College Physics - Serway Vuille - GCE Guide

SOLUTION MANUAL

I NSTRUCTOR ’ S S OLUTIONS M ANUAL FOR

S ERWAY

AND

V UILLE ’ S

C O L L E G E P H YS I C S N INTH E DITION , V OLUME 1

Charles Teague

Emeritus, Eastern Kentucky University

Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States

56157_00_fm_pi-viii.indd i

10/13/10 9:54:04 AM

© 2012 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher.

For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions Further permissions questions can be emailed to [email protected]

ISBN-13: 978 0 8400 6871 9 ISBN-10: 0 8400 6871 9 Brooks/Cole 20 Channel Center Street Boston, MA 02210 USA Cengage Learning is a leading provider of customized learning solutions with oﬃce locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local oﬃce at: www.cengage.com/global Cengage Learning products are represented in Canada by Nelson Education, Ltd. To learn more about Brooks/Cole, visit www.cengage.com/brookscole Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com

NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED, OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN. READ IMPORTANT LICENSE INFORMATION Dear Professor or Other Supplement Recipient: Cengage Learning has provided you with this product (the “Supplement”) for your review and, to the extent that you adopt the associated textbook for use in connection with your course (the “Course”), you and your students who purchase the textbook may use the Supplement as described below. Cengage Learning has established these use limitations in response to concerns raised by authors, professors, and other users regarding the pedagogical problems stemming from unlimited distribution of Supplements. Cengage Learning hereby grants you a nontransferable license to use the Supplement in connection with the Course, subject to the following conditions. The Supplement is for your personal, noncommercial use only and may not be reproduced, posted electronically or distributed, except that portions of the Supplement may be provided to your students IN PRINT FORM ONLY in connection with your instruction of the Course, so long as such students are advised that they may not copy or distribute any portion of

the Supplement to any third party. You may not sell, license, auction, or otherwise redistribute the Supplement in any form. We ask that you take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution. Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement. If you do not accept these conditions, you must return the Supplement unused within 30 days of receipt. All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/ or its licensors. The Supplement is furnished by Cengage Learning on an “as is” basis without any warranties, express or implied. This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s conﬂict of law rules. Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement. We trust you ﬁnd the Supplement a useful teaching tool.

Printed in the United States of America 1 2 3 4 5 6 7 14 13 12 11 10

68719_00_fm_pi-viii.indd ii

1/7/11 3:07:37 PM

TABLE OF CONTENTS Acknowledgements Preface

v vii

Part 1 – Mechanics Chapter 1 – Introduction Chapter 2 – Motion in One Dimension Chapter 3 – Vectors and Two-Dimensional Motion Chapter 4 – The Laws of Motion Chapter 5 – Energy Chapter 6 – Momentum and Collisions Chapter 7 – Rotational Motion and the Law of Gravity Chapter 8 – Rotational Equilibrium and Rotational Dynamics Chapter 9 – Solids and Fluids

1 17 49 89 129 169 209 243 288

Part 2 – Thermodynamics Chapter 10 – Thermal Physics Chapter 11 – Energy in Thermal Processes Chapter 12 – The Laws of Thermodynamics

322 344 371

Part 3 – Vibrations and Waves Chapter 13 – Vibrations and Waves Chapter 14 – Sound

400 426

iii

56157_00_fm_pi-viii.indd iii

10/13/10 9:54:06 AM

TABLE OF CONTENTS Acknowledgements Preface

v vii

Part 4 – Electricity and Magnetism Chapter 15 – Electric Forces and Electric Fields Chapter 16 – Electrical Energy and Capacitance Chapter 17 – Current and Resistance Chapter 18 – Direct-Current Circuits Chapter 19 – Magnetism Chapter 20 – Induced Voltages and Inductance Chapter 21 – Alternating Current Circuits and Electromagnetic Waves

1 30 62 83 120 148 171

Part 5 – Light and Optics Chapter 22 – Reflection and Refraction of Light Chapter 23 – Mirrors and Lenses Chapter 24 – Wave Optics Chapter 25 – Optical Instruments

198 223 256 282

Part 6 – Modern Physics Chapter 26 – Relativity Chapter 27 – Quantum Physics Chapter 28 – Atomic Physics Chapter 29 – Nuclear Physics Chapter 30 – Nuclear Physics and Elementary Particles

305 324 341 360 377

iii

68719_00_fm_pi-viii.indd iii

1/7/11 3:07:38 PM

ACKNOWLEDGEMENTS The author would like to thank everyone who has contributed to this work. In particular, thanks go to the support staff at Cengage Learning for their excellent guidance and support in all phases of this project. Special mention goes to Physics Acquisitions Editor, Charles Hartford; Development Editor, Ed Dodd; Associate Content Project Manager, Holly Schaff; Associate Development Editor, Brandi Kirksey; and Editorial Assistant, Brendan Killion. Susan English of Durham Technical Community College served as accuracy reviewer for this manual. Her contributions are deeply appreciated. Any remaining errors in this work are the responsibility of the author alone. I would like to acknowledge the staff of MPS Limited, a Macmillan Company for their excellent work in assembling and typing this manual and preparing diagrams and page layouts. Finally, the author would like to thank his wife, Carol, for her patience, understanding, and great support during this effort.

v

56157_00_fm_pi-viii.indd v

10/13/10 9:54:06 AM

PREFACE This manual is written to accompany College Physics, Ninth Edition, by Raymond A. Serway and Chris Vuille. For each chapter in that text, the manual includes solutions to all end-of-chapter problems, more detailed answers to Quick Quizzes and Multiple Choice Questions than available in the main text, and answers to the even-numbered Conceptual Questions. Considerable effort has been made to insure that the solutions and answers given in this manual comply with the rules on significant figures and rounding given in the Chapter 1 of the textbook. This means that intermediate answers are rounded to the proper number of significant figures when written, and that rounded value is used in all subsequent calculations. Users should not be concerned if their answers differ slightly in the last digit from the answers given here. Most often, this will be caused by choosing to round intermediate answers at different stages of the solution. You are encouraged to keep this manual out of the hands of students as instructors in many colleges throughout the country use this textbook, and many of them use graded problem assignments as part of the final course grade. Additionally, even when the problems are not used in such a direct fashion, it is advantageous for students to struggle with some problems in order to improve their problem-solving skills. Feel free to post answers and solutions to selected questions and problems, but please preserve the manual as a whole. You may also encourage students to purchase a copy of the Student Solutions Manual with Study Guide, which provides chapter summaries as well as detailed solutions to selected problems in the main text. Attempting to keep the manual of manageable size, and recognizing that the primary users will be instructors well versed in the field, answers and solutions are kept fairly brief. Answers to conceptual questions have been shortened by not offering detailed arguments that lead to the answer. Problem solutions often omit commentary, intermediate steps, as well as initial steps that could be necessary for clear understanding by students. On occasions where selected problem solutions are to be shared with students, you may wish to supply intermediate steps and additional comments as needed. An electronic version of this manual can be obtained by requesting the Instructor’s Power Lecture CD from your local Cengage Learning Sales Representative. Contact information for your sales representative is available under the “Find Your Rep” tab found at the bottom of the page at www.academic.cengage.com. We welcome your comments on the accuracy of the solutions as presented here, as well as suggestions for alternative approaches. Charles Teague

vii

56157_00_fm_pi-viii.indd vii

10/13/10 9:54:06 AM

56157_00_fm_pi-viii.indd viii

10/13/10 9:54:06 AM

1 Introduction ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

Using a calculator to multiply the length by the width gives a raw answer of 6783 m 2 , but this answer must be rounded to contain the same number of signiﬁcant ﬁgures as the least accurate factor in the product. The least accurate factor is the length, which contains either 2 or 3 signiﬁcant ﬁgures, depending on whether the trailing zero is signiﬁcant or is being used only to locate the decimal point. Assuming the length contains 3 signiﬁcant ﬁgures, answer (c) correctly expresses the area as 6.78 × 10 3 m 2 . However, if the length contains only 2 signiﬁcant ﬁgures, answer (d) gives the correct result as 6.8 × 10 3 m 2 .

2.

Both answers (d) and (e) could be physically meaningful. Answers (a), (b), and (c) must be meaningless since quantities can be added or subtracted only if they have the same dimensions.

3.

According to Newton’s second law, Force = mass × acceleration . Thus, the units of Force must be the product of the units of mass (kg) and the units of acceleration ( m s 2 ). This yields kg ⋅ m s 2, which is answer (a).

4.

The calculator gives an answer of 57.573 for the sum of the 4 given numbers. However, this sum must be rounded to 58 as given in answer (d) so the number of decimal places in the result is the same (zero) as the number of decimal places in the integer 15 (the term in the sum containing the smallest number of decimal places).

5.

The required conversion is given by: ⎛ 1 000 mm ⎞ ⎛ 1.00 cubitus ⎞ h = ( 2.00 m ) ⎜ ⎟⎜ ⎟ = 4.49 cubiti ⎝ 1.00 m ⎠ ⎝ 445 mm ⎠ This result corresponds to answer (c).

6.

The given area (1 420 ft 2 ) contains 3 signiﬁcant ﬁgures, assuming that the trailing zero is used only to locate the decimal point. The conversion of this value to square meters is given by: 2

1.00 m ⎞ 2 2 2 A = (1.42 × 10 3 ft 2 ) ⎛⎜ ⎟ = 1.32 × 10 m = 132 m ⎝ 3.281 ft ⎠ Note that the result contains 3 signiﬁcant ﬁgures, the same as the number of signiﬁcant ﬁgures in the least accurate factor used in the calculation. This result matches answer (b). 7.

You cannot add, subtract, or equate a number apples and a number of days. Thus, the answer is yes for (a), (c), and (e). However, you can multiply or divide a number of apples and a number of days. For example, you might divide the number of apples by a number of days to ﬁnd the number of apples you could eat per day. In summary, the answers are (a) yes, (b) no, (c) yes, (d) no, and (e) yes.

1

56157_01_ch01_p001-016.indd 1

10/12/10 1:26:50 PM

Chapter 1

2

8.

The given Cartesian coordinates are x = −5.00, and y = 12.00 , with the least accurate containing 3 signiﬁcant ﬁgures. Note that the speciﬁed point (with x < 0 and y > 0 ) is in the second quadrant. The conversion to polar coordinates is then given by: r = x 2 + y 2 = ( −5.00 ) + (12.00 ) = 13.0 2

tan θ =

y 12.00 = = −2.40 x −5.00

2

and θ = tan −1 ( −2.40 ) = −67.3° + 180° = 113°

Note that 180° was added in the last step to yield a second quadrant angle. The correct answer is therefore (b) (13.0, 113°). 9.

Doing dimensional analysis on the ﬁrst 4 given choices yields: (a)

(c)

[ v] ⎡⎣t ⎤⎦ 2

=

LT L = 3 T2 T

⎡⎣ v 2 ⎤⎦ ( L T )2 L2 T 2 L2 = = = 3 T T T [t ]

(b)

(d)

[ v] ⎡⎣ x ⎤⎦ 2

=

LT = L−1T −1 2 L

⎡⎣ v 2 ⎤⎦ ( L T )2 L2 T 2 L = = = 2 L L T [ x]

Since acceleration has units of length divided by time squared, it is seen that the relation given in answer (d) is consistent with an expression yielding a value for acceleration. 10.

The number of gallons of gasoline she can purchase is # gallons =

total expenditure 33 Euros ≈ cost per gallon ⎛ Euros ⎞ ⎛ 1 L ⎞ ⎟ ⎜ 1.5 ⎟⎜ L ⎠ ⎜⎝ 1 quart ⎟⎠ ⎝

⎛ 4 quarts ⎜⎜ ⎝ 1 gal

⎞ ⎟⎟ ⎠

≈ 5 gal

so the correct answer is (b). 11.

The situation described is shown in the drawing at the right. h From this, observe that tan 26° = , or 45 m

h

h = ( 45 m ) tan 26° = 22 m 26⬚

Thus, the correct answer is (a). 12.

45 m

Note that we may write 1.365 248 0 × 10 7 as 136.524 80 × 10 5. Thus, the raw answer, including the uncertainty, is x = (136.524 80 ± 2) × 10 5. Since the ﬁnal answer should contain all the digits we are sure of and one estimated digit, this result should be rounded and displayed as 137 × 10 5 = 1.37 × 10 7 (we are sure of the 1 and the 3, but have uncertainty about the 7). We see that this answer has three signiﬁcant ﬁgures and choice (d) is correct.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

56157_01_ch01_p001-016.indd 2

Atomic clocks are based on the electromagnetic waves that atoms emit. Also, pulsars are highly regular astronomical clocks.

10/12/10 1:26:51 PM

Introduction

4.

(a)

~ 0.5 lb ≈ 0.25 kg or ~10 −1 kg

(b)

~ 4 lb ≈ 2 kg or ~10 0 kg

(c)

~ 4000 lb ≈ 2000 kg or ~10 3 kg

3

6.

Let us assume the atoms are solid spheres of diameter 10−10 m. Then, the volume of each atom is of the order of 10−30 m3. (More precisely, volume = 4π r 3 3 = π d 3 6 .) Therefore, since 1 cm 3 = 10 −6 m 3, the number of atoms in the 1 cm3 solid is on the order of 10 −6 10 −30 = 10 24 atoms. A more precise calculation would require knowledge of the density of the solid and the mass of each atom. However, our estimate agrees with the more precise calculation to within a factor of 10.

8.

Realistically, the only lengths you might be able to verify are the length of a football ﬁeld and the length of a houseﬂy. The only time intervals subject to veriﬁcation would be the length of a day and the time between normal heartbeats.

10.

In the metric system, units differ by powers of ten, so it’s very easy and accurate to convert from one unit to another.

ANSWERS TO EVEN NUMBERED PROBLEMS L T2

(a)

4.

All three equations are dimensionally incorrect.

6.

(a)

kg ⋅ m s

(b)

Ft = p

8.

(a)

22.6

(b)

22.7

(c)

22.6 is more reliable

10.

(a)

3.00 × 108 m s

(b)

2 .997 9 × 108 m s

(c)

2.997 925 × 108 m s

12.

(a)

346 m 2 ± 13 m 2

(b)

66.0 m ± 1.3 m

14.

(a)

797

(b)

1.1

(c)

17.66

16.

3.09 cm s

18.

(a)

5.60 × 10 2 km = 5.60 × 10 5 m = 5.60 × 10 7 cm

(b)

0.491 2 km = 491.2 m = 4.912 × 10 4 cm

(c)

6.192 km = 6.192 × 10 3 m = 6.192 × 10 5 cm

(d)

2.499 km = 2.499 × 10 3 m = 2.499 × 10 5 cm

20.

10.6 km L

22.

9.2 nm s

24.

2 .9 × 10 2 m 3 = 2 .9 × 108 cm 3

26.

2 .57 × 10 6 m 3

56157_01_ch01_p001-016.indd 3

(b)

L

2.

10/12/10 1:26:51 PM

4

Chapter 1

28.

∼ 108 steps

30.

~108 people with colds on any given day

32.

(a)

4.2 × 10 −18 m 3

(b)

~10 −1 m 3

34.

(a)

∼ 10 29 prokaryotes

(b)

~1014 kg

(c)

~1016 cells

(c) The very large mass of prokaryotes implies they are important to the biosphere. They are responsible for ﬁxing carbon, producing oxygen, and breaking up pollutants, among many other biological roles. Humans depend on them! 36.

2.2 m

38.

8.1 cm

40.

Δs = r12 + r22 − 2r1r2 cos (θ1 − θ2 )

42.

2.33 m

44.

(a)

46.

8.60 m

48.

(a) and (b)

(c)

1.50 m

(b)

2.60 m

y x = tan 12.0°, y ( x − 1.00 km ) = tan 14.0°

(d)

1.44 × 10 3 m

(c)

16 km h

d ⋅ tan θ ⋅ tan φ tan φ − tan θ

50.

y=

52.

(a)

54.

Assumes population of 300 million, average of 1 can week per person, and 0.5 oz per can.

1.609 km h

(b)

88 km h

(a)

∼ 1010 cans yr

(b)

∼ 10 5 tons yr

56.

(a)

7.14 × 10 −2 gal s

(b)

2.70 × 10 −4 m 3 s

58.

(a)

A2 A1 = 4

(b)

V2 V1 = 8

60.

(a)

∼ 10 2 yr

(b)

∼ 10 4 times

62.

∼ 10 4 balls yr. Assumes 1 lost ball per hitter, 10 hitters per inning, 9 innings per game, and 81 games per year.

56157_01_ch01_p001-016.indd 4

(c)

1.03 h

10/12/10 1:26:52 PM

Introduction

5

PROBLEM SOLUTIONS 1.1

Substituting dimensions into the given equation T = 2π sionless constant, we have

[T ] =

[ ] [ g]

T=

or

L = L T2

g , and recognizing that 2π is a dimen-

T2 = T

Thus, the dimensions are consistent . 1.2

(a)

From x = Bt2, we ﬁnd that B = [ B] =

(b)

x . Thus, B has units of t2

[ x] L = 2 2 [t ] T

If x = A sin ( 2π ft ), then [ A] = [ x ] [sin ( 2π ft )] But the sine of an angle is a dimensionless ratio. Therefore, [ A] = [ x ] = L

1.3

(a)

The units of volume, area, and height are: [V ] = L3, [ A] = L2 , and [h] = L We then observe that L3 = L2 L or [V ] = [ A][h] Thus, the equation V = Ah is dimensionally correct .

(b)

Vcylinder = π R 2 h = (π R 2 ) h = Ah , where A = π R 2 Vrectangular box = wh = ( w ) h = Ah, where A = w = length × width 2

1.4

(a)

(b)

(c)

2

L ML m v 2 = 12 m v02 + mgh, [ m v 2 ] = [ m v02 ] = M ⎛⎜ ⎞⎟ = 2 T T ⎝ ⎠ 1 2 L M L while ⎡⎣ mgh ⎤⎦ = M ⎛⎜ 2 ⎞⎟ L = . Thus, the equation is dimensionally incorrect . T ⎝T ⎠ In the equation

1 2

L L but [at 2 ] = [a][t 2 ] = ⎛⎜ 2 ⎞⎟ ( T 2 ) = L. Hence, this equation T ⎝T ⎠ is dimensionally incorrect . In v = v0 + at 2, [ v] = [ v0 ] =

L In the equation ma = v 2, we see that [ ma] = [ m][a] = M ⎛⎜ 2 T ⎝

2

2 ⎞ = ML , while [ v 2 ] = ⎛ L ⎞ = L . ⎟ ⎜ ⎟ 2 T2 ⎠ T ⎝T⎠

Therefore, this equation is also dimensionally incorrect . 1.5

From the universal gravitation law, the constant G is G = Fr 2 Mm. Its units are then

[G ] =

56157_01_ch01_p001-016.indd 5

[ F ] ⎡⎣r 2 ⎤⎦ ( kg ⋅ m s2 ) ( m 2 ) m3 = = kg ⋅ kg kg ⋅ s 2 [ M ][ m ]

10/12/10 1:26:52 PM

Chapter 1

6

1.6

(a)

p2 for the momentum, p, gives p = 2 m ( KE ) where the numeral 2 is a 2m dimensionless constant. Dimensional analysis gives the units of momentum as:

Solving KE =

[ p] = [ m ][ KE ] =

M ( M ⋅ L2 T 2 ) = M 2 ⋅ L2 T 2 = M ( L T )

Therefore, in the SI system, the units of momentum are kg ⋅ ( m s ) . (b)

Note that the units of force are kg ⋅ m s 2 or [ F ] = M ⋅ L T 2 . Then, observe that

[ F ][ t ] = ( M ⋅ L T 2 ) ⋅ T = M ( L T ) = [ p ] From this, it follows that force multiplied by time is proportional to momentum: Ft = p . (See the impulse–momentum theorem in Chapter 6, F ⋅ Δt = Δp , which says that a constant force F multiplied by a duration of time ∆t equals the change in momentum, ∆p.) 1.7

Area = ( length ) × ( width ) = ( 9.72 m )( 5.3 m ) = 52 m 2

1.8

(a)

Computing

( 8) (b)

3

( 8)

3

without rounding the intermediate result yields

= 22.6 to three signiﬁcant ﬁgures.

Rounding the intermediate result to three signiﬁcant ﬁgures yields 8 = 2.8284 → 2.83 Then, we obtain

1.9

1.10

1.11

( 8)

3

= ( 2.83) = 22.7 to three signiﬁcant ﬁgures. 3

(c)

The answer 22.6 is more reliable because rounding in part (b) was carried out too soon.

(a)

78.9 ± 0.2 has 3 significant figures with the uncertainty in the tenths position.

(b)

3.788 ×10 9 has 4 significant figures

(c)

2.46 ×10 −6 has 3 significant figures

(d)

0.003 2 = 3.2 × 10 −3 has 2 significant figures . The two zeros were originally included only to position the decimal.

c = 2 .997 924 58 × 108 m s (a)

Rounded to 3 signiﬁcant ﬁgures: c = 3.00 × 108 m s

(b)

Rounded to 5 signiﬁcant ﬁgures: c = 2 .997 9 × 108 m s

(c)

Rounded to 7 signiﬁcant ﬁgures: c = 2 .997 925 ×108 m s

Observe that the length = 5.62 cm, the width w = 6.35 cm, and the height h = 2.78 cm all contain 3 signiﬁcant ﬁgures. Thus, any product of these quantities should contain 3 signiﬁcant ﬁgures. (a) (b)

w = ( 5.62 cm )( 6.35 cm ) = 35.7 cm 2 V = ( w ) h = ( 35.7 cm 2 ) ( 2.78 cm ) = 99.2 cm 3

continued on next page

56157_01_ch01_p001-016.indd 6

10/12/10 1:26:53 PM

Introduction

7

wh = ( 6.35 cm )( 2.78 cm ) = 17.7 cm 2

(c)

V = ( wh ) = (17.7 cm 2 ) ( 5.62 cm ) = 99.5 cm 3 (d)

1.12

In the rounding process, small amounts are either added to or subtracted from an answer to satisfy the rules of signiﬁcant ﬁgures. For a given rounding, different small adjustments are made, introducing a certain amount of randomness in the last signiﬁcant digit of the ﬁnal answer. 2 2 2 A = π r 2 = π (10.5 m ± 0.2 m ) = π ⎡⎣(10.5 m ) ± 2 (10.5 m )( 0.2 m ) + ( 0.2 m ) ⎤⎦

(a)

Recognize that the last term in the brackets is insigniﬁcant in comparison to the other two. Thus, we have A = π ⎡⎣110 m 2 ± 4.2 m 2 ⎤⎦ = 346 m 2 ± 13 m 2 C = 2π r = 2π (10.5 m ± 0.2 m ) = 66.0 m ± 1.3 m

(b) 1.13

The least accurate dimension of the box has two signiﬁcant ﬁgures. Thus, the volume (product of the three dimensions) will contain only two signiﬁcant ﬁgures. V = ⋅ w ⋅ h = ( 29 cm )(17.8 cm )(11.4 cm ) = 5.9 × 10 3 cm 3

1.14

1.15

(a)

The sum is rounded to 797 because 756 in the terms to be added has no positions beyond the decimal.

(b)

0.003 2 × 356.3 = ( 3.2 × 10 − 3 ) × 356.3 = 1.14016 must be rounded to 1.1 because 3.2 × 10 −3 has only two signiﬁcant ﬁgures.

(c)

5.620 × π must be rounded to 17.66 because 5.620 has only four signiﬁcant ﬁgures.

5 280 ft ⎞ ⎛ 1 fathom ⎞ 8 d = ( 250 000 mi ) ⎛⎜ ⎟⎜ ⎟ = 2 × 10 fathoms ⎝ 1.000 mi ⎠ ⎝ 6 ft ⎠ The answer is limited to one signiﬁcant ﬁgure because of the accuracy to which the conversion from fathoms to feet is given.

1.16

v=

t

giving

=

186 furlongs 1 fortnight

⎛ 1 fortnight ⎜⎜ ⎝ 14 days

⎞ ⎛ 1 day ⎞ ⎛ 220 yds ⎞ ⎛ 3 ft ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ 4 ⎠ ⎝ 8.64 × 10 s ⎠ ⎝ 1 furlong ⎠ ⎝ 1 yd

⎞ ⎛ 100 cm ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ 3.281 ft ⎠

v = 3.09 cm s

1.17

⎛ 9 gal 6.00 firkins = 6.00 firkins ⎜⎜ ⎝ 1 firkin

1.18

(a)

⎞ ⎛ 3.786 L ⎟⎟ ⎜⎜ ⎠ ⎝ 1 gal

⎞ ⎛ 10 3 cm 3 ⎞ ⎛ 1 m 3 ⎞ = 0.204 m 3 ⎟⎟ ⎜ ⎟⎜ 6 3 ⎟ 1 L 10 cm ⎝ ⎠ ⎝ ⎠ ⎠

1.609 km ⎞ 2 5 7 = ( 348 mi ) ⎛⎜ ⎟ = 5.60 × 10 km = 5.660 × 10 m = 5.60 × 10 cm ⎝ 1.000 mi ⎠

(b)

⎛ 1.609 km ⎞ 4 h = (1 612 ft ) ⎜ ⎟ = 0.491 2 km = 491.22 m = 4.912 × 10 cm ⎝ 5 280 ft ⎠

(c)

⎛ 1.609 km ⎞ 3 5 h = ( 20 320 ft ) ⎜ ⎟ = 6.192 km = 6.192 × 10 m = 6.192 × 10 cm ⎝ 5 280 ft ⎠

continued on next page

56157_01_ch01_p001-016.indd 7

10/12/10 1:26:53 PM

8

Chapter 1

(d)

⎛ 1.609 km ⎞ 3 5 d = (8 200 ft ) ⎜ ⎟ = 2 .499 km = 2 .499 × 10 m = 2 .499 × 10 cm 5 280 ft ⎝ ⎠

In (a), the answer is limited to three signiﬁcant ﬁgures because of the accuracy of the original data value, 348 miles. In (b), (c), and (d), the answers are limited to four signiﬁcant ﬁgures because of the accuracy to which the kilometers-to-feet conversion factor is given. m ⎛ 1 km ⎞ ⎛ 1 mi ⎞ ⎛ 3 600 s ⎞ ⎟⎜ ⎟ = 85.0 mi h ⎜ ⎟⎜ s ⎝ 10 3 m ⎠ ⎝ 1.609 km ⎠ ⎝ 1 h ⎠ Yes, the driver is exceeding the speed limit by 10.0 mi h .

1.19

v = 38.0

1.20

efficiency = 25.0

1.21

(a)

r=

(b)

A = 4π r 2 = 4π ( 6.81 cm ) = 5.83 × 10 2 cm 2

(c)

V=

1.22

mi ⎛ 1 km ⎞ ⎛ 1 gal ⎞ ⎟ = 10.6 km L ⎜ ⎟⎜ gal ⎝ 0.621 mi ⎠ ⎜⎝ 3.786 L ⎟⎠

diameter 5.36 in ⎛ 2.54 cm ⎞ = ⎜ ⎟ = 6.81 cm 2 2 ⎝ 1 in ⎠ 2

4 3 4 3 π r = π ( 6.81 cm ) = 1.32 × 10 3 cm 3 3 3

⎛ 1 in ⎞ ⎛ 1 day rate = ⎜ ⎜ 32 day ⎟⎟ ⎜⎜ 24 h ⎝ ⎠⎝

⎞ ⎛ 1 h ⎞ ⎛ 2.54 cm ⎞ ⎛ 10 9 nm ⎞ ⎟⎟ ⎜ ⎟⎜ ⎟⎜ 2 ⎟ = 9.2 nm s ⎠ ⎝ 3 600 s ⎠ ⎝ 1.00 in ⎠ ⎝ 10 cm ⎠

This means that the proteins are assembled at a rate of many layers of atoms each second! 1.23

⎛ m ⎞ ⎛ 3 600 s ⎞ ⎛ 1 km ⎞ ⎛ 1 mi ⎞ 8 c = ⎜ 3.00 × 10 8 ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 6.71 × 10 mi h s ⎠ ⎝ 1 h ⎠ ⎝ 10 3 m ⎠ ⎝ 1.609 km ⎠ ⎝

1.24

⎛ 2 .832 × 10 − 2 m3 ⎞ Volume of house = ( 50.0 ft )( 26 ft )(8.0 ft ) ⎜ ⎟⎠ 1 ft 3 ⎝ 3

⎛ 100 cm ⎞ = 2 .9 × 10 2 m3 = ( 2 .9 × 10 2 m3 ) ⎜ = 2 .9 × 10 8 cm3 ⎝ 1 m ⎟⎠ 1.25 1.26

2 2 ⎛ 1 m ⎞ ⎡⎛ 43 560 ft ⎞ ⎛ 1 m ⎞ ⎤ ⎢ ⎥ = 3.08 × 10 4 m3 Volume = 25.0 acre ft ⎜ ⎜ ⎟ ⎝ 3.281 ft ⎟⎠ ⎢⎝ 1 acre ⎠ ⎜⎝ 3.281 ft ⎟⎠ ⎥ ⎣ ⎦ 1 Volume of pyramid = ( area of base )( height ) 3

(

)

=

1 ⎡(13.0 acres )( 43 560 ft 2 acre ) ⎤ ( 481 ft ) = 9.08 × 10 7 f ⎦ 3⎣

⎛ 2 .832 × 10 − 2 m3 ⎞ 6 3 = ( 9.08 × 10 7 ft 3 ) ⎜ ⎟⎠ = 2 .557 × 10 m 1 ft3 ⎝ 1.27

Volume of cube = L 3 = 1 quart (Where L = length of one side of the cube.) ⎛ 1 gallon ⎞ ⎛ 3.786 liiter ⎞ ⎛ 1000 cm3 ⎞ = 947 cm3 Thus, L 3 = 1 quart ⎜ ⎟ ⎜⎝ 4 quarts ⎟⎠ ⎜⎝ 1 gallon ⎟⎠ ⎜⎝ 1 liter ⎟⎠

(

and

56157_01_ch01_p001-016.indd 8

)

L = 3 947 cm3 = 9.82 cm

10/18/10 1:21:48 PM

Introduction

1.28

9

We estimate that the length of a step for an average person is about 18 inches, or roughly 0.5 m. Then, an estimate for the number of steps required to travel a distance equal to the circumference of the Earth would be N= or

1.29.

2π ( 6.338 × 10 6 m ) Circumference 2π RE = ≈ ≈ 8 × 10 7 steps 0.5 m step Step Length Step Length

N ∼ 108 steps

We assume an average respiration rate of about 10 breaths/minute and a typical life span of 70 years. Then, an estimate of the number of breaths an average person would take in a lifetime is ⎛ 3.156 × 10 7 s ⎞ ⎛ 1 min ⎛ breaths ⎞ n = ⎜ 10 ⎟⎟ ⎜ ⎟ ( 70 yr ) ⎜⎜ min ⎠ 1 yr ⎝ ⎠ ⎝ 60 s ⎝ or

1.30

⎞ 8 ⎟ = 4 × 10 breaths ⎠

n ∼ 108 breaths

We assume that the average person catches a cold twice a year and is sick an average of 7 days (or 1 week) each time. Thus, on average, each person is sick for 2 weeks out of each year (52 weeks). The probability that a particular person will be sick at any given time equals the percentage of time that person is sick, or probability of sickness =

2 weeks 1 = 52 weeks 26

The population of the Earth is approximately 7 billion. The number of people expected to have a cold on any given day is then 1 Number sick = ( population )(( probability of sickness ) = ( 7 × 10 9 ) ⎛⎜ ⎞⎟ = 3 × 108 or ∼ 108 ⎝ 26 ⎠ 1.31

(a)

Assume that a typical intestinal tract has a length of about 7 m and average diameter of 4 cm. The estimated total intestinal volume is then ⎛ πd 2 ⎞ π ( 0.04 m ) Vtotal = A = ⎜ ( 7 m ) = 0.009 m 3 ⎟ = 4 4 ⎝ ⎠ 2

The approximate volume occupied by a single bacterium is Vbacteria ∼ ( typical length scale ) = (10 −6 m ) = 10 −18 m 3 3

3

If it is assumed that bacteria occupy one hundredth of the total intestinal volume, the estimate of the number of microorganisms in the human intestinal tract is 3 Vtotal 100 ( 0.009 m ) 100 n= = = 9 × 1013 or n ∼ 1014 −18 3 10 m Vbacteria

1.32

(b)

The large value of the number of bacteria estimated to exist in the intestinal tract means that they are probably not dangerous. Intestinal bacteria help digest food and provide important nutrients. Humans and bacteria enjoy a mutually beneﬁcial symbiotic relationship.

(a)

Vcell =

(b)

Consider your body to be a cylinder having a radius of about 6 inches (or 0.15 m) and a height of about 1.5 meters. Then, its volume is

3 4 3 4 π r = π (1.0 × 10 −6 m ) = 4.2 × 10 −18 m 3 3 3

Vbody = Ah = (π r 2 ) h = π ( 0.15 m ) (1.5 m ) = 0.11 m 3 or ∼ 10 −1 m 3 2

continued on next page

56157_01_ch01_p001-016.indd 9

10/12/10 1:26:54 PM

10

Chapter 1

(c)

The estimate of the number of cells in the body is then n=

1.33

Vcell

=

0.11 m 3 = 2.6 × 1016 or ∼ 1016 4.2 × 10 −18 m 3

A reasonable guess for the diameter of a tire might be 3 ft, with a circumference (C = 2π r = π D = distance travels per revolution) of about 9 ft. Thus, the total number of revolutions the tire might make is n=

1.34

Vbody

total distance traveled ( 50 000 mi )( 5 280 ft mi ) = 3 × 10 7 rev, or ~ 10 7 rev = distance per revolution 9 ft rev

Answers to this problem will vary, dependent on the assumptions one makes. This solution assumes that bacteria and other prokaryotes occupy approximately one ten-millionth (10−7) of the Earth’s volume, and that the density of a prokaryote, like the density of the human body, is approximately equal to that of water (103 kg/m3). (a)

estimated number = n =

Vtotal Vsingle

(10 )V ∼

(c)

1.35

Vsingle

prokaryote

3

3

6

−6

3

∼ 10 29

The very large mass of prokaryotes implies they are important to the biosphere. They are responsible for ﬁxing carbon, producing oxygen, and breaking up pollutants, among many other biological roles. Humans depend on them!

) y = r sinθ = ( 2 .5 m ) sin 35° = 1.4 m (

x = r cosθ = 2 .5 m cos 35° = 2.0 m

The x coordinate is found as

The x distance out to the ﬂy is 2.0 m and the y distance up to the ﬂy is 1.0 m. Thus, we can use the Pythagorean theorem to ﬁnd the distance from the origin to the ﬂy as

( 2.0 m ) + (1.0 m ) 2

d = x 2 + y2 = 1.37

−7

3 Earth

3 kg ⎞ ⎛ ⎞ ⎛ mtotal = ( density )( total volume) ∼ ρwater ⎜ nVsingle ⎟ = ⎜ 10 3 3 ⎟ (10 29 )(10 −6 m ) ∼ 1014 kg ⎝ prokaryote ⎠ ⎝ m ⎠

and the y coordinate 1.36

(10 )(10 m ) ∼ ∼ (length scale) (10 m ) −7

Earth

prokaryote

(b)

(10 ) R

−7

2

=

2.2 m

The distance from the origin to the ﬂy is r in polar coordinates, and this was found to be 2.2 m in Problem 36. The angle θ is the angle between r and the horizontal reference line (the x axis in this case). Thus, the angle can be found as tan θ =

y 1.0 m = = 0.50 x 2.0 m

and

θ = tan −1 ( 0.50 ) = 27°

The polar coordinates are r = 2.2 m and θ = 27 ° 1.38

The x distance between the two points is Δx = x2 − x1 = −3.0 cm − 5.0 cm = 8.0 cm and the y distance between them is Δy = y2 − y1 = 3.0 cm − 4.0 cm = 1.0 cm. The distance between them is found from the Pythagorean theorem: d=

1.39

Δx + Δy = (8.0 cm ) + (1.0 cm ) = 2

2

2

2

65 cm 2 = 8.1 cm

Refer to the Figure given in Problem 1.40 below. The Cartesian coordinates for the two given points are: x1 = r1 cos θ1 = ( 2.00 m ) cos 50.0° = 1.29 m

x2 = r2 cos θ2 = ( 5.00 m ) cos ( −50.0°) = 3.21 m

y1 = r1 sin θ1 = ( 2.00 m ) sin 50.0° = 1.53 m

y2 = r2 sin θ2 = ( 5.00 m ) sin ( −50.0°) = −3.83 m

continued on next page

56157_01_ch01_p001-016.indd 10

10/12/10 1:26:55 PM

Introduction

11

The distance between the two points is then: Δs = ( Δx ) + ( Δy ) = (1.29 m − 3.21 m ) + (1.53 m + 3.83 m ) = 5.69 m 2

1.40

2

2

2

Consider the Figure shown at the right. The Cartesian coordinates for the two points are: x1 = r1 cos θ1

x2 = r2 cos θ2

y1 = r1 sin θ1

y2 = r2 sin θ2

r1

The distance between the two points is the length of the hypotenuse of the shaded triangle and is given by Δs = ( Δx ) + ( Δy ) = 2

2

(x1, y1)

y

q1

( x1 − x2 )2 + ( y1 − y2 )2

Δs

Δy⫽ y1⫺ y2

(x2, y2) r2 Δx⫽ x1⫺ x2 q2 x

or Δs =

(r

2 1

cos 2 θ1 + r22 cos 2 θ2 − 2r1r2 cos θ1 cos θ2 ) + ( r12 sin 2 θ1 + r22 sin 2 θ2 − 2r1r2 sin θ1 sin θ2 )

= r12 ( cos 2 θ1 + sin 2 θ1 ) + r22 ( cos 2 θ2 + sin 2 θ2 ) − 2r1r2 ( cos θ1 cos θ2 + sin θ1 siin θ2 ) Applying the identities cos 2 θ + sin 2 θ = 1 and cos θ1 cos θ2 + sin θ1 sin θ2 = cos (θ1 − θ2 ) , this reduces to Δs = r12 + r22 − 2r1r2 ( cos θ1 cos θ2 + sin θ1 sin θ2 ) = 1.41

(a)

With a = 6.00 m and b being two sides of this right triangle having hypotenuse c = 9.00 m, the Pythagorean theorem gives the unknown side as b = c2 − a2 =

(b)

1.42

r12 + r22 − 2r1r2 cos (θ1 − θ2 )

tan θ =

( 9.00 m )2 − ( 6.00 m )2 = 6.71 m

a 6.00 m = = 0.894 b 6.71 m

(c)

sin φ =

b 6.71 m = = 0.746 c 9.00 m

From the diagram, cos ( 75.0°) = d L

L⫽9

.00 m

Thus, d = L cos ( 75.0°) = ( 9.00 m ) cos ( 75.0°) = 2.33 m

75.0⬚

d

56157_01_ch01_p001-016.indd 11

10/12/10 1:26:55 PM

12

1.43

Chapter 1

The circumference of the fountain is C = 2π r , so the radius is C 15.0 m = = 2.39 m 2π 2π h h Thus, tan ( 55.0°) = = which gives r 2.39 m r=

h = ( 2.39 m ) tan ( 55.0°) = 3.41 m 1.44

1.45

1.46

(a)

sin θ =

opposite side so, opposite side = ( 3.00 m ) sin ( 30.0° ) = 1.50 m hypotenuse

(b)

cos θ =

adjacent side so, adjacent side = ( 3.00 m ) cos ( 30.0° ) = 2 .60 m hypotenuse

(a)

The side opposite θ = 3.00

(c)

cos θ =

4.00 = 0.800 5.00

(e)

tan φ =

4.00 = 1.33 3.00

(b)

The side adjacent to φ = 3.00

(d)

sin φ =

4.00 = 0.800 5.00

Using the diagram at the right, the Pythagorean theorem yields 5.00 m

c = ( 5.00 m ) + ( 7.00 m ) = 8.60 m 2

2

c q 7.00 m

1.47

From the diagram given in Problem 1.46 above, it is seen that 5.00 = 0.714 7.00

tan θ = 1.48

(a) and (b) (c)

and θ = tan −1 ( 0.714 ) = 35.5°

See the Figure given at the right.

Applying the deﬁnition of the tangent function to the large right triangle containing the 12.0° angle gives: y x = tan 12.0°

[1]

Also, applying the deﬁnition of the tangent function to the smaller right triangle containing the 14.0° angle gives: y = tan 14.0° x − 1.00 km (d)

From Equation [1] above, observe that

[2] x = y tan 12.0°

Substituting this result into Equation [2] gives y ⋅ tan 12.0° = tan 14.0° y − (1.00 km ) tan 12.0° continued on next page

56157_01_ch01_p001-016.indd 12

10/15/10 9:53:57 AM

Introduction

13

Then, solving for the height of the mountain, y, yields y= 1.49

(1.00 km ) tan 12.0° tan 14.0° tan 14.0° − tan 12.0°

= 1.44 km = 1.44 × 10 3 m

Using the sketch at the right: w = tan 35.0° , or 100 m

w

w = (100 m ) tan 35.0° = 70.0 m 1.50

The ﬁgure at the right shows the situation described in the problem statement. Applying the deﬁnition of the tangent function to the large right triangle containing the angle θ in the Figure, one obtains y x = tan θ

[1]

Also, applying the deﬁnition of the tangent function to the small right triangle containing the angle φ gives y = tan φ x−d

[2]

Solving Equation [1] for x and substituting the result into Equation [2] yields y = tan φ y tan θ − d The last result simpliﬁes to

y ⋅ tan θ = tan φ y − d ⋅ tan θ

or

y ⋅ tan θ = y ⋅ tan φ − d ⋅ tan θ ⋅ tan φ

Solving for y: y ( tan θ − tan φ ) = −d ⋅ tan θ ⋅ tan φ y=− 1.51

(a)

or

d ⋅ tan θ ⋅ tan φ d ⋅ tan θ ⋅ tan φ = tan θ − tan φ tan φ − tan θ

Given that a ∝ F m , we have F ∝ ma . Therefore, the units of force are those of ma, [ F ] = [ ma] = [ m][a] = M ( L T 2 ) = M L T-2

1.52

56157_01_ch01_p001-016.indd 13

(b)

L [ F ] = M ⎛⎜ 2 ⎝T

(a)

1

(b)

vmax = 55

(c)

Δvmax = 65

⎞ = M⋅L ⎟ T2 ⎠

so

newton =

kg ⋅ m s2

mi ⎛ mi ⎞ ⎛ 1.609 km ⎞ km = ⎜1 ⎟⎜ ⎟ = 1.609 h ⎝ h ⎠ ⎝ 1 mi ⎠ h mi ⎛ mi ⎞ ⎛ 1.609 km h ⎞ km = ⎜ 55 ⎟⎜ ⎟ = 88 h ⎝ h ⎠ ⎝ 1 mi h ⎠ h mi mi ⎛ mi ⎞ ⎛ 1.609 km h ⎞ km − 55 = ⎜ 10 ⎟⎜ ⎟ = 16 h h ⎝ h ⎠ ⎝ 1 mi h ⎠ h

10/15/10 9:53:59 AM

14

1.53

Chapter 1

(a)

Since 1 m = 10 2 cm , then 1 m 3 = (1 m ) = (10 2 cm ) = (10 2 ) cm 3 = 10 6 cm 3, giving 3

3

⎛ 1.0 × 10 −3 kg ⎞ 3 mass = density volume = ⎜ ⎟ 1.0 m 3 ⎝ 1.0 cm ⎠

)(

(

(

)

3

)

⎛ 10 6 cm3 ⎞ ⎛ kg ⎞ 3 = ⎜ 1.0 × 10 −3 3 ⎟ 1.0 m 3 ⎜ ⎟ = 1.0 × 10 kg 3 ⎝ cm ⎠ ⎝ 1m ⎠

(

)

As a rough calculation, treat each of the following objects as if they were 100% water. (b)

3 kg 4 cell: mass = density × volume = ⎛⎜ 10 3 3 ⎞⎟ π ( 0.50 × 10 −6 m ) = 5.2 × 10 −16 kg m ⎠3 ⎝

(c)

3 4 kg 4 kidney: mass = density × volume = ρ ⎛⎜ π r 3 ⎞⎟ = ⎛⎜ 10 3 3 ⎞⎟ π ( 4.0 × 10 −2 m ) = 0.27 kg m ⎠3 ⎝3 ⎠ ⎝

(d)

ﬂy: mass = density × volume = ( density ) (π r 2 h ) kg = ⎛⎜ 10 3 3 m ⎝

1.54

⎞ π 1.0 × 10 −3 m 2 4.0 × 10 −3 m = 1.3 × 10 −5 kg )( ) ⎟ ( ⎠

Assume an average of 1 can per person each week and a population of 300 million. (a)

number cans person ⎞ number cans year = ⎛⎜ ⎟ ( population )( weeks year ) week ⎝ ⎠ ≈ ⎛⎜ 1 ⎝

can person ⎞ 8 ⎟ ( 3 × 10 people ) ( 52 weeks yr ) week ⎠

≈ 2 × 1010 cans yr , or ~10 10 cans yr (b)

number of tons = ( weight can )( number cans year ) ⎡ oz ⎞ ⎛ 1 lb ⎞ ⎛ 1 ton ⎞⎤ ⎛ 10 can ⎞ ≈ ⎢⎛⎜ 0.5 ⎟⎜ ⎟⎜ ⎟⎥ ⎜ 2 ×10 ⎟ can 16 oz 2 000 lb yr ⎠ ⎠⎝ ⎠⎝ ⎠⎦ ⎝ ⎣⎝ ≈ 3 × 10 5 ton yr , or ~10 5 ton yr Assumes an average weight of 0.5 oz of aluminum per can.

1.55

The term s has dimensions of L, a has dimensions of LT−2, and t has dimensions of T. Therefore, the equation, s = k a m t n with k being dimensionless, has dimensions of L = ( LT −2 ) ( T ) m

n

or

L1T 0 = L m T n−2 m

The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m =1 Likewise, equating powers of T, we see that n − 2 m = 0, or n = 2 m = 2 Dimensional analysis cannot determine the value of k , a dimensionless constant. 1.56

(a)

The rate of ﬁlling in gallons per second is rate =

30.0 gal ⎛ 1 min ⎞ −2 ⎜ ⎟ = 7.14 × 10 gal s 7.00 min ⎝ 60 s ⎠

continued on next page

56157_01_ch01_p001-016.indd 14

10/12/10 1:26:57 PM

Introduction

(b)

3 1L ⎞ Note that 1 m 3 = (10 2 cm ) = (10 6 cm 3 ) ⎛⎜ 3 = 10 3 L. Thus, 3 ⎟ ⎝ 10 cm ⎠

rate = 7.14 × 10 −2 (c)

1.57

gal ⎛ 3.786 L ⎞ ⎛ 1 m 3 ⎞ −4 3 ⎜ ⎟⎜ ⎟ = 2.70 × 10 m s s ⎝ 1 gal ⎠ ⎝ 10 3 L ⎠

Vfilled 1.00 m 3 = = 3.70 × 10 3 rate 2.70 × 10 −4 m 3 s

t=

⎛ 1h s⎜ ⎝ 3 600

⎞ ⎟ = 1.03 h s⎠

The volume of paint used is given by V = Ah, where A is the area covered and h is the thickness of the layer. Thus, h=

1.58

15

(a)

V 3.79 × 10 −3 m 3 = = 1.52 × 10 −4 m = 152 × 10 −6 m = 152 μ m 25.0 m 2 A

For a sphere, A = 4π R 2 . In this case, the radius of the second sphere is twice that of the ﬁrst, or R2 = 2 R1. A2 4π R 22 R 22 ( 2 R1 ) = = = = 4 A1 4π R 12 R 12 R12 2

Hence, (b)

For a sphere, the volume is

V=

4 3 πR 3

V2 ( 4 3) π R 32 R 32 ( 2 R1 ) = = = = 8 V1 ( 4 3) π R 13 R 13 R 13 3

Thus,

1.59

The estimate of the total distance cars are driven each year is d = ( cars in use ) ( distance traveled per car ) = (100 × 10 6 cars )(10 4 mi car ) = 1 × 1012 mi At a rate of 20 mi/gal, the fuel used per year would be V1 =

d 1 × 1012 mi = = 5 × 1010 gal rate1 20 mi gal

If the rate increased to 25 mi gal, the annual fuel consumption would be V2 =

d 1 × 1012 mi = = 4 × 1010 gal rate2 25 mi gal

and the fuel savings each year would be savings = V1 − V2 = 5 × 1010 gal − 4 × 1010 gal = 1 × 1010 gal 1.60

(a)

The amount paid per year would be dollars ⎞ ⎛ 8.64 × 10 4 s ⎞ ⎛ 365.25 days ⎞ 10 dollars annual amount = ⎛⎜ 1 000 ⎟⎜ ⎟⎜ ⎟ = 3.16 × 10 s ⎠ ⎝ 1.00 day ⎠ ⎝ yr yr ⎝ ⎠ Therefore, it would take

(b)

10 × 10 12 dollars = 3 × 10 2 yr, 3.16 × 10 10 dollars yr

or

~10 2 yr

The circumference of the Earth at the equator is

(

)

C = 2π r = 2π 6.378 × 10 6 m = 4.007 × 10 7 m

continued on next page

56157_01_ch01_p001-016.indd 15

10/12/10 1:26:57 PM

16

Chapter 1

The length of one dollar bill is 0.155 m, so the length of ten trillion bills is m ⎞ 12 12 = ⎛⎜ 0.155 ⎟ (10 × 10 dollars ) = 1×10 m. Thus, the ten trillion dollars would dollar ⎠ ⎝ encircle the Earth 1 × 1012 m n= = = 2 × 10 4 , or ~10 4 times C 4.007 × 10 7 m 1.61

(a) (b)

⎛ 365.2 days ⎞ ⎛ 8.64 × 10 4 s ⎞ 7 1 yr = (1 yr ) ⎜ ⎟⎟ ⎜ ⎟ = 3.16 × 10 s ⎜ y 1 yr 1 da ⎝ ⎠ ⎝ ⎠ Consider a segment of the surface of the Moon which has an area of 1 m2 and a depth of 1 m. When ﬁlled with meteorites, each having a diameter 10−6 m, the number of meteorites along each edge of this box is n=

length of an edge 1m = −6 = 10 6 meteorite diameter 10 m

The total number of meteorites in the ﬁlled box is then

( )

N = n 3 = 10 6 3 = 10 18 At the rate of 1 meteorite per second, the time to ﬁll the box is 1y ⎞ = 3 × 10 10 yr, or t = 1018 s = (1018 s ) ⎛⎜ ⎟ 7 ⎝ 3.16 × 10 s ⎠ 1.62

~1010 yr

We will assume that, on average, 1 ball will be lost per hitter, that there will be about 10 hitters per inning, a game has 9 innings, and the team plays 81 home games per season. Our estimate of the number of game balls needed per season is then number of balls needed = ( number lost per hitter ) ( number hitters/game )( home games/year ) ⎡⎛ games ⎞ hitters ⎞ ⎛ innings ⎞⎤ ⎛ = (1 ball per hitter ) ⎢⎜ 10 ⎟⎥ ⎜ 81 ⎟ ⎟ ⎜9 year ⎠ inning ⎠ ⎝ game ⎠⎦ ⎝ ⎣⎝ balls year

= 7300

1.63

~10 4

or

balls year

The volume of the Milky Way galaxy is roughly ⎛ πd2 ⎞ π 21 VG = At = ⎜ ⎟ t ≈ 4 10 m ⎝ 4 ⎠

(

) (10 m ) 2

19

orr VG ∼10 61 m3

If, within the Milky Way galaxy, there is typically one neutron star in a spherical volume of radius r = 3 × 1018 m, then the galactic volume per neutron star is V1 =

3 4 3 4 π r = π ( 3 × 1018 m ) = 1 × 10 56 m 3 3 3

or

V1 ∼ 10 56 m 3

The order of magnitude of the number of neutron stars in the Milky Way is then n=

56157_01_ch01_p001-016.indd 16

VG 10 61 m 3 ∼ V1 10 56 m 3

or

n ∼ 10 5 neutron stars

10/12/10 1:26:58 PM

2 Motion in One Dimension QUICK QUIZZES 1.

(a)

200 yd

2.

(a)

False. The car may be slowing down, so that the direction of its acceleration is opposite the direction of its velocity.

(b)

True. If the velocity is in the direction chosen as negative, a positive acceleration causes a decrease in speed.

(c)

True. For an accelerating particle to stop at all, the velocity and acceleration must have opposite signs, so that the speed is decreasing. If this is the case, the particle will eventually come to rest. If the acceleration remains constant, however, the particle must begin to move again, opposite to the direction of its original velocity. If the particle comes to rest and then stays at rest, the acceleration has become zero at the moment the motion stops. This is the case for a braking car—the acceleration is negative and goes to zero as the car comes to rest.

3.

(b)

0

(c)

0

The velocity-vs.-time graph (a) has a constant slope, indicating a constant acceleration, which is represented by the acceleration-vs.-time graph (e). Graph (b) represents an object whose speed always increases, and does so at an ever increasing rate. Thus, the acceleration must be increasing, and the acceleration-vs.-time graph that best indicates this behavior is (d). Graph (c) depicts an object which ﬁrst has a velocity that increases at a constant rate, which means that the object’s acceleration is constant. The motion then changes to one at constant speed, indicating that the acceleration of the object becomes zero. Thus, the best match to this situation is graph (f).

4.

Choice (b). According to graph b, there are some instants in time when the object is simultaneously at two different x-coordinates. This is physically impossible.

5.

(a)

The blue graph of Figure 2.14b best shows the puck’s position as a function of time. As seen in Figure 2.14a, the distance the puck has traveled grows at an increasing rate for approximately three time intervals, grows at a steady rate for about four time intervals, and then grows at a diminishing rate for the last two intervals.

(b)

The red graph of Figure 2.14c best illustrates the speed (distance traveled per time interval) of the puck as a function of time. It shows the puck gaining speed for approximately three time intervals, moving at constant speed for about four time intervals, then slowing to rest during the last two intervals.

17

56157_02_ch02_p017-048.indd 17

10/12/10 1:27:27 PM

18

Chapter 2

(c)

The green graph of Figure 2.14d best shows the puck’s acceleration as a function of time. The puck gains velocity (positive acceleration) for approximately three time intervals, moves at constant velocity (zero acceleration) for about four time intervals, and then loses velocity (negative acceleration) for roughly the last two time intervals.

6.

Choice (e). The acceleration of the ball remains constant while it is in the air. The magnitude of its acceleration is the free-fall acceleration, g = 9.80 m/s2.

7.

Choice (c). As it travels upward, its speed decreases by 9.80 m/s during each second of its motion. When it reaches the peak of its motion, its speed becomes zero. As the ball moves downward, its speed increases by 9.80 m/s each second.

8.

Choices (a) and (f). The ﬁrst jumper will always be moving with a higher velocity than the second. Thus, in a given time interval, the ﬁrst jumper covers more distance than the second, and the separation distance between them increases. At any given instant of time, the velocities of the jumpers are deﬁnitely different, because one had a head start. In a time interval after this instant, however, each jumper increases his or her velocity by the same amount, because they have the same acceleration. Thus, the difference in velocities stays the same.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

Once the arrow has left the bow, it has a constant downward acceleration equal to the freefall acceleration, g. Taking upward as the positive direction, the elapsed time required for the velocity to change from an initial value of 15.0 m s upward ( v0 = +15.0 m s ) to a value of 8.00 m s downward ( v f = −8.00 m s ) is given by Δt =

Δv v f − v0 −8.00 m s − ( +15.0 m s ) = = = 2.35 s a −g − 9.80 m s 2

Thus, the correct choice is (d). 2.

In Figure MCQ2.2, there are ﬁve spaces separating adjacent oil drops, and these spaces span a distance of Δx = 600 meters. Since the drops occur every 5.0 s, the time span of each space is 5.0 s and the total time interval shown in the ﬁgure is Δt = 5 ( 5.0 s ) = 25 s. The average speed of the car is then v=

Δx 600 m = = 24 m s Δt 25 s

making (b) the correct choice. 3.

56157_02_ch02_p017-048.indd 18

The derivation of the equations of kinematics for an object moving in one dimension (Equations 2.6 through 2.10 in the textbook) was based on the assumption that the object had a constant acceleration. Thus, (b) is the correct answer. An object having constant acceleration would have constant velocity only if that acceleration had a value of zero, so (a) is not a necessary condition. The speed (magnitude of the velocity) will increase in time only in cases when the velocity is in the same direction as the constant acceleration, so (c) is not a correct response. An object projected straight upward into the air has a constant acceleration. Yet its position (altitude) does not always increase in time (it eventually starts to fall back downward) nor is its velocity always directed downward (the direction of the constant acceleration). Thus, neither (d) nor (e) can be correct.

10/12/10 1:27:28 PM

Motion in One Dimension

4.

19

The bowling pin has a constant downward acceleration ( a = − g = −9.80 m s 2 ) while in ﬂight. The velocity of the pin is directed upward on the upward part of its ﬂight and is directed downward as it falls back toward the juggler’s hand. Thus, only (d) is a true statement.

5.

The initial velocity of the car is v0 = 0 and the velocity at time t is v. The constant acceleration is therefore given by a = Δv Δt = ( v − v0 ) t = ( v − 0 ) t = v t and the average velocity of the car is v = ( v + v0 ) 2 = ( v + 0 ) 2 = v 2. The distance traveled in time t is Δx = vt = vt 2. In the special case where a = 0 ( and hence v = v0 = 0 ) , we see that statements (a), (b), (c), and (d) are all correct. However, in the general case ( a ≠ 0, and hence v ≠ 0 ), only statements (b) and (c) are true. Statement (e) is not true in either case.

6.

The motion of the boat is very similar to that of a object thrown straight upward into the air. In both cases, the object has a constant acceleration which is directed opposite to the direction of the initial velocity. Just as the object thrown upward slows down and stops momentarily before it starts speeding up as it falls back downward, the boat will continue to move northward for some time, slowing uniformly until it comes to a momentary stop. It will then start to move in the southward direction, gaining speed as it goes. The correct answer is (c).

7.

In a position versus time graph, the velocity of the object at any point in time is the slope of the line tangent to the graph at that instant in time. The speed of the particle at this point in time is simply the magnitude (or absolute value) of the velocity at this instant in time. The displacement occurring during a time interval is equal to the difference in x-coordinates at the ﬁnal and initial times of the interval Δx = x t f − x ti .

)

(

The average velocity during a time interval is the slope of the straight line connecting the points on the curve corresponding to the initial and ﬁnal times of the interval ⎡ v = Δx Δt = ( x f − xi ) ( t f − ti ) ⎤ . Thus, we see how the quantities in choices (a), (e), (c), and (d) ⎣ ⎦ can all be obtained from the graph. Only the acceleration, choice (b), cannot be obtained from the position versus time graph. 8.

From Δx = v0 t + 12 at 2, the distance traveled in time t, starting from rest ( v0 = 0 ) with constant acceleration a, is Δx = 12 at 2 . Thus, the ratio of the distances traveled in two individual trials, one of duration t1 = 6 s and the second of duration t 2 = 2 s, is 2

2 Δx2 12 at 22 ⎛ t 2 ⎞ ⎛ 2 s ⎞ 1 = 1 2 =⎜ ⎟ =⎜ ⎟ = Δx1 2 at1 ⎝ t1 ⎠ ⎝ 6 s ⎠ 9 and the correct answer is (c).

9.

The distance an object moving at a uniform speed of v = 8.5 m s will travel during a time interval of Δt = 1 1 000 s = 1.0 × 10 −3 s is given by Δx = v ( Δt ) = (8.5 m s ) (1.0 × 10 −3 s ) = 8.5 × 10 −3 m = 8.5 mm so the only correct answer to this question is choice (d).

10.

Once either ball has left the student’s hand, it is a freely falling body with a constant acceleration a = − g (taking upward as positive). Therefore, choice (e) cannot be true. The initial velocities of the red and blue balls are given by viR = + v0 and viB = − v0 , respectively. The velocity of either ball when it has a displacement from the launch point of Δy = −h (where h is the height of the building) is found from v 2 = vi2 + 2a ( Δy ) as follows: vR = − viR2 + 2a ( Δy ) R = −

56157_02_ch02_p017-048.indd 19

( + v0 ) 2 + 2 ( − g ) ( − h ) = −

v02 + 2 gh

10/12/10 1:27:28 PM

20

Chapter 2

and vB = − viB2 + 2a ( Δy ) B = −

( − v0 ) 2 + 2 ( − g ) ( − h ) = −

v02 + 2 gh

Note that the negative sign was chosen for the radical in both cases since each ball is moving in the downward direction immediately before it reaches the ground. From this, we see that choice (c) is true. Also, the speeds of the two balls just before hitting the ground are vR = − v02 + 2 gh = v02 + 2 gh > v0 and vB = − v02 + 2 gh = v02 + 2 gh > v0 Therefore, vR = vB , so both choices (a) and (b) are false. However, we see that both ﬁnal speeds exceed the initial speed and choice (d) is true. The correct answer to this question is then (c) and (d). 11.

At ground level, the displacement of the rock from its launch point is Δy = −h , where h is the height of the tower and upward has been chosen as the positive direction. From v 2 = vo2 + 2a ( Δy ) , the speed of the rock just before hitting the ground is found to be v = ± v02 + 2a ( Δy ) = v02 + 2 ( − g ) ( −h ) =

(12 m s )2 + 2 ( 9.8 m s2 ) ( 40.0 m ) = 30 m s

Choice (b) is therefore the correct response to this question. 12.

Once the ball has left the thrower’s hand, it is a freely falling body with a constant, non-zero, acceleration of a = − g . Since the acceleration of the ball is not zero at any point on its trajectory, choices (a) through (d) are all false and the correct response is (e).

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

Yes. The particle may stop at some instant, but still have an acceleration, as when a ball thrown straight up reaches its maximum height.

4.

(a)

No. They can be used only when the acceleration is constant.

(b)

Yes. Zero is a constant.

(a)

In Figure (c), the images are farther apart for each successive time interval. The object is moving toward the right and speeding up. This means that the acceleration is positive in Figure (c).

(b)

In Figure (a), the ﬁrst four images show an increasing distance traveled each time interval and therefore a positive acceleration. However, after the fourth image, the spacing is decreasing, showing that the object is now slowing down (or has negative acceleration).

(c)

In Figure (b), the images are equally spaced, showing that the object moved the same distance in each time interval. Hence, the velocity is constant in Figure (b).

(a)

At the maximum height, the ball is momentarily at rest (i.e., has zero velocity). The acceleration remains constant, with magnitude equal to the free-fall acceleration g and directed downward. Thus, even though the velocity is momentarily zero, it continues to change, and the ball will begin to gain speed in the downward direction.

(b)

The acceleration of the ball remains constant in magnitude and direction throughout the ball’s free ﬂight, from the instant it leaves the hand until the instant just before it strikes the

6.

8.

56157_02_ch02_p017-048.indd 20

10/12/10 1:27:29 PM

Motion in One Dimension

21

ground. The acceleration is directed downward and has a magnitude equal to the freefall acceleration g. 10.

(a)

Successive images on the ﬁlm will be separated by a constant distance if the ball has constant velocity.

(b)

Starting at the right-most image, the images will be getting closer together as one moves toward the left.

(c)

Starting at the right-most image, the images will be getting farther apart as one moves toward the left.

(d)

As one moves from left to right, the balls will ﬁrst get farther apart in each successive image, then closer together when the ball begins to slow down.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a)

2 ×10 4 mi

(b)

Δx 2 RE = 2.4

4.

(a)

10.04 m s

(b)

7.042 m s

6.

(a)

5.00 m s

(b)

1.25 m s

(d)

−3.33 m s

(e)

0

(a)

+4.0 m s

(b)

−0.50 m s

(d)

0

10.

(a)

2.3 min

(b)

64 mi

12.

(a)

L t1

(b)

−L t 2

(d)

2 L ( t1 + t 2 )

14.

(a)

1.3 × 10 2 s

16.

(a) The trailing runner’s speed must be greater than that of the leader, and the leader’s distance from the ﬁnish line must be great enough to give the trailing runner time to make up the deﬁcient distance.

8.

(b) 18.

56157_02_ch02_p017-048.indd 21

(a)

(b)

t = d ( v1 − v2 )

(c)

(c)

−2.50 m s

(c)

−1.0 m s

(c)

0

13 m

d2 = v2 d ( v1 − v2 )

Some data points that can be used to plot the graph are as given below:

x (m)

5.75

16.0

35.3

68.0

119

192

t (s)

1.00

2.00

3.00

4.00

5.00

6.00

(b)

41.0 m s , 41.0 m s , 41.0 m s

(c)

17.0 m s , much smaller than the instantaneous vellocity at t = 4.00 s

10/12/10 1:27:29 PM

22

Chapter 2

20.

(a)

22.

0.391 s

24.

(i)

(a)

(ii)

(a)

(b)

263 m

0

(b)

1.6 m s 2

(c)

0.80 m s 2

0

(b)

1.6 m s 2

(c)

0

The curves intersect at t = 16.9 s.

500 x (m)

26.

20.0 m s , 5.00 m s

car

250

police officer

0 0

28.

a = 2.74 × 10 5 m s 2 = ( 2.79 × 10 4 ) g

30.

(a)

(b)

v 2f = vi2 + 2a ( Δx )

4.00 8.00 12.0 16.0 20.0 t (s)

(c)

a = ( v 2f − vi2 ) 2 ( Δx )

(d)

1.25 m s 2

(b)

13.5 m

(c)

13.5 m

(e)

8.00 s

(a)

13.5 m

(d)

22.5 m

34.

(a)

20.0 s

(b)

No, it cannot land safely on the 0.800 km runway.

36.

(a)

5.51 km

(b)

20.8 m s, 41.6 m s, 20.8 m s, 38.7 m s

38.

(a)

107 m

(b)

1.49 m s 2

40.

(a)

v = a1t1

(b)

Δx = 12 a1t12

(c)

Δxtotal = 12 a1t12 + a1t1t 2 + 12 a2 t 22

32.

42.

95 m

44.

29.1 s

46.

1.79 s

48.

(a) Yes.

(b)

vtop = 3.69 m s

(c)

Δv downward = 2.39 m s

(d) No, Δv upward = 3.71 m s. The two rocks have the same acceleration, but the rock thrown downward has a higher average speed between the two levels, and is accelerated over a smaller time interval.

56157_02_ch02_p017-048.indd 22

10/12/10 1:27:30 PM

Motion in One Dimension

50.

(a)

21.1 m s

(b)

19.6 m

52.

(a)

v = − v0 − gt = v0 + gt

(b)

d = 12 gt 2

(c)

v = v0 − gt , d = 12 gt 2

54.

(a)

29.4 m s

(b)

44.1 m

56.

(a)

−202 m s 2

(b)

198 m

58.

(a)

4.53 s

(b)

14.1 m s

60.

(a)

vi = h t + gt 2

(b)

v = h t − gt 2

62.

See Solutions Section for Motion Diagrams.

64.

Yes. The minimum acceleration needed to complete the 1 mile distance in the allotted time is amin = 0.032 m s 2 , considerably less than what she is capable of producing.

66.

(a)

y1 = h − v0 t − 12 gt 2 , y2 = h + v0 t − 12 gt 2

(c)

v1 f = v2 f = − v02 + 2 gh (d)

(c)

(b)

23

18.1 m s, 19.6 m

t 2 − t1 = 2 v0 g

y2 − y1 = 2 v0 t as long as both balls are still in the air.

68.

3.10 m s

70.

(a)

3.00 s

(c)

v1 = −31.4 m s, v2 = −34.8 m s

v0 ,2 = −15.2 m s

(b)

72.

(a)

2.2 s

(b)

−21 m s

74.

(a)

only if acceleration = 0

(b)

Yes, for all initial velocities and accelerations.

PROBLEM SOLUTIONS 2.1

We assume that you are approximately 2 m tall and that the nerve impulse travels at uniform speed. The elapsed time is then Δt =

2.2

(a)

Δx 2m = = 2 × 10 −2 s = 0.02 s v 100 m s

At constant speed, c = 3 × 108 m s, the distance light travels in 0.1 s is Δx = c ( Δt ) = ( 3 × 108 m s ) ( 0.1 s ) ⎛ 1 mi ⎞ ⎛ 1 km ⎞ 4 = ( 3 × 10 7 m ) ⎜ ⎟ = 2 × 10 mi ⎟⎜ 3 1.609 k m 10 m ⎝ ⎠⎝ ⎠

(b)

Comparing the result of part (a) to the diameter of the Earth, DE, we ﬁnd Δx Δx 3.0 × 10 7 m = = ≈ 2.4 DE 2 RE 2 ( 6.38 × 10 6 m )

56157_02_ch02_p017-048.indd 23

( with RE = Earth’s radius )

10/12/10 1:27:30 PM

24

2.3

Chapter 2

Distances traveled between pairs of cities are Δx1 = v1 ( Δt1 ) = (80.0 km h ) ( 0.500 h ) = 40.0 km Δx2 = v2 ( Δt 2 ) = (100 km h ) ( 0.200 h ) = 20.0 km Δx3 = v3 ( Δt3 ) = ( 40.0 km h ) ( 0.750 h ) = 30.0 km Thus, the total distance traveled is Δx = ( 40.0 + 20.0 + 30.0 ) km = 90.0 km, and the elapsed time is Δt = 0.500 h + 0.200 h + 0.750 h + 0.250 h = 1.70 h.

2.4

v=

(b)

Δx = 90.0 km (see above)

(a)

v=

(b) 2.5

(a)

(b)

2.6

Δx 90.0 km = = 52.9 km h Δt 1.70 h

(a)

Δx 2.000 × 10 2 m = = 10.04 m s Δt 19.92 s

⎛ 1.609 km ⎞ ⎛ 10 3 m ⎞ ⎜ ⎟⎜ ⎟ = 7.042 m s ⎝ 1 mi ⎠ ⎝ 1 km ⎠ Boat A requires 1.0 h to cross the lake and 1.0 h to return, total time 2.0 h. Boat B requires 2.0 h to cross the lake at which time the race is over. Boat A wins, being 60 km ahead of B when the race ends. v=

Δx 1.000 mi = Δt 228.5 s

Average velocity is the net displacement of the boat divided by the total elapsed time. The winning boat is back where it started, its displacement thus being zero, yielding an average velocity of zero .

The average velocity over any time interval is Δx x f − xi = Δt t f − ti Δx 10.0 m − 0 v= = = 5.00 m s Δt 2.00 s − 0 v= (a)

2.7

(b)

v=

Δx 5.00 m − 0 = = 1.25 m s Δt 4.00 s − 0

(c)

v=

Δx 5.00 m − 10.0 m = = − 2.50 m s Δt 4.00 s − 2.00 s

(d)

v=

Δx − 5.00 m − 5.00 m = = − 3.33 m s Δt 7.00 s − 4.00 s

(e)

v=

0−0 Δx x2 − x1 = = = 0 Δt t 2 − t1 8.00 s − 0

(a)

1h ⎞ Displacement = Δx = (85.0 km h ) ( 35.0 min ) ⎛⎜ ⎟ + 130 km = 180 km 60 . ⎝ 0 min ⎠

(b)

1h ⎞ The total elapsed time is Δt = ( 35.0 min + 15.0 min ) ⎛⎜ ⎟ + 2.00 h = 2.83 h ⎝ 60.0 min ⎠ so,

56157_02_ch02_p017-048.indd 24

v=

Δx 180 km = = 63.6 km h Δt 2.84 h

10/12/10 1:27:31 PM

Motion in One Dimension

2.8

25

The average velocity over any time interval is Δx x f − xi = Δt t f − ti Δx 4.0 m − 0 v= = = + 4.0 m s Δt 1.0 s − 0 Δx − 2 .0 m − 0 v= = = − 0.50 m s Δt 4.0 s − 0 v= (a) (b)

2.9

(c)

v=

Δx 0 − 4.0 m = = − 1.0 m s Δt 5.0 s − 1.0 s

(d)

v=

Δx 0−0 = = 0 Δt 5.0 s − 0

The plane starts from rest ( v0 = 0 ) and maintains a constant acceleration of a = +1.3 m s 2 . Thus, we ﬁnd the distance it will travel before reaching the required takeoff speed ( v = 75 m s ) , from v 2 = v02 + 2a ( Δx ) , as v 2 − v02 ( 75 m s ) − 0 = = 2.2 × 10 3 m = 2.2 km 2a 2 (1.3 m s 2 ) 2

Δx =

Since this distance is less than the length of the runway, the plane takes off safely. 2.10

(a)

The time for a car to make the trip is t =

Δx . Thus, the difference in the times for the two v

cars to complete the same 10 mile trip is Δt = t1 − t 2 = (b)

Δx Δx ⎛ 10 mi 10 mi ⎞ ⎛ 60 min ⎞ − =⎜ − ⎟ = 2.3 min ⎟⎜ v1 v2 ⎝ 55 mi h 70 mi h ⎠ ⎝ 1 h ⎠

When the faster car has a 15.0 min lead, it is ahead by a distance equal to that traveled by the slower car in a time of 15.0 min. This distance is given by Δx1 = v1 ( Δt ) = ( 55 mi h ) (15 min ). The faster car pulls ahead of the slower car at a rate of vrelative = 70 mi h − 55 mi h = 15 mi h Thus, the time required for it to get distance Δx1 ahead is Δt =

Δx1 = vrelative

( 55 mi h ) (15 min ) 15.0 mi h

= 55 min

Finally, the distance the faster car has traveled during this time is Δx2 = v2 ( Δt ) = 2.11

(a)

( 70 mi h ) ( 55 min ) ⎛⎜

1h ⎞ ⎟ = 64 mi ⎝ 60 min ⎠

From v 2f = vi2 + 2a ( Δx ) , with vi = 0 , v f = 72 km h , and ∆x = 45 m, the acceleration of the cheetah is found to be ⎡⎛ km ⎞ ⎛ 10 3 m ⎞ ⎛ 1 h ⎢⎜ 72 ⎟⎜ ⎟⎜ 2 2 h ⎠ ⎝ 1 km ⎠ ⎝ 3 600 v − vi ⎝ a= f =⎣ 2 ( Δx ) 2 ( 45 m )

2

⎞⎤ ⎟⎥ − 0 s ⎠⎦ = 4.4 m s 2

continued on next page

56157_02_ch02_p017-048.indd 25

10/12/10 1:27:31 PM

26

Chapter 2

(b)

The cheetah’s displacement 3.5 s after starting from rest is 1 1 2 Δx = vi t + at 2 = 0 + ( 4.4 m s 2 ) ( 3.5 s ) = 27 m 2 2

2.12

(a)

v1 =

( Δx )1 + L = = + L t1 ( Δt )1 t1

(b)

v2 =

( Δx )2 − L = = − L t2 ( Δt )2 t 2

(c) (d)

2.13

(a)

( Δx ) total ( Δx )1 + ( Δx )2 + L − L 0 = = = 0 = t1 + t 2 t1 + t 2 t1 + t 2 ( Δt ) total +L + −L total distance traveled ( Δx )1 + ( Δx )2 2L = = = ( ave. speed )trip = t1 + t 2 t1 + t 2 t1 + t 2 ( Δt ) total vtotal =

The total time for the trip is t total = t1 + 22 .0 min = t1 + 0.367 h , where t1 is the time spent traveling at v1 = 89.5 km h. Thus, the distance traveled is Δx = v1 t1 = vt total, which gives

(89.5 km h ) t1 = ( 77.8 km h ) ( t1 + 0.367 h ) = ( 77.8 km h ) t1 + 28.5 km (89.5 km h − 77.8 km h ) t1 = 28.5 km

or,

From which, t1 = 2 .44 h for a total time of t total = t1 + 0.367 h = 2.81 h (b)

The distance traveled during the trip is Δx = v1 t1 = vt total, giving Δx = v ttotal = ( 77.8 km h ) ( 2.81 h ) = 219 km

2.14

(a)

At the end of the race, the tortoise has been moving for time t and the hare for a time t − 2 .0 min = t − 120 s. The speed of the tortoise is vt = 0.100 m s, and the speed of the hare is vh = 20 vt = 2 .0 m s. The tortoise travels distance xt, which is 0.20 m larger than the distance xh traveled by the hare. Hence, xt = xh + 0.20 m which becomes or

vt t = vh ( t − 120 s ) + 0.20 m

( 0.100 m s ) t = ( 2 .0 m s ) ( t − 120 s ) + 0.20 m

This gives the time of the race as (b) 2.15

t = 1.3 × 10 2 s

xt = vt t = ( 0.100 m s ) (1.3 × 10 2 s ) = 13 m

The maximum allowed time to complete the trip is t total =

total distance 1600 m ⎛ 1 km h ⎞ = ⎜ ⎟ = 23.0 s required average speed 250 km h ⎝ 0.278 m s ⎠

The time spent in the ﬁrst half of the trip is t1 =

half distance 800 m ⎛ 1 km h ⎞ = ⎜ ⎟ = 12 .5 s v1 230 km h ⎝ 0.278 m s ⎠

continued on next page

56157_02_ch02_p017-048.indd 26

10/12/10 1:27:32 PM

Motion in One Dimension

27

Thus, the maximum time that can be spent on the second half of the trip is t 2 = t total − t1 = 23.0 s − 12 .5 s = 10.5 s and the required average speed on the second half is v2 = 2.16

⎛ 1 km h ⎞ half distance 800 m = = 76.2 m s ⎜ ⎟ = 274 km h t2 10.5 s ⎝ 0.278 m s ⎠

(a)

In order for the trailing athlete to be able to catch the leader, his speed (v1) must be greater than that of the leading athlete (v2), and the distance between the leading athlete and the ﬁnish line must be great enough to give the trailing athlete sufﬁcient time to make up the deﬁcient distance, d.

(b)

During a time t the leading athlete will travel a distance d2 = v2 t and the trailing athlete will travel a distance d1 = v1t . Only when d1 = d2 + d (where d is the initial distance the trailing athlete was behind the leader) will the trailing athlete have caught the leader. Requiring that this condition be satisﬁed gives the elapsed time required for the second athlete to overtake the ﬁrst: d1 = d2 + d giving

(c)

or

v1t − v2 t = d

v1t = v2 t + d or

t = d ( v1 − v2 )

In order for the trailing athlete to be able to at least tie for ﬁrst place, the initial distance D between the leader and the ﬁnish line must be greater than or equal to the distance the leader can travel in the time t calculated above (i.e., the time required to overtake the leader). That is, we must require that D ≥ d2 = v2 t = v2 ⎡⎣d ( v1 − v2 ) ⎤⎦

2.17

D≥

v2 d v1 − v2

The instantaneous velocity at any time is the slope of the x vs. t graph at that time. We compute this slope by using two points on a straight segment of the curve, one point on each side of the point of interest. (a)

vt=1.00 s =

(b)

vt=3.00 s =

(c) (d)

56157_02_ch02_p017-048.indd 27

or

10.0 m − 0 = 5.00 m s 2 .00 s − 0

( 5.00 − 10.0 ) m = − 2 .50 m s ( 4.00 − 2 .00 ) s ( 5.00 − 5.00 ) m vt=4.50 s = = 0 ( 5.00 − 4.00 ) s 0 − ( − 5.00 m ) vt=7.50 s = = 5.00 m s (8.00 − 7.00 ) s

10/12/10 1:27:32 PM

28

2.18

Chapter 2

(a)

A few typical values are t (s) 1.00 2.00 3.00 4.00 5.00 6.00

(b)

x (m) 5.75 16.0 35.3 68.0 119 192

We will use a 0.400 s interval centered at t = 400 s. We ﬁnd at t = 3.80 s, x = 60.2 m and at t = 4.20 s, x = 76.6 m. Therefore, v=

Δx 16.4 m = = 41.0 m s Δt 0.400 s

Using a time interval of 0.200 s, we ﬁnd the corresponding values to be: at t = 3.90 s, x = 64.0 m and at t = 4.10 s, x = 72.2 m. Thus, v=

Δx 8.20 m = = 41.0 m s Δt 0.200 s

For a time interval of 0.100 s, the values are: at t = 3.95 s, x = 66.0 m, and at t = 4.05 s, x = 70.1 m. Therefore, v= (c)

Δx 4.10 m = = 41.0 m s Δt 0.100 s

At t = 4.00 s, x = 68.0 m. Thus, for the ﬁrst 4.00 s, v=

Δx 68.0 m − 0 = = 17.0 m s Δt 4.00 s − 0

This value is much less than the instantaneous velocity at t = 4.00 s. 2.19

Choose a coordinate axis with the origin at the ﬂagpole and east as the positive direction. Then, using x = x0 + v0 t + 12 at 2 with a = 0 for each runner, the x-coordinate of each runner at time t is x A = − 4.0 mi + ( 6.0 mi h ) t

and

x B = 3.0 mi + ( − 5.0 mi h ) t

When the runners meet,

xA = xB

giving

− 4.0 mi + ( 6.0 mi h ) t = 3.0 mi + ( − 5.0 mi h ) t

or

( 6.0 mi h + 5.0 mi h ) t = 3.0 mi + 4.0 mi

This gives the elapsed time when they meet as t = (7.0 mi) (11.0 mi h) = 0.64 h. At this time, x A = x B = −0.18 mi. Thus, they meet 0.18 mi west of the flagpole .

56157_02_ch02_p017-048.indd 28

10/12/10 1:27:33 PM

Motion in One Dimension

2.20

From the ﬁgure at the right, observe that the motion of this particle can be broken into three distinct time intervals, during each of which the particle has a constant acceleration. These intervals and the associated accelerations are

ax(m/s2) 2 1 0 ⫺1

0 ≤ t < 10.0 s, a = a1 = +2.00 m s 2

5

10

15

t(s) 20

⫺2 ⫺3

10 ≤ t < 15.0 s, a = a2 = 0 15.0 ≤ t < 20.0 s, a = a3 = −3.00 m s 2

and (a)

29

Applying v f = vi + a ( Δt ) to each of the three time intervals gives for 0 ≤ t < 10.0 s,

v10 = v0 + a1 ( Δt1 ) = 0 + ( 2.00 m s 2 ) (10.0 s ) = 20.0 m s

for 10.0 s ≤ t < 15.0 s , v15 = v10 + a2 ( Δt 2 ) = 20.0 m s + 0 = 20.0 m s for 15.0 s ≤ t < 20.0 s, v20 = v15 + a3 ( Δt3 ) = 20.0 m s + ( −3.00 m s 2 ) ( 5.00 s ) = 5.00 m s (b)

Applying Δx = vi ( Δt ) + 12 a ( Δt ) to each of the time intervals gives 2

for 0 ≤ t < 10.0 s, 1 1 2 2 Δx1 = v0 Δt1 + a1 ( Δt1 ) = 0 + ( 2.00 m s 2 ) (10.0 s ) = 1.00 × 10 2 m 2 2 for 10.0 s ≤ t < 15.0 s, 1 2 Δx2 = v10 Δt 2 + a2 ( Δt 2 ) = ( 20.0 m s ) ( 5.00 s ) + 0 = 1.00 × 10 2 m 2 for 15.0 s ≤ t < 20.0 s , 1 2 Δx3 = v15 Δt3 + a3 ( Δt3 ) 2 = ( 20.0 m s ) ( 5.00 s ) +

1 ( −3.00 m s2 ) (5.00 s)2 = 62.5 m 2

Thus, the total distance traveled in the ﬁrst 20.0 s is Δxtotal = Δx1 + Δx2 + Δx3 = 100 m + 100 m + 62.5 m = 263 m 2.21

We choose the positive direction to point away from the wall. Then, the initial velocity of the ball is vi = −25.0 m s and the ﬁnal velocity is v f = +22.0 m s. If this change in velocity occurs over a time interval of Δt = 3.50 ms (i.e., the interval during which the ball is in contact with the wall), the average acceleration is a=

2.22

Δv v f − vi +22.0 m s − ( −25.0 m s ) = = = 1.34 × 10 4 m s 2 3.50 × 10 −3 s Δt Δt

From a = Δv Δt , the required time is Δt =

2.23

56157_02_ch02_p017-048.indd 29

From a =

⎞ ⎛ 0.447 m s ⎞ Δv ⎛ 60.0 mi h − 0 ⎞ ⎛ 1g =⎜ ⎟⎜ ⎟ = 0.391 s 2 ⎟ ⎜ a ⎝ 7g ⎠ ⎝ 9.80 m s ⎠ ⎝ 1 mi h ⎠ Δv Δv ( 60 − 55) mi h ⎛ 0.447 m s ⎞ , we have Δt = = ⎜ ⎟ = 3.7 s Δt a 0.60 m s 2 ⎝ 1 mi h ⎠

10/12/10 1:27:33 PM

30

2.24

Chapter 2

(i)

(a)

From t = 0 to t = 5.0 s, a=

(b)

2.25

(a)

−8.0 m s − ( −8.0 m s ) = 0 5.0 s − 0

8.0 m s − ( −8.0 m s ) = 1.6 m s 2 15 s − 5.0 s 8.0 m s − ( −8.0 m s ) = 0.80 m s 2 20 s − 0

At any instant, the instantaneous acceleration equals the slope of the line tangent to the v vs. t graph at that point in time. (a)

At t = 2.0 s, the slope of the tangent line to the curve is 0 .

(b)

At t = 10 s, the slope of the tangent line is 1.6 m s 2 .

(c)

At t = 18 s, the slope of the tangent line is 0 .

a= or

Δv 175 mi h − 0 = = 70.0 mi h ⋅ s Δt 2.5 s ⎛ mi ⎞ ⎛ 1609 m ⎞ ⎛ 1 h ⎞ 2 a = ⎜ 70.0 ⎟⎜ ⎟ = 31.3 m s ⎟⎜ h ⋅ s ⎠ ⎝ 1 mi ⎠ ⎝ 3600 s ⎠ ⎝

Alternatively,

(b)

=

From t = 0 to t = 20 s, a=

(ii)

t f − ti

From t = 5.0 s to t = 15 s, a=

(c)

v f − vi

⎞ m ⎛ 1g a = ⎛⎜ 31.3 2 ⎞⎟ ⎜ = 3.19 g 2 ⎟ s ⎠ ⎝ 9.80 m s ⎠ ⎝

If the acceleration is constant, Δx = v0 t + 12 at 2 1 m 2 Δx = 0 + ⎛⎜ 31.3 2 ⎞⎟ ( 2.50 s ) = 97.8 m 2⎝ s ⎠ or

As in the algebraic solution to Example 2.5, we let t represent the time the trooper has been moving.

500

We graph

250

x (m)

2.26

3.281 ft ⎞ Δx = ( 97.8 m ) ⎛⎜ ⎟ = 321 ft ⎝ 1m ⎠

car police officer

xcar = 24.0 m + ( 24.0 m s ) t 0

and

0

4.00 8.00 12.0 16.0 20.0 t (s)

xtrooper = (1.50 m s 2 ) t 2 The curves intersect at t = 16.9 s

56157_02_ch02_p017-048.indd 30

10/12/10 1:27:33 PM

Motion in One Dimension

2.27

31

Apply Δx = v0 + 12 at 2 to the 2.00-second time interval during which the object moves from xi = 3.00 cm to x f = − 5.00 cm. With v0 = 12.0 cm s, this yields an acceleration of 2 ⎡( x f − xi ) − v0 t ⎤⎦ 2 ⎡⎣( −5.00 − 3.00 ) cm − (12.0 cm s ) ( 2.00 s ) ⎤⎦ = a= ⎣ t2 ( 2.00 s )2 or

2.28

a = −16.0 cm s 2

From v 2 = v02 + 2 a ( Δx ), we have ( 10.97 × 10 3 m s ) 2 = 0 + 2 a ( 220 m ) so that a=

3 2 v 2 − v02 ( 10.97 × 10 m s ) − 0 = = 2.74 × 10 5 m s 2 2 ( Δx ) 2 ( 220 m )

⎞ ⎛ 1g = ( 2.74 × 10 5 m s 2 ) ⎜ = 2.79 × 10 4 times g! 2 ⎟ ⎝ 9.80 m s ⎠ 2.29

(a)

Δx = v ( Δt ) = ⎡⎣( v + v0 ) 2 ⎤⎦ Δt becomes ⎛ 2 .80 m s + v0 ⎞ 40.0 m = ⎜ ⎟ (8.50 s ) 2 ⎝ ⎠ which yields

(b) 2.30

2.31

a=

v0 =

2 ( 40.0 m ) − 2 .80 m s = 6.61 m s 8.50 s

v − v0 2 .80 m s − 6.61 m s = = − 0.448 m s 2 Δt 8.50 s

(a)

(b)

The known quantities are initial velocity, ﬁnal velocity, and displacement. The kinematics equation that relates these quantities to acceleration is v 2f = vi2 + 2a ( Δx )

(c)

a=

(d)

a=

(e)

Using a = Δv Δt , we ﬁnd that

(a)

With v = 120 km h , v 2 = v02 + 2a ( Δx ) yields

v 2f − vi2 2 ( Δx ) v 2f − vi2 2 ( Δx )

=

( 30.0 m s )2 − ( 20.0 m s )2 2 ( 2.00 × 10 2 m )

Δt =

= 1.25 m s 2

Δv v f − vi 30.0 m s − 20.0 m s = = = 8.00 s a a 1.25 m s 2

2 2 v 2 − v02 ⎡⎣(120 km h ) − 0 ⎤⎦ ⎛ 0.278 m s ⎞ 2 a= = ⎜ ⎟ = 2.32 m s 2 ( Δx ) 2 ( 240 m ) ⎝ 1 km h ⎠

(b)

56157_02_ch02_p017-048.indd 31

The required time is

Δt =

v − v0 (120 km h − 0 ) ⎛ 0.278 m s ⎞ = ⎜ ⎟ = 14.4 s 2 .32 m s 2 ⎝ 1 km h ⎠ a

10/12/10 1:27:34 PM

32

2.32

Chapter 2

(a)

From v 2f = vi2 + 2a ( Δx ), with vi = 6.00 m s and v f = 12.0 m s, we ﬁnd Δx =

(b)

v 2f − vi2 2a

=

(12.0 m s )2 − ( 6.00 m s )2 2 ( 4.00 m s 2 )

= 13.5 m

In this case, the object moves in the same direction for the entire time interval and the total distance traveled is simply the magnitude or absolute value of the displacement. That is, d = Δx = 13.5 m

(c)

Here, vi = −6.00 m s and v f = 12.0 m s, and we ﬁnd Δx =

(d)

v 2f − vi2 2a

= 13.5 m

[the same as in part (a)]

In this case, the object initially slows down as it travels in the negative x-direction, stops momentarily, and then gains speed as it begins traveling in the positive x-direction. We ﬁnd the total distance traveled by ﬁrst ﬁnding the displacement during each phase of this motion. While coming to rest ( vi = −6.00 m s , v f = 0 ), Δx1 =

v 2f − vi2 2a

=

( 0 )2 − ( −6.00 m s )2 2 ( 4.00 m s 2 )

= −4.50 m

After reversing direction ( vi = 0, v f = 12.0 m s ), Δx2 =

v 2f − vi2 2a

=

(12.0 m s )2 − ( 0 )2 2 ( 4.00 m s 2 )

= 18.0 m

Note that the net displacement is Δx = Δx1 + Δx2 = −4.50 m + 18.0 m = 13.5 m, as found in part (c) above. However, the total distance traveled in this case is d = Δx1 + Δx2 = −4.50 m + 18.0 m = 22.5 m 2.33

2.34

v − v0 24.0 m s 2 − 0 = = 8.14 m s 2 Δt 2.95 s

(a)

a=

(b)

From a = Δv Δt , the required time is Δt =

(c)

Yes. For uniform acceleration, the change in velocity ∆v generated in time ∆t is given by Δv = a ( Δt ). From this, it is seen that doubling the length of the time interval ∆t will always double the change in velocity ∆v. A more precise way of stating this is: “When acceleration is constant, velocity is a linear function of time.”

(a)

The time required to stop the plane is

(b)

The minimum distance needed to stop is

v f − vi

t=

a

=

20.0 m s − 10.0 m s = 1.23 s 8.14 m s 2

v − v0 0 − 100 m s = = 20.0 s a − 5.00 m s 2

⎛ v + v0 ⎞ ⎛ 0 + 100 m s ⎞ 20.0 s = 1000 m = 1.00 km Δ x = vt = ⎜ ) ⎟( ⎟ t=⎜ 2 ⎝ ⎠ ⎝ 2 ⎠ Thus, the plane requires a minimum runway length of 1.00 km. It cannot land safely on a 0.800 km runway.

56157_02_ch02_p017-048.indd 32

10/12/10 1:27:34 PM

Motion in One Dimension

2.35

33

We choose x = 0 and t = 0 at the location of Sue’s car when she ﬁrst spots the van and applies the brakes. Then, the initial conditions for Sue’s car are x0 S = 0 and v0 S = 30.0 m s. Her constant acceleration for t ≥ 0 is aS = −2.00 m s 2. The initial conditions for the van are x0V = 155 m, v0V = 5.00 m s , and its constant acceleration is aV = 0. We then use Δx = x − x0 = v0 t + 12 at 2 to write an equation for the x-coordinate of each vehicle for t ≥ 0. This gives 1 ( −2.00 m s2 ) t 2 2 1 xV − 155 m = ( 5.00 m s ) t + ( 0 ) t 2 2

xS − 0 = ( 30.0 m s ) t +

Sue’s Car: Van:

or

xS = ( 30.0 m s ) t − (1.00 m s 2 ) t 2

or

xV = 155 m + ( 5.00 m s ) t

In order for a collision to occur, the two vehicles must be at the same location ( i.e., xS = xV ). Thus, we test for a collision by equating the two equations for the x-coordinates and see if the resulting equation has any real solutions.

( 30.0 m s ) t − (1.00 m s2 ) t 2 = 155 m + ( 5.00 m s ) t

⇒

xS = xV

(1.00

or

m s 2 ) t 2 − ( 25.00 m s ) + 155 m = 0

Using the quadratic formula yields t=

− ( −25.00 m s ) ±

( −25.00 m s )2 − 4 (1.00 m s2 ) (155 m ) 2 (1.00 m s 2 )

= 13.6 s or 11.4 s

The solutions are real, not imaginary, so a collision will occur . The smaller of the two solutions is the collision time. (The larger solution tells when the van would pull ahead of the car again if the vehicles could pass harmlessly through each other.) The x-coordinate where the collision occurs is given by xcollision = xS 2.36

t =11.4 s

= xV

t =11.4 s

= 155 m + ( 5.00 m s ) (11.4 s ) = 212 m

The velocity at the end of the ﬁrst interval is v = v0 + at = 0 + (2 .77 m s) (15.0 s ) = 41.6 m s This is also the constant velocity during the second interval and the initial velocity for the third interval. Also, note that the duration of the second interval is t 2 = ( 2.05 min ) ( 60.0 s 1 min ) = 123 s. (a)

1 From Δx = v0 t + at 2, the total displacement is 2

( Δx ) total = ( Δx )1 + ( Δx )2 + ( Δx )3 1 2 = ⎡⎢0 + ( 2 .77 m s 2 ) (15.0 s ) ⎤⎥ + ⎡⎣( 41.6 m s ) (123 s ) + 0 ⎤⎦ 2 ⎣ ⎦ 1 2 + ⎡⎢( 41.6 m s ) ( 4.39 s ) + ( − 9.47 m s 2 ) ( 4.39 s ) ⎤⎥ 2 ⎣ ⎦ or ( Δx ) total = 312 m + 5.12 × 10 3 m + 91.4 m = 5.52 × 10 3 m = 5. 52 km

continued on next page

56157_02_ch02_p017-048.indd 33

10/12/10 1:27:35 PM

34

Chapter 2

(b)

v1 = v2 = v = 3

( Δx )1 t1

( Δx )2 t2

( Δx )3 t3

=

312 m = 20.8 m s 15.0 s

=

5.12 × 10 3 m = 41.6 m s 123 s

=

91.4 m = 20.8 m s , and the average velocity for the 4.39 s

total trip is vtotal = 2.37

( Δx )total t total

=

5.52 ×10 3 m = 38.8 m s (15.0 + 123 + 4.39 ) s

Using the uniformly accelerated motion equation Δx = v0 t + 12 at 2 for the full 40 s interval yields 2 Δx = ( 20 m s ) ( 40 s ) + 12 ( −1.0 m s 2 ) ( 40 s ) = 0 , which is obviously wrong. The source of the error is found by computing the time required for the train to come to rest. This time is v − v0 0 − 20 m s = = 20 s a −1.0 m s 2 Thus, the train is slowing down for the ﬁrst 20 s and is at rest for the last 20 s of the 40 s interval. t=

The acceleration is not constant during the full 40 s. It is, however, constant during the ﬁrst 20 s as the train slows to rest. Application of Δx = v0 t + 12 at 2 to this interval gives the stopping distance as Δx = ( 20 m s ) ( 20 s ) + 2.38

1 ( −1.0 m s2 ) ( 20 s )2 = 200 m 2

mi ⎞ ⎛ 0.447 m s ⎞ v0 = 0 and v f = ⎛⎜ 40.0 ⎟⎜ ⎟ = 17.9 m s h ⎠ ⎝ 1 mi h ⎠ ⎝ (a) To ﬁnd the distance traveled, we use ⎛ v + v0 ⎞ ⎛ 17.9 m s + 0 ⎞ 12 .0 s = 107 m Δx = v t = ⎜ f ) ⎟( ⎟ t= ⎜ 2 2 ⎝ ⎠ ⎝ ⎠ (b)

2.39

The constant acceleration is

a=

v f − v0 t

=

17.9 m s − 0 = 1.49 m s 2 12.0 s

At the end of the acceleration period, the velocity is v = v0 + ataccel = 0 + (1.5 m s 2 ) ( 5.0 s ) = 7.5 m s This is also the initial velocity for the braking period. v f = v + at brake = 7.5 m s + ( −2.0 m s 2 ) ( 3.0 s ) = 1.5 m s

(a)

After braking,

(b)

The total distance traveled is v + v0 ⎞ ⎛ vf + v ⎞ Δxtotal = ( Δx )accel + (Δ x ) brake = ( v t )accel + (v t ) brake = ⎛⎜ ⎟ taccel + ⎜ ⎟ t brake ⎝ 2 ⎠ ⎝ 2 ⎠ 7.5 m s + 0 ⎞ ⎛ 1.5 m s + 7.5 m s ⎞ 3.0 s = 32 m Δxtotal = ⎛⎜ ) ⎟ ( 5.0 s ) + ⎜ ⎟( 2 2 ⎝ ⎠ ⎝ ⎠

2.40

For the acceleration period, the parameters for the car are: initial velocity = via = 0, acceleration = aa = a1, elapsed time = ( Δt )a = t1, and final velocity = v fa. For the braking period, the parameters are: initial velocity = vib = final velocity of acceleration period = v fa, acceleration = ab = a2, and elapsed time = ( Δt )b = t 2.

continued on next page

56157_02_ch02_p017-048.indd 34

10/12/10 1:27:35 PM

Motion in One Dimension

(a)

To determine the velocity of the car just before the brakes are engaged, we apply v f = vi + a ( Δt ) to the acceleration period and ﬁnd vib = v fa = via + aa ( Δt )a = 0 + a1t1

(b)

vib = a1t1

or

We may use Δx = vi ( Δt ) + 12 a ( Δt ) to determine the distance traveled during the acceleration period (i.e., before the driver begins to brake). This gives 2

1 2

( Δx )a = via ( Δt )a + 12 aa ( Δt )a2 = 0 + a1t12 (c)

35

( Δx )a =

or

1 2 a1t1 2

The displacement occurring during the braking period is 1 2

( Δx )b = vib ( Δt )b + 12 ab ( Δt )b2 = ( a1t1 ) t 2 + a2t 22 Thus, the total displacement of the car during the two intervals combined is

( Δx ) total = ( Δx )a + ( Δx )b = 2.41

1 2 1 a1t1 + a1t1t 2 + a2 t 22 2 2

The time the Thunderbird spends slowing down is Δt1 =

2 ( 250 m ) Δx1 2 ( Δx1 ) = = 6.99 s = v1 v + v0 0 + 71.5 m s

The time required to regain speed after the pit stop is Δt 2 =

2 ( 350 m ) Δx2 2 ( Δx2 ) = = 9.79 s = v2 v + v0 71.5 m s + 0

Thus, the total elapsed time before the Thunderbird is back up to speed is Δt = Δt1 + 5.00 s + Δt 2 = 6.99 s + 5.00 s + 9.79 s = 21.8 s During this time, the Mercedes has traveled (at constant speed) a distance Δx M = v0 ( Δt ) = ( 71.5 m s ) ( 21.8 s ) = 1 559 m and the Thunderbird has fallen behind a distance d = Δx M − Δx1 − Δx2 = 1 559 m − 250 m − 350 m = 959 m 2.42

The car is distance d from the dog and has initial velocity v0 when the brakes are applied, giving it a constant acceleration a. Apply v = Δx Δt = ( v f + v0 ) 2 to the entire trip (for which Δx = d + 4.0 m, Δt = 10 s, and v f = 0) to obtain d + 4.0 m 0 + v0 = 10 s 2

or

v0 =

d + 4.0 m 5.0 s

[1]

Then, applying v 2f = v02 + 2a ( Δx ) to the entire trip yields 0 = v02 + 2a ( d + 4.0 m ). Substitute for v0 from Equation [1] to ﬁnd that 0=

( d + 4.0 m )2 25 s 2

+ 2a ( d + 4.0 m )

and

a=−

d + 4.0 m 50 s 2

[2] continued on next page

56157_02_ch02_p017-048.indd 35

10/15/10 9:54:54 AM

36

Chapter 2

Finally, apply Δx = v0 t + 12 at 2 to the ﬁrst 8.0 s of the trip (for which Δx = d ). d = v0 (8.0 s ) + 12 a ( 64 s 2 )

This gives

[3]

Substitute Equations [1] and [2] into Equation [3] to obtain ⎛ d + 4.0 m ⎞ 1 ⎛ d + 4.0 m ⎞ 2 d =⎜ ⎟ (8.0 s ) + ⎜ − ⎟ ( 64 s ) = 0.96d + 3.8 m 2⎝ 50 s 2 ⎠ ⎝ 5.0 s ⎠ d = 3.8 m 0.04 = 95 m

which yields 2.43

(a)

Take t = 0 at the time when the player starts to chase his opponent. At this time, the opponent is distance d = (12 m s ) ( 3.0 s ) = 36 m in front of the player. At time t > 0, the displacements of the players from their initial positions are 1 1 Δxplayer = ( v0 )player t + aplayer t 2 = 0 + ( 4.0 m s 2 ) t 2 2 2 1 2 Δxopponent = ( v0 )opponent t + aopponent t = (12 m s ) t + 0 2

and

When the players are side-by-side, Δxplayer = Δxopponent + 36 m

[1] [2] [3]

Substituting Equations [1] and [2] into Equation [3] gives 1 ( 4.0 m s2 ) t 2 = (12 m s ) t + 36 m 2

or

t 2 + ( − 6.0 s ) t + ( −18 s 2 ) = 0

Applying the quadratic formula to this result gives t=

− ( − 6.0 s ) ±

( − 6.0 s )2 − 4 (1) ( −18 s2 ) 2 (1)

which has solutions of t = −2.2 s and t = +8.2 s. Since the time must be greater than zero, we must choose t = 8.2 s as the proper answer. (b) 2.44

1 1 2 Δxplayer = ( v0 )player t + aplayer t 2 = 0 + ( 4.0 m s 2 ) (8..2 s ) = 1.3 × 10 2 m 2 2

The initial velocity of the train is v0 = 82 .4 km h and the ﬁnal velocity is v = 16.4 km h. The time required for the 400 m train to pass the crossing is found from Δ x = v t = [ (v + v0 ) 2] t as t=

2.45

(a)

2 ( Δx ) 2 ( 0.400 km ) 3600 s ⎞ = = (8.10 × 10 −3 h ) ⎛⎜ ⎟ = 29.1 s v + v0 (82 .4 + 16.4 ) km h ⎝ 1h ⎠

From v 2 = v02 + 2a ( Δy ) with v = 0, we have

( Δy )max =

v 2 − v02 0 − ( 25.0 m s ) = = 31.9 m 2a 2 ( −9.80 m s 2 ) 2

(b)

The time to reach the highest point is

(c)

v − v0 0 − 25.0 m s = 2.55 s = a − 9.80 m s 2 The time required for the ball to fall 31.9 m, starting from rest, is found from t up =

1 Δy = ( 0 ) t + at 2 as t = 2

2 ( Δy ) = a

2 ( − 31.9 m ) = 2.55 s − 9.80 m s 2

continued on next page

56157_02_ch02_p017-048.indd 36

10/15/10 9:54:55 AM

Motion in One Dimension

(d)

37

The velocity of the ball when it returns to the original level (2.55 s after it starts to fall from rest) is v = v0 + at = 0 + ( − 9.80 m s 2 ) ( 2 .55 s ) = − 25.0 m s

2.46

We take upward as the positive y-direction and y = 0 at the point where the ball is released. Then, v0 y = −8.00 m s , ay = − g = −9.80 m s 2, and Δy = −30.0 m when the ball reaches the ground. From vy2 = v02 y + 2ay ( Δy ) , the velocity of the ball just before it hits the ground is vy = − v02 y + 2ay ( Δy ) = −

(8.00 m s )2 + 2 ( −9.80 m s2 ) ( −30.0 m ) = −25.5 m s

Then, vy = v0 y + ay t gives the elapsed time as t= 2.47

(a)

v y − v0 y ay

=

−25.5 m s − ( −8.00 m s ) = 1.79 s −9.80 m s 2

The velocity of the object when it was 30.0 m above the ground can be determined by applying Δy = v0 t + 12 at 2 to the last 1.50 s of the fall. This gives 1 m 2 −30.0 m = v0 (1.50 s ) + ⎛⎜ −9.80 2 ⎞⎟ (1.50 s ) 2⎝ s ⎠

(b)

or

v0 = −12.7 m s

The displacement the object must have undergone, starting from rest, to achieve this velocity at a point 30.0 m above the ground is given by v 2 = v02 + 2a ( Δy ) as

( Δy )1 =

v 2 − v02 ( −12.7 m s ) − 0 = = −8.23 m 2a 2 ( −9.80 m s 2 ) 2

The total distance the object drops during the fall is then

( Δy )total = (−8.23 m ) + (− 30.0 m ) = 38.2 m 2.48

(a)

Consider the rock’s entire upward ﬂight, for which v0 = +7.40 m s, v f = 0 , a = − g = −9.80 m s 2, yi = 1.55 m (taking y = 0 at ground level), and y f = hmax = maximum altitude reached by rock. Then applying v 2f = vi2 + 2a ( Δy ) to this upward ﬂight gives 0 = ( 7.40 m s ) + 2 ( −9.80 m s 2 ) ( hmax − 1.55 m ) 2

Solving for the maximum altitude of the rock gives hmax = 1.55 m+

( 7.40 m s )2

2 ( 9.80 m s 2 )

= 4.34 m

Since hmax > 3.65 m ( height of the wall ), the rock does reach the top of the wall . (b)

To ﬁnd the velocity of the rock when it reaches the top of the wall, we use v 2f = vi2 + 2a ( Δy ) and solve for v f when y f = 3.65 m (starting with vi = +7.40 m s at yi = 1.55 m). This yields v f = vi2 + 2a ( y f − yi ) =

( 7.40 m s )2 + 2 ( −9.80 m s2 ) ( 3.65 m − 1.55 m ) = 3.69 m s

continued on next page

56157_02_ch02_p017-048.indd 37

10/12/10 1:27:37 PM

38

Chapter 2

(c)

A rock thrown downward at a speed of 7.40 m s ( vi = −7.40 m s ) from the top of the wall undergoes a displacement of ( Δy ) = y f − yi = 1.55 m − 3.65 m = −2.10 m before reaching the level of the attacker. Its velocity when it reaches the attacker is v f = − vi2 + 2a ( Δy ) = −

( −7.40 m s )2 + 2 ( −9.80 m s2 ) ( −2.10 m ) = −9.79 m s

so the change in speed of this rock as it goes between the 2 points located at the top of the wall and the attacker is given by Δ ( speed )down = v f − vi = −9.79 m s − −7.40 m s = 2.39 m s (d)

Observe that the change in speed of the ball thrown upward as it went from the attacker to the top of the wall was Δ ( speed ) up = v f − vi = 3.69 m s − 7.40 m s = 3.71 m s The two rocks do not undergo the same magnitude speed change. The rocks have the same acceleration, but the rock thrown downward has a higher average speed between the two levels, and is accelerated over a smaller time interval.

2.49

The velocity of the child’s head just before impact (after falling a distance of 0.40 m, starting from rest) is given by v 2 = v02 + 2a ( Δy ) as vI = − v02 + 2a ( Δy ) = − 0 + 2 ( −9.8 m s 2 ) ( −0.40 m ) = −2.8 m s If, upon impact, the child’s head undergoes an additional displacement Δy = −h before coming to rest, the acceleration during the impact can be found from v 2 = v02 + 2a ( Δy ) to be a = ( 0 − vI2 ) 2 ( −h ) = vI2 2h . The duration of the impact is found from v = v0 + at as t = Δv a = − vI ( vI2 2h ), or t = −2h vI . Applying these results to the two cases yields vI2 ( −2.8 m s ) = = 2.0 × 10 3 m s 2 2h 2 ( 2.0 × 10 −3 m ) 2

Hardwood Floor (h = 2.0 × 10 −3 m): a =

and

t=

−3 −2h −2 ( 2.0 × 10 m ) = 1.4 × 10 −3 s = 1.4 ms = vI −2.8 m s

vI2 ( −2.8 m s ) = = 3.9 × 10 2 m s 2 2h 2 (1.0 × 10 −2 m ) 2

Carpeted Floor (h = 1.0 × 10 −2 m): a =

and 2.50

(a)

t=

−2 −2h −2 (1.0 × 10 m ) = = 7.1 × 10 −3 s = 7.1 ms vI −2.8 m s

After 2.00 s, the velocity of the mailbag is vbag = v0 + at = −1.50 m s + ( −9.80 m s 2 ) ( 2 .00 s ) = − 21.1 m s The negative sign tells us that the bag is moving downward and the magnitude of the velocity gives the speed as 21.1 m s .

(b)

The displacement of the mailbag after 2.00 s is

( Δy )bag = ⎛⎜ ⎝

v + v0 2

⎞ ⎡ −21.1 m s + ( −1.50 m s ) ⎤ ⎟t = ⎢ ⎥ ( 2 .00 s ) = − 22 .6 m 2 ⎠ ⎣ ⎦

continued on next page

56157_02_ch02_p017-048.indd 38

10/12/10 1:27:38 PM

Motion in One Dimension

39

During this time, the helicopter, moving downward with constant velocity, undergoes a displacement of 1 2

( Δy )copter = v0 t + at 2 = ( −1.5 m s )( 2 .00 s ) + 0 = − 3.00 m The distance separating the package and the helicopter at this time is then d = ( Δy ) p − ( Δy )h = −22.6 m − ( −3.00 m ) = −19.6 m = 19.6 m (c)

Here, ( v0 ) bag = ( v0 )copter = + 1.50 m s and abag = −9.80 m s 2 while acopter = 0. After 2.00 s, the velocity of the mailbag is vbag = 1.50

m ⎛ m m m + ⎜ −9.80 2 ⎞⎟ ( 2.00 s ) = − 18.1 and its speed is vbag = 18.1 s ⎝ s ⎠ s s

In this case, the displacement of the helicopter during the 2.00 s interval is Δycopter = ( +1.50 m s ) ( 2.00 s ) + 0 = +3.00 m Meanwhile, the mailbag has a displacement of ⎛ vbag + v0 2 ⎝

( Δy )bag = ⎜

⎞ ⎡ −18.1 m s + 1.50 m s ⎤ ⎟t = ⎢ ⎥⎦ ( 2 .00 s ) = −16.6 m 2 ⎠ ⎣

The distance separating the package and the helicopter at this time is then d = ( Δy ) p − ( Δy )h = −16.6 m − ( +3.00 m ) = −19.6 m = 19.6 m 2.51

(a)

From the instant the ball leaves the player’s hand until it is caught, the ball is a freely falling body with an acceleration of a = − g = −9.80 m s 2 = 9.80 m s 2 ( downward )

(b)

At its maximum height, the ball comes to rest momentarily and then begins to fall back downward. Thus, vmax = 0 . height

(c)

Consider the relation Δy = v0 t + 12 at 2 with a = − g . When the ball is at the thrower’s hand, the displacement is Δy = 0, giving 0 = v0 t − 12 gt 2 This equation has two solutions, t = 0, which corresponds to when the ball was thrown, and t = 2 v0 g, corresponding to when the ball is caught. Therefore, if the ball is caught at t = 2 .00 s, the initial velocity must have been 2 gt ( 9.80 m s ) ( 2 .00 s ) v0 = = = 9.80 m s 2 2

(d)

From v 2 = v02 + 2a ( Δy ), with v = 0 at the maximum height,

( Δy )max =

56157_02_ch02_p017-048.indd 39

v 2 − v02 0 − ( 9.80 m s ) = = 4.90 m 2a 2 ( − 9.80 m s 2 ) 2

10/12/10 1:27:38 PM

40

2.52

Chapter 2

(a)

Let t = 0 be the instant the package leaves the helicopter, so the package and the helicopter have a common initial velocity of vi = − v0 ( choosing upward as positive ) . At times t > 0, the velocity of the package (in free-fall with constant acceleration a p = − g) is given by v = vi + at as v p = − v0 − gt = − ( v0 + gt ) and speed = v p = v0 + gt .

(b)

After an elapsed time t, the downward displacement of the package from its point of release will be

( Δy ) p = vi t + a p t 2 = − v0 t − gt 2 = − ⎛⎜ v0 t + gt 2 ⎞⎟ 2 2 2 1

1

1

⎝

⎠

and the downward displacement of the helicopter (moving with constant velocity, or acceleration ah = 0) from the release point at this time is 1 2

( Δy )h = vi t + ah t 2 = − v0 t + 0 = − v0 t The distance separating the package and the helicopter at this time is then

(c)

1 1 d = ( Δy ) p − ( Δy )h = − ⎛⎜ v0 t + gt 2 ⎞⎟ − ( − v0 t ) = gt 2 2 2 ⎝ ⎠ If the helicopter and package are moving upward at the instant of release, then the common initial velocity is vi = + v0. The accelerations of the helicopter (moving with constant velocity) and the package (a freely falling object) remain unchanged from the previous case ( a p = − g and ah = 0 ). In this case, the package speed at time t > 0 is v p = vi + a p t = v0 − gt . At this time, the displacements from the release point of the package and the helicopter are given by 1 2

1 2

( Δy ) p = vi t + a p t 2 = v0 t − gt 2

and

1 2

( Δy )h = vi t + ah t 2 = v0 t + 0 = + v0 t

The distance separating the package and helicopter at time t is now given by d = ( Δy ) p − ( Δy )h = v0 t − 2.53

1 2 1 gt − v0 t = gt 2 2 2

(the same as earlier!)

(a)

After its engines stop, the rocket is a freely falling body. It continues upward, slowing under the inﬂuence of gravity until it comes to rest momentarily at its maximum altitude. Then it falls back to Earth, gaining speed as it falls.

(b)

When it reaches a height of 150 m, the speed of the rocket is v = v02 + 2a ( Δy ) =

( 50.0 m s )2 + 2 ( 2 .00 m s2 ) (150 m ) = 55.7 m s

After the engines stop, the rocket continues moving upward with an initial velocity of v0 = 55.7 m s and acceleration a = − g = −9.80 m s 2 . When the rocket reaches maximum height, v = 0. The displacement of the rocket above the point where the engines stopped (that is, above the 150 m level) is v 2 − v02 0 − ( 55.7 m s ) = = 158 m 2a 2 ( − 9.80 m s 2 ) 2

Δy =

The maximum height above ground that the rocket reaches is then given by hmax = 150 m + 158 m = 308 m continued on next page

56157_02_ch02_p017-048.indd 40

10/12/10 1:27:39 PM

Motion in One Dimension

(c)

41

The total time of the upward motion of the rocket is the sum of two intervals. The ﬁrst is the time for the rocket to go from v0 = 50.0 m s at the ground to a velocity of v = 55.7 m s at an altitude of 150 m. This time is given by t1 =

( Δy )1 v1

=

( Δy )1

( v + v0 )

2

=

2 (150 m ) = 2 .84 s ( 55.7 + 50.0 ) m s

The second interval is the time to rise 158 m starting with v0 = 55.7 m s and ending with v = 0. This time is t2 =

( Δy )2 v2

=

( Δy )2

( v + v0 )

2

=

2 (158 m ) = 5.67 s 0 + 55.7 m s

The total time of the upward ﬂight is then (d)

The time for the rocket to fall 308 m back to the ground, with v0 = 0 and acceleration 1 a = − g = −9.80 m s 2 , is found from Δy = v0 t + at 2 as 2 t down =

2 ( Δy ) = a

2 ( − 308 m ) = 7.93 s − 9.80 m s 2

so the total time of the ﬂight is 2.54

(a)

t up = t1 + t 2 = ( 2 .84 + 5.67 ) s = 8.51 s

t flight = t up + t down = (8.51 + 7.93) s = 16.4 s

For the upward ﬂight of the ball, we have vi = v0 , v f = 0, a = − g , and Δt = 3.00 s. Thus, v f = vi + a ( Δt ) gives the initial velocity as vi = v f − a ( Δt ) = v f + g ( Δt ) or v0 = 0 + ( 9.80 m s 2 ) ( 3.00 s ) = +29.4 m s

(b)

The vertical displacement of the ball during this 3.00-s upward ﬂight is ⎛ vi + v f ⎝ 2

( Δy )max = h = v ( Δt ) = ⎜ 2.55

⎞ ⎛ 29.4 m s + 0 ⎞ 3.00 s = 44.1 m ) ⎟( ⎟ ( Δt ) = ⎜ 2 ⎝ ⎠ ⎠

During the 0.600 s required for the rig to pass completely onto the bridge, the front bumper of the tractor moves a distance equal to the length of the rig at constant velocity of v = 100 km h. Therefore the length of the rig is ⎡ km ⎛ 0.278 m s ⎞⎤ Lrig = vt = ⎢100 ⎜ ⎟⎥ ( 0.600 s ) = 16.7 m h ⎝ 1 km h ⎠⎦ ⎣ While some part of the rig is on the bridge, the front bumper moves a distance Δx = Lbridge + Lrig = 400 m + 16.7 m . With a constant velocity of v = 100 km h, the time for this to occur is t=

2.56

(a)

Lbridge + Lrig v

=

400 m + 16.7 m ⎛ 1 km h ⎞ ⎜ ⎟ = 15.0 s 100 km h ⎝ 0.278 m s ⎠

The acceleration experienced as he came to rest is given by v = v0 + at as mi ⎞ ⎛ 0.447 m s ⎞ 0 − ⎛⎜ 632 ⎟⎜ ⎟ h ⎠ ⎝ 1 mi h ⎠ v − v0 ⎝ a= = = −202 m s 2 t 1.40 s

continued on next page

56157_02_ch02_p017-048.indd 41

10/12/10 1:27:39 PM

42

Chapter 2

(b)

The distance traveled while stopping is found from ⎡ ⎛ mi ⎞ ⎛ 0.447 m s ⎞ ⎤ ⎟ ⎢ 0 + ⎜⎝ 632 ⎥ h ⎠ ⎜⎝ 1 mi h ⎟⎠ ⎦ ⎛ v + v0 ⎞ ⎣ = Δx = vt = ⎜ t (1.40 s ) = 198 m ⎝ 2 ⎟⎠ 2

2.57

(a)

The acceleration of the bullet is v 2 − v02 ( 300 m/s ) − ( 400 m/s ) 5 2 = = − 3.50 × 10 m s 2 ( Δx ) 2 ( 0.100 m ) 2

a= (b)

The time of contact with the board is t=

2.58

(a)

2

v − v0 ( 300 − 400 ) m s −4 = = 2.86 × 10 s 5 2 a − 3.50 × 10 m s

From Δx = v0 t + 12 at 2, we have 100 m = ( 30.0 m s ) t +

1 −3.50 m s 2 ) t 2 ( 2

This reduces to 3.50 t 2 + ( − 60.0 s ) t + ( 200 s 2 ) = 0 , and the quadratic formula gives − ( −60.0 s ) ± ( −60.0 s ) − 4 ( 3.50 ) ( 200 s 2 ) 2

t=

2 ( 3.50 )

The desired time is the smaller solution of t = 4.53 s . The larger solution of t = 12 .6 s is the time when the boat would pass the buoy moving backwards, assuming it maintained a constant acceleration. (b)

The velocity of the boat when it ﬁrst reaches the buoy is v = v0 + at = 30.0 m s+ ( − 3.50 m s 2 ) ( 4.53 s ) = 14.1 m s

2.59

(a)

The keys have acceleration a = − g = −9.80 m s 2 from the release point until they are caught 1.50 s later. Thus, Δy = v0 t + 12 at 2 gives 2 Δy − at 2 2 ( + 4.00 m ) − ( − 9.80 m s ) (1.50 s ) 2 = = +10.0 m s t 1.50 s 2

v0 = or (b)

v0 = 10.0 m s upward

The velocity of the keys just before the catch was v = v0 + at = 10.0 m s + ( − 9.80 m s 2 ) (1.50 s ) = − 4.70 m s or

2.60

(a)

v = 4.70 m s downward

The keys, moving freely under the inﬂuence of gravity ( a = − g ), undergo a vertical displacement of Δy = +h in time t. We use Δy = vi t + 12 at 2 to ﬁnd the initial velocity as h = vi t +

1 (−g) t 2 2

giving

vi =

h + gt 2 2 h gt = + t t 2

continued on next page

56157_02_ch02_p017-048.indd 42

10/18/10 3:27:59 PM

Motion in One Dimension

(b)

43

The velocity of the keys just before they were caught ( at time t ) is given by v = vi + at as h gt h gt h gt v = ⎛⎜ + ⎞⎟ + ( − g ) t = + − gt = − t 2 t 2 ⎝t 2 ⎠

2.61

(a)

From v 2 = v02 + 2a ( Δy ), the insect’s velocity after straightening its legs is v = v02 + 2a ( Δy ) = 0 + 2 ( 4 000 m s 2 ) ( 2.0 × 10 −3 m ) = 4.0 m s

(b)

The time to reach this velocity is v − v0 4.0 m s − 0 = = 1.0 × 10 −3 s = 1.0 ms a 4 000 m s 2 The upward displacement of the insect between when its feet leave the ground and it comes to rest momentarily at maximum altitude is t=

(c)

− ( 4.0 m s ) v 2 − v02 0 − v02 = = = 0.82 m 2a 2 ( − g ) 2 ( −9.8 m s 2 ) 2

Δy =

2.62

(a)

= reading order = velocity

(b)

= acceleration

(c)

(d)

(e)

(f)

2.63

If the speed did not change at a constant rate, the drawings would have less regularity than those given above.

The falling ball moves a distance of (15 m − h ) before they meet, where h is the height above the ground where they meet. Apply Δy = v0 t + 12 at 2 , with a = − g, to obtain − (15 m − h ) = 0 −

1 2 gt 2

or

h = 15 m −

1 2 gt 2

[1]

1 Applying Δy = v0 t + at 2 to the rising ball gives 2 h = ( 25 m s ) t −

1 2 gt 2

[2]

Combining Equations [1] and [2] gives 1 2

1 2

( 25 m s ) t − gt 2 = 15 m − gt 2 or

56157_02_ch02_p017-048.indd 43

t=

15 m = 0.60 s 25 m s

10/12/10 1:27:40 PM

44

2.64

Chapter 2

The constant speed the student has maintained for the ﬁrst 10 minutes, and hence her initial speed for the ﬁnal 500 yard dash, is v0 =

Δx10 1.0 mi − 500 yards ( 5280 ft − 1500 ft ) ⎛ 1 m ⎞ = = ⎜ ⎟ = 1.9 m s Δt 10 min 6000 s ⎝ 3.281 ft ⎠

With an initial speed of v0 = 1.9 m s, the minimum constant acceleration that would be needed to complete the last 500 yards (1 500 ft) in the remaining 2.0 min (120 s) of her allotted time is found from Δx = vi t + 12 at 2 as amin =

2 [ Δx − v0 t ] t

2

2 ⎡(1 500 ft )(1 m 3.281 ft ) − (1.9 m s ) (120 s ) ⎤⎦ = ⎣ = 0.032 m s 2 (120 s )2

Since this acceleration is considerably smaller than the acceleration of 0.15 m s 2 that she is capable of producing, she should be able to easily meet the requirement of running 1.0 mile in 12 minutes. 2.65

2.66

Once the gymnast’s feet leave the ground, she is a freely falling body with constant acceleration a = − g = −9.80 m s 2 . Starting with an initial upward velocity of v0 = 2.80 m s, the vertical displacement of the gymnast’s center of mass from its starting point is given as a function of time by Δy = v0 t + 12 at 2. (a)

At t = 0.100 s,

Δy = ( 2.80 m s ) ( 0.100 s ) − ( 4.90 m s 2 ) ( 0.100 s ) = 0.231 m

(b)

At t = 0.200 s,

Δy = ( 2.80 m s ) ( 0.200 s ) − ( 4.90 m s 2 ) ( 0.200 s ) = 0.364 m

(c)

At t = 0.300 s,

Δy = ( 2.80 m s ) ( 0.300 s ) − ( 4.90 m s 2 ) ( 0.300 s ) = 0.399 m

(d)

At t = 0.500 s,

Δy = ( 2.80 m s ) ( 0.500 s ) − ( 4.90 m s 2 ) ( 0.500 s ) = 0.175 m

(a)

While in the air, both balls have acceleration a1 = a2 = − g (where upward is taken as positive). Ball 1 (thrown downward) has initial velocity v01 = − v0 , while ball 2 (thrown upward) has initial velocity v02 = + v0. Taking y = 0 at ground level, the initial y-coordinate of each ball is y01 = y02 = +h. Applying Δy = y − yi = vi t + 12 at 2 to each ball gives their ycoordinates at time t as:

(b)

2

2

2

2

Ball 1:

y1 − h = − v0 t +

1 (−g) t 2 2

or

y1 = h − v0 t −

1 2 gt 2

Ball 2:

y2 − h = + v0 t +

1 (−g) t 2 2

or

y2 = h + v0 t −

1 2 gt 2

At ground level, y = 0. Thus, we equate each of the equations found above to zero and use the quadratic formula to solve for the times when each ball reaches the ground. This gives: Ball 1:

0 = h − v0 t1 −

so

t1 =

1 2 gt1 2

−2 v0 ±

→

gt12 + ( 2 v0 ) t1 + ( −2h ) = 0

( 2v0 )2 − 4 ( g ) ( −2h ) 2g

2

=−

⎛ v ⎞ 2h v0 ± ⎜ 0⎟ + g g ⎝ g ⎠ 2

Using only the positive solution gives

t1 = −

⎛ v ⎞ 2h v0 + ⎜ 0⎟ + g g ⎝ g ⎠

continued on next page

56157_02_ch02_p017-048.indd 44

10/12/10 1:27:41 PM

Motion in One Dimension

Ball 2:

0 = h + v0 t 2 − t2 =

and

1 2 gt 2 2

45

gt 22 + ( −2 v0 ) t 2 + ( −2h ) = 0

→

− ( −2 v0 ) ±

( −2v0 )2 − 4 ( g ) ( −2h ) 2g

2

⎛ v ⎞ 2h v =+ 0 ± ⎜ 0 ⎟ + g g ⎝ g ⎠ 2

⎛ v ⎞ 2h v Again, using only the positive solution t2 = 0 + ⎜ 0 ⎟ + g g ⎝ g ⎠ Thus, the difference in the times of ﬂight of the two balls is 2 2 ⎛ v0 ⎞ 2h ⎛⎜ v0 ⎛ v0 ⎞ 2h ⎞⎟ 2 v0 v0 Δt = t 2 − t1 = + ⎜ ⎟ + − − + ⎜ ⎟ + = g g g ⎜ g g ⎟ ⎝ g ⎠ ⎝ g ⎠ ⎝ ⎠

(c)

(d)

Realizing that the balls are going downward ( v < 0 ) as they near the ground, we use v 2f = vi2 + 2a ( Δy ) with Δy = −h to ﬁnd the velocity of each ball just before it strikes the ground: Ball 1:

v1 f = − v12i + 2a1 ( −h ) = −

( − v0 ) 2 + 2 ( − g ) ( − h ) =

− v02 + 2 gh

Ball 2:

v2 f = − v22i + 2a2 ( −h ) = −

( + v0 ) 2 + 2 ( − g ) ( − h ) =

− v02 + 2 gh

While both balls are still in the air, the distance separating them is ⎛ 1 ⎞ ⎛ 1 ⎞ d = y2 − y1 = ⎜ h + v0 t − gt 2 ⎟ − ⎜ h − v0 t − gt 2 ⎟ = 2v0 t 2 ⎠ ⎝ 2 ⎠ ⎝

2.67

(a)

The ﬁrst ball is dropped from rest ( v01 = 0 ) from the height h of the window. Thus, v 2f = v02 + 2a ( Δy ) gives the speed of this ball as it reaches the ground (and hence the initial 2 velocity of the second ball) as v f = v01 + 2a1 ( Δy1 ) = 0 + 2 ( − g ) ( −h ) = 2 gh . When ball 2 is thrown upward at the same time that ball 1 is dropped, their y-coordinates at time t during the ﬂights are given by y − yo = v0 t + 12 at 2 as: Ball 1:

y1 − h = ( 0 ) t + 12 ( − g ) t 2

Ball 2:

y2 − 0 =

(

)

2 gh t + 12 ( − g ) t 2

or

y1 = h − 12 gt 2

or

y2 =

When the two balls pass, y1 = y2 , or giving

(b)

t=

h = 2 gh

h = 2g

(

)

2 gh t − 12 gt 2

h − 12 gt 2 =

(

)

2 gh t − 12 gt 2

28.7 m = 1.21 s 2 ( 9.80 m s 2 )

When the balls meet, 2

t=

h 2g

and

y1 = h −

1 ⎛ h ⎞ h 3h g⎜ ⎟ =h− = 2 ⎜⎝ 2 g ⎟⎠ 4 4

Thus, the distance below the window where this event occurs is d = h − y1 = h −

56157_02_ch02_p017-048.indd 45

3h h 28.7 m = = = 7.18 m 4 4 4

10/12/10 1:27:41 PM

46

2.68

Chapter 2

We do not know either the initial velocity or the ﬁnal velocity (that is, velocity just before impact) for the truck. What we do know is that the truck skids 62.4 m in 4.20 s while accelerating at −5.60 m s 2. We have v = v0 + at and Δ x = v t = [ (v + v0 ) 2] t. Applied to the motion of the truck, these yield v − v0 = at = ( −5.60 m s 2 ) ( 4.20 s )

or

v − v0 = −23.5 m s

[1]

or

v + v0 = 29.7 m s

[2]

and v + v0 =

2 ( Δx ) 2 ( 62.4 m ) = t 4.20 s

Adding Equations [1] and [2] gives the velocity just before impact as 2 v = ( −23.5 + 29.7 ) m s 2.69

or

v = 3.10 m s

When released from rest ( v0 = 0 ), the bill falls freely with a downward acceleration due to gravity ( a = − g = −9.80 m s2 ). Thus, the magnitude of its downward displacement during David’s 0.2 s reaction time will be 1 1 2 Δy = v0 t + at 2 = 0 + ( −9.80 m s 2 ) ( 0.2 s ) = 0.2 m = 20 cm 2 2 This is over twice the distance from the center of the bill to its top edge ( ≈ 8 cm ), so David will be unsuccessful .

2.70

(a)

The velocity with which the ﬁrst stone hits the water is 2

m⎞ m⎞ m ⎛ 2 v1 = − v01 + 2 a ( Δy ) = − ⎛⎜ − 2 .00 ⎟ + 2 ⎜ − 9.80 2 ⎟ ( −50.0 m ) = −31.4 s ⎠ s ⎠ s ⎝ ⎝ The time for this stone to hit the water is t1 = (b)

v1 − v01 ⎡⎣−31.4 m s − ( −2 .00 m s ) ⎦⎤ = = 3.00 s a −9.80 m s 2

Since they hit simultaneously, the second stone, which is released 1.00 s later, will hit the water after an ﬂight time of 2.00 s. Thus, 2 Δy − at 22 2 −50.0 m − ( −9.80 m s ) ( 2 .00 s ) 2 v02 = = = − 15.2 m s t2 2 .00 s 2

(c)

From part (a), the ﬁnal velocity of the ﬁrst stone is v1 = − 31.4 m s . The ﬁnal velocity of the second stone is v2 = v02 + at 2 = −15.2 m s + ( −9.80 m s 2 ) ( 2.00 s ) = − 34.8 m s

56157_02_ch02_p017-048.indd 46

10/12/10 1:27:42 PM

Motion in One Dimension

2.71

(a)

47

The sled’s displacement, Δx1, after accelerating at a1 = +40 ft s 2 for time t1, is Δx1 = ( 0 ) t1 + 12 a1t12 = ( 20 ft s 2 ) t12

Δx1 = ( 20 ft s 2 ) t12

or

[1]

At the end of time t1, the sled had achieved a velocity of v = v0 + a1t1 = 0 + ( 40 ft s 2 ) t1

v = ( 40 ft s 2 ) t1

or

[2]

The displacement of the sled while moving at constant velocity v for time t2 is Δx2 = vt 2 = ⎡⎣( 40 ft s 2 ) t1 ⎤⎦ t 2

Δx2 = ( 40 ft s 2 ) t1t 2

or

[3]

It is known that Δx1 + Δx2 = 17 500 ft, and substitutions from Equations [1] and [3] give

( 20 ft s ) t + ( 40 ft s ) t t 2

2 1

2

1 2

= 17 500 ft

or

Also, it is known that

t12 + 2t1t 2 = 875 s 2

[4]

t1 + t 2 = 90 s

[5]

Solving Equations [4] and [5] simultaneously yields t12 + 2t1 ( 90 s − t1 ) = 875 s 2

or

t12 + ( −180 s ) t1 + 875 s 2 = 0 − ( −180 s ) ± ( −180 s ) − 4 (1) (875 s 2 ) 2

The quadratic formula then gives with solutions t1 = 5.00 s

t1 =

(and t 2 = 90 s − 5.0 s = 85 s )

2 (1)

or t1 = 175 s

(and t 2 = −85 s )

Since it is necessary that t 2 > 0 , the valid solutions are t1 = 5.0 s and t 2 = 85 s . v = ( 40 ft s 2 ) t1 = ( 40 ft s 2 ) ( 5.0 s ) = 200 ft s

(b)

From Equation [2] above,

(c)

The displacement Δx3 of the sled as it comes to rest (with acceleration a3 = −20 ft s 2) is 0 − v 2 − ( 200 ft s ) = 1 000 ft = 2a3 2 ( −20 ft s 2 ) 2

Δx3 =

Thus, the total displacement for the trip (measured from the starting point) is Δxtotal = ( Δx1 + Δx2 ) + Δx3 = 17 500 ft + 1 000 ft = 18 500 ft (d)

The time required to come to rest from velocity v (with acceleration a3) is t3 =

0 − v −200 ft s = = 10 s −20 ft s 2 a3

so the duration of the entire trip is t total = t1 + t 2 + t3 = 5.0 s + 85 s + 10 s = 100 s 2.72

(a)

1 From Δy = v0 t + at 2 with v0 = 0 , we have 2 t=

56157_02_ch02_p017-048.indd 47

2 ( Δy ) = a

2 ( −23 m ) = 2.2 s − 9.80 m s 2

continued on next page

10/12/10 1:27:42 PM

48

Chapter 2

(b)

The ﬁnal velocity is v = 0 + ( − 9.80 m s 2) (2.2s ) = − 22 m s

(c)

The time it takes for the sound of the impact to reach the spectator is Δy 23 m = = 6.8 × 10 −2 s vsound 340 m s

tsound =

so the total elapsed time is t total = 2 .2 s + 6.8 × 10 −2 s ≈ 2.3 s . 2.73

(a)

Since the sound has constant velocity, the distance it traveled is Δx = vsound t = (1100 ft s ) ( 5.0 s ) = 5.5 × 10 3 ft

(b)

The plane travels this distance in a time of 5.0 s + 10 s = 15 s, so its velocity must be vplane =

(c)

Δx 5.5 × 10 3 ft = = 3.7 × 10 2 ft s t 15 s

The time the light took to reach the observer was t light =

Δx 5.5 × 10 3 ft ⎛ 1 m s ⎞ −6 = ⎜ ⎟ = 5.6 × 10 s 8 vlight 3.00 × 10 m s ⎝ 3.281 ft s ⎠

During this time the plane would only travel a distance of 0.002 ft. 2.74

The distance the glider moves during the time Δt d is given by Δx = = v0 ( Δt d ) + 12 a ( Δt d ) , where v0 is the glider’s velocity when the ﬂag ﬁrst enters the photogate and a is the glider’s acceleration. Thus, the average velocity is 2

v ( Δt d ) + 12 a ( Δt d ) 1 = 0 = v0 + a ( Δt d ) Δt d Δt d 2 2

vd = (a)

The glider’s velocity when it is halfway through the photogate in space

( i.e., when Δx = 2 ) is found from v 2 = v02 + 2a ( Δx ) as

v1 = v02 + 2a ( 2 ) = v02 + a = v02 + a ⎡⎣ vd ( Δt d ) ⎤⎦ = v02 + avd ( Δt d ) Note that this is not equal to vd unless a = 0 , in which case v1 = vd = v0 . (b)

The speed v2when the glider is halfway through the photogate in time (i.e., when the elapsed time is t 2 = Δt d 2) is given by v = v0 + at as 1 v = v0 + at 2 = v0 + a ( Δt d 2 ) = v0 + a ( Δt d ) 2 which is equal to vd for all possible values of v0 and a.

2.75

The time required for the stuntman to fall 3.00 m, starting from rest, is found from Δy = v0 t + 12 at 2 as −3.00 m = 0 + (a)

1 −9.80 m s 2 ) t 2 ( 2

so

t=

2 ( 3.00 m ) = 0.782 s 9.80 m s 2

With the horse moving with constant velocity of 10.0 m s, the horizontal distance is Δx = vhorse t = (10.0 m s ) ( 0.782 s ) = 7.82 m

(b)

56157_02_ch02_p017-048.indd 48

The required time is t = 0.782 s as calculated above.

10/12/10 1:27:43 PM

3 Vectors and Two-Dimensional Motion QUICK QUIZZES 1.

2.

Choice (c). The largest possible magnitude of the resultant occurs when the two vectors are in the same direction. In this case, the magnitude of the resultant is the sum of the magnitudes of A and B: R = A + B = 20 units. The smallest possible magnitude of the resultant occurs when the two vectors are in opposite directions, and the magnitude is the difference of the magnitudes of A and B: R = | A – B | = 4 units. Vector

x-component

y-component

A

−

+

B

+

−

A + B

−

−

3.

Vector B. The range of the inverse tangent function includes only the ﬁrst and fourth quadrants (i.e., angles in the range − π 2 < θ < π 2). Only vector B has an orientation in this range.

4.

Choice (b). If velocity is constant, the acceleration (rate of change in velocity) is zero. An object may have constant speed (magnitude of velocity) but still be accelerating due to a change in direction of the velocity. If an object is following a curved path, it is accelerating because the velocity is changing in direction.

5.

Choice (a). Any change in the magnitude and/or direction of the velocity is an acceleration. The gas pedal and the brake produce accelerations by altering the magnitude of the velocity. The steering wheel produces accelerations by altering the direction of the velocity.

6.

Choice (c). A projectile has constant horizontal velocity. Thus, if the runner throws the ball straight up into the air, the ball maintains the horizontal velocity it had before it was thrown (that is, the velocity of the runner). In the runner’s frame of reference, the ball appears to go straight upward and come straight downward. To a stationary observer, the ball follows a parabolic trajectory, moving with the same horizontal velocity as the runner and staying above the runner’s hand.

7.

Choice (b). The velocity is always tangential to the path while the acceleration is always directed vertically downward. Thus, the velocity and acceleration are perpendicular only where the path is horizontal. This only occurs at the peak of the path.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

Choose coordinate system with north as the positive y-direction and east as the positive x-direction. The velocity of the cruise ship relative to Earth is v CE = 4.50 m s due north, with 49

56157_03_ch03_p049-088.indd 49

10/12/10 1:28:10 PM

50

Chapter 3

( )

( )

components of vCE x = 0 and vCE y = 4.50 m s. The velocity of the patrol boat relative to Earth is vPE = 5.20 m s at 45.0° north of west, with components of

(v ) (v ) PE

and

PE

(

)(

)

x

= − vPE cos 45.0° = − 5.20 m s 0.707 = −3.68 m s

y

= + vPE sin 45.0° = 5.20 m s 0.707 = 3.68 m s

(

)(

)

Thus, the velocity of the cruise ship relative to the patrol boat is vCP = vCE − vPE , which has components of

(v ) = (v ) − (v ) CP

and

CE

x

x

PE

x

( vCP )y = ( vCE )y − ( vPE )y

(

)

= 0 − −3.68 m s = +3.68 m s = 4.50 m s − 3.68 m s = +0.82 m s

Choice (a) is seen to be the correct answer. 2.

The skier has zero initial velocity in the vertical direction ( v0 y = 0 ) and undergoes a vertical displacement of Δy = −3.20 m. The constant acceleration in the vertical direction is ay = − g, so we use Δy = v0 y t + 12 ay t 2 to ﬁnd the time of ﬂight as −3.20 m = 0 +

1 ( −9.80 m s2 ) t 2 2

t=

or

2 ( −3.20 m ) = 0.808 s −9.80 m s 2

During this time, the object moves with constant horizontal velocity vx = v0 x = 22.0 m s. The horizontal distance traveled during the ﬂight is Δx = vx t = ( 22.0 m s ) ( 0.808 s ) = 17.8 m, 3.

which is choice (d).

At maximum height ( Δy = hmax ), the vertical velocity of the stone will be zero. Thus, vy2 = v02 y + 2ay ( Δy ) gives hmax =

vy2 − v02 y 2ay

0 − v02 sin 2 θ − ( 45 m s ) sin 2 ( 55.0°) = = 69.3 m 2 (−g) 2 ( −9.80 m s 2 ) 2

=

and we see that choice (c) is the correct answer. 4.

For vectors in the x-y plane, their components have the signs indicated in the following table: Quadrant of Vector First

Second

Third

Fourth

x-component

Positive

Negative

Negative

Positive

y-component

Positive

Positive

Negative

Negative

Thus, a vector having components of opposite sign must lie in either the second or fourth quadrants and choice (e) is the correct answer. 5.

56157_03_ch03_p049-088.indd 50

Whether on Earth or the Moon, a golf ball is in free fall with a constant downward acceleration of magnitude determined by local gravity from the time it leaves the tee until it strikes the ground or other object. Thus, the vertical component of velocity is constantly changing while the horizontal component of velocity is constant. Note that the speed (or magnitude of the velocity)

10/13/10 10:24:22 AM

Vectors and Two-Dimensional Motion

51

v = v = vx2 + vy2 will change in time since vy changes in time. Thus, the only correct choices for this question are (b) and (d). 6.

Consider any two very closely spaced points on a circular path and draw vectors of the same length (to represent a constant velocity magnitude or speed) tangent to the path at each of these points as shown in the leftmost diagram below. Now carefully move the velocity vector v f at the second point down so its tail is at the ﬁrst point as shown in the rightmost diagram. Then, draw the vector difference Δv = v f − vi , and observe that if the start of this vector were located on the circular path midway between the two points, its direction would be inward toward the center of the circle.

Thus, for an object following the circular path at constant speed, its instantaneous acceleration, a = lim ( Δ v Δt ) , at the point midway between your initial and end points is directed toward the Δt → 0

center of the circle, and the only correct choice for this question is (d). 7.

The path followed (and distance traveled) by the athlete is shown in the sketch, along with the vectors for the initial position, ﬁnal position, and change in position. The average speed for the elapsed time interval Δt is vav =

d Δt

and the magnitude of the average velocity for this time interval is v av =

Δr Δt

The sketch clearly shows that d > Δr in this case, meaning that vav > v av and that (a) is the correct choice. 8.

At maximum altitude, the projectile’s vertical component of velocity is zero. The time for the projectile to reach its maximum height is found from vy = v0 y + ay t as t max =

56157_03_ch03_p049-088.indd 51

vy

Δy = hmax

ay

− v0 y

=

0 − v0 sin θ0 v0 sin θ0 = −g g

10/12/10 1:28:11 PM

52

Chapter 3

Since the acceleration of gravity on the Moon is one-sixth that of Earth, we see that (with v0 and θ0 kept constant) t max

Moon

=

⎛ v sin θ0 ⎞ v0 sin θ0 v0 sin θ0 = = 6⎜ 0 ⎟ = 6 t max gMoon gEarth 6 ⎝ gEarth ⎠

Earth

and the correct answer for this question is (e). 9.

The boat moves with a constant horizontal velocity (or its velocity relative to Earth has components of ( vBE )x = v0 = constant, and ( vBE )y = 0), where the y-axis is vertical and the x-axis is parallel to the keel of the boat. Once the wrench is released, it is a projectile whose velocity relative to Earth has components of

(v ) WE

x

= v0 x + a x t = v0 + 0 = v0

and

( vWE )y

= v0 y + ay t = 0 − gt = − gt

The velocity of the wrench relative to the boat (vWB = vWE − vBE ) has components of

( vWB )x = ( vWE )x − ( vBE )x

= v0 − v0 = 0

and

(v ) = (v ) − (v ) WB

y

WE

y

BE

y

= − gt − 0 = − gt

Thus, the wrench has zero horizontal velocity relative to the boat and will land on the deck at a point directly below where it was released (i.e., at the base of the mast). The correct choice is (b). 10.

While in the air, the baseball is a projectile whose velocity always has a constant horizontal component ( vx = v0 x ) and a vertical component that changes at a constant rate (Δvy Δt = ay = − g). At the highest point on the path, the vertical velocity of the ball is momentarily zero. Thus, at this point, the resultant velocity of the ball is horizontal and its acceleration continues to be directed downward (ax = 0, ay = − g). The only correct choice given for this question is (c).

11.

Note that for each ball, v0 y = 0 . Thus, the vertical velocity of each ball when it reaches the ground (Δy = −h) is given by vy2 = v02u + 2ay (Δy) as vy = − 0 + 2 ( − g ) ( −h ) = − 2 gh and the time required for each ball to reach the ground is given by vy = v0 y + ay t as t=

v y − v0 y ay

=

− 2 gh − 0 = −g

2h g

The speeds ( i.e., magnitudes of total velocities ) of the balls at ground level are:

(

)

2

= v02 + 2 gh

(

)

2

= 0 + 2 gh = 2 gh

Red Ball:

vR = vx2 + vy2 = v02 x + − 2 gh

Blue Ball:

vB = vx2 + vy2 = v02 x + − 2 gh

Therefore, we see that the two balls reach the ground at the same time but with different speeds ( vR > vB ), so only choice (b) is correct. 12.

When the apple ﬁrst comes off the tree, it is moving forward with the same horizontal velocity as the truck. Since, while in free fall, the apple has zero horizontal acceleration, it will maintain this constant horizontal velocity as it falls. Also, while in free fall, the apple has constant downward acceleration (ay = − g), so its downward speed increases uniformly in time. (i)

56157_03_ch03_p049-088.indd 52

As the truck moves left to right past an observer stationary on the ground, this observer will see both the constant velocity horizontal motion and the uniformly accelerated downward motion of the apple. The curve that best describes the path of the apple as seen by this observer is (a).

10/12/10 1:28:12 PM

Vectors and Two-Dimensional Motion

(ii)

13.

53

An observer on the truck moves with the same horizontal motion as does the apple. This observer does not detect any horizontal motion of the apple relative to him. However, this observer does detect the uniformly accelerated vertical motion of the apple. The curve best describing the path of the apple as seen by the observer on the truck is (b).

Of the choices listed, the quantities which have magnitude or size, but no direction, associated with them (i.e., scalar quantities) are (b) temperature, (c) volume, and (e) height. The other quantities, (a) velocity of a sports car and (d) displacement of a tennis player who moves from the court’s backline to the net, have both magnitude and direction associated with them, and are both vector quantities.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The components of a vector will be equal in magnitude if the vector lies at a 45° angle with the two axes along which the components lie.

4.

(a)

6.

(a)

The balls will be closest at the instant the second ball is projected.

(b)

The ﬁrst ball will always be going faster than the second ball.

(c)

There will be a one second time interval between their collisions with the ground.

(d)

The two move with the same acceleration in the vertical direction. Thus, changing their horizontal velocities can never make them hit at the same time.

8.

10.

(b)

The equations of projectile motion are only valid for objects moving freely under the inﬂuence of gravity. The only acceleration such an object has is the free-fall acceleration, g, directed vertically downward. Of the objects listed, only a and d meet this requirement. (a)

The passenger sees the ball go into the air and come back in the same way he would if he were at rest on Earth. An observer by the tracks would see the ball follow the path of a projectile.

(b)

If the train were accelerating, the ball would fall behind the position it would reach in the absence of the acceleration.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a)

6.1 units at θ = + 113°

(b)

4.

(a)

See Solution.

(b) The sum does not depend on the order of adding.

6.

(a)

484 km

(b)

(c)

Because of Earth’s curvature, the plane does not follow straight lines.

8.

R = 9.5 units at 57° above the +x -axis

10.

1.31 km northward, 2.81 km eastward

56157_03_ch03_p049-088.indd 53

15 units at θ = + 23°

18.1° N of W

10/12/10 1:28:13 PM

54

Chapter 3

12.

(a)

10.0 m

14.

(a)

R = 358 m at 2.00° south of east

(b)

(c)

15.7 m

0

(b) The vector sum is independent of the order in which they are added. 16.

42.7 yards

18.

788 mi at 48.0° N of E

20.

(a)

22.

2.68 ft

24.

(a)

185 N at 77.8° from the x-axis

76.0°

(b)

185 N at 258° from the x-axis

(b)

Rmax = 2.13R

(c) The result is independent of the acceleration of gravity and valid on all planets. 26.

(a) 28.2 m s

(b)

4.07 s

(b)

v0 x = 7.52 m s , v0 y = −2.74 m s

(c) v0 and t would increase 28.

(a)

x = 0, y = y0

(c)

x = ( 7.52 m s ) t , y = y0 − ( 2.74 m s ) t − ( 4.90 m s 2 ) t 2

(d)

22.6 m

(e)

y0 = 52.3 m

(f )

1.18 s

(b)

1.09 m

(c)

2.8 m

30.

x = 7.22 km, y = 1.69 km

32.

18.6 m

34.

(a) 18.1 m s

36.

(a) 57.7 km h at 60.0° west of vertical (b) 28.9 km h vertically downward 27.7° east of north

38.

(a)

40.

(a) 1.81 m s

42.

(a)

(b)

47.6 min

(b) Yes, the required initial speed is < 6.26 m s.

44.

(a)

d v − vs 2d tb = v t woman = L v1

46.

(a)

hmax = 57.4 m, t = 1.22 s

(b)

Same as for part (a).

48.

(a)

t = 2h g

(b)

vi = d g 2h

(c)

v = d 2 g 2h + 2 gh

(d) θ = tan −1 ( −2h d )

(a)

0.345 m

(b)

(d)

50.

56157_03_ch03_p049-088.indd 54

t up =

d v + vs

(b)

t down =

(e)

t a > tb as long as vs > 0

(b)

t man =

(c)

See Solution.

L v1 + v2

2.35 m

10/12/10 1:28:13 PM

Vectors and Two-Dimensional Motion

52.

(a)

0.85 m s

(b) 2.1 m s

54.

(a)

0° to the vertical

(b) 8.25 m s

(c)

straight up and down

(d)

55

parabolic arc opening downward

(e) 12.6 m s at 41.0° above the horizontal eastward line 56.

(a) 14.7 m s

58.

10.7 m s

60.

(a)

d male = 132 cm at 69.6° and d female = 111 cm at 70.0°

(b)

d′male = 146 cm at 69.6° and d′female = 132 cm at 70.0°, Δd ′ = 14 cm at 66°

(a)

⎛ ⎛v ⎞ g ⎞ y = ⎜ − 2 ⎟ x2 + ⎜ 0y ⎟ x + 0 ⎝ 2 v0 x ⎠ ⎝ v0 x ⎠

(b)

a = − g 2 v02 x , b = v0 y v0 x , and c = 0

64.

(a)

26.6°

66.

3.95 m s

68.

(a)

26 knots at 50° S of E

70.

(a)

0.519 m, too narrow for a pedestrian walkway

(b)

0.217 m s

62.

72.

4.14 m

74.

x = 18.8 m, y = −17.3 m

(b)

29.4 m s

(b)

t1 bounce t no bounce = 0.950

(b)

20 knots due south

PROBLEM SOLUTIONS 3.1

We are given that R = A + B . When two vectors are added graphically, the second vector is positioned with its tail at the tip of the first vector. The resultant then runs from the tail of the first vector to the tip of the second vector. In this case, vector A will be positioned with its tail at the origin and its tip at the point (0, 29). The resultant is then drawn, starting at the origin (tail of first vector) and going 14 units in the negative y-direction to the point (0, −14). The second vector, B , must then start from the tip of A at point (0, 29) and end on the tip of R at point (0, −14) as shown in the sketch at the right. From this, it is seen that B is 43 units in the negative y-direction

continued on next page

56157_03_ch03_p049-088.indd 55

10/12/10 1:28:14 PM

56

3.2

3.3

Chapter 3

(a)

Using graphical methods, place the tail of vector B at the head of vector A. The new vector A + B has a magnitude of 6.1 units at 113° from the positive x-axis.

(b)

The vector difference A − B = A + (− B) is found by placing the negative of vector B (a vector of the same magnitude as B , but opposite direction) at the head of vector A. The resultant vector A − B has magnitude 15 units at 23° from the positive x-axis.

(a)

In your vector diagram, place the tail of vector B at the tip of vector A. The vector sum, A + B, is then found as shown in the vector diagram and should be A + B = 5.0 units at –53°

(b)

To ﬁnd the vector difference A − B = A + (− B), form the vector −B (same magnitude as B, opposite direction) and add it to vector A as shown in the diagram. You should ﬁnd that A − B = 5.0 units at +53°

3.4

(a)

The three diagrams shown below represent the graphical solutions for the three vector sums: R1 = A + B + C, R 2 = = B + C + A, and R 3 = C + B + A . 100 m C

B

A

A

B R1

R2

A

C

R3

C B

(b) 3.5

We observe that R1 = R 2 = R 3 , illustrating that the sum of a set of vectors is not affected by the order in which the vectors are added.

Your sketch should be drawn to scale, and be similar to that pictured below. The length of R and the angle θ can be measured to ﬁnd, with use of your scale factor, the magnitude and direction of the resultant displacement. The result should be approximately 421 ft at 3° below the horizonntal .

→

56157_03_ch03_p049-088.indd 56

10/12/10 1:28:15 PM

Vectors and Two-Dimensional Motion

3.6

(a)

57

The distance d from A to C is d = x 2 + y2 where x = 200 km + ( 300 km ) cos 30.0° = 460 km and y = 0 + ( 300 km ) sin 30.0° = 150 km ∴ d = (460 km)2 + (150 km)2 = 484 km

(b) (c)

3.7

y 150 km ⎞ φ = tan −1 ⎛⎜ ⎞⎟ = tan −1 ⎛⎜ ⎟ = 18.1° N of W ⎝x⎠ ⎝ 460 km ⎠ Because of the curvature of the Earth, the planee doesn’t travel along straight lines. Thus, the answer computed above is only approximately correct.

Using a vector diagram, drawn to scale, like that shown at the right, the displacement from Lake B back to base camp is given by the vector D. Measuring the length of this vector and multiplying by the chosen scale factor should give the magnitude of this displacement as 310 km. Measuring the angle θ should yield a value of 57°. Thus, the displacement from B to the base camp is

D

D = 310 km at θ = 57° S of W

Find the resultant R = F1 + F2 graphically by placing the tail of F2 at the head of F1 in a scale drawing as shown at the right. The resultant is the vector drawn from the tail of F1 to the head of F2 to close up the triangle. Measuring the length and orientation of this vector shows that

5.00 units

3.8

R= F1

+F

2

R = 9.5 units at 57° above the +x -axis

0 6.0

its

un

30.0⬚

3.9

The displacement vectors A = 8.00 m westward and B = 13.0 m north can be drawn to scale as at the right. The vector C represents the displacement that the man in the maze must undergo to return to his starting point. The scale used to draw the sketch can be used to ﬁnd C to be 15 m at 58° S of E

56157_03_ch03_p049-088.indd 57

10/12/10 1:28:16 PM

58

3.10

Chapter 3

The person undergoes a displacement A = 3.10 km at 25.0° north of east. Choose a coordinate system with origin at the starting point, positive x-direction oriented eastward, and positive ydirection oriented northward. N

Then the components of her displacement are

y

Ax = A cos θ = ( 3.10 km ) cos 25.0° = 2.81 km eastward and 3.11

m 0k 25.0⬚

3.1

Ay = A sin θ = ( 3.10 km ) sin 25.0° = 1.31 km northwardd

Ax

Ay x

E

The x- and y-components of vector A are its projections on lines parallel to the x- and y-axes, respectively, as shown in the sketch. The magnitude of these components can be computed using the sine and cosine functions as shown below: Ax = A cos 325° = + A cos 35° = ( 35.0 ) cos 35° = 28.7 units and Ay = A sin 325° = − A sin 35° = − ( 35.0 ) sin 35° = −20.1 units

3.12

(a)

The skater’s displacement vector, d, extends in a straight line from her starting point A to the end point B. When she has coasted half way around a circular path as shown in the sketch at the right, the displacement vector coincides with the diameter of the circle and has magnitude B

C

r ⫽5.00 m d

A

d = 2r = 2 ( 5.00 m ) = 10.0 m (b)

The actual distance skated, s, is one half the circumference of the circular path of radius r. Thus s=

3.13

1 ( 2π r ) = π ( 5.00 m ) = 15.7 m 2

(c)

When the skater skates all the way around the circular path, her end point, B, coincides with the start point, A. Thus, the displacement vector has zero length, or d = 0 .

(a)

Her net x (east-west) displacement is − 3.00 + 0 + 6.00 = + 3.00 blocks, while her net y (north-south) displacement is 0 + 4.00 + 0 = + 4.00 blocks. The magnitude of the resultant displacement is R = ( Σx ) + ( Σy ) = ( 3.00 ) + ( 4.00 ) = 5.00 blocks 2

2

2

2

and the angle the resultant makes with the x-axis (eastward direction) is Σy 4.00 ⎞ −1 θ = tan −1 ⎛⎜ ⎞⎟ = tan −1 ⎛⎜ ⎟ = tan (1.33) = 53.1° ⎝ Σx ⎠ ⎝ 3.00 ⎠ The resultant displacement is then 5.00 blocks at 53.1° N of E . (b)

56157_03_ch03_p049-088.indd 58

The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks .

10/12/10 1:28:16 PM

Vectors and Two-Dimensional Motion

3.14

(a)

59

We choose a coordinate system with the positive x-axis eastward and the positive y-axis northward. The hiker then undergoes four successive displacements, given below along with their x- and y-components: A = 75.0 m due north

Ax = 0

B = 250.0 m due east

Bx = + 250 m

C = 125.0 m at 30.0° N of E D = 150 m due south

Ay = + 75.0 m By = 0

C x = +108 m

Dx = 0

C y = + 62.5 m

Dy = −150 m

The components of the resultant displacement are then Rx = Ax + Bx + C x + Dx = +358 m

and

Ry = Ay + By + C y + Dy = −12.5 m

The magnitude of the resultant is R = Rx2 + Ry2 = ( 358 m ) + ( −12.5 m ) = 358 m 2

⎛R ⎞ −12.5 m ⎞ θ = tan −1 ⎜ y ⎟ = tan −1 ⎛⎜ ⎟ = −2.00° ⎝ 358 m ⎠ ⎝ Rx ⎠

and its direction is giving (b) 3.15

2

R = 358 m at 2.00° south of east

No. Because of the commutative property of vector addition, the net displacement is the same regardless of the order in which the individual displacements are executed.

Ax = −25.0

Ay = 40.0

A = Ax2 + Ay2 = ( −25.0 ) + ( 40.0 ) = 47.2 units 2

2

From the triangle, we ﬁnd that ⎛ A ⎞ ⎛ 40.0 ⎞ φ = tan −1 ⎜ y ⎟ = tan −1 ⎜ ⎟ = 58.0°, so θ = 180° − φ = 122° ⎜ A ⎟ ⎝ 25.0 ⎠ ⎝ x ⎠ Thus, A = 47.2 units at 122° counterclockwise from thee +x -axis . 3.16

Let A be the vector corresponding to the 10.0 yd run, B to the 15.0 yd run, and C to the 50.0 yd pass. Also, we choose a coordinate system with the +y-direction downﬁeld, and the +x-direction toward the sideline to which the player runs. The components of the vectors are then Ax = 0

Ay = −10.0 yds

Bx = 15.0 yds

By = 0

Cx = 0

C y = + 50.0 yds

From these, Rx = Σx = 15.0 yds, and Ry = Σy = 40.0 yds, and

56157_03_ch03_p049-088.indd 59

R = Rx2 + Ry2 =

(15.0 yds )2 + ( 40.0 yds )2 = 42.7 yardss

10/12/10 1:28:17 PM

60

3.17

Chapter 3

After 3.00 h moving at 41.0 km/h, the hurricane is 123 km at 60.0° N of W from the island. In the next 1.50 h, it travels 37.5 km due north. The components of these two displacements are: Displacement

x-component (eastward)

y-component (northward)

123 km

–61.5 km

+107 km

37.5 km

0

+37.5 km

Resultant

–61.5 km

144 km

Therefore, the eye of the hurricane is now R = (− 61.5 km ) + (144 km ) = 157 km from the islaand 2

3.18

2

Choose the positive x-direction to be eastward and positive y as northward. Then, the components of the resultant displacement from Dallas to Chicago are Rx = Σx = ( 730 mi ) cos 5.00° − ( 560 mi ) sin 21.0° = 527 mi and

Ry = Σy = ( 730 mi ) sin 5.00° + ( 560 mi ) cos 21.0° = 586 mi

R = Rx2 + Ry2 = ( 527 mi ) + ( 586 mi ) = 788 mi 2

2

Σy θ = tan −1 ⎛⎜ ⎞⎟ = tan −1 (1.11) = 48.0° ⎝ Σx ⎠ Thus, the displacement from Dallas to Chicago is R = 788 mi at 48.0° N of E 3.19

The components of the displacements a, b, and c are ax = a ⋅ cos 30.0° = + 152 km

(north)

bx = b ⋅ cos 110° = − 51.3 km cx = c ⋅ cos180° = − 190 km and

ay = a ⋅ sin 30.0° = + 87.5 km by = b ⋅ sin110° = + 141 km (east)

cy = c ⋅ sin180° = 0 Thus, Rx = ax + bx + cx = − 89.3 km , and Ry = ay + by + cy = + 229 km so

(

)

R = Rx2 + Ry2 = 246 km, and θ = tan −1 Rx Ry = tan −1 ( 0.390 ) = 21.3°

City C is 246 km at 21.3° W of N from the starting point. 3.20

(a)

F1 = 120 N

F1x = (120 N ) cos 60.0° = 60.0 N

F1y = (120 N ) sin 60.0° = 104 N

F2 = 80.0 N

F2 x = − (80.0 N ) cos 75.0° = − 20.7 N

F2 y = (80.0 N ) sin 75.0° = 77.3 N

FR =

( ΣFx )2 + (ΣFy )

2

= ( 39.3 N ) + (181 N ) = 185 N 2

2

181 N ⎞ −1 and θ = tan −1 ⎛⎜ ⎟ = tan ( 4.61) = 77.8° 39 . 3 N ⎝ ⎠ The resultant force is FR = 185 N at 77.8° from the x-axis . continued on next page

56157_03_ch03_p049-088.indd 60

10/12/10 1:28:18 PM

Vectors and Two-Dimensional Motion

(b)

61

To have zero net force on the mule, the resultant above must be cancelled by a force equal in magnitude and oppositely directed. Thus, the required force is 185 N at 258° from the x -axis

3.21

The single displacement required to sink the putt in one stroke is equal to the resultant of the three actual putts used by the novice. Taking east as the positive x-direction and north as the positive y-direction, the components of the three individual putts and their resultant are Ax = 0

Ay = + 4.00 m

Bx = ( 2.00 m ) cos 45.0° = +1.41 m

By = ( 2.00 m ) sin 45.0° = +1.41 m

C x = − (1.00 m ) sin 30.0° = −0.500 m

C y = − (1.00 m ) cos 30.0° = −0.866 m

Rx = Ax + Bx + C x = + 0.910 m

Ry = Ay + By + C y = +4.55 m

The magnitude and direction of the desired resultant is then R = Rx2 + Ry2 = 4.64 m

⎛ Ry ⎞ θ = tan −1 ⎜ ⎟ = +78.7° ⎝ Rx ⎠

R = 4.64 m at 78.7° north of east

Thus, 3.22

and

⎛ 0.447 m s ⎞ 1m ⎞ = 45.1 m s and Δx = ( 60.5 ft ) ⎛⎜ v0 x = (101 mi h ) ⎜ ⎟ = 18.4 m ⎝ 1 mi h ⎟⎠ ⎝ 3.281 ft ⎠ The time to reach home plate is t =

Δx 18.4 m = = 0.408 s. v0 x 45.1 m s

In this time interval, the vertical displacement is Δy = v0 y t +

1 2 1 2 ay t = 0 + ( −9.80 m s 2 ) ( 0.408 s ) = − 0.817 m 2 2

Thus, the ball drops vertically 0.817 m ( 3.281 ft 1 m ) = 2.68 ft . 3.23

(a)

With the origin chosen at point O as shown in Figure P3.23, the coordinates of the original position of the stone are x0 = 0 and y0 = +50.0 m .

(b)

The components of the initial velocity of the stone are v0 x = +18.0 m s and v0 y = 0 .

(c)

The components of the stone’s velocity during its ﬂight are given as functions of time by vx = v0 x + ax t = 18.0 m s + ( 0 ) t and

(d)

v y = v0 y + a y t = 0 + ( − g ) t

or

or

vx = 18.0 m s

vy = − ( 9.80 m s 2 ) t

The coordinates of the stone during its ﬂight are x = x 0 + v0 x t +

1 2 1 ax t = 0 + (18.0 m s ) t + ( 0 ) t 2 2 2

or

x = (18.0 m s ) t

and y = y0 + v0 y t +

1 2 1 ay t = 50.0 m + ( 0 ) t + ( − g ) t 2 2 2

or

y = 50.0 m − ( 4.90 m s 2 ) t 2 continued on next page

56157_03_ch03_p049-088.indd 61

10/12/10 1:28:19 PM

62

Chapter 3

(e)

We ﬁnd the time of fall from Δy = v0 y t + t=

(f)

2 ( Δy ) = a

1 2 ay t with v0 y = 0 : 2

2 ( − 50.0 m ) = 3.19 s − 9.80 m s 2

At impact, vx = v0 x = 18.0 m s, and the vertical component is vy = v0 y + ay t = 0 + ( − 9.80 m s 2 ) ( 3.19 s ) = − 31.3 m s

3.24

(a)

(18.0 m s )2 + ( −31.3 m s )2 = 36.1 m s

Thus,

v = vx2 + vy2 =

and

⎛ vy ⎞ ⎛ −31.3 ⎞ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ = −60.1° ⎝ 18.0 ⎟⎠ ⎝ vx ⎠

or

v = 36.1 m s at 60.1° below the horizontal

At y = hmax, the vertical component of velocity is zero, so vy = v0 y + ay t gives the time to reach the peak as

y = hmax’ t = tp y = 0, t = 2tp

hmax v0 R

tp =

0 − v0 y ay

=

− v0 sin θ v0 sin θ = −g g

Then, Δy = ( vy )av t = ⎡⎣( vy + v0 y ) 2 ⎤⎦ t gives the maximum vertical displacement as 2 2 ⎛ 0 + v0 y ⎞ ⎛ v sin θ ⎞ ⎛ v0 sin θ ⎞ v0 sin θ hmax = ⎜ tp = ⎜ 0 ⎟⎜ ⎟= ⎟ 2g ⎝ 2 ⎠⎝ g ⎠ ⎝ 2 ⎠

The total time of ﬂight is t total = 2t p = 2 v0 sin θ g, so the horizontal range is ⎛ 2 v sin θ ⎞ 2 v02 sin θ cos θ R = v0 x t total = ( v0 cos θ ) ⎜ 0 ⎟= g g ⎝ ⎠ v02 sin 2 θ 2 v02 sin θ cos θ = 2g g

If we are to have hmax = R, it is necessary that This requirement reduces to

sin θ = 2 cos θ 2

which gives the required launch angle as (b)

tan θ = 4

or

θ = tan −1 ( 4 ) = 76.0°

For maximum range, the launch angle would be θ = 45°, so Rmax =

2 v02 sin 45° cos 45° v02 = g g

The ratio of Rmax to the range in part (a) ( where θ = 76.0°) is

continued on next page

56157_03_ch03_p049-088.indd 62

Rmax v02 g 1 = = 2 R ⎡⎣2 v0 sin ( 76.0°) cos ( 76.0°) ⎤⎦ g 0.469

or

Rmax = 2.13 R

10/12/10 1:28:19 PM

Vectors and Two-Dimensional Motion

(c)

3.25

63

Note that in the calculation of the answer to part (a), the acceleration of gravity canceled out. Thus, the result is valid on every planet independent of the local acceleration of gravity.

At the maximum height vy = 0, and the time to reach this height is found from v y = v0 y + a y t

t=

as

v y − v0 y ay

=

0 − v0 y −g

=

v0 y g

The vertical displacement that has occurred during this time is ⎛ v y + v0 y 2 ⎝

( Δy )max = ( vy )av t = ⎜

2 ⎞ ⎛ 0 + v0 y ⎞ ⎛ v0 y ⎞ v0 y t = = ⎟ ⎜ 2 ⎟ ⎜ g ⎟ 2g ⎠ ⎝ ⎠⎝ ⎠

If ( Δy )max = 3.7 m , we ﬁnd v0 y = 2 g ( Δy )max = 2 ( 9.80 m s 2 ) ( 3.7 m ) = 8.5 m s and if the angle of projection is θ = 45°, the launch speed is v0 = 3.26

(a)

v0 y sin θ

=

8.5 m s = 12 m s sin 45°

When a projectile is launched with speed v0 at angle θ0 above the horizontal, the initial velocity components are v0 x = v0 cos θ0 and v0 y = v0 sin θ0 . Neglecting air resistance, the vertical velocity when the projectile returns to the level from which it was launched (in this case, the ground) will be vy = − v0 y. From this information, the total time of ﬂight is found from vy = v0 y + ay t to be t total =

vyf − v0 y ay

=

− v0 y − v0 y −g

=

2 v0 y

or t total =

g

2 v0 sin θ0 g

Since the horizontal velocity of a projectile with no air resistance is constant, the horizontal distance it will travel in this time (i.e., its range) is given by ⎛ 2 v sin θ0 R = v0 x t total = ( v0 cos θ0 ) ⎜ 0 g ⎝

v02 sin ( 2θ0 ) ⎞ v02 2 s i n θ cos θ = = ( ) ⎟ 0 0 g ⎠ g

Thus, if the projectile is to have a range of R = 81.1 m when launched at an angle of θ0 = 45.0°, the required initial speed is v0 = (b)

(81.1 m ) ( 9.80 m s2 ) = 28.2 m s sin ( 90.0°)

With v0 = 28.2 m s and θ0 = 45.0°, the total time of ﬂight (as found above) will be t total =

(c)

Rg = sin ( 2θ0 )

2 v0 sin θ0 2 ( 28.2 m s ) sin ( 45.0°) = = 4.07 s 9.80 m s 2 g

Note that at θ0 = 45.0°, sin ( 2θ0 ) = 1 and that sin ( 2θ0 ) will decrease as θ0 is increased above this optimum launch angle. Thus, if the range is to be kept constant while the launch angle is increased above 45.0°, we see from v0 = Rg sin ( 2θ0 ) that the required initial velocity will increase.. continued on next page

56157_03_ch03_p049-088.indd 63

10/12/10 1:28:20 PM

64

Chapter 3

Observe that for θ0 < 90°, the function sin θ0 increases as θ0 is increased. Thus, increasing the launch angle above 45.0° while keeping the range constant means that both v0 and sin θ0 will increase. Considering the expression for t total given above, we see that the total time of flight will increase. 3.27

(a)

The time required for the ball to travel Δx = 36.0 m horizontally to the goal is found from Δx = v0 x t = ( v0 cos θ ) t as t = Δx v0 cos θ . At this time, the vertical displacement of the ball is given by Δy = v0 y t + 12 ay t 2. The vertical distance by which the ball clears the bar is d = Δy − h , or ⎛ 36.0 m ⎞ 1 ⎛ ⎞ 36.0 m d = v0 sin 53.0° ⎜ ⎟ + ( −9.80 m s 2 ) ⎜ ⎟ − 3.05 m ⎜ v0 cos 53.0° ⎟ 2 ⎝ ( 20.0 m s ) cos 53.0° ⎠ ⎝ ⎠ 2

yielding d = + 0.89 m. Thus, the ball clears the crossbar by 0.89 m . (b)

The ball reaches the plane of the goal post at t = Δx v0 cos θ , or t=

36.0 m

( 20.0 m s ) cos 53.0°

= 2.99 s

At this time, its vertical velocity is given by v y = v0 y + ay t as v y = ( 20.0 m s ) sin 53.0° + ( −9.80 m s 2 ) ( 2.99 s ) = −13.3 m s Since v y < 0, the ball has passed the peak of its arc and is descending when it crosses the crossbar. 3.28

(a)

With the origin at ground level directly below the window, the original coordinates of the ball are ( x , y) = (0, y0 ) .

(b)

v0 x = v0 cos θ0 = (8.00 m s ) cos ( −20.0°) = +7.52 m s v0 y = v0 sin θ0 = (8.00 m s ) sin ( −20.0°) = −2.74 m s

(c)

x = x 0 + v0 x t +

1 2 1 ax t = 0 + ( 7.52 m s ) t + ( 0 ) t 2 2 2

y = y0 + v0 y t +

1 2 1 ay t = y0 + ( −2.74 m s ) t + ( −9.80 m s 2 ) t 2 2 2

x = ( 7.52 m s ) t

y = y0 − ( 2.74 m s ) t − ( 4.90 m s 2 ) t 2

or (d)

orr

Since the ball hits the ground at t = 3.00 s, the x-coordinate at the landing site is xlanding = x t =3.00 s = ( 7.52 m s ) ( 3.00 s ) = 22.6 m

(e)

Since y = 0 when the ball reaches the ground at t = 3.00 s, the result of (c) above gives m⎞ ⎛ m ⎞ 2⎤ m⎞ m⎞ ⎡ 2 ⎛ y0 = ⎢ y + ⎛⎜ 2.74 = 0 + ⎛⎜ 2.74 ⎟ t + ⎜ 4.90 2 ⎟ t ⎥ ⎟ ( 3.00 s ) + ⎜ 4.90 2 ⎟ ( 3.00 s ) s s s s ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ t =3.00 s or

y0 = 52.3 m

continued on next page

56157_03_ch03_p049-088.indd 64

10/12/10 1:28:20 PM

Vectors and Two-Dimensional Motion

(f)

65

When the ball has a vertical displacement of Δy = −10.0 m, it will be moving downward with a velocity given by vy2 = v02 y + 2ay ( Δy ) as vy = − v02 y + 2ay ( Δy ) = −

( −2.74 m s )2 + 2 ( −9.80 m s2 ) ( −10.0 m ) = −14.3 m s

The elapsed time at this point is then t= 3.29

v y − v0 y ay

=

−14.3 m s − ( −2.74 m s ) = 1.18 s −9.80 m s 2

We choose our origin at the initial position of the projectile. After 3.0 s, it is at ground level, so the vertical displacement is Δy = − H . To ﬁnd H, we use Δy = v0 y t +

1 2 ay t , which becomes 2

1 2 − H = ⎡⎣(15 m s ) sin 25°⎤⎦ ( 3.0 s ) + ( −9.80 m s 2 ) ( 3.00 s ) , or H = 25 m 2 3.30

The initial velocity components of the projectile are v0 x = ( 300 m s ) cos 55.0° = 172 m s

and

while the constant acceleration components are

v0 y = ( 300 m s ) sin 55.0° = 246 m s ax = 0

and

ay = − g = −9.80 m s 2

The coordinates of where the shell strikes the mountain at t = 42.0 s are 1 x = v0 x t + ax t 2 = (172 m s ) ( 42.0 s ) + 0 = 7.22 × 10 3 m = 7.22 km 2 and 1 y = v0 y t + a y t 2 2 = ( 246 m s ) ( 42.0 s ) +

3.31

1 2 −9.80 m s 2 ) ( 42.00 s ) = 1.69 × 10 3 m = 1.69 km ( 2

The speed of the car when it reaches the edge of the cliff is v = v02 + 2a ( Δx ) = 0 + 2 ( 4.00 m s 2 ) ( 50.0 m ) = 20.0 m s

Now, consider the projectile phase of the car’s motion. The vertical velocity of the car as it reaches the water is vy = − v02 y + 2ay ( Δy ) = − ⎡⎣ − ( 20.0 m s ) sin 24.0°⎤⎦ + 2 ( − 9.80 m s 2 ) ( − 30.0 m ) 2

or

vy = − 25.6 m s

(b)

The time of ﬂight is t=

(a)

v y − v0 y ay

=

− 25.6 m s − ⎡⎣ − ( 20.0 m s ) sin 24.0°⎤⎦ = 1.78 s − 9.80 m s 2

The horizontal displacement of the car during this time is Δx = v0 x t = ⎡⎣( 20.0 m s ) cos 24.0°⎤⎦ (1.78 s ) = 32 .5 m

56157_03_ch03_p049-088.indd 65

10/12/10 1:28:21 PM

66

3.32

Chapter 3

The components of the initial velocity are v0 x = ( 40.0 m s ) cos 30.0° = 34.6 m s and h

v0 y = ( 40.0 m s ) sin 30.0° = 20.0 m s The time for the water to reach the building is t=

Δx 50.0 m = = 1.44 s v0 x 34.6 m

The height of the water at this time is h = Δy = v0 y t + 12 ay t 2 , or h = ( 20.0 m s )(1.44 s ) + 3.33

(a)

1 ( − 9.80 m s2 )(1.44 s)2 = 18.6 m 2

At the highest point of the trajectory, the projectile is moving horizontally with velocity components of vy = 0 and vx = v0 x = v0 cos θ = ( 60.0 m/s ) cos 30.0° = 52.0 m s

(b)

The horizontal displacement is Δx = v0 x t = (52.0 m s)(4.00 s) = 208 m and, from Δy = ( v0 sin θ )t + 12 ay t 2 , the vertical displacement is 1 2 Δy = ( 60.0 m s ) ( sin 30.0°)( 4.00 s ) + ( −9.80 m s 2 ) ( 4.00 s ) = 41.6 m 2 The straight line distance is d = ( Δx ) + ( Δy ) = ( 208 m ) + ( 41.6 m ) = 212 m

3.34

(a)

2

2

v0 53.0⬚

6.00 m

The horizontal displacement of the ball at time t is Δx = v x t = ( v0 cos 53.0°) t . Thus, if the ball travels 24.0 m horizontally in 2.20 s, the initial speed of the ball must be

2

7.00 m

2

24.0 m

v0 = (b)

Δx 24.0 m = = 18.1 m s t ⋅ cos 53.0° ( 2.20 s ) cos 53.0°

The vertical displacement of the ball at the instant it passes over the top of the wall (at t = 2.20 s, and 24.0 m horizontally from the launch point) is given by Δy = v0 y t + 12 ay t 2 , with v0 y = v0 sin 53.0°, as Δy = (18.1 m s ) sin 53.0° ( 2.20 s ) +

1 2 −9.80 m s 2 ) ( 2.20 s ) = 8.09 m ( 2

Thus, the ball clears the top of the wall by: (c)

d = 8.09 m − 7.00 m = 1.09 m

The times when the ball is 6.00 m above ground level are found from Δy = v0 y t + 12 ay t 2 , using Δy = 6.00 m, v0 y = v0 sin 53.0°, and ay = − g: 6.00 m = ⎡⎣(18.1 m s ) sin 53.0°⎤⎦ t − ( 4.90 m s 2 ) t 2

continued on next page

56157_03_ch03_p049-088.indd 66

10/12/10 1:28:21 PM

Vectors and Two-Dimensional Motion

or

67

( 4.90 ) t 2 − (14.5 s ) t + ( 6.00 s2 ) = 0

This has solutions of t = 0.497 s and t = 2.46 s. The ﬁrst solution is when the ball passes the 6.00 m level on the way up and the second solution is the time when it lands on the roof. Note that this is Δt = 2.46 s − 2.20 s = 0.26 s after it crosses over the wall. Thus, the ball lands a horizontal distance beyond the wall given by Δx = v x ( Δt ) = ( v0 cos 53.0°) Δt = (18.1 m s ) cos 53.0° ( 0.26 s ) = 2.8 m 3.35

(a)

The jet moves at 3.00 × 10 2 mi h due east relative to the air. Choosing a coordinate system with the positive x-direction eastward and the positive y-direction northward, the components of this velocity are

( v JA ) x = 3.00 ×10 2 (b)

( v JA ) y = 0

and

The velocity of the air relative to Earth is 1.00 × 10 2 mi h at 30.0° north of east. Using the coordinate system adopted in (a) above, the components of this velocity are

( v AE ) x = v AE and (c)

mi h

cos θ = (1.00 × 10 2 mi h ) cos30.0° = 86.6 mi h

( v AE ) y = v AE sin θ = (1.00 × 10 2

mi h ) sin30.0° = 50.0 mi h

Carefully observe the pattern of the subscripts in Equation 3.16 of the textbook. There, two objects (cars A and B) both move relative to a third object (Earth, E). The velocity of object A relative to object B is given in terms of the velocities of these objects relative to E as v AB = v AE − v BE . In the present case, we have two objects, a jet (J) and the air (A), both moving relative to a third object, Earth (E). Using the same pattern of subscripts as that in Equation 3.16, the velocity of the jet relative to the air is given by v JA = v JE − v AE

(d)

From the expression for v JA found in (c) above, the velocity of the jet relative to the ground is v JE = v JA + v AE . Its components are then

( v JE ) x = ( v JA ) x + ( v AE ) x = 3.00 × 10 2 and

mi h + 86.6 mi h = 3.87 × 10 2 mi h

( v JE ) y = ( v JA ) y + ( v AE ) y = 0 + 50.0

mi h = 50.0 mi h

This gives the magnitude and direction of the jet’s motion relative to Earth as v JE = and

2 y

⎛ ( v JE ) y θ = tan −1 ⎜ ⎜ ( v JE ) x ⎝

Therefore, 3.36

2

v JE x + v JE

=

(3.87 × 10

2

mi h ) + ( 50.0 mi h ) = 3.90 × 10 2 mi h 2

2

⎞ ⎛ 50.0 mi h ⎞ ⎟ = tan −1 ⎜ ⎟ = 7.36° 2 ⎟ ⎝ 3.87 × 10 mi h ⎠ ⎠

v JE = 3.90 × 10 2 mi h at 7.36° north of east

Choose a reference system with the positive x-axis in the eastward direction and the positive y-axis vertically upward. Then, the velocities of the car and the raindrops relative to Earth are: v CE = 50.0 km h in the + x -direction , and v RE = vrain in the negative y-direction . We also know continued on next page

56157_03_ch03_p049-088.indd 67

10/12/10 1:28:22 PM

68

Chapter 3

that the velocity of the rain relative to the car, v RC , is directed downward at θ = 60.0° from the vertical. From Equation 3.16 in the text, these relative velocities are related by v RC = v RE − v CE

v RE = v CE + v RC

or

Thus, these relative velocities form a 90°-vector triangle as shown below: +

We then have tan θ = and

(a)

vrain v RC = and

(b)

50.0 km h = vrain

v RE

VRE

VCE q

+x

VRC

50.0 km h = = 28.9 km h tan 60.0° 2

v CE + v RE

2

=

( 50.0 km h )2 + ( 28.9 km h )2 = 57.7 km h

v RC = 57.7 km h at 60.0° west of vertical

v RE = vrain in the negative y-direction and

3.37

v CE

v RE = 28.9 km h vertically downward

Choose a reference system with the positive x-axis in the northward direction and the positive y-axis vertically upward. Then, the accelerations of the car and the bolt (in free-fall) relative to Earth are: a CE = 2.50 m s 2 in the + x -direction , and a BE = 9.80 m s 2 in the negative y-direction . Similar to Equation 3.16 in the text for relative velocities, these accelerations are related to the acceleration of the bolt relative to the car, a BC , by a BC = a BE − a CE

or

a BE = a CE + a BC

Thus, these relative accelerations form a 90°-vector triangle as shown below: (a)

We see that a BC has components of

( a BC ) x = − ( a CE ) x = −2.50

(b)

+ (upward) aCE

m s2 aBE

and

( a BC ) y = − ( a BE ) y = −9.80

so

a BC = 2.50 m s 2 southward and 9.80 m s 2 downward

m s2

+x (northward) aBC

The acceleration of the bolt relative to Earth is that of a freely falling body, namely 9.80 m s 2 downward .

(c)

56157_03_ch03_p049-088.indd 68

Observers ﬁxed on Earth see the bolt follow a parabolic path with a vertical axis , the same as any freely falling body having a horizontal initial velocity.

10/12/10 1:28:22 PM

69

Vectors and Two-Dimensional Motion

3.38

North

As shown in the ﬁgure at the right, we choose a coordinate system with the y-axis lying at 15.0° east of north (i.e., along the direction of the original displacement, d 0, of the ship from the cutter. During the time t required for the cutter to intercept the ship, the ship undergoes displacement d S (directed at 40.0° east of north and 25.0° east of our y-axis) and the cutter undergoes displacement d C (directed at angle θ east of north or angle α relative to our y-axis).

15.0⬚

ds 25.0⬚

40.0⬚

dc

d0 15.0⬚ ␣

The magnitude of d 0 is given to be 20.0 km while the magnitudes of d S and d C will be dS = ( 26.0 km h ) t and dC = ( 50.0 km h ) t , respectively.

␪

East 15.0⬚ x

(a)

Equating the x-components in the vector triangle formed by the three displacements d 0 , d S, and d C, we see that dC sin α = dS sin 25.0° yielding

sin α =

or

( 50.0 km h ) t sin α = ( 26.0 km h ) t sin 25.0°

( 26.0 km h ) sin 25.0° 50.0 km h

= 0.220

or

α = 12.7°

or to intercept the ship, the cutter must steer a course directed at

θ = α + 15.0° = 12.7° + 15.0° = 27.7° east of north (b)

To ﬁnd the time required to intercept the ship, we equate the y-components in the vector triangle to ﬁnd ⎡⎣( 50.0 km h ) cos 12.7°⎤⎦ t = 20.0 km + ⎡⎣( 26.0 km h ) cos 25.0°⎤⎦ t or

[( 48.8 − 23.6)

yielding 3.39

t=

km h ] t = 20.0 km 20.0 km

( 48.8 − 23.6 ) km h

= 0.794 h = 47.6 min

During the trip of duration t, the displacement of the plane relative to the ground, d PG , is to have a magnitude of 750 km and be directed due north. We choose the positive y-axis to be directed northward and the positive x-axis directed eastward. During the trip, the plane’s displacement relative to the air d PA has magnitude d PA = ( 630 km h ) t and is directed at some angle α relative to the y-axis. The displacement of the air relative to the ground, d AG , has magnitude d AG = ( 35.0 km h ) t and is assumed to be at angle β from the y-axis.

(North) dAG ␤ dPG dPA

␣ x (East)

continued on next page

56157_03_ch03_p049-088.indd 69

10/12/10 1:28:23 PM

70

Chapter 3

Since these relative displacements are related by d PA = d PG − d AG , or d PG = d PA + d AG , they form a vector triangle as shown above. Equating the x-components in the vector triangle gives − d PA sin α + d AG sin β = 0, or − ( 630 km h ) ⋅ t ⋅ sin α + ( 35.0 km h ) ⋅ t ⋅ sin β = 0

and

35.0 ⎞ sin α = ⎛⎜ ⎟ sin β ⎝ 630 ⎠

[1]

Equating y-components in the vector triangle gives d PA cos α + d AG cos β = d PG , or ⎡⎣( 630 km h ) cos α + ( 35.0 km h ) cos β ⎤⎦ t = 750 km h (a)

The wind blows toward the south ( β = 180°) and is a headwind for the plane (α = 0°). Then, Equation [2] gives

[ 630 (b)

km h − 35.0 km h ] t = 750 km h

or

t=

750 km h = 1.26 h 595 km h

The wind blows northward as a tailwind (α = β = 0°), and Equation [2] yields

[ 630 (c)

[2]

km h + 35.0 km h ] t = 750 km h

or

t=

750 km h = 1.13 h 665 km h

The wind blows due East, so β = 90.0°. Then Equation [1] requires that 35.0 ⎞ sin α = ⎛⎜ ⎟ sin 90.0° = 0.056 ⎝ 630 ⎠

and

α = 3.18°

or the plane must ﬂy 3.18° W of N relative to the air to maintain a due north heading relative to the ground. Finally, Equation [2] for this case gives ⎡⎣( 630 km h ) cos 3.18° + ( 35.0 km h ) cos 90.0°⎤⎦ t = 750 km h or 3.40

(a)

t=

750 km h = 1.19 h ( 629 km h + 0 )

If the salmon (a projectile) is to have vy = 0 when Δy = +1.50 m, the required initial velocity in the vertical direction is given by vy2 = v02 y + 2ay Δy as v0 y = + vy2 − 2ay Δy = 0 − 2 ( −9.80 m s 2 ) ( +1.50 m ) = 5.42 m s The elapsed time for the upward ﬂight will be Δt =

v y − v0 y ay

=

0 − 5.42 m s = 0.553 s −9.80 m s 2

If the horizontal displacement at this time is to be Δx = +1.00 m, the required constant horizontal component of the salmon’s velocity must be v0 x = (b)

Δx 1.00 m = = 1.81 m s Δt 0.553 s

The speed with which the salmon must leave the water is then v0 = v02 x + v02 y =

(1.81 m s )2 + (5.42

m s ) = 5.71 m s 2

Yes , since v0 < 6.26 m s, the salmon is capable of making this jump.

56157_03_ch03_p049-088.indd 70

10/15/10 9:56:05 AM

Vectors and Two-Dimensional Motion

3.41

(a)

71

Both the student (S) and the water (W) move relative to Earth (E). The velocity of the student relative to the water is given by v SW = v SE − v WE , where v SE and v WE are the velocities of the student relative to Earth and the water relative to Earth, respectively. If we choose downstream as the positive direction, then v WE = + 0.500 m s, v SW = −1.20 m s when the student is going up stream, and v SW = +1.20 m s when the student moves downstream. The velocity of the student relative to Earth for each leg of the trip is

(v ) (v )

SE upstream

and

SE downstream

= v WE + ( v SW ) upstream = 0.500 m s + ( −1.20 m s ) = −0.700 m s = v WE + ( v SW )downstream = 0.500 m s + ( +1.20 m s ) = +1.70 m s

The distance (measured relative to Earth) for each leg of the trip is d = 1.00 km = 1.00 × 10 3 m. The times required for each of the two legs are t upstream = and

t downstream =

d v SE

=

upstream

d v SE

downstream

1.00 × 10 3 m = 1.43 × 10 3 s 0.700 m s =

1.00 × 10 3 m = 5.88 × 10 2 s 1.70 m s

so the time for the total trip is t total = t upstream + t downstream = 1.43 × 10 3 s + 5.88 × 10 2 s = 2.02 × 10 3 s (b)

If the water had been still ( v WE = 0 ), the speed of the student relative to Earth would have been the same for each leg of the trip, v SE = v SE upstream = v SE downstream = 1.20 m s. In this case, the time for each leg and the total time would have been d 1.00 × 10 3 m = = 8.33 × 10 2 s, and v SE 1.20 m s

t leg = (c)

3.42

(a)

The time savings going downstream with the curreent is always less than the extra time required to go the same distance against the current. The speed of the student relative to shore is vup = v − vs while swimming upstream and vdown = v + vs while swimming downstream. The time required to travel distance d upstream is then t up =

(b)

d d = vup v − vs

The time required to swim the same distance d downstream is t down =

(c)

d vdown

=

56157_03_ch03_p049-088.indd 71

d ( v + vs ) + d ( v − vs ) d d 2d v 2d v + = = 2 = 2 v − vs v + vs v − vs 1 − vs2 v 2 ( v − vs ) ( v + vs )

In still water, vs = 0 and the time for the complete trip is seen to be 2d v 2d = 1 − 0 v2 v 2d 2d v 2d v Note that tb = . Thus, when there is a current ( vs > 0 ), = 2 and that t a = 2 v v v − vs2 it is always true that t a > tb . tb = t a

(e)

d v + vs

The total time for the trip is therefore t a = t up + t down =

(d)

t total = 2t leg = 1.67 × 10 3 s

vs = 0

=

10/12/10 1:28:24 PM

72

3.43

Chapter 3

(a)

The bomb starts its fall with v0 y = 0 and v0 x = vplane = 275 m s . Choosing the origin at the location of the plane when the bomb is released and upward as positive, the y-coordinate of the bomb at ground level is y = −h = −3.00 × 10 3 m. The time required for the bomb to fall 2 is given by y = y0 + v0 y t + 12 ay t 2 as 0 = y0 + 0 + 12 ( − g ) t fall or t fall = 2 y0 g . With ax = 0, the horizontal distance the bomb travels during this time is d = v0 x t fall = v0 x

2 ( 3.00 × 10 3 m ) 2 y0 = ( 275 m s ) = 6.80 × 10 3 m = 6.80 km 9.80 m s 2 g

(b)

While the bomb is falling, the plane travels in the same horizontal direction with the same constant horizontal speed, vx = v0 x = vplane, as the bomb. Thus, the plane remains directly above the bomb as the bomb falls to the ground. When impact occurs, the plane is directly over the impact point, at an altituude of 3.00 km.

(c)

The angle, measured in the forward direction from the vertical, at which the bombsight must have been set is d 6.80 km ⎞ θ = tan −1 ⎛⎜ ⎞⎟ = tan −1 ⎛⎜ ⎟ = 66.2° ⎝h⎠ ⎝ 3.00 km ⎠

3.44

(a)

The time required for the woman, traveling at constant speed v1 relative to the ground, to travel distance L relative to the ground is t woman = L v1 .

(b)

With both the walkway (W) and the man (M) moving relative to Earth (E), we know that the velocity of the man relative to the moving walkway is v MW = v ME − v WE . His velocity relative to Earth is then v ME = v MW + v WE . Since all of these velocities are in the same direction, his speed relative to Earth is v ME = v MW + v WE = v2 + v1 . The time required for the man to travel distance L relative to the ground is then t man =

3.45

L L = v ME v1 + v2

Choose the positive direction to be the direction of each car’s motion relative to Earth. The velocity of the faster car relative to the slower car is given by v FS = v FE − v SE , where v FE = + 60.0 km h is the velocity of the faster car relative to Earth, and v SE = 40.0 km h is the velocity of the slower car relative to Earth. Thus, v FS = + 60.0 km h − 40.0 km h = + 20.0 km h and the time required for the faster car to move 100 m (0.100 km) closer to the slower car is t=

3.46

d 0.100 km 3600 s ⎞ = = 5.00 × 10 −3 h ⎛⎜ ⎟ = 18.0 s vFS 20.0 km h ⎝ 1h ⎠

The vertical displacement from the launch point (top of the building) to the top of the arc may be found from vy2 = v02 y + 2ay Δy with vy = 0 at the top of the arc. This yields Δy = and

vy2 − v02 y 2ay

0 − (12.0 m s ) = +7.35 m 2 ( −9.80 m s 2 ) 2

=

Δy = ymax − y0 gives ymax = y0 + Δy = y0 + 7.35 m

continued on next page

56157_03_ch03_p049-088.indd 72

10/12/10 1:28:24 PM

Vectors and Two-Dimensional Motion

(a)

73

If the origin is chosen at the top of the building, then y0 = 0 and ymax = 7.35 m. Thus, the maximum height above the ground is hmax = 50.0 m + ymax = 50.0 m + 7.35 m = 57.4 m The elapsed time from the point of release to the top of the arc is found from vy = v0 y + ay t as v y − v0 y

t= (b)

ay

=

0 − 12.0 m s = 1.22 s −9.80 m s 2

If the origin is chosen at the base of the building (ground level), then y0 = +50.0 m and hmax = ymax, giving hmax = y0 + Δy = 50.0 m + 7.35 m = 57.4 m The calculation for the time required to reach maximum height is exactly the same as that given above. Thus, t = 1.22 s

3.47

(a)

The known parameters for this jump are: θ0 = −10.0°, Δx = 108 m, Δy = −55.0 m, ax = 0, and ay = − g = −9.80 m s. Since ax = 0, the horizontal displacement is Δx = v0 x t = ( v0 cos θ0 ) t , where t is the total time of the ﬂight. Thus, t = Δx ( v0 cos θ0 ). The vertical displacement during the ﬂight is given by Δy = v0 y t +

1 2 gt 2 ay t = ( v0 sin θ0 ) t − 2 2

or

⎛ ⎞ g ⎛ Δx ⎞ ⎡ g ( Δx )2 ⎤ 1 Δx Δy = v0 sin θ0 ⎜ Δ x tan θ − = ⎟− ⎜ ( ) ⎥ 2 ⎢ 0 2 ⎜ v0 cos θ0 ⎟ 2 ⎝ v0 cos θ0 ⎟⎠ ⎣ 2 cos θ0 ⎦ v0 ⎝ ⎠

Thus,

( ) [ Δy − ( Δx ) tan θ0 ] = − ⎢ 2 cos 2 θ

(

2

)

⎡ g Δx ⎣

⎤ 1 ⎥ 2 0 ⎦ v0

2

− g ( Δx ) = 2 [ Δy − ( Δx ) tan θ0 ] cos 2 θ0 2

or

v0 =

yielding (b)

3.48

v0 =

− ( 9.80 m s 2 ) (108 m )

2

2 [ −55.0 m − (108 m ) tan ( −10.0°)] cos 2 ( −10.0°)

−1.143 × 10 5 m 3 s 2 = 40.5 m s −69.75 m

Rather than falling like a rock, the skier glides through the air much like a bird, prolonging the jump.

The cup leaves the counter with initial velocity components of ( v0 x = vi , v0 y = 0 ), and has acceleration components of ( ax = 0, ay = − g ) while in ﬂight. (a)

Applying Δy = v0 y t + 12 ay t 2 from when the cup leaves the counter until it reaches the ﬂoor gives −h = 0 +

(−g) 2

t2

so the time of the fall is

t = 2h g . continued on next page

56157_03_ch03_p049-088.indd 73

10/12/10 1:28:25 PM

74

Chapter 3

(b)

If the cup travels a horizontal distance d while falling to the ﬂoor, Δx = v0 x t gives vi = v0 x =

(c)

Δx = t

d 2h g

vi = d

or

The components of the cup’s velocity just before hitting the ﬂoor are vx = v0 x = vi = d

g 2h

v y = v0 y + a y t = 0 − g

and

v = vx2 + vy2 =

Thus, the total speed at this point is (d)

2h = − 2 gh g d2g + 2 gh 2h

The direction of the cup’s motion just before it hits the ﬂoor is ⎛v θ = tan −1 ⎜ y ⎝ vx

3.49

g 2h

⎛ ⎛ 2h ⎞ ⎞ ⎛ − 2 gh ⎞ ⎞ −1 −1 ⎜ −1 −1 ⎛ −2h ⎞ tan tan 2 g h = = ⎜⎜ ⎟⎟ ⎟ = tan ⎜ ⎜ ⎟ ⎟ ⎟ ⎜ d g 2h ⎟ ⎜ ⎟ d g ⎝ d ⎠ ⎠ ⎝ ⎠ ⎝ ⎠⎠ ⎝

AL = v1t = ( 90.0 km h ) ( 2.50 h ) = 225 km BD = AD − AB = AL cos 40.0° − 80.0 km = 92.4 km

From the triangle BLD,

or

BL =

( BD ) + ( DL )

BL =

( 92.4 km )2 + ( AL sin 40.0°) = 172 km

2

2

2

Since Car 2 travels this distance in 2.50 h, its constant speed is v2 = 3.50

(a)

172 km = 68.8 km h 2.50 h

After leaving the ledge, the water has a constant horizontal component of velocity. vx = v0 x = 1.50 m s 2 Thus, when the speed of the water is v = 3.00 m s, the vertical component of its velocity will be vy = − v 2 − vx2 = −

( 3.00 m s )2 − (1.50 m s )2 = −2.60 m s

The vertical displacement of the water at this point is Δy =

v y2 − v02 y 2a y

( −2.60 =

m s) − 0 = − 0.345 m 2 ( −9.80 m s 2 ) 2

or the water is 0.345 m below the ledge. (b)

If its speed leaving the water is 6.26 m/s, the maximum vertical leap of the salmon is Δyleap =

0 − v02 y 2ay

0 − ( 6.26 m s ) = 2.000 m 2 ( −9.80 m s ) 2

=

Therefore, the maximum height waterfall the salmon can clear is hmax = Δyleap + 0.345 m = 2.35 m

56157_03_ch03_p049-088.indd 74

10/12/10 1:28:25 PM

Vectors and Two-Dimensional Motion

3.51

75

The distance, s, moved in the ﬁrst 3.00 seconds is given by s = v0 t +

1 2 1 2 at = (100 m s ) ( 3.00 s ) + ( 30.0 m s 2 ) ( 3.000 s ) = 435 m 2 2

Choosing the origin at the point where the rocket was launched, the coordinates of the rocket at the end of powered ﬂight are x1 = s ( cos 53.0°) = 262 m

y1 = s ( sin 53.0°) = 347 m

and

The speed of the rocket at the end of powered ﬂight is v1 = v0 + at = 100 m s + ( 30.0 m s 2 ) ( 3.00 s ) = 190 m s

so the initial velocity components for the free-fall phase of the ﬂight are v0 x = v1 cos 53.0° = 114 m s (a)

v0 y = v1 sin 53.0° = 152 m s

and

When the rocket is at maximum altitude, vy = 0. The rise time during the free-fall phase can be found from vy = v0 y + ay t as t rise =

0 − v0 y ay

=

0 − 152 m = 15.5 s − 9.80 m s 2

The vertical displacement occurring during this time is 0 + 152 m s ⎞ ⎛ v + v0 y ⎞ 3 Δy = ⎜ y t rise = ⎛⎜ ⎟ (15.5 s ) = 1.18 × 10 m ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ The maximum altitude reached is then H = y1 + Δy = 347 m + 1.18 × 10 3 m = 1.53 × 10 3 m (b)

After reaching the top of the arc, the rocket falls 1.53 × 10 3 m to the ground, starting with zero vertical velocity ( v0 y = 0 ). The time for this fall is found from Δy = v0 y t + 12 ay t 2 as t fall =

2 ( Δy ) = ay

2 ( −1.53 × 10 3 m ) −9.80 m s 2

= 17.7 s

The total time of ﬂight is t = t powered + t rise + t fall = ( 3.00 + 15.5 + 17.7 ) s = 36.2 s (c)

The free-fall phase of the ﬂight lasts for t 2 = t rise + t fall = (15.5 + 17.7 ) s = 33.2 s The horizontal displacement occurring during this time is Δx = v0 x t 2 = (114 m s ) ( 33.2 s ) = 3.78 × 10 3 m and the full horizontal range is R = x1 + Δx = 262 m + 3.78 × 10 3 m = 4.04 × 10 3 m

3.52

Taking downstream as the positive direction, the velocity of the water relative to shore is v WS = +vWS, where vWS is the speed of the ﬂowing water. Also, if vCW is the common speed of the two canoes relative to the water, their velocities relative to the water are

( v CW )downstream = +vCW

and

( v CW )upstream = −vCW continued on next page

56157_03_ch03_p049-088.indd 75

10/12/10 1:28:26 PM

76

Chapter 3

The velocity of a canoe relative to the water can also be expressed as v CW = v CS − v WS. Applying this to the canoe moving downstream gives + vCW = + 2.9 m s − vWS

[1]

and for the canoe going upstream − vCW = −1.2 m s − vWS (a)

Adding Equations [1] and [2] gives 2 vWS = 1.7 m s,

(b)

(a)

so

vWS = 0.85 m s

Subtracting Equation [2] from [1] yields 2 vCW = 4.1 m s,

3.53

[2]

or

vCW = 2.1 m s

The time of ﬂight is found from Δy = v0 y t + 12 ay t 2 with Δy = 0, as t = 2 v0 y g. 2 v0 x v0 y

This gives the range as

R = v0 x t =

On Earth this becomes

REarth = 2 v0 x v0 y gEarth

and on the Moon,

RMoon = 2 v0 x v0 y gMoon

g

Dividing RMoon by REarth, we ﬁnd RMoon = ( gEarth gMoon ) REarth. With gMoon = gEarth 6, this gives RMoon = 6 REarth = 6 ( 3.0 m ) = 18 m

(b) 3.54

Similarly, RMars = ( gEarth gMars ) REarth = 3.0 m 0.38 = 7.9 m

(a)

Since the can returns to the same spot on the truck bed that it was thrown from, the can must have zero horizontal velocity relative to the truck. This means that, in the reference frame of the truck, the can was thrown vertically upward or at 0° to the vertical .

(b)

While the can was in the air, the truck moved 16.0 m, at a constant speed of 9.50 m s, relative to the ground. The time of ﬂight of the can was therefore t flight = (16.0 m ) ( 9.50 m s ). The time for the can to go from the truck bed to the top of its arc was then t up =

1 1 ⎛ 16.0 m ⎞ t flight = ⎜ ⎟ = 0.842 s 2 2 ⎝ 9.50 m s ⎠

Thus, v y = v0 y + ay t ( with v y = 0 at t = t up ) gives the initial vertical velocity, and hence the initial speed of the can relative to the truck, as v0 y = v y − ay t = 0 − ( −9.80 m s 2 ) ( 0.842 s ) = 8.25 m s (c)

Since the boy is stationary in the reference frame of the truck, he sees the can go straight upward and fall straight back doownward.

(d)

In a frame of reference ﬁxed on the ground, the can has a constant horizontal velocity v x = vtruck = 9.50 m s as it rises upward and falls back to the truck bed. Thus, wnward in this reference frame. the can follows a parabolic path opening dow

continued on next page

56157_03_ch03_p049-088.indd 76

10/15/10 9:56:26 AM

Vectors and Two-Dimensional Motion

(e)

77

The initial velocity of the can as seen by an observer on the ground is v0 = v02 x + v02 y = ( 9.50 ) + (8.25) 2

m s = 12.6 m s

2

directed at θ = tan −1 ( v0 y v0 x ) = tan −1 (8.25 9.50 ) = 41.0° above the horizontal or 3.55

(a)

v 0 = 12.6 m s at 41.0° above the horizontal eaastward line

The time for the ball to reach the fence is t=

Δx 130 m 159 m = = v0 x v0 cos 35° v0

At this time, the ball must be Δy = 21 m − 1.0 m = 20 m above its launch position, so Δy = v0 y t + 12 ay t 2 gives ⎛ 159 m ⎞ ⎛ 159 m ⎞ 20 m = v0 sin 35° ⎜ ⎟ − 4.90 m s 2 ) ⎜ ⎟ ⎜ v0 ⎟ ( ⎝ v0 ⎠ ⎝ ⎠

(

or

)

( 4.90 (159 m ) sin 35° − 20 m = v0 =

From which,

( 4.90

m s 2 ) (159 m )

2

2

v02

m s 2 ) (159 m )

2

(159 m ) sin 35° − 20 m

= 42 m s

(b)

From above, t = (159 m ) v0 = (159 m ) ( 42 m s ) = 3.8 s

(c)

When the ball reaches the wall (at t = 3.8 s), vx = v0 x = ( 42 m s ) cos 35° = 34 m s vy = v0 y + ay t = ( 42 m s ) sin 35° − ( 9.80 m s 2 ) ( 3.8 s ) = − 13 m s v = vx2 + vy2 =

3.56

(a)

(34

m s ) + ( −13 m s ) = 36 m s 2

When a projectile returns to the level it was launched from, the time to reach the top of the arc is one half of the total time of ﬂight. Thus, the elapsed time when the ﬁrst ball reaches maximum height is t = ( 3.00 s ) 2 = 1.50 s. Also, at this time, vy = 0, and vy = v0 y + ay t gives 0 = v0 y − ( 9.80 m s 2 ) (1.50 s )

(b)

or

v0 y = 14.7 m s

In order for the second ball to reach the same vertical height as the ﬁrst, the second must have the same initial vertical velocity as the ﬁrst. Thus, we ﬁnd v0 as v0 =

3.57

2

v0 y sin 30.0°

=

14.7 m s = 29.4 m s 0.500

The time of ﬂight of the ball is given by Δy = v0 y t + 12 ay t 2, with Δy = 0, as 1 ( −9.80 m s2 ) t 2 2 yielding a non-zero solution of t = 2.0 s 0 = ⎡⎣( 20 m s ) sin 30°⎤⎦ t +

The horizontal distance the football moves in this time is Δx = v0 x t = ⎡⎣( 20 m s ) cos 30°⎤⎦ ( 2.0 s ) = 35 m continued on next page

56157_03_ch03_p049-088.indd 77

10/12/10 1:28:27 PM

78

Chapter 3

(a)

Since Δx > 20 m, the receiver must run away from the quarterback, in the direction the ball was thrown , if he is to catch the ball.

(b)

The receiver has a time of 2.0 s to run a distance of d = Δx − 20 m = 15 m, so the required speed is v=

3.58

d 15 m = = 7.5 m s t 2.0 s

The horizontal component of the initial velocity is v0 x = v0 cos 40° = 0.766 v0 and the time required for the ball to move 10.0 m horizontally is t=

Δx 10.0 m 13.1 m = = v0 x 0.766 v0 v0

At this time, the vertical displacement of the ball must be Δy = y − y0 = 3.05 m − 2.00 m = 1.05 m Thus, Δy = v0 y t + 12 ay t 2 becomes

(

1.05 m = v0 sin 40.0°

which yields v0 = 3.59

2

2

2 0

0

1.05 m = 8.42 m −

or

) 13v.1 m + 12 ( − 9.80 m s ) (13.v1 m )

841 m 3 s 2 v02

841 m 3 s 2 = 10.7 m s 8.42 m − 1.05 m

Choose an origin where the projectile leaves the gun and let the y-coordinates of the projectile and the target at time t be labeled y p and yT , respectively. Then,

Target

( Δy ) p = y p − 0 = ( v0 sin θ0 ) t − ( g 2 ) t 2, and ( Δy )T = yT − h = 0 − ( g 2 ) t or yT = h − ( g 2 ) t 2

h 2

The time when the projectile will have the same x-coordinate as the target is t=

q0 x0

x0 Δx = v0 x v0 cos θ0

For a collision to occur, it is necessary that y p = yT at this time, or

(v

0

⎛ ⎞ g 2 x0 g sin θ0 ⎜ ⎟ − t = h − t2 ⎜ v0 cos θ0 ⎟ 2 2 ⎝ ⎠

)

which reduces to

tan θ0 =

h x0

This requirement is satisﬁed provided that the gun is aimed at the initial location of the target. ms the gun in this manner. Thus, a collision is guaranteed if the shooter aim

56157_03_ch03_p049-088.indd 78

10/12/10 1:28:28 PM

Vectors and Two-Dimensional Motion

3.60

(a)

79

The components of the vectors are Vector

x-component (cm)

y-component (cm)

d1m

0

104

d 2m

46.0

19.5

d1f

0

84.0

d 2f

38.0

20.2

The sums d m = d1m + d 2m and d f = d1f + d 2f are computed as 104 + 19.5 ⎞ 2 2 dm = ( 0 + 46.0 ) + (104 + 19.5) = 132 cm and θ = tan −1 ⎛⎜ ⎟ = 69.6° ⎝ 0 + 46.0 ⎠ 84.0 + 20.2 ⎞ 2 2 df = ( 0 + 38.0 ) + (84.0 + 20.2 ) = 111 cm and θ = tan −1 ⎛⎜ ⎟ = 70.0° ⎝ 0 + 38.0 ⎠ or d m = 132 cm at 69.6° and d f = 111 cm at 70.0° (b)

To normalize, multiply each component in the above calculation by the appropriate scale factor. The scale factor required for the components of d1m and d 2m is sm =

200 cm = 1.11 180 cm

and the scale factor needed for components of d1f and d 2f is sf =

200 cm = 1.19 168 cm

After using these scale factors and recalculating the vector sums, the results are d′m = 146 cm at 69.6° and d′f = 132 cm at 70.0° The difference in the normalized vector sums is Δd′ = d′m – d′f . Vector

x-component (cm)

y-component (cm)

d′m

50.9

137

−d′f

–45.1

–124

Δd′

Σx = 5.8

Σy = 13

Therefore, Δd ′ =

( Σx )2 + ( Σy )2 = ( 5.8 )2 + (13)2 cm = 14 cm, and

⎛ Σy ⎞ ⎛ 13 ⎞ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ = 66°, or Δd ′ = 14 cm at 66° ⎝ Σx ⎠ ⎝ 5.8 ⎟⎠

56157_03_ch03_p049-088.indd 79

10/12/10 1:28:28 PM

80

3.61

Chapter 3

To achieve maximum range, the projectile should be launched at 45° above the horizontal. In this case, the initial velocity components are v0 x = v0 y =

v0

v0 2

45° R

The time of ﬂight may be found from vy = v0 y − gt by recognizing that when the projectile returns to the original level, vy = − v0 y . Thus, the time of ﬂight is t=

− v0 y − v0 y −g

=

2 v0 y g

=

2 ⎛ v0 ⎞ v0 2 ⎜ ⎟= g⎝ 2 ⎠ g

2 ⎛ v ⎞ ⎛ v 2 ⎞ v0 The maximum horizontal range is then R = v0 x t = ⎜ 0 ⎟ ⎜ 0 ⎟= ⎝ 2 ⎠⎝ g ⎠ g

[1]

Now, consider throwing the projectile straight upward at speed v0 . At maximum height, vy = 0, and the time required to reach this height is found from vy = v0 y − gt as v0 y = 0 , which yields t = v0 g. Therefore, the maximum height the projectile will reach is 0 + v0 ⎝ 2

( Δy )max = ( vy )av t = ⎛⎜

2 ⎞ ⎛ v0 ⎞ v0 ⎟⎜ ⎟ = ⎠ ⎝ g ⎠ 2g

Comparing this result with the maximum range found in Equation [1] above reveals that

( Δy )max = R 2 provided the projectile is given the same initial speed in the two tosses.

If the girl takes a step when she makes the horizontal throw, she can likely give a higher initial speed for that throw than for the vertical throw. 3.62

(a)

x = v0 x t , so the time may be written as t = x v0 x . Thus, y = v0 y t − 12 gt 2 becomes ⎛ x y = v0 y ⎜ ⎝ v0 x or

(b)

⎞ 1 ⎛ x ⎞ ⎟− g⎜ ⎟ ⎠ 2 ⎝ v0 x ⎠

⎛ ⎛v ⎞ g ⎞ y = ⎜ − 2 ⎟ x2 + ⎜ 0y ⎟ x + 0 ⎝ 2 v0 x ⎠ ⎝ v0 x ⎠

Note that this result is of the general form y = ax 2 + bx + c with ⎛ g ⎞ a = ⎜− 2 ⎟, ⎝ 2 v0 x ⎠

3.63

2

⎛v ⎞ b = ⎜ 0y ⎟ , ⎝ v0 x ⎠

andd

c=0

In order to cross the river in minimum time, the velocity of the boat relative to the water (v BW) must be perpendicular to the banks (and hence perpendicular to the velocity v WS of the water relative to shore).

continued on next page

56157_03_ch03_p049-088.indd 80

10/15/10 9:57:03 AM

Vectors and Two-Dimensional Motion

81

The velocity of the boat relative to the water is v BW = v BS − v WS , where v BS is the velocity of the boat relative to shore. Note that this vector equation can be rewritten as v BS = v BW + v WS. Since v BW and v WS are to be perpendicular in this case, the vector diagram for this equation is a right triangle with v BS as the hypotenuse. Hence, velocity of the boat relative to shore must have magnitude 2 2 vBS = vBW + vWS =

(12 km h )2 + ( 5.0 km h )2 = 13 km h

and be directed at ⎛v ⎞ ⎛ 12 km h ⎞ θ = tan −1 ⎜ BW ⎟ = tan −1 ⎜ ⎟ = 67° v ⎝ 5.0 km h ⎠ ⎝ WS ⎠ to the direction of the current in the river (which is the same as the line of the riverbank). The minimum time to cross the river is t=

width of river 1.5 km ⎛ 60 min ⎞ = ⎜ ⎟ = 7.5 min vBW 12 km h ⎝ 1 h ⎠

During this time, the boat drifts downstream a distance of 1 h ⎞ ⎛ 10 3 m ⎞ 2 d = vWS t = ( 5.0 km h ) ( 7.5 min ) ⎛⎜ ⎟⎜ ⎟ = 6.3 × 10 m ⎝ 60 min ⎠ ⎝ 1 km ⎠ 3.64

For the ball thrown at 45.0°, the time of ﬂight is found from g ⎛ v ⎞ 0 = ⎜ 0 ⎟ t1 − t12 2 ⎝ 2⎠ which has the single non-zero solution of Δy = v0 y t +

t1 =

1 2 ay t 2

as

v0 2 g

The horizontal range of this ball is 2 ⎛ v ⎞ ⎛ v 2 ⎞ v0 R1 = v0 x t1 = ⎜ 0 ⎟ ⎜ 0 = ⎟ ⎝ 2 ⎠⎝ g ⎠ g Now consider the ﬁrst arc in the motion of the second ball, started at angle θ with initial speed v0 . Applied to this arc, Δy = v0 y t + 12 ay t 2 becomes

0 = ( v0 sin θ ) t 21 −

g 2 t 21 2

with non-zero solution t 21 =

2 v0 sin θ g

Similarly, the time of ﬂight for the second arc (started at angle θ with initial speed v0 2) of this ball’s motion is found to be t 22 =

2 ( v0 2 ) sin θ v0 sin θ = g g

The horizontal displacement of the second ball during the ﬁrst arc of its motion is ⎛ 2 v sin θ ⎞ v02 ( 2 sin θ cos θ ) v02 sin ( 2 θ ) R21 = v0 x t 21 = ( v0 cos θ ) ⎜ 0 = ⎟= g g g ⎝ ⎠ continued on next page

56157_03_ch03_p049-088.indd 81

10/12/10 1:28:29 PM

82

Chapter 3

Similarly, the horizontal displacement during the second arc of this motion is R22 =

( v0 2)2 sin ( 2 θ ) = 1 v02 sin ( 2 θ ) g

g

4

The total horizontal distance traveled in the two arcs is then R2 = R21 + R22 = (a)

5 v02 sin ( 2 θ ) g 4

Requiring that the two balls cover the same horizontal distance (that is, requiring that R2 = R1) gives 5 v02 sin ( 2 θ ) v02 = g g 4 This reduces to sin ( 2 θ ) = 4 5, which yields 2 θ = 53.1°, so θ = 26.6° is the required projection angle for the second ball.

(b)

The total time of ﬂight for the second ball is t 2 = t 21 + t 22 =

2 v0 sin θ v0 sin θ 3 v0 sin θ + = g g g

Therefore, the ratio of the times of ﬂight for the two balls is t 2 ( 3 v0 sin θ ) g 3 = sin θ = t1 2 v0 2 g

(

)

With θ = 26.6° as found in (a), this becomes t2 3 sin ( 26.6°) = 0.950 = t1 2 3.65

The initial velocity components for the daredevil are v0 x = v0 cos 45° and v0 y = v0 sin 45°, or v0 x = v0 y =

v0 25.0 m s = 2 2

The time required to travel 50.0 m horizontally is t=

Δx ( 50.0 m ) 2 = =2 2 s v0 x 25.0 m s

The vertical displacement of the daredevil at this time, and the proper height above the level of the cannon to place the net, is Δy = v0 y t + 3.66

1 2 ⎛ 25.0 m ay t = ⎜ ⎝ 2 2

s⎞ 1 2 ⎟⎠ 2 2 s − ( 9.80 m s ) 2 2 s 2

(

)

(

)

2

= 10.8 m

The vertical component of the salmon’s velocity as it leaves the water is v0 y = + v0 sin θ = + ( 6.26 m s ) sin 45.0° = + 4.43 m s When the salmon returns to water level at the end of the leap, the vertical component of velocity will be vy = − v0 y = − 4.43 m s. The time the salmon is out of the water is given by t1 =

v y − v0 y ay

=

− 4.43 m s − 4.43 m s = 0.904 s −9.80 m s 2

continued on next page

56157_03_ch03_p049-088.indd 82

10/12/10 1:28:30 PM

Vectors and Two-Dimensional Motion

83

The horizontal distance traveled during the leap is L = v0 x t1 = ( v0 cos θ ) t1 = ( 6.26 m s ) cos 45.0° ( 0.904 s ) = 4.00 m To travel this same distance underwater, at speed v = 3.58 m s, requires a time of t2 =

L 4.00 m = = 1.12 s v 3.58 m s

The average horizontal speed for the full porpoising maneuver is then vav = 3.67

Δxtotal 2L 2 ( 4.00 m ) = = = 3.95 m s Δt total t1 + t 2 0.904 s + 1.12 s

(a) and (b) Since the shot leaves the gun horizontally, v0 x = v0 and the time required to reach the target is t=

Δx x = v0 x v0

The vertical displacement occurring in this time is 1 2 1 ⎛ x ⎞ ay t = 0 − g ⎜ ⎟ 2 2 ⎝ v0 ⎠ which gives the drop as

2

Δy = − y = v0 y t +

2

y=

(c)

1 ⎛ x ⎞ g g ⎜ ⎟ = Ax 2 with A = 2 , where v0 is the muzzle velocity 2 ⎝ v0 ⎠ 2 v0

If x = 3.00 m, and y = 0.210 m, then A=

and

3.68

y 0.210 m −2 = m −1 2 = 2.33 × 10 x 2 ( 3.00 m )

v0 =

g = 2A

9.80 m s 2 = 14.5 m s 2 ( 2 .33 × 10 −2 m −1 )

The velocity of the wind relative to the boat, v WB, is given by v WB = v WE − v BE where v WE and v BE are the velocities of the wind and the boat relative to Earth, respectively. Choosing the positive x-direction as east and positive y as north, these relative velocities have components of

( v WE ) x = +17 knots

( v WE ) y = 0 knots

( v BE ) x = 0

( v BE ) y = +20 knots

so

( v WB ) x = ( v WE ) x − ( v BE ) x = +17 knots

( v WB ) y = ( v WE ) y − ( v BE ) y = −20 knots

(a)

The velocity of the wind relative to the boat has magnitude and direction of v WB =

( v WB )2x + ( v WB )2y

= (17 knots ) + ( −20 knots ) = 26 knots

and

⎛ (v ) θ = tan −1 ⎜ WB y ⎜ (v ) ⎝ WB x

or

v WB = 26 knots at 50° south of east

2

2

⎞ ⎛ −20 knots ⎞ = −50° ⎟ = tan −1 ⎜ ⎟ ⎟ ⎝ 17 knots ⎠ ⎠

continued on next page

56157_03_ch03_p049-088.indd 83

10/12/10 1:28:30 PM

84

3.69

Chapter 3

(b)

The component of this velocity parallel to the motion of the boat (that is, parallel to a northsouth line) is ( v WB ) y = −20 knots or 20 knots south .

(a)

Take the origin at the point where the ball is launched. Then x0 = y0 = 0 , and the coordinates of the ball at time t later are x = v0 x t = ( v0 cos θ0 ) t

y = v0 y t +

and

g 1 2 ay t = ( v0 sin θ0 ) t − ⎛⎜ ⎞⎟ t 2 2 ⎝2⎠

When the ball lands at x = R = 240 m, the y-coordinate is y = 0 and the elapsed time is found from g 0 = ( v0 sin θ0 ) t − ⎛⎜ ⎞⎟ t 2 ⎝2⎠ for which the non-zero solution is t=

2 v0 sin θ0 g

Substituting this time into the equation for the x-coordinate gives ⎛ 2 v sin θ0 x = 240 m = ( v0 cos θ0 ) ⎜ 0 g ⎝ Thus, if v0 = 50.0 m s, we must have sin ( 2θ0 ) =

( 240 m ) g v02

=

and

2θ0 = 180° − 70.2° = 109.8°

So, the two possible launch angles are θ0 = 35.1° (b)

⎞ ⎟ sin ( 2θ0 ) ⎠

( 240 m ) ( 9.80 m s2 ) = +0.941 ( 50.0 m s )2

2θ0 = 70.2°

with solutions of

⎛ v02 ⎞ ⎛ v02 ⎞ ⎟ = ⎜ ⎟ ( 2 sin θ0 cos θ0 ) = ⎜ ⎝ g ⎠ ⎝ g ⎠

and

θ0 = 54.9°

At maximum height, vy = 0, and the elapsed time is given by t peak =

(v )

y peak

− v0 y

ay

=

0 − v0 sin θ0 −g

or

t peak =

v0 sin θ0 g

The y-coordinate of the ball at this time will be ⎛ v sin θ0 ⎞ ⎛ g ⎞ v02 sin 2 θ0 v02 sin 2 θ0 g 2 ymax = ( v0 sin θ0 ) t peak − ⎛⎜ ⎞⎟ t peak = ( v0 sin θ0 ) ⎜ 0 = ⎟−⎜ ⎟ g g2 2g ⎝2⎠ ⎝ ⎠ ⎝2⎠ The maximum heights corresponding to the two possible launch angles are

and 3.70

(a)

( ymax )1 =

( 50.0 m s )2 sin 2 ( 35.1°)

( ymax )2 =

( 50.0 m s )2 sin 2 ( 54.9°)

2 ( 9.80 m s 2 )

2 ( 9.80 m s 2 )

= 42.2 m = 85.4 m

Consider the falling water to consist of droplets, each following a projectile trajectory. If the origin is chosen at the level of the pool and directly beneath the end of the channel, the parameters for these projectiles are: x0 = 0

y0 = h = 2.35 m

v0 x = 0.75 m s

v0 y = 0

ax = 0

ay = − g

continued on next page

56157_03_ch03_p049-088.indd 84

10/12/10 1:28:30 PM

Vectors and Two-Dimensional Motion

85

The elapsed time when the droplet reaches the pool is found from y − y0 = v0 y t + 12 ay t 2 as 0−h=0−

g 2 tp 2

tp =

or

2h g

The distance from the wall where the water lands is then R = xmax = v0 x t p = v0 x

2 ( 2.35 m ) 2h = ( 0.750 m s ) = 0.519 m g 9.80 m s 2

This space is too narrow for a pedestrian walkway. (b)

It is desired to build a model in which linear dimensions, such as the height hmodel and horizontal range of the water Rmodel, are one-twelfth the corresponding dimensions in the actual waterfall. If vmodel is to be the speed of the water ﬂow in the model, then we would have 2hmodel g

Rmodel = vmodel ( t p )model = vmodel or

vmodel = Rmodel

g 2hmodel

=

Ractual 12

g

2 ( hactual 12 )

=

1 12

⎛ g ⎜⎜ Ractual 2 h actual ⎝

⎞ vactual ⎟⎟ = 12 ⎠

and the needed speed of ﬂow in the model is vmodel = 3.71

(a)

vactual 0.750 m s = = 0.217 m s 12 12

Applying Δy = v0 y t +

1 2 ay t to the vertical motion of the ﬁrst snowball gives 2

0 = ⎡⎣( 25.0 m s ) sin 70.0°⎤⎦ t1 +

1 ( − 9.80 m s2 ) t12 2

which has the non-zero solution of t1 =

2 ( 25.0 m s ) sin 70.0° = 4.79 s 9.80 m s 2

as the time of ﬂight for this snowball. The horizontal displacement this snowball achieves is Δx = v0 x t1 = ⎡⎣( 25.0 m s ) cos 70.0°⎤⎦ ( 4.79 s ) = 41.0 m Now consider the second snowball, also given an initial speed of v0 = 25.0 m s, thrown at angle θ , and in the air for time t 2 . Applying Δy = v0 y t + 12 ay t 2 to its vertical motion yields 0 = ⎡⎣( 25.0 m s ) sin θ ⎤⎦ t 2 +

1 −9.80 m s 2 ) t 22 ( 2

which has a non-zero solution of t2 =

2 ( 25.0 m s ) sin θ = ( 5.10 s ) sin θ 9.80 m s 2

We require the horizontal range of this snowball be the same as that of the ﬁrst ball, namely Δx = v0 x t 2 = ⎡⎣( 25.0 m s ) cos θ ⎤⎦ [ ( 5.10 s ) sinθ ] = 41.0 m continued on next page

56157_03_ch03_p049-088.indd 85

10/12/10 1:28:31 PM

86

Chapter 3

This yields the equation sinθ cos θ =

41.0 m = 0.321 ( 25.0 m s ) ( 5.10 s )

From the trigonometric identity sin 2θ = 2 sinθ cos θ , this result becomes sin 2θ = 2 ( 0.321) = 0.642

so

2θ = 39.9°

and the required angle of projection for the second snowball is

θ = 20.0° above the horizontal (b)

From above, the time of ﬂight for the ﬁrst snowball is t1 = 4.79 s and that for the second snowball is t 2 = ( 5.10 s ) sin θ = ( 5.10 s ) sin 20.0° = 1.74 s Thus, if they are to arrive simultaneously, the time delay between the ﬁrst and second snowballs should be Δt = t1 − t 2 = 4.79 s − 1.74 s = 3.05 s

3.72

First, we determine the velocity with which the dart leaves the gun by using the data collected when the dart was ﬁred horizontally (v0 y = 0) from a stationary gun. In this case, Δy = v0 y t + 12 ay t 2 gives the time of ﬂight as t=

2 ( −1.00 m s ) = 0.452 s −9.80 m s 2

2Δy = ay

Thus, the initial speed of the dart relative to the gun is vDG = v0 x =

Δx 5.00 m = = 11.1 m s t 0.452 s

At the instant when the dart is ﬁred horizontally from a moving gun, the velocity of the dart relative to the gun may be expressed as v DG = v DE − v GE, where v DE and v GE are the velocities of the dart and gun relative to Earth, respectively. The initial velocity of the dart relative to Earth is therefore v 0 = v DE = v DG + v GE From the vector diagram, observe that v0 y = − vGE sin 45.0° = − ( 2.00 m s ) sin 45.0° = −1.41 m s and

v0 x = vDG + vGE cos 45.0° = 11.1 m s + ( 2.00 m s ) cos 45.00° = 12.5 m s

The vertical velocity of the dart after dropping 1.00 m to the ground is vy = − v02 y + 2ay Δy = −

( −1.41 m s )2 + 2 ( −9.80 m s2 ) ( −1.00 m ) = −4.65 m s

and the time of ﬂight is t=

v y − v0 y ay

=

−4.65 m s − ( −1.41 m s ) = 0.331 s −9.80 m s 2

continued on next page

56157_03_ch03_p049-088.indd 86

10/12/10 1:28:32 PM

87

Vectors and Two-Dimensional Motion

The displacement during the ﬂight is Δx = v0 x t = (12.5 m s ) ( 0.331 s ) = 4.14 m 3.73

(a)

First, use Δx = v0 x t + 12 ax t 2 to ﬁnd the time for the coyote to travel 70 m, starting from rest with constant acceleration ax = 15 m s 2. 2 ( 70 m ) = 3.1 s 15 m s 2

2Δx = ax

t1 =

The minimum constant speed the roadrunner must have to reach the edge in this time is v= (b)

Δx 70 m = = 23 m s t1 3.1 s

The initial velocity of the coyote as it goes over the edge of the cliff is horizontal and equal to v0 = v0 x = 0 + ax t1 = (15 m s 2 ) ( 3.1 s ) = 46 m s From Δy = v0 y t + 12 ay t 2, the time for the coyote to drop 100 m, with v0 y = 0, is t2 =

2 ( −100 m ) = 4.52 s −9.80 m s 2

2Δy = ay

The horizontal displacement of the coyote during his fall is Δx = v0 x t 2 + 12 ax t 22 = ( 46 m s ) ( 4.52 s ) + 3.74

1 2

(15 m s ) ( 4.52 s) 2

The known parameters for the ﬂight of the melon are:

v0

2

= 3.6 × 10 2 m

x

v0 x = v0 = 10.0 m s v0 y = 0

2

ax = 0

= (16.0 m)x

ay = − g = −9.80 m s 2 x0 = y0 = 0 At impact: y 2 = (16.0 m ) x The y-coordinate of the melon as a function of time is given by Δy = v0 y t + 12 ay t 2 as y−0=0+

1 (−g) t 2 2

or

y=−

gt 2 2

[1]

and the x-coordinate as a function of time is x = v0 x t = v0 t . Thus, the elapsed time may be expressed in terms of the x-coordinate as t = x v0 , so Equation [1] for the y-coordinate becomes 2

⎛ g ⎞ g⎛ x ⎞ y = − ⎜ ⎟ = − ⎜ 2 ⎟ x2 2 ⎝ v0 ⎠ ⎝ 2 v0 ⎠

and

⎛ g2 ⎞ y2 = ⎜ 4 ⎟ x 4 ⎝ 4 v0 ⎠

[2]

Since, at impact, we must have y 2 = (16.0 m ) x , Equation [2] says that when impact occurs we must have ⎛ g2 4 ⎝ 4 v0

(16.0 m ) x = ⎜

⎞ 4 ⎟x ⎠

or

⎛ ⎞ g2 x ⎜ 16.0 m − 4 x 3 ⎟ = 0 4 v0 ⎠ ⎝

[3] continued on next page

56157_03_ch03_p049-088.indd 87

10/15/10 9:57:26 AM

88

Chapter 3

Equation [3] has a trivial solution x = 0, but the x-coordinate of the melon at impact is a solution of the equation 16.0 m −

g2 3 x =0 4 v04

or

x3 =

(16.0 m ) ( 4 v04 ) g2

Thus, at impact 13

⎡ ( 64.0 m ) v04 ⎤ x=⎢ ⎥ g2 ⎣ ⎦ and the y-coordinate is

56157_03_ch03_p049-088.indd 88

13

⎡ ( 64.0 m ) (10.0 m s )4 ⎤ ⎥ =⎢ 2 2 ⎢ ⎥ 9 . 80 m s ( ) ⎣ ⎦

= 18.8 m

y = − (16.0 m ) x = − (16.0 m )(18.8 m ) = −17.3 m

10/12/10 1:28:32 PM

4 The Laws of Motion QUICK QUIZZES 1.

Newton’s second law says that the acceleration of an object is directly proportional to the resultant (or net) force acting on. Recognizing this, consider the given statements one at a time. (a)

True – If the resultant force on an object is zero (either because no forces are present or the vector sum of the forces present is zero), the object can still move with constant velocity.

(b)

False – An object that remains at rest has zero acceleration. However, any number of external forces could be acting on it, provided that the vector sum of these forces is zero.

(c)

True – When a single force acts, the resultant force cannot be zero and the object must accelerate.

(d)

True – When an object accelerates, a set containing one or more forces with a non-zero resultant must be acting on it.

(e)

False – Many external forces could be acting on an object with zero acceleration, provided that the vector sum of these forces is zero.

(f)

False – If the net force is in the positive x-direction, the acceleration will be in the positive x-direction. However, the velocity of an object does not have to be in the same direction as its acceleration (consider the motion of a projectile).

2.

Choice (b). The newton is a unit of weight, and the quantity (or mass) of gold that weighs 1 newton is m = 1 N g. Because the value of g is smaller on the Moon than on the Earth, someone possessing 1 newton of gold on the Moon has more gold than does a person having 1 newton of gold on Earth.

3.

(a)

False – When on an orbiting space station the astronaut is farther from the center of Earth than when on the surface of the Earth. Thus, the astronaut experiences a reduced gravitational force, but that force is not zero.

(b)

True – On or above the surface of Earth, the acceleration of gravity is inversely proportional to the square of the distance from the center of Earth. Therefore, when this distance is increased by a factor of 3, the acceleration of gravity decreases by a factor of 9.

(c)

False – The acceleration of gravity on the surface of a planet of mass M and volume V is 13 g = Fg m = GM R 2, where R = ⎡⎣3V 4π ⎤⎦ . If the mass and volume are doubled (as when 13 two identical planets coalesce), the radius becomes R′ = [ 3 ( 2V ) 4π ] = 21 3 R and the surface gravity becomes

( )

(

g ′ = GM ′ R ′ 2 = G ( 2M ) 21 3 R (d)

)

2

= 22 3 g

False – Unlike the weight of an object, its mass is independent of the local acceleration of gravity. Thus, one kilogram of gold contains the same amount of gold on the Moon as it does on Earth. 89

56157_04_ch04_p089-107.indd 89

10/15/10 3:31:26 PM

90

Chapter 4

4.

Choices (c) and (d). Newton’s third law states that the car and truck will experience equal magnitude (but oppositely directed) forces. Newton’s second law states that acceleration is inversely proportional to mass when the force is constant. Thus, the lower mass vehicle (the car) will experience the greater acceleration.

5.

Choice (c). In case (i), the scale records the tension in the rope attached to its right end. The section of rope in the man’s hands has zero acceleration, and hence, zero net force acting on it. This means that the tension in the rope pulling to the left on this section must equal the force F the man exerts toward the right on it. The scale reading in this case will then be F. In case (ii), the person on the left can be modeled as simply holding the rope tightly while the person on the right pulls. Thus, the person on the left is doing the same thing that the wall does in case (i). The resulting scale reading is the same whether there is a wall or a person holding the left side of the scale.

6.

Choice (c). The tension in the rope has a vertical component that supports part of the total weight of the woman and sled. Thus, the upward normal force exerted by the ground is less than the total weight.

7.

Choice (b). Friction forces are always parallel to the surfaces in contact, which, in this case, are the wall and the cover of the book. This tells us that the friction force is either upward or downward. Because the tendency of the book is to fall downward due to gravity, the friction force must be in the upward direction.

8.

Choice (b). The static friction force between the bottom surface of the crate and the surface of the truck bed is the net horizontal force on the crate that causes it to accelerate. It is in the same direction as the acceleration, toward the east.

9.

Choice (b). It is easier to attach the rope and pull. In this case, there is a component of your applied force that is upward. This reduces the normal force between the sled and the snow. In turn, this reduces the friction force between the sled and the snow, making it easier to move. If you push from behind, with a force having a downward component, the normal force is larger, the friction force is larger, and the sled is harder to move.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

Newton’s second law gives the net force acting on the crate as

(

)

Fnet = 95.0 N − fk = ( 60.0 kg ) 1.20 m s 2 = 72.0 N This gives the kinetic friction force as fk = 23.0 N, so choice (a) is correct. 2.

As the block slides down the frictionless incline, there is a constant net force directed down the incline (i.e., the tangential component of the weight of the block) acting on it. This force will give the block a constant acceleration down the incline, meaning that its speed down the incline will increase at a constant rate. Thus, the only correct choice is (c).

3.

From Newton’s law of universal gravitation, the force Earth exerts on an object on its surface is Fg = GM E m RE2 = mg, or the acceleration of gravity at Earth’s surface is g = GM E RE2 . If both the mass and radius of the Earth should double, so M E′ = 2M E and RE′ = 2RE , the acceleration of gravity at the surface would then be g′ = G

M E′

( RE′ )

2

⎛ 2M E ⎞ 1 ⎛ M E ⎞ g 9.80 m s 2 = G⎜ = 4.90 m s 2 = ⎜G 2 ⎟ = = 2 ⎟ 4R R 2 2 2 ⎝ ⎝ E ⎠ E ⎠

meaning that (b) is the correct answer.

56157_04_ch04_p089-107.indd 90

10/15/10 3:31:28 PM

The Laws of Motion

91

4.

If a net external force acts on an object, Newton’s second law (Fnet = ma) tells us that the object will have a non-zero acceleration, and choice (d) is correct. Since the net force is constant, the acceleration will be constant and choice (c) is false. With a non-zero acceleration, the velocity of the object will undergo either a change in magnitude (could be an increase or a decrease) and/or a change in direction. Thus, neither choice (b) nor (e) is necessarily true. While an object with a non-zero acceleration may come to rest momentarily, it will immediately start moving again, so choice (a) is correct. Thus, we see that choices (a) and (d) are true and all others are false.

5.

The tension, Fupper , in the vine at the point above the upper monkey must support the weight ⎛ ⎞ of both monkeys ⎜ i.e., Fupper = 2 Fg single ⎟ . However, the tension in the vine between the two ⎝ monkey ⎠ ⎛ ⎞ monkeys supports only the weight of the lower monkey ⎜ Flower = Fg single ⎟ , meaning that ⎝ monkey ⎠ Fupper Flower = 2 and choice (d) is correct.

( )

( )

y

6.

When the crate is held in equilibrium on the incline as shown in the sketch, Newton’s second law requires that ΣFx = ΣFy = 0. From ΣFx = + F g x − f s = 0, the magnitude of the friction force equals the component of gravitational force acting down the incline, or choice (e) is correct. Note that f s = f s max = m s n only when the crate is on the verge of starting to slide.

→

n

→

fs x

q →

q

Fg

7.

As stated by Newton’s third law of motion, “For every force of action, there is a force of reaction having equal magnitude but opposite direction.” Thus, the force exerted on the locomotive by the wall has the same magnitude as the force exerted on the wall by the locomotive, and (b) is the correct choice.

8.

Constant velocity means zero acceleration. From Newton’s second law, ΣF = ma, so the total (or resultant) force acting on the object must be zero if it moves at constant velocity. This means that (d) is the correct choice.

9.

Choose a coordinate system with the positive x-direction being east and the positive y-direction being north. Then the components of the four given forces are Ax = + 40 N, Ay = 0

Bx = 0, By = + 50 N

C x = − 70 N, C y = 0

Dx = 0, Dy = − 90 N

The components of the resultant (or net) force are Rx = Ax + Bx + C x + Dx = −30 N and Ry = Ay + By + C y + Dy = −40 N. Therefore, the magnitude of the net force acting on the object is R = Rx2 + Ry2 = 10.

11.

56157_04_ch04_p089-107.indd 91

( −30 N )2 + ( −40 N )2 = 50 N, or choice (a) is correct.

An object in equilibrium has zero acceleration ( a = 0 ), so both the magnitude and direction of the object’s velocity must be constant. Also, Newton’s second law states that the net force acting on an object in equilibrium is zero. The only untrue statement among the given choices is (d), untrue because the value of the velocity’s constant magnitude need not be zero. The box will accelerate in the direction of the resultant force acting on it. The only horizontal forces present are the force exerted by the manager and the friction force. Since the acceleration is given to be in the direction of the force applied by the manager, the magnitude of this force must exceed that of the opposing friction force; however, the friction force is not necessarily zero. Thus, choice (b) is correct.

10/15/10 3:31:29 PM

92

Chapter 4

12.

As the trailer leaks sand at a constant rate, the total mass of the vehicle (truck, trailer and remaining sand) decreases at a steady rate. Then, with a constant net force present, Newton’s second law states that the magnitude of the vehicle’s acceleration ( a = Fnet m ) will steadily increase. Choice (b) is the correct answer.

13.

When the truck accelerates forward, its natural tendency is to slip from beneath the crate, leaving the crate behind. However, friction between the crate and the bed of the truck acts in such a manner as to oppose this relative motion between truck and crate. Thus, the friction force acting on the crate will be in the forward horizontal direction and tend to accelerate the crate forward. The crate will slide only when the coefﬁcient of static friction is inadequate to prevent slipping. The correct response to this question is (c).

14.

Assuming that the cord connecting m1 and m2 has constant length, the two masses are a ﬁxed distance (measured along the cord) apart. Thus, their speeds must always be the same, which means that their accelerations must have equal magnitudes. The magnitude of the downward acceleration of m2 is given by Newton’s second law as a2 =

ΣFy m2

=

⎛ T ⎞ m2 g − T = g−⎜ m2 34.

(a) TA > TB

(b) a1 = a2

(c) Yes, by Newton’s 3rd law. 36.

(a) a1 = a2 = 6.53 m s2

(b) T = 32.7 N

38.

(a) 36.8 N

(b) 2.44 m s2

40.

(a) a = 0

(b) 0.70 m s2

42.

(a) See Solution.

(b) static friction between blocks

(c) 1.22 m

(c) a = F 4m 44.

(a) 3.43 m s2

(b) 3.14 m s2

46.

(a) ax = − m k g

(b) Δx = v02 2m k g

48.

(a) 0.368 m s2

(b) 1.28 m s2 down the incline

50.

(a) 98.6 m

(b) 16.4 m

52.

(a) 15.0 lb

(b) 5.00 lb

56157_04_ch04_p089-107.indd 94

(c) zero

10/15/10 3:31:34 PM

The Laws of Motion

54.

(a) 0.125 m s2

56.

21.4 N

58.

2.6 m s 2 northward

60.

(b) 39.7 N

(c) 0.235

(a) 50.0 N

(b) m s ≥ 0.500

(c) 25.0 N

62.

(a) See Solution.

(b) 55.2°

(c) 167 N

64.

(a) 1.63 m s2

(b) T1 = 57.0 N, T2 = 24.5 N

66.

1.18 × 10 3 N

68.

(a) a1 = a2 2

95

(b) T1 = 4m1m2 g ( m1 + 4m2 ), T2 = 2m1m2 g ( m1 + 4m2 ) (c) 70.

72.

a1 = m1 g ( m1 + 4m2 ) , a2 = 2m1 g ( m1 + 4m2 )

(a) See Solution.

(b) No. For q ≤ q c , f ≤ fs,max = m s n.

(c) See Solution.

(d) m k = tanq c′

(a) See Solution.

(b) a = F ( mb + mr )

(c) Tleft = ⎡⎣ mb ( mb + mr ) ⎤⎦ F end (d) The tension is uniform throughout a light cord. 74.

173 lb

76.

(a) 7.25 × 103 N

78.

104 N

80.

6.00 cm

82.

5.10 × 10 3 N

84.

(a) a1 = 2a2

(b) 4.57 m s2

(b) a2 = m2 g ( 4m1 + m2 ) ≈ 12.7 N ( 4m1 + 1.30 kg ) (c) 9.80 m s2 (d) a2 → 0 (e) T = m2 g 2 = 6.37 N (f) Yes. If m1

56157_04_ch04_p089-107.indd 95

m2 , m2 is in near free-fall; If m1

m2 , m2 is held almost stationary.

10/15/10 3:31:37 PM

96

Chapter 4

PROBLEM SOLUTIONS

4.1

⎛ 2000 lbs ⎞ ⎛ 4.448 N ⎞ 4 w = 2 tons ⎜ ⎜ ⎟ = 2 × 10 N ⎝ 1 ton ⎟⎠ ⎝ 1 lb ⎠

4.2

From v = v0 + at, the acceleration given to the football is

(

)

aav =

v − v0 10 m s − 0 = = 50 m s 2 0.20 s t

Then, from Newton’s second law, we ﬁnd

(

)

Fav = m aav = ( 0.50 kg ) 50 m s 2 = 25 N 4.3

4.4

4.5

(

)

(a)

ΣFx = max = ( 6.0 kg ) 2.0 m s 2 = 12 N

(b)

ax =

(a)

Action: The hand exerts a force to the right on the spring. Reaction: The spring exerts an equal magnitude force to the left on the hand. Action: The wall exerts a force to the left on the spring. Reaction: The spring exerts an equal magnitude force to the right on the wall. Action: Earth exerts a downward gravitational force on the spring. Reaction: The spring exerts an equal magnitude gravitational force upward on the Earth.

(b)

Action: The handle exerts a force upward to the right on the wagon. Reaction: The wagon exerts an equal magnitude force downward to the left on the handle. Action: Earth exerts an upward contact force on the wagon. Reaction: The wagon exerts an equal magnitude downward contact force on the Earth. Action: Earth exerts a downward gravitational force on the wagon. Reaction: The wagon exerts an equal magnitude gravitational force upward on the Earth.

(c)

Action: The player exerts a force upward to the left on the ball. Reaction: The ball exerts an equal magnitude force downward to the right on the player. Action: Earth exerts an downward gravitational force on the ball. Reaction: The ball exerts an equal magnitude gravitational force upward on the Earth.

(d)

Action: M exerts a gravitational force to the right on m. Reaction: m exerts an equal magnitude gravitational force to the left on M.

(e)

Action: The charge +Q exerts an electrostatic force to the right on the charge –q. Reaction: The charge – q exerts an equal magnitude electrostatic force to the left on the charge + Q.

(f)

Action: The magnet exerts a force to the right on the iron. Reaction: The iron exerts an equal magnitude force to the left on the magnet.

ΣFx 12 N = = 3.0 m s 2 4.0 kg m

The weight of the bag of sugar on Earth is ⎛ 4.448 N ⎞ wE = mgE = ( 5.00 lbs ) ⎜ = 22.2 N ⎝ 1 lb ⎟⎠ If gM is the free-fall acceleration on the surface of the Moon, the ratio of the weight of an object on the Moon to its weight when on Earth is w M wE = mgM mgE = gM gE, so w M = wE ( gM gE ). Hence, the weight of the bag of sugar on the Moon is

⎛ 1⎞ w M = ( 22.2 N ) ⎜ ⎟ = 3.70 N ⎝ 6⎠ continued on next page

56157_04_ch04_p089-107.indd 96

10/15/10 3:31:41 PM

The Laws of Motion

97

On Jupiter, its weight would be ⎛g ⎞ wJ = wE ⎜ J ⎟ = ( 22.2 N )( 2.64 ) = 58.6 N ⎝ gE ⎠ The mass is the same at all three locations, and is given by m=

4.6

a=

wE ( 5.00 lb ) ( 4.448 N lb ) = = 2.27 kg gE 9.80 m s 2

ΣF 7.5 × 10 5 N = = 5.0 × 10 −2 m s2 m 1.5 × 10 7 kg

and v = v0 + at gives t= 4.7

v − v0 80 km h − 0 ⎛ 0.278 m s ⎞ ⎛ 1 min ⎞ = ⎜ ⎟ = 7.4 min 5.0 × 10 −2 m s2 ⎜⎝ 1 km h ⎟⎠ ⎝ 60 s ⎠ a

If F = 825 N is the upward force exerted on the man by the scales, the upward acceleration of the man (and hence, the acceleration of the elevator) is ay =

4.8

4.9

ΣFy mman

=

825 N − mman g 825 N 825 N = −g= − 9.8 m s 2 = 1.2 m s 2 mman 75 kg mman

(a)

The sphere has a larger mass than the feather. Hence, the sphere experiences a larger gravitational force Fg = mg than does the feather.

(b)

The time of fall is less for the sphere than for the feather. This is because air resistance affects the motion of the feather more than that of the sphere.

(c)

In a vacuum, the time of fall is the same for the sphere and the feather. In the absence of air resistance, both objects have the free-fall acceleration g.

(d)

In a vacuum, the total force on the sphere is greater than that on the feather. In the absence of air resistance, the total force is just the gravitational force, and the sphere weighs more than the feather.

(a)

From v 2y = v02 y + 2a y ( Δy ), with Δy = ( 2 3) (1.50 m ) = 1.00 m, we ﬁnd ay =

(b)

v 2y − v02 y 2 ( Δy )

=

(6.00

m s ) − ( 3.00 m s ) = 13.5 m s 2 (1.00 m ) 2

2

We apply Newton’s second law to the vertical motion of the ﬁsh with F being the upward force exerted by the tail ﬁn. This gives

ΣFy = may or

(

⇒

F − mg = ma y

)

(

)

F = m a y + g = ( 61.0 kg ) 13.5 m s2 + 9.80 m s 2 = 1.42 × 10 3 N v 2 − v02 ( 320 m s ) − 0 = 2 ( 0.82 m ) 2 ( Δx ) 2

4.10

The acceleration of the bullet is given by Then,

56157_04_ch04_p089-107.indd 97

a=

⎡ ( 320 m s )2 ⎤ 2 ΣF = ma = 5.0 × 10 −3 kg ⎢ ⎥ = 3.1 × 10 N 2 0.82 m ( ) ⎢⎣ ⎥⎦

(

)

10/15/10 3:31:42 PM

98

4.11

Chapter 4

(a)

y

From the second law, the acceleration of the boat is a=

ΣF 2 000 N − 1 800 N = = 0.200 m s 2 m 1 000 kg

f 1 800 N F 2 000 N

x →

F

(b)

The distance moved is →

(

)

1 1 2 Δx = v0 t + at 2 = 0 + 0.200 m s 2 (10.0 s ) = 10.0 m 2 2

)

v = v0 + at = 0 + 0.200 m s 2 (10.0 s ) = 2.00 m s

Choose the positive y-axis in the forward direction. We resolve the forces into their components as y-component

400 N

200 N

346 N

450 N

–78.1 N

443 N

Resultant

ΣFx = 122 N

ΣFy = 789 N

0N

x-component

F 2 45

Force

y 10° N

(a)

30°

00

The ﬁnal velocity is

4

(c)

F1

4.12

(

f

x

The magnitude and direction of the resultant force is FR =

( ΣFx )2 + ( ΣFy )

2

= 798 N

⎛ ΣF ⎞ q = tan −1 ⎜ x ⎟ = 8.79° to right of y-axis ⎝ ΣFy ⎠

Thus, FR = 798 N at 8.79° to the right of the forward direction . (b)

The acceleration is in the same direction as FR and has magnitude a=

4.13

FR 798 N = = 0.266 m s 2 m 3 000 kg

Taking eastward as the positive x-direction, the average horizontal acceleration of the car is ax =

v x − v0 x 25.0 m s − 0 = = +5.00 m s 2 5.00 s Δt

Thus, the average horizontal force acting on the car during this 5.00-s period is

ΣFx = max = ( 970 kg ) ( + 5.00 m s2 ) = + 4.85 × 103 N = 4.85 kN eastward 4.14

(a)

With the wind force being horizontal, the only vertical force acting on the object is its own weight, mg. This gives the object a downward acceleration of ay =

ΣFy m

=

−mg = −g m

The time required to undergo a vertical displacement Δy = −h, starting with initial vertical velocity v0 y = 0, is found from Δy = v0 y t + 12 a y t 2 as −h = 0 −

g 2 t 2

or

t=

2h g

continued on next page

56157_04_ch04_p089-107.indd 98

10/15/10 3:31:45 PM

The Laws of Motion

99

(b)

The only horizontal force acting on the object is that due to the wind, so ΣFx = F and the ΣFx F horizontal acceleration will be ax = = m m

(c)

With v0 x = 0, the horizontal displacement the object undergoes while falling a vertical distance h is given by Δx = v0 x t + 12 ax t 2 as 2

Δx = 0 + (d)

1 ⎛ F ⎞ ⎛ 2h ⎞ Fh = ⎜ ⎟ 2 ⎝ m ⎠ ⎜⎝ g ⎟⎠ mg

The total acceleration of this object while it is falling will be a = ax2 + a y2 =

4.15

( F m )2 + ( −g )2

( F m )2 + g 2

=

Starting with v0 y = 0 and falling 30 m to the ground, the velocity of the ball just before it hits is

(

)

v1 = − v02 y + 2a y Δy = − 0 + 2 −9.80 m s 2 ( −30 m ) = −24 m s On the rebound, the ball has v y = 0 after a displacement Δy = +20 m. Its velocity as it left the ground must have been

(

)

v2 = + v 2y − 2a y Δy = + 0 − 2 −9.80 m s 2 ( 20 m ) = +20 m s Thus, the average acceleration of the ball during the 2.0-ms contact with the ground was aav =

v2 − v1 +20 m s − ( −24 m s ) = = + 2.2 × 10 4 m s2 Δt 2.0 × 10 −3 s

upward

The average resultant force acting on the ball during this time interval must have been

(

)

Fnet = maav = ( 0.50 kg ) +2.2 × 10 4 m s2 = +1.1 × 10 4 N or 4.16

F net = 1.1 × 10 4 N upward

Since the two forces are perpendicular to each other, their resultant is FR = a=

Thus, 4.17

(a)

⎛ 390 N ⎞ q = tan −1 ⎜ = 65.2° N of E ⎝ 180 N ⎟⎠

(180 N )2 + ( 390 N )2 = 430 N, at

FR 430 N = = 1.59 m s 2 m 270 kg

or

a = 1.59 m s 2 at 65.2° N of E y

Since the burglar is held in equilibrium, the tension in the vertical cable equals the burglar’s weight of 600 N . Now, consider the junction in the three cables: ΣFy = 0, giving T2 sin 37.0° − 600 N = 0 or

T2 =

→ T2 → T1

37.0° x w 600 N

600 N = 997 N in the inclined cable sin 37.0°

Also, ΣFx = 0, which yields T2 cos 37.0° − T1 = 0 or

T1 = ( 997 N ) cos 37.0° = 796 N in the horizontal cable continued on next page

56157_04_ch04_p089-107.indd 99

10/15/10 3:31:47 PM

100

Chapter 4

(b)

If the left end of the originally horizontal cable was attached to a point higher up the wall, the tension in this cable would then have an upward component. This upward component would support part of the weight of the cat burglar, thus decreasing the tension in the cable on the right .

4.18

Using the reference axis shown in the sketch at the right, we see that

y x 14.0°

ΣFx = T cos14.0° − T cos14.0° = 0 and

14.0°

ΣFy = −T sin14.0° − T sin14.0° = −2T sin14.0°

T

T

Thus, the magnitude of the resultant force exerted on the tooth by the wire brace is R= or 4.19

( ΣFx )2 + ( ΣFy )

2

= 0 + ( −2T sin14.0° ) = 2T sin14.0° 2

R = 2 (18.0 N ) sin14.0° = 8.71 N

From ΣFx = 0, T1 cos 30.0° − T2 cos 60.0° = 0 or

T2 = (1.73) T1

y → T2

[1]

→

T1

60.0°

The tension in the vertical cable is the full weight of the feeder, or Tvertical = 150 N

30.0° 150 N

x

Then ΣFy = 0 becomes T1 sin 30.0° + (1.73 T1 ) sin 60.0° − 150 N = 0 which gives T1 = 75.1 N in the right side cable . Finally, Equation [1] above gives T2 = 130 N in the left side cable . 4.20

If the hip exerts no force on the leg, the system must be in equilibrium with the three forces shown in the freebody diagram.

y 11

0N

40°

Thus ΣFx = 0 becomes w2 cosa = (110 N ) cos 40°

[1]

→ w2

α 220 N

x

From ΣFy = 0, we ﬁnd w2 sina = 220 N − (110 N ) sin 40°

[2]

Dividing Equation [2] by Equation [1] yields ⎛ 220 N − (110 N ) sin 40° ⎞ a = tan −1 ⎜ = 61° (110 N ) cos 40° ⎟⎠ ⎝ Then, from either Equation [1] or [2],

56157_04_ch04_p089-107.indd 100

w2 = 1.7 × 10 2 N

10/15/10 3:31:51 PM

The Laws of Motion

4.21

(a)

Force diagrams of the two blocks are shown at the right. Note that each block experiences a downward gravitational force

(

T1

101

T2

)

Fg = ( 3.50 kg ) 9.80 m s 2 = 34.3 N T2

Also, each has the same upward acceleration as the elevator, in this case a y = +1.60 m s 2 .

Fg

Fg

Applying Newton’s second law to the lower block: ΣFy = ma y or

⇒

T2 − Fg = ma y

(

)

T2 = Fg + ma y = 34.3 N + ( 3.50 kg ) 1.60 m s 2 = 39.9 N

Next, applying Newton’s second law to the upper block: ΣFy = ma y or (b)

⇒

T1 − T2 − Fg = ma y

(

)

T1 = T2 + Fg + ma y = 39.9 N + 34.3 N + ( 3.50 kg ) 1.60 m s 2 = 79.8 N

As the acceleration of the system increases, we wish to ﬁnd the value of a y when the upper string reaches its breaking point (i.e., when T1 = 85.0 N). Making use of the general relationships derived in (a) above gives

(

)

T1 = T2 + Fg + ma y = Fg + ma y + Fg + ma y = 2Fg + 2ma y or

4.22

(a)

ay =

T1 − 2Fg 2m

=

85.0 N − 2 ( 34.3 N ) = 2.34 m s 2 2 ( 3.50 kg )

Force diagrams of the two blocks are shown at the right. Note that each block experiences a downward gravitational force Fg = mg.

T1

T2

Upper Block m = 3.50 kg

Also, each has the same upward acceleration as the elevator, a y = + a.

T2

Lower Block m = 3.50 kg

Fg

Fg

Applying Newton’s second law to the lower block: ΣFy = ma y

⇒

T2 − Fg = ma y

or

T2 = mg + ma = m ( g + a )

Next, applying Newton’s second law to the upper block: ΣFy = ma y or

56157_04_ch04_p089-107.indd 101

⇒

T1 − T2 − Fg = ma y

T1 = T2 + Fg + ma y = ( mg + ma ) + mg + ma = 2 ( mg + ma ) = 2T2

(b)

Note that T1 = 2T2 , so the upper string breaks first as the acceleration of the system increases.

(c)

When the upper string breaks, both blocks will be in free-fall with a = −g. Then, using the results of part (a), T2 = m ( g + a ) = m ( g − g ) = 0 and T1 = 2T2 = 0 .

10/15/10 3:31:54 PM

102

4.23

Chapter 4

m = 1.00 kg and mg = 9.80 N

25.0 m a

⎛ 0.200 m ⎞ a = tan −1 ⎜ = 0.458° ⎝ 25.0 m ⎟⎠

→

0.200 m

T

25.0 m a →

T

→

mg

Since a y = 0, this requires that ΣFy = T sina + T sina − mgg = 00, 2T sina = mg

giving 4.24

9.80 N = 613 N 2sina

In each case, the scale is measuring the tension in the cord connecting to it. This tension can be determined by applying Newton’s second law (with a = 0 for equilibrium) to the object attached to the end of this cord. (a)

ΣFy = 0 or

⇒

(

2

)=

49.0 N

The solution to part (a) is also the solution to part (b).

(c)

Neglecting the mass of the pulley and the cord supporting the two objects of mass m, we ﬁnd

or (d)

m mg

(b)

⇒

T

T − mg = 0

T = mg = ( 5.00 kg ) 9.80 m s

ΣFy = 0

T

T − mg − mg = 0

(

)

mg

In the ﬁgure at the right, n is the normal force exerted on the block by the frictionless incline.

or

x

n T

⇒ T − mg sinq = 0

T = 24.5 N

q q

mg

The forces on the bucket are the tension in the rope and the weight of the bucket, mg = ( 5.0 kg ) 9.80 m s 2 = 49 N. Choose the positive direction upward and use Newton’s second law:

(

mg

y

T = mg sinq = ( 5.00 kg ) ( 9.80 m s 2 ) sin 30.0°

and

m

m

T = 2mg = 2 ( 5.00 kg ) 9.80 m s 2 = 98.0 N

ΣFx = 0

4.25

T=

or

)

→

T

ΣFy = ma y

(

T − 49 N = ( 5.0 kg ) 3.0 m s 2

→

)

a

T = 64 N →

mg

4.26

With the truck accelerating in a forward direction on a horizontal roadway, the acceleration of the crate is the same as that of the truck as long as the cord does not break. Applying Newton’s second law to the horizontal motion of the block gives ΣFx = max Thus

56157_04_ch04_p089-107.indd 102

amax =

⇒

T = ma or a = T m

Tmax 68 N = = 2.1 m s2 32 kg m

y n m m

x T mg

10/15/10 3:31:57 PM

The Laws of Motion

4.27

We choose reference axes that are parallel to and perpendicular to the incline as shown in the force diagrams at the right. Since both blocks are in equilibrium, ax = a y = 0 for each block. Then, applying Newton’s second law to each block gives

n1 T1

or

⇒

+y

m

q

For Block 1 (mass m): ΣFx = max

103

+x T2

mg

− T1 + T2 + mg sinq = 0

T1 = T2 + mg sinq

+y

n2

[1] 2m

T2

For Block 2 (mass 2m): ΣFx = max

⇒

+x

− T2 + 2mg sinq = 0 q

or T2 = 2mg sinq

4.28

[2]

(a)

Substituting Equation [2] into Equation [1] gives

(b)

From Equation [2] above, we have

T1 = 3mg sinq

T2 = 2mg sinq

Let m1 = 10.0 kg, m2 = 5.00 kg, and q = 40.0°. (a)

→

a

→

T

Applying the second law to each object gives m1a = m1 g − T and

→

→

n

T

[1]

m2 a = T − m2 g sinq

2 mg

→

a

m2

m1

[2]

→

m1g

q →

m2g

Adding these equations yields m1a + m2 a = m1 g − T + T − m2 g sinq

so (b)

or

⎛ m − m2 sinq ⎞ a=⎜ 1 g ⎝ m1 + m2 ⎟⎠

⎛ 10.0 kg − ( 5.00 kg ) sin 40.0° ⎞ a=⎜ 9.80 m s 2 = 4.43 m s2 ⎟ 15.0 kg ⎝ ⎠

(

)

Then, Equation [1] yields T = m1 ( g − a ) = (10.0 kg ) ⎡⎣( 9.80 − 4.43) m s2 ⎤⎦ = 53.7 N

4.29

(a)

The resultant external force acting on this system, consisting of all three blocks having a total mass of 6.0 kg, is 42 N directed horizontally toward the right. Thus, the acceleration produced is a=

(b)

ΣF 42 N = = 7.0 m s2 horizontally to the right m 6.0 kg

Draw a free body diagram of the 3.0-kg block and apply Newton’s second law to the horizontal forces acting on this block: ΣFx = max ⇒

(c)

(

The force accelerating the 2.0-kg block is the force exerted on it by the 1.0-kg block. Therefore, this force is given by

(

)

F = ma = ( 2.0 kg ) 7.0 m s 2 ,

56157_04_ch04_p089-107.indd 103

)

42 N − T = ( 3.0 kg ) 7.0 m s 2 , and therefore T = 21 N

or

F = 14 N horizontally to the right

10/15/10 3:32:03 PM

104

4.30

Chapter 4

The ﬁgure at the right shows the forces acting on the block. The incline is tilted at q = 25°, the mass of the block is m = 5.8 kg, while the applied force pulling the block up the incline is F = 32 N. Since a y = 0 for this block,

⫹y

n

(a)

m

mg

q

n = mg cosq

q

Since the incline is considered frictionless for this part, we take the friction force to be fk = 0 and ﬁnd F ΣFx = F − mg sinq = max or ax = − g sinq m 32 N giving ax = − 9.8 m s 2 sin 25° = 1.4 m s 2 5.8 kg

(

(b)

⫹x

fk

ΣFy = n − mg cosq = 0 and the normal force is

F

)

If the coefﬁcient of kinetic friction between the block and the incline is m k, then the friction force is fk = m k n = m k mg cosq , and ΣFx = F − fk − mg sinq = F − mg ( m k cosq + sinq ) = max

4.31

(a)

Thus,

ax =

F − g ( m k cosq + sinq ) m

and

ax =

32 N − 9.8 m s 2 5.8 kg

(

) [( 0.10 ) cos 25° + sin 25°] =

Assuming frictionless pulleys, the tension is uniform through the entire length of the rope. Thus, the tension at the point where the rope attaches to the leg is the same as that at the 8.00 kg block. Part (a) of the sketch at the right gives a force diagram of the suspended block. Recognizing that the block has zero acceleration, Newton’s second law gives

T

8.00 kg

(a)

(

F

T 70.0° T

mg

ΣFy = T − mg = 0

(b)

)

T = mg = (8.00 kg ) 9.80 m s 2 = 78.4 N

or (b)

0.49 m s 2

Part (b) of the sketch above gives a force diagram of the pulley near the foot. Here, F is the magnitude of the force the foot exerts on the pulley. By Newton’s third law, this is the same as the magnitude of the force the pulley exerts on the foot. Applying the second law gives ΣFx = T + T cos 70.0° − F = max = 0 or

4.32

F = T (1 + cos 70.0° ) = ( 78.4 N ) (1 + cos 70.0° ) = 105 N

(a)

→

→

n1

m1

n2

→

→

P

F

→

m1g

→

m2

P

→

m2g

continued on next page

56157_04_ch04_p089-107.indd 104

10/15/10 3:32:06 PM

The Laws of Motion

105

(b)

Note that the blocks move on a horizontal surface with a y = 0. Thus, the net vertical force acting on each block and on the combined system of both blocks is zero. The net horizontal force acting on the combined system consisting of both m1 and m2 is ΣFx = F − P + P = F .

(c)

Looking at just m1, ΣFy = 0 as explained above, while ΣFx = F − P .

(d)

Looking at just m2, we again have ΣFy = 0, while ΣFx = +P .

(e)

For m1 :

ΣFx = max

⇒

F − P = m1a

For m2 :

ΣFx = max

⇒

P = m2 a

(f)

Substituting the second of the equations found in (e) above into the ﬁrst gives m1a = F − P = F − m2 a

( m1 + m2 ) a = F

or

a = F ( m1 + m2 )

and

Then substituting this result into the second equation from (e), we have ⎛ F ⎞ P = m2 a = m2 ⎜ ⎝ m1 + m2 ⎟⎠ (g)

⎛ m2 ⎞ P=⎜ F ⎝ m1 + m2 ⎟⎠

or

Realize that applying the force to m2 rather than m1 would have the effect of interchanging the roles of m1 and m2 . We may easily ﬁnd the results for that case by simply interchanging the labels m1 and m2 in the results found in (f) above. This gives a = F ( m2 + m1 ) ⎛ m1 ⎞ F. (the same result as in the ﬁrst case) and P = ⎜ ⎝ m2 + m1 ⎟⎠ We see that the contact force, P, is larger in this case because m1 > m2 .

4.33

(a)

First, we consider the glider, plane, and connecting rope to be a single unit having mass

n1 mtotal glider + plane + rope

mtotal = 276 kg + 1 950 kg = 2 226 kg For this system, the tension in the rope is an internal force and is not included in an application of Newton’s second law. Applying the second law to the horizontal motion of this combined system gives ΣFx = mtotal ax (b)

(

⇒

mtotal g

)

F = ( 2 226 kg ) 2.20 m s2 = 4.90 × 10 3 N = 4.90 kN

To determine the tension in the rope connecting the glider and the plane, we consider a system consisting of the glider alone. For this system, the rope is a external agent and the tension force it exerts on our system (glider) is included in a second law calculation. ΣFx = mglider ax

56157_04_ch04_p089-107.indd 105

⇒

F(thrust)

(

)

T = ( 276 kg ) 2.20 m s 2 = 607 N

n2

mglider

T

mglider g

10/15/10 3:32:09 PM

106

Chapter 4

4.34

(a)

First, consider a system consisting of the two blocks combined, with mass m1 + m2 . For this system, the only external horizontal force is the tension in cord A pulling to the right. The tension in cord B is a force one part of our system exerts on another part of our system, and is therefore an internal force.

a m2

B

2

m1 1

A

Applying Newton’s second law to this system (including only external forces, as we should) gives ΣFx = ma

⇒

TA = ( m1 + m2 ) a

[1]

Now, consider a system consisting of only m2. For this system, the tension in cord B is an external force since it is a force exerted on block 2 by block 1 (which is not part of this system). Applying Newton’s second law to this system gives ΣFx = ma

⇒

TB = m2 a

[2]

Comparing Equations [1] and [2], and realizing that the acceleration is the same in both cases (see part (b) below), it is clear that Cord A exerts a larger force on block 1 than cord B exerts on block 2 .

4.35

(b)

Since cord B connecting the two blocks is taut and unstretchable, the two blocks stay a ﬁxed distance apart, and the velocities of the two blocks must be equal at all times. Thus, the rates at which the velocities of the two blocks change in time are equal, or the two blocks must have equal accelerations .

(c)

Yes. Block 1 exerts a forward force on Cord B, so Newton’s third law tells us that Cord B exerts a force of equal magnitude in the backward direction on Block 1.

(a)

When the acceleration is upward, the total upward force T must exceed the total downward force w = mg = (1 500 kg ) 9.80 m s 2 = 1.47 × 10 4 N.

(b)

When the velocity is constant, the acceleration is zero. The total upward force T and the total downward force w must be equal in magnitude .

(c)

If the acceleration is directed downward, the total downward force w must exceed the total upward force T.

(d)

ΣFy = ma y

(

)

(

)

⇒ T = mg + ma y = (1 500 kg ) 9.80 m s 2 + 2.50 m s 2 = 1.85 × 10 4 N

Yes , T > w. (e)

ΣFy = ma y

(

)

⇒ T = mg + ma y = (1 500 kg ) 9.80 m s 2 + 0 = 1.47 × 10 4 N

Yes , T = w. (f)

ΣFy = ma y

(

)

⇒ T = mg + ma y = (1 500 kg ) 9.80 m s 2 − 1.50 m s 2 = 1.25 × 10 4 N

Yes , T < w. 4.36

Note that if the cord connecting the two blocks has a ﬁxed length, the accelerations of the blocks must have equal magnitudes, even though they differ in directions. Also, observe from the diagrams, we choose the positive direction for each block to be in its direction of motion.

continued on next page

56157_04_ch04_p089-107.indd 106

10/15/10 3:32:13 PM

The Laws of Motion

First consider the block moving along the horizontal. The only force in the direction of movement is T. Thus, ΣFx = max ⇒

⫹x

107

→

T

→

T = ( 5.00 kg ) a

[1]

T

5.00 kg

⫹y

10.0 kg

Next consider the block which moves vertically. The forces on it are the tension T and its weight, 98.0 N.

mg ⫽ 98.0 N (a)

ΣFy = ma y ⇒ 98.0 N − T = (10.0 kg ) a [2]

(b)

Equations [1] and [2] can be solved simultaneously to give (a)

98.0 N − ( 5.00 kg ) a = (10.0 kg ) a

(b)

Then, Equation [1] yields

a=

or

98.0 N = 6.53 m s2 15.0 kg

(

)

T = ( 5.00 kg ) 6.53 m s 2 = 32.7 N

4.37 Trailer 300 kg →

T

→ wT

→

nT

Car 1 000 kg

→

→

→

nc

nc

q

→

→

→

Rcar

F

T

wc

→

F

Choosing the +x-direction to be horizontal and forward, the +y-direction vertical and upward, the common acceleration of the car and trailer has components of ax = +2.15 m s 2 and a y = 0 . (a)

The net force on the car is horizontal and given by

( ΣFx )car = F − T = mcar ax = (1 000 kg) ( 2.15 m s2 ) = (b)

2.15 × 10 3 N forward

The net force on the trailer is also horizontal and given by

( ΣFx )trailer = +T = mtrailer ax = (300 kg) ( 2.15

)

m s2 = 645 N forward

(c)

Consider the free-body diagrams of the car and trailer. The only horizontal force acting on the trailer is T = 645 N forward, and this is exerted on the trailer by the car. Newton’s third law then states that the force the trailer exerts on the car is 645 N toward the rear .

(d)

The road exerts two forces on the car. These are F and nc shown in the free-body diagram of the car. From part (a),

( )

Also, ΣFy

car

F = T + 2.15 × 10 3 N = 645 N + 2.15 × 10 3 N = + 2.80 × 10 3 N = nc − wc = mcar a y = 0, so nc = wc = mcar g = 9.80 × 10 3 N

The resultant force exerted on the car by the road is then Rcar = F 2 + nc2 =

( 2.80 × 10 N ) + (9.80 × 10 N ) 3

2

3

2

= 1.02 × 10 4 N

at q = tan −1 ( nc F ) = tan −1 ( 3.50 ) = 74.1° above the horizontal and forward. Newton’s third law then states that the resultant force exerted on the road by the car is 1.02 × 10 4 N at 74.1° below the horizontal and rearward .

56157_04_ch04_p089-107.indd 107

10/15/10 3:32:16 PM

108

4.38

Chapter 4

First, consider the 3.00-kg rising mass. The forces on it are the tension, T, and its weight, 29.4 N. With the upward direction as positive, the second law becomes T − 29.4 N = ( 3.00 kg ) a

→

m1 3.00 kg

The forces on the falling 5.00-kg mass are its weight and T, and its acceleration has the same magnitude as that of the rising mass. Choosing the positive direction down for this mass, gives

(a)

m2 5.00 kg

w2 49.0 N

[2]

or

1.60 T = 58.8 N

T = 36.8 N

Equation [2] then gives the acceleration as 49.0 N − 36.8 N = 2.44 m s 2 5.00 kg

a=

Consider the 3.00-kg mass. We have 1 2 1 2 a y t = 0 + ( 2.44 m s 2 )(1.00 s ) = 1.22 m 2 2

Δy = v0 y t + 4.39

Falling Mass

Solving Equation [2] for a and substituting into [1] gives

and the tension is

(c)

w1 29.4 N

⎛ 3.00 kg ⎞ T − 29.4 N = ⎜ ( 49.0 N − T ) ⎝ 5.00 kg ⎟⎠

(b)

T

Rising Mass

[1]

49.0 N − T = ( 5.00 kg ) a

→

T

When the block is on the verge of moving, the static friction force has a magnitude fs = ( fs )m ax = m s n. Since equilibrium still exists and the applied force is 75 N, we have

ΣFx = 75 N − fs = 0

or

(f ) s

m ax

= 75 N

In this case, the normal force is just the weight of the crate, or n = mg. Thus, the coefficient of static friction is ms =

(f ) s

m ax

n

=

(f ) s

m ax

mg

=

75 N = 0.38 ( 20 kg) (9.80 m s2 )

After motion exists, the friction force is that of kinetic friction, fk = m k n. Since the crate moves with constant velocity when the applied force is 60 N, we find that ΣFx = 60 N − fk = 0 or fk = 60 N. Therefore, the coefficient of kinetic friction is mk =

56157_04_ch04_p108-128.indd 108

fk f 60 N = k = = 0.31 n mg ( 20 kg ) ( 9.80 m s 2 )

10/15/10 3:33:45 PM

The Laws of Motion

4.40

(a)

109

The static friction force attempting to prevent motion may reach a maximum value of

(f ) s

m ax

= m s n1 = m s m1 g = ( 0.50 )(10 kg ) ( 9.80 m s 2 ) = 49 N

This exceeds the force attempting to move the system, w2 = m2 g = 39 N. Hence, the system remains at rest and the acceleration is a = 0 . (b)

Once motion begins, the friction force retarding the motion is fk = m k n1 = m k m1 g = ( 0.30 )(10 kg ) ( 9.80 m s 2 ) = 29 N This is less than the force trying to move the system, w2 = m2 g. Hence, the system gains speed at the rate 2 Fnet m2 g − m k m1 g ⎡⎣ 4.0 kg − 0.30 (10 kg ) ⎤⎦( 9.80 m s ) a= = = = 0.70 m s 2 m1 + m2 mtotal 4.0 kg + 10 kg

4.41

(a)

Since the crate has constant velocity, ax = a y = 0. Applying Newton’s second law: ΣFx = F cos 20.0° − fk = max = 0 and ΣFy = n − F sin 20.0° − w = 0

or

fk = ( 300 N ) cos 20.0° = 282 N

or

n = ( 300 N ) sin 20.0° + 1 000 N = 1.10 × 10 3 N

The coefficient of friction is then m k = (b)

fk 282 N = = 0.256 . n 1.10 × 10 3 N

In this case, ΣFy = n + F sin 20.0° − w = 0

so

n = w − F sin 20.0° = 897 N

fk = m k n = ( 0.256 )(897 N ) = 230 N

The friction force now becomes

Therefore, ΣFx = F cos 20.0° − fk = max = ( w g ) ax and the acceleration is a= 4.42

( F cos 20.0° − f ) g = [(300 N ) cos 20.0° − 230 N ](9.80

(a)

k

w

n1

1 000 N

−f

m s2 )

= 0.509 m s 2

−n1 F

m

3m

f mg

3mg

n2

(b)

The static friction force, f , exerted on the upper block (mass m) by the lower block causes the upper block to accelerate toward the right.

(c)

As long as slipping does not occur between the two blocks, both have acceleration a directed toward the right. Apply Newton’s second law to the horizontal motion of the upper block [see the leftmost diagram in part (a)] to find ΣFx = max

⇒ f = ma continued on next page

56157_04_ch04_p108-128.indd 109

10/15/10 3:33:48 PM

110

Chapter 4

Now, we make use of this result as we apply Newton’s second law to the horizontal motion of the lower block (mass 3m): ΣFx = max

F = ma + 3ma = 4ma

Thus, 4.43

⇒ F − f = ( 3m ) a

a = F 4m

or

When the load is on the verge of sliding forward on the bed of the slowing truck, the rearward-directed static friction force has its maximum value

(f ) s

F = f + 3ma

or

mload g→

→

fs →

n m ax

= m s n = m s mload g

Since slipping is not yet occurring, this single horizontal force must give the load an acceleration equal to that the truck. Thus, ΣFx = max (a)

⇒

− m s mload g = mload ( atruck )m ax

(a ) truck

m ax

= − ms g

If slipping is to be avoided, the maximum allowable rearward acceleration of the truck is seen to be ( atruck )m ax = − m s g, and vx2 = v02 x + 2ax ( Δx ) gives the minimum stopping distance as

( Δx )m in =

0 − v02 x v2 = 0x 2 ( atruck )m ax 2m s g

If v0 x = 12 m s and m s = 0.500, then

4.44

or

( Δx )m in =

(12.0

m s)

2

2 ( 0.500 )( 9.80 m s 2 )

= 14.7 m

(b)

Examining the calculation of part (a) shows that neither mass is necessary .

(a)

The free-body diagram of the crate is shown at the right. Since the crate has no vertical acceleration a y = 0 , we see that

(

ΣFy = ma y

⇒

n − mg = 0

)

or

n = mg

The only horizontal force present is the friction force exerted on the crate by the truck bed. Thus, ΣFx = max

⇒

f = ma

If the crate is not to slip (i.e., the static case is to prevail), it is necessary that the required friction force not exceed the maximum possible static friction force, ( fs )m ax = m s n. From this, we find the maximum allowable acceleration as am ax = (b)

Once slipping has started, the kinetic friction case prevails and f = fk = m k n. The acceleration of the crate in this case will be a=

56157_04_ch04_p108-128.indd 110

fm ax ( fs )m ax m s n m s ( mg ) = = = = m s g = ( 0.350 )( 9.80 m s 2 ) = 3.43 m s 2 m m m m

m n m ( mg ) f = k = k = m k g = ( 0.320 )( 9.80 m s 2 ) = 3.14 m s 2 m m m

10/15/10 3:33:51 PM

The Laws of Motion

4.45

→

The acceleration of the system is found from

a

1 1 2 Δy = v0 y t + at 2 , or 1.00 m = 0 + a (1.20 s ) 2 2

→

→

→

n m1g →

which gives a = 1.39 m s . 2

f

m1 m1g

Using the force diagram of m2 , the second law gives m s 2 ) − T = ( 5.00 kg ) (1.39 m s 2 )

or

T

→

m2

→ →

(5.00 kg) (9.80

111

T

a

→

m2g

T = 42.1 N

Then applying the second law to the horizontal motion of m1 , 42.1 N − f = (10.0 kg ) (1.39 m s 2 ) Since n = m1 g = 98.0 N, we have m k = 4.46

(a)

or

f = 28.2 N

f 28.2 N = = 0.288 n 98.0 N

Since the puck is on a horizontal surface, the normal force is vertical. With a y = 0, we see that ΣFy = ma y

⇒

n − mg = 0

or

n = mg

Once the puck leaves the stick, the only horizontal force is a friction force in the negative x-direction (to oppose the motion of the puck). The acceleration of the puck is ax = (b)

Then vx2 = v02 x + 2ax ( Δx ) gives the horizontal displacement of the puck before coming to rest as Δx =

4.47

ΣFx − fk − m k n − m k ( mg ) = = = = − mk g m m m m

vx2 − v02 x 0 − v02 v02 = = 2ax 2 ( − m k g ) 2m k g

The crate does not accelerate perpendicular to the incline. Thus, →

ΣF⊥ = ma⊥ = 0

⇒ n = F + mg cosq

F

→

n →

fs

The net force tending to move the crate down the incline is ΣF = mg sinq − fs , where fs is the force of static friction between the crate and the incline. If the crate is in equilibrium, then

q 35.0° →

mg sinq − fs = 0

so

fs = Fg sinq

mg

But, we also know fs ≤ m s n = m s ( F + mg cosq ) Therefore, we may write or

56157_04_ch04_p108-128.indd 111

mg sinq ≤ m s ( F + mg cosq )

⎛ sinq ⎞ ⎛ sin 35.0° ⎞ F ≥ mg ⎜ − cosq ⎟ = ( 3.00 kg ) ( 9.80 m s 2 ) ⎜ − cos35.0°⎟ = 32.1 N ⎝ 0.300 ⎠ ⎝ ms ⎠

10/15/10 3:33:54 PM

112

Chapter 4

4.48

(a)

y

Find the normal force n on the 25.0 kg box: →

ΣFy = n + (80.0 N ) sin 25.0° − 245 N = 0

n

F

→

f

0N

80.

25.0° x

n = 211 N

or

mg 245 N

Now find the friction force, f, as f = m k n = 0.300 ( 211 N ) = 63.3 N From the second law, we have ΣFx = ma, or

(80.0 N ) cos 25.0° − 63.3 N = ( 25.0 kg ) a which yields a = 0.368 m s2 (b)

When the box is on the incline,

y →

ΣFy = n + (80.0 N ) sin 25.0° − ( 245 N ) cos10.0° = 0

n →

f

n = 207 N

The friction force is f = m k n = 0.300 ( 207 N ) = 62.1 N. The net force parallel to the incline is then

245 N

giving

N .0 80 F 25.0° x

10.0°

ΣFx = (80.0 N ) cos 25.0° − ( 245 N ) sin10.0° − 62.1 N = −32.1 N Thus, a =

4.49

(a)

ΣFx −32.1 N = = −1.28 m s 2 25.0 kg m

1.28 m s 2 down the incline

or

( mg − bv ) ,

The object will fall so that ma = mg − bv, or a = where the downward direction is taken as positive.

m

Equilibrium ( a = 0 ) is reached when v = vterm inal =

4.50

→

→

f bv

→

m

mg ( 50 kg ) ( 9.80 m s = b 15 kg s

2

)=

33 m s

v

→

mg

(b)

If the initial velocity is less than 33 m/s, then a ≥ 0 and 33 m/s is the largest velocity attained by the object. On the other hand, if the initial velocity is greater than 33 m/s, then a ≤ 0 and 33 m/s is the smallest velocity attained by the object. Note also that if the initial velocity is 33 m/s, then a = 0 and the object continues falling with a constant speed of 33 m/s.

(a)

The force of friction is found as

f = m k n = m k ( mg )

Choose the positive direction of the x-axis in the direction of motion and apply Newton’s second law. We have ΣFx = − f = max

or

ax =

−f = − mk g m

continued on next page

56157_04_ch04_p108-128.indd 112

10/15/10 3:33:58 PM

The Laws of Motion

113

From v 2 = v02 + 2a ( Δx ) , with v = 0, v0 = 50.0 km h = 13.9 m s, we find 0 = (13.9 m s ) + 2 ( − m k g )( Δx ) 2

With m k = 0.100, this gives Δx =

4.51

(13.9 Δx =

or

(13.9

m s)

m s) 2m k g

[1]

2

2 ( 0.100 )( 9.80 m s 2 )

= 98.6 m

(13.9

(b)

With m k = 0.600, Equation [1] above gives Δx =

(a)

Δx = v0 t +

(b)

Considering forces parallel to the incline, Newton’s second law yields

m s)

2

2 ( 0.600 )( 9.80 m s 2 )

2

fk = 9.36 N

y →

)

kn

x

ΣFy = n − ( 29.4 N ) cos30.0° = 0 mk =

m f k

n

30.0°

Perpendicular to the plane, we have equilibrium, so

Then,

= 16.4 m

1 1 2 ( Δx ) 2 ( 2.00 m ) ax t 2 = 0 + ax t 2 gives ax = = = 1.78 m s 2 2 2 2 t2 (1.50 s )

ΣFx = ( 29.4 N ) sin 30.0° − fk = ( 3.00 kg ) (1.78 m s or

2

or

n = 25.5 N

30.0°

w mg 29.4 N

fk 9.36 N = = 0.367 n 25.5 N fk = 9.36 N

(c)

From part (b) above,

(d)

Finally, v 2 = v02 + 2ax ( Δx ) gives v = v02 + 2ax ( Δx ) = 0 + 2 (1.78 m s 2 )( 2.00 m ) = 2.67 m s

4.52

(a)

(

ΣFy = n − mg = 0 (b)

⇒

56157_04_ch04_p108-128.indd 113

n = mg = 15.0 lb

When a 10.0-lb object hangs in equilibrium on one end of the rope, the tension in the rope must be T = 10.0 lb. Thus, the 15.0-lb block has now has three vertical forces acting on it: an upward-directed tension force, T = 10.0 lb, a downward gravitational force of magnitude mg = 15.0 lb, and the upward normal force, n, exerted by the floor. Since the block is in equilibrium, a y = 0 and we find ΣFy = n + T − mg = 0

(c)

)

When the block is resting in equilibrium a y = 0 on a horizontal floor with only two vertical forces acting on it, the upward normal force exerted on the block by the floor must equal the downward gravitational force. That is,

or

n = mg − T = 15.0 lb − 10.0 lb = 5.00 lb

If the hanging object on the end of the rope is heavier than the 15.0-lb block, the system will not remain in equilibrium. Rather, the hanging object will accelerate downward, lifting the 15.0-lb block off the floor. As soon as the block leaves the floor, the normal force exerted on the block by the floor becomes zero .

10/15/10 3:34:02 PM

114

4.53

Chapter 4

When a person walks in a forward direction on a level floor, the force propelling them is a forward reaction force equal in magnitude to the rearward static friction force, fs , their foot exerts on the floor. The normal force exerted on the person by the level floor is n = mg. If the person is to have maximum acceleration (in order to travel distance d = 3.00 m in minimum time), the static friction force must have its maximum value, ( fs )m ax = m s n = m s mg, so the maximum acceleration possible is am ax =

(f ) s

m ax

m

=

m s mg = ms g m

The minimum time required to travel distance d (starting from rest) is then given by Δx = v0 t + 12 at 2 as 3.00 m = 0 +

4.54

1 ( m s g ) tm2 in 2

or

6.00 m ms g

tm in =

(a)

If m s = 0.500,

tm in =

6.00 m = 1.11 s 0.500 ( 9.80 m s 2 )

(b)

If m s = 0.800 ,

tm in =

6.00 m = 0.875 s 0.800 ( 9.80 m s 2 )

(a)

Both objects start from rest and have accelerations of the same magnitude, a. This magnitude can be determined by applying Δy = v0 y t + 12 a y t 2 to the motion of m1:

a= (b)

2 ( Δy ) 2 (1.00 m ) = = 0.125 m s 2 t2 ( 4.00s )2

Consider the free-body diagram of m1 and apply Newton’s second law: ΣFy = ma y or

(c)

T

⇒ T − m1 g = m1 ( +a )

m1

T = m1 ( g + a ) = ( 4.00 kg ) ( 9.80 m s + 0.125 m s 2

2

)=

→

m1g

39.7 N

Considering the free-body diagram of m2 : ΣFy = ma y so

⇒ n − m2 g cosq = 0

or

→

n = m2 g cosq

n = ( 9.00 kg ) ( 9.80 m s 2 ) cos 40.0° = 67.6 N ΣFx = max ⇒ m2 g sinq − T − fk = m2 ( + a )

y

T

→

n

m

→

2

fk q

x →

m2g

Then

fk = m2 ( g sinq − a ) − T

or

fk = ( 9.00 kg ) ⎡⎣( 9.80 m s 2 ) sin 40.0° − 0.125 m s 2 ⎤⎦ − 39.7 N = 15.9 N

The coefficient of kinetic friction is m k =

56157_04_ch04_p108-128.indd 114

→

fk 15.9 N = = 0.235 . n 67.6 N

10/15/10 3:34:06 PM

The Laws of Motion

y

4.55

115

y

nfeet mg/2 85.0 lb 22.0° 22.0° →

→ F2

n tip

→ F1

→

fs x

45.8

lb

x

F

mg 170 lb

Free-Body Diagram of Person

22.0°

Free-Body Diagram of Crutch Tip

From the free-body diagram of the person, ΣFx = F1 sin ( 22.0° ) − F2 sin ( 22.0° ) = 0, which gives or F1 = F2 = F Then, ΣFy = 2F cos 22.0° + 85.0 lbs − 170 lbs = 0 yields F = 45.8 lb. (a)

Now consider the free-body diagram of a crutch tip. ΣFx = fs − ( 45.8 lb )sin 22.0° = 0, or

fs = 17.2 lb

ΣFy = ntip − ( 45.8 lb ) cos 22.0° = 0, which gives

ntip = 42.5 lb

For minimum coefficient of friction, the crutch tip will be on the verge of slipping, so fs = ( fs )m ax = m s ntip (b)

and m s =

fs 17.2 lb = = 0.405 ntip 42.5 lb

As found above, the compression force in each crutch is F1 = F2 = F = 45.8 lb

4.56

The acceleration of the ball is found from

→

v 2 − v02 ( 20.0 m s ) − 0 a= = = 133 m s 2 2 (1.50 m ) 2 ( Δy )

F

2

From the second law, ΣFy = F − w = ma y , so

m 0.150 kg

F = w + ma y = 1.47 N + ( 0.150 kg ) (133 m s 2 ) = 21.4 N

→

4.57

→

T1

(a)

w 1.47 N

a2

→ T2

→

n →

a1

→

m1

T1

4.00 kg

→

T2

m2

m3

→ a3

2.00 kg

1.00 kg → →

→

m2g

fk

→

m3g

m1g

continued on next page

56157_04_ch04_p108-128.indd 115

10/15/10 3:34:10 PM

116

Chapter 4

(b)

Note that the suspended block on the left, m1, is heavier than that on the right, m3. Thus, if the system overcomes friction and moves, the center block will move right to left with each block’s acceleration being in the directions shown above . First, consider the center block, m2, which has no vertical acceleration. Then, ΣFy = n − m2 g = 0

or

This means the friction force is:

n = m2 g = (1.00 kg ) ( 9.80 m s 2 ) = 9.80 N fk = m k n = ( 0.350 )( 9.80 N ) = 3.43 N

Assuming the cords do not stretch, the speeds of the three blocks must always be equal. Thus, the magnitudes of the blocks’ accelerations must have a common value, a. a1 = a 2 = a 3 = a Taking the indicated direction of the acceleration as the positive direction of motion for each block, we apply Newton’s second law to each block as follows: For m1:

m1 g − T1 = m1 a

or

T1 = m1 ( g − a ) = ( 4.00 kg ) ( g − a )

[1]

For m2:

T1 − T2 − fk = m2 a

or

T1 − T2 = (1.00 kg ) a + 3.43 N

[2]

For m3:

T2 − m3 g = m3 a

or

T2 = m3 ( g + a ) = ( 2.00 kg ) ( g + a )

[3]

Substituting Equations [1] and [3] into Equation [2], and solving for a yields

( 4.00 kg) ( g − a ) − ( 2.00 kg) ( g + a ) = (1.00 kg) a + 3.43 N a=

so

( 4.00 kg − 2.00 kg) (9.80

m s 2 ) − 3.43 N

4.00 kg + 2.00 kg + 1.00 kg

= 2.31 m s 2

2 a1 = 2.31 m s 2 downward , a 2 = 2.31 m s to the left ,

and a 3 = 2.31 m s 2 upward (c)

Using this result in Equations [1] and [3] gives the tensions in the two cords as T1 = ( 4.00 kg ) ( g − a ) = ( 4.00 kg )( 9.80 − 2.31) m s 2 = 30.0 N and

(d)

4.58

T2 = ( 2.00 kg ) ( g + a ) = ( 2.00 kg )( 9.80 + 2.31) m s 2 = 24.2 N

From the final calculation in part (b), observe that if the friction force had a value of zero (rather than 3.53 N), the acceleration of the system would increase in magnitude. Then, observe from Equations [1] and [3] that this would mean T1 would decrease while T2 would increase .

The sketch at the right gives an edge view of the sail (heavy line) as seen from above. The velocity of the wind, v wind , is directed to the east and the force the wind exerts on the sail is perpendicular to the sail. The magnitude of this force is ⎛ N ⎞ v wind Fsail = ⎜ 550 m s ⎟⎠ ⎝

north

→ Fsail

30°

30° →

⊥

where v wind

vwind ⊥

is the component of the wind velocity perpendicular to the sail.

east

continued on next page

56157_04_ch04_p108-128.indd 116

10/15/10 3:34:13 PM

The Laws of Motion

117

When the sail is oriented at 30° from the north-south line and the wind speed is vwind = 17 knots, we have ⎛ N ⎞ v wind Fsail = ⎜ 550 m s ⎟⎠ ⎝

⊥

⎤ ⎛ ⎛ 0.514 m s ⎞ N ⎞⎡ = ⎜ 550 cos30° ⎥ = 4.2 × 10 3 N ⎢ 17 knots ⎜ ⎟ ⎟ ⎝ 1 knot ⎠ m s⎠ ⎣ ⎝ ⎦

(

)

The eastward component of this force will be counterbalanced by the force of the water on the keel of the boat. Before the sailboat has significant speed (that is, before the drag force develops), its acceleration is provided by the northward component of Fsail. Thus, the initial acceleration is

(a)

( 4.2 × 10 N ) sin 30° = 3

north

m

=

800 kg

2.6 m s 2

N

Rx = ΣFx = ( 60.0 N ) cos 45.0° − ( 60.0 N ) cos 45.0° = 0

y .0 60

The horizontal component of the resultant force exerted on the light by the cables is

45.0°

N

4.59

Fsail

60 .0

a=

45.0° x

The resultant y-component is Ry = ΣFy = ( 60.0 N ) sin 45.0° + ( 60.0 N ) sin 45.0° = 84.9 N Hence, the resultant force is 84.9 N vertically upward .

4.60

(b)

The forces on the traffic light are the weight, directed downward, and the 84.9 N vertically upward force exerted by the cables. Since the light is in equilibrium, the resultant of these forces must be zero. Thus, w = 84.9 N downward .

(a)

For the suspended block, ΣFy = T − 50.0 N = 0, so the tension in the rope is T = 50.0 N. Then, considering the horizontal forces on the 100-N block, we find ΣFx = T − fs = 0

(b)

fs = T = 50.0 N

If the system is on the verge of slipping, fs = ( fs )m ax = m s n . Therefore, the minimum acceptable coefficient of friction is ms =

(c)

or

fs 50.0 N = = 0.500 n 100 N

If m k = 0.250, then the friction force acting on the 100-N block is fk = m k n = ( 0.250 )(100 N ) = 25.0 N Since the system is to move with constant velocity, the net horizontal force on the 100-N block must be zero, or ΣFx = T − fk = T − 25.0 N = 0. The required tension in the rope is T = 25.0 N. Now, considering the forces acting on the suspended block when it moves with constant velocity, ΣFy = T − w = 0, giving the required weight of this block as w = T = 25.0 N .

4.61

On the level surface, the normal force exerted on the sled by the ice equals the total weight, or n = 600 N. Thus, the friction force is fk = m k n = ( 0.050 )( 600 N ) = 30 N

continued on next page

56157_04_ch04_p108-128.indd 117

10/15/10 3:34:16 PM

118

Chapter 4

Hence, Newton’s second law yields ΣFx = − fk = max , or ax =

2 − fk − fk − ( 30 N )( 9.80 m s ) = = = − 0.49 m s 2 m wg 600 N

The distance the sled travels on the level surface before coming to rest is 0 − ( 7.0 m s ) vx2 − v02 x = = 50 m 2 ax 2 ( − 0.49 m s 2 ) 2

Δx =

4.62

+y

(a)

F = 35.0 N

n

q

m 20.0 kg ƒk = 20.0 N

(b)

⇒

( 35.0 N ) cosq − 20.0 N = 0

20.0 N = 0.571 35.0 N

cosq =

or

and

q = 55.2°

We find the normal force the ground exerts on the suitcase by applying Newton’s second law to the vertical motion: ⇒ n + ( 35.0 N ) sinq − w = 0

ΣFy = ma y = 0

n = w − ( 35.0 N ) sinq = 196 N − ( 35.0 N ) sin 55.2° = 167 N

and 4.63

w = mg = 196 N

Since the suitcase moves with constant velocity, a = 0 and ax = a y = 0. Applying Newton’s second law to the horizontal motion of the suitcase gives ΣFx = max = 0

(c)

+x

(a)

The force that accelerates the box is the friction force between the box and truck.

(b)

We assume the truck is on level ground. Then, the normal force exerted on the box by the truck equals the weight of the box, n = mg. The maximum acceleration the truck can have before the box slides is found by considering the maximum static friction force the truck bed can exert on the box:

(f ) s

m ax

= m s n = m s ( mg )

Thus, from Newton’s second law, am ax = 4.64

(f ) s

m ax

m

=

m s ( mg ) = m s g = ( 0.300 )( 9.80 m s 2 ) = 2.94 m s 2 m

Let m1 = 5.00 kg, m2 = 4.00 kg, and m3 = 3.00 kg. Let T1 be the tension in the string between m1 and m2 , and T2 be the tension in the string between m2 and m3 . Note that the three objects have the same magnitude acceleration: a1 = a directed upward, a 2 = a 3 = a directed downward. We take the positive direction for each object to be the direction of its acceleration.

continued on next page

56157_04_ch04_p108-128.indd 118

10/15/10 3:34:20 PM

The Laws of Motion

(a)

119

We may apply Newton’s second law to each of the masses. for m1:

m1 a = T1 − m1 g

[1]

for m2:

m2 a = T2 + m2 g − T1

[2]

for m3:

m3 a = m3 g − T2

[3]

Adding these equations yields ( m1 + m2 + m3 ) a = ( −m1 + m2 + m3 ) g, so ⎛ −m1 + m2 + m3 ⎞ ⎛ 2.00 kg ⎞ 9.80 m s 2 ) = 1.63 m s 2 g=⎜ a=⎜ ( ⎟ ⎟ 12.0 kg m + m + m ⎠ ⎝ ⎝ 1 2 3 ⎠ (b)

From Equation [1], T1 = m1 ( a + g ) = ( 5.00 kg ) (11.4 m s 2 ) = 57.0 N , and from Equation [3], T2 = m3 ( g − a ) = ( 3.00 kg ) (8.17 m s 2 ) = 24.5 N

4.65

When an object of mass m is on this frictionless incline, the only force acting parallel to the incline is the parallel component of weight, mg sinq , directed down the incline. The acceleration is then a= (a)

mg sinq = g sinq = ( 9.80 m s 2 ) sin 35.0° = 5.62 m s 2 (directed down the incline) m

Taking up the incline as positive, the time for the sled projected up the incline to come to rest is given by t=

v − v0 0 − 5.00 m s = = 0.890 s − 5.62 m s 2 a

The distance the sled travels up the incline in this time is ⎛ v + v0 ⎞ ⎛ 0 + 5.00 m s ⎞ t=⎜ Δs = vav t = ⎜ ⎟⎠ ( 0.890 s ) = 2.23 m ⎝ ⎝ 2 ⎟⎠ 2 (b)

The time required for the first sled to return to the bottom of the incline is the same as the time needed to go up, that is, t = 0.890 s. In this time, the second sled must travel down the entire 10.0 m length of the incline. The needed initial velocity is found from 1 Δs = v0 t + at 2 as 2 2 Δs at −10.0 m ( −5.62 m s )( 0.890 s ) v0 = − = − = −8.74 m s t 2 0.890 s 2

or 4.66

8.74 m s down the incline

Before she enters the water, the diver is in free-fall with an acceleration of 9.80 m s 2 downward. Taking downward as the positive direction, her velocity when she reaches the water is given by v = v02 + 2a ( Δy ) = 0 + 2 ( 9.80 m s 2 )(10.0 m ) = 14.0 m s This is also her initial velocity for the 2.00 s after hitting the water. Her average acceleration during this 2.00 s interval is aav =

v − v0 0 − (14.0 m s ) = = − 7.00 m s 2 t 2.00 s continued on next page

56157_04_ch04_p108-128.indd 119

10/15/10 3:34:23 PM

120

Chapter 4

Continuing to take downward as the positive direction, the average upward force by the water is found as ΣFy = Fav + mg = maav , or Fav = m ( aav − g ) = ( 70.0 kg ) ⎡⎣( −7.00 m s 2 ) − 9.80 m s 2 ⎤⎦ = −1.18 × 10 3 N Fav = 1.18 × 10 3 N upward

or 4.67

(a)

Free-body diagrams for the two blocks are given at the right. The coefficient of kinetic friction for aluminum on steel is m1 = 0.47 while that for copper on steel is m 2 = 0.36. Since a y = 0 for each block, n1 = w1

and

+y →

→

n1

a

→

T

Aluminum 2.00 kg

n2 = w2 cos30.0°

+x

→

f1

Thus,

f1 = m1 n1 = 0.47 (19.6 N ) = 9.2 N

and

f2 = m 2 n2 = 0.36 ( 58.8 N ) cos30.0° = 18 N

w1 19.6 N +y

→

a

→

T

→

n2

For the aluminum block: →

ΣFx = max ⇒ T − f1 = m ( +a ) or T = f1 + ma T = 9.2 N + ( 2.00 kg ) a

giving

f2

Co 6.0 pper 0k g x

[1] 30°

For the copper block: ΣFx = max ⇒

w2 58.8 N

( 58.8 N ) sin 30.0° − T − 18 N = ( 6.00 kg ) a

11 N − T = ( 6.00 kg ) a

or

[2]

Substituting Equation [1] into Equation [2] gives 11 N − 9.2 N − ( 2.00 kg ) a = ( 6.00 kg ) a

4.68

or

a=

1.8 N = 0.23 m s 2 8.00 kg

T = 9.2 N + ( 2.00 kg ) ( 0.23 m s 2 ) = 9.7 N

(b)

From Equation [1] above,

(a)

If mass m1 moves 1 unit downward, pulley P2 must move 1 unit to the right. Also, mass m2 must get 1 unit closer to P2 (to provide an additional unit of length in the cord between P2 and the wall). This means that during the time m1 moved 1 unit, m2 has moved 2 units. Therefore, m2 is always moving twice as fast as m1, giving v1 = v2 2 and a1 = a2 2 .

(b)

Draw force diagrams of m1 , m2 , and P2 as shown below: a2

T1

a1

n m1

a1

m2

T2

T2 P2 T2

m1 g

m2 g

T1 m≈0

continued on next page

56157_04_ch04_p108-128.indd 120

10/15/10 3:34:26 PM

The Laws of Motion

121

Now, we apply Newton’s second law to the motion of each object, using the fact that a1 = a2 2 from part (a), and taking the direction of each object’s acceleration as the positive direction for that object. For m1:

m1 g − T1 = m1 a1 = m1 ( a2 2 )

or

m1 ( g − a2 2 ) = T1

[1]

For m2:

T2 = m2 a2

or

a2 = T2 m2

[2]

For P2:

T1 − 2T2 ≈ 0

or

T1 = 2T2

[3]

Substitute Equations [2] and [3] into Equation [1] and simplify to obtain T2 =

(c)

2m1 m2 g m1 + 4m2

and, from Equation [3],

4m1 m2 g m1 + 4m2 a2 =

From Equation [2], and the answer for T2 from part (b), And, from the answer of part (a),

4.69

T1 =

a1 =

2m1 g m1 + 4m2

m1 g m1 + 4m2

Figure 1 is a free-body diagram for the system consisting of both blocks. The friction forces are

→

a →

f1 = m k n1 = m k ( m1 g )

and

f 2 = m k ( m2 g )

n2

m1 →

For this system, the tension in the connecting rope is an internal force and is not included in second law calculations. The second law gives

→

n1

f1

50 N

m2 →

m1g

→

f2

→

m2g

FIGURE 1

ΣFx = 50 N − f1 − f2 = ( m1 + m2 ) a which reduces to

a=

50 N − mk g m1 + m2

[1]

Figure 2 gives a free-body diagram of m1 alone. For this system, the tension is an external force and must be included in the second law. We find ΣFx = T − f1 = m1 a, or T = m1 ( a + m k g ) (a)

If the surface is frictionless, m k = 0. Then, Equation [1] gives a=

FIGURE 2

50 N 50 N −0= = 1.7 m s 2 m1 + m2 30 kg

and Equation [2] yields (b)

[2]

T = (10 kg ) (1.7 m s 2 + 0 ) = 17 N

If m k = 0.10, Equation [1] gives the acceleration as 50 N − ( 0.10 )( 9.80 m s 2 ) = 0.69 m s 2 30 kg while Equation [2] gives the tension as a=

T = (10 kg ) ⎡⎣ 0.69 m s 2 + ( 0.10 )( 9.80 m s 2 ) ⎤⎦ = 17 N

56157_04_ch04_p108-128.indd 121

10/15/10 3:34:29 PM

122

Chapter 4

4.70

(a)

(b)

No. In general, the static friction force is less than the maximum value of ( fs )m ax = m s n . It is equal to this maximum value only when the coin is on the verge of slipping, or at the critical angle q c . For q ≤ q c , f ≤ ( fs )m ax = m s n .

(c)

Recognize that when the y-axis is chosen perpendicular to the incline as shown above, a y = 0 and we find ΣFy = n − mg cosq = ma y = 0

n = mg cosq

or

Also, when static conditions still prevail, but the coin is on the verge of slipping, we have ax = 0, q = q c , and f = ( fs )m ax = m s n = m s mg cosq c. Then, Newton’s second law becomes ΣFx = mg sinq c − m s mg cosq c = max = 0 m s mg cosq c = mg sinq c

and (d)

yielding

ms =

sinq c = tanq c cosq c

Once the coin starts to slide, kinetic conditions prevail and the friction force is f = fk = m k n = m k mg cosq At q = q c′ < q c, the coin slides with constant velocity, and ax = 0 again. Under these conditions, Newton’s second law gives ΣFx = mg sinq c′ − m k mg cosq c′ = max = 0 m k mg cosq c′ = mg sinq c′

and

4.71

(a)

yielding

mk =

sinq c′ = tanq c′ cosq c′

When the pole exerts a force downward and toward the rear on the lake bed, the lake bed exerts an oppositely directed force of equal magnitude, F, on the end of the pole. As the boat floats on the surface of the lake, its vertical acceleration is a y = 0. Thus, Newton’s second law gives the magnitude of the buoyant force, FB , as ΣFy = FB + F cosq − mg = 0 and, with q = 35.0°, or

FB = mg − F cosq = ( 370 kg ) ( 9.80 m s 2 ) − ( 240 N ) cos35.0°

FB = 3.43 × 10 3 N = 3.43 kN

continued on next page

56157_04_ch04_p108-128.indd 122

10/15/10 3:34:33 PM

The Laws of Motion

(b)

123

Applying Newton’s second law to the horizontal motion of the boat gives ΣFx = F sinq − Fdrag = max

ax =

or

( 240 N ) sin 35.0° − 47.5 N 370 kg

= 0.244 m s 2

After an elapsed time t = 0.450 s , vx = v0 x + ax t gives the velocity of the boat as vx = 0.857 m s + ( 0.244 m s 2 )( 0.450 s ) = 0.967 m s (c)

4.72

If angle q increased while the magnitude of F remained constant, the vertical component of this force would decrease. The buoyant force would have to increase to support more of the weight of the boat and its contents. At the same time, the horizontal component of F would increase, which would increase the acceleration of the boat. →

(a)

n →

→

T

mb

→

mr

T

F

→

mbg

(b)

Applying Newton’s second law to the rope yields ΣFx = max

⇒

F − T = mr a

or

T = F − mr a

[1]

Then, applying Newton’s second law to the block, we find ΣFx = max (c)

⇒ T = mb a

or

F − mr a = mb a which gives a =

Substituting the acceleration found above back into Equation [1] gives the tension at the left end of the rope as ⎛ mb + mr − mr ⎞ ⎛ F ⎞ T = F − mr a = F − mr ⎜ = F⎜ ⎟ ⎟ mb + mr ⎝ mb + mr ⎠ ⎝ ⎠

(d)

4.73

⎛ mb ⎞ T =⎜ F ⎝ mb + mr ⎟⎠

or

From the result of (c) above, we see that as mr approaches zero, T approaches F. Thus, the tension in a cord of negligible mass is constant along its length . y

Choose the positive x-axis to be down the incline and the y-axis perpendicular to this as shown in the free-body diagram of the toy. The acceleration of the toy then has components of Δv x + 30.0 m s a y = 0, and ax = = = + 5.00 m s 2 6.00 s Δt Applying the second law to the toy gives (a)

ΣFx = mg sinq = max and

(b)

F mb + mr

⇒

→

T →

a x

q

m

→ mg q

sinq = max mg = ax g,

⎛a ⎞ ⎛ 5.00 m s 2 ⎞ = 30.7° q = sin −1 ⎜ x ⎟ = sin −1 ⎜ ⎝ g⎠ ⎝ 9.80 m s 2 ⎟⎠

ΣFy = T − mg cosq = ma y = 0 , or T = mg cosq = ( 0.100 kg ) ( 9.80 m s 2 ) cos 30.7° = 0.843 N

56157_04_ch04_p108-128.indd 123

10/15/10 3:34:36 PM

124

4.74

Chapter 4

The sketch at the right gives the force diagram of the person. The scale simply reads the magnitude of the normal force exerted on the student by the seat. From Newton’s second law, we obtain ΣFy = ma y = 0 or

4.75

⇒

n − mg cos 30.0° = 0

n = mg cosq = ( 200 lb ) cos 30.0° = 173 lb

The acceleration the car has as it is coming to a stop is v 2 − v02 0 − ( 35 m s ) = = − 0.61 m s 2 2 ( Δx ) 2 (1 000 m ) 2

a=

Thus, the magnitude of the total retarding force acting on the car is ⎛ w⎞ ⎛ 8 820 N ⎞ F=ma =⎜ ⎟ a =⎜ ( 0.61 m s2 ) = 5.5 × 102 N ⎝ g⎠ ⎝ 9.80 m s 2 ⎟⎠ 4.76

(a)

8 000 N

In the vertical direction, we have ΣFy = (8 000 N ) sin 65.0° − w = ma y = 0

65.0°

w = (8 000 N ) sin 65.0° = 7.25 ×10 3 N

so

→

→

w mg

(b)

Along the horizontal, Newton’s second law yields ⎛ w⎞ ΣFx = (8 000 N ) cos 65.0° = max = ⎜ ⎟ ax ⎝ g⎠ or

4.77

ax =

g ⎡⎣(8 000 N ) cos 65.0° ⎤⎦ w

=

(9.80

m s 2 ) (8 000 N ) cos 65.0° 7.25 × 10 3 N

Since the board is in equilibrium, ΣFx = 0, and we see that the normal forces must have the same magnitudes on both sides of the board. Also, if the minimum normal forces (compression forces) are being applied, the board is on the verge of slipping and the friction force on each side is f = ( fs )m ax = m s n.

= 4.57 m s 2

→

→

f

f

→

→

n

n

The board is also in equilibrium in the vertical direction, so ΣFy = 2 f − w = 0, or f =

w 2

w 95.5 N

The minimum compression force needed is then n= 4.78

f w 95.5 N = = = 72.0 N m s 2 m s 2 ( 0.663)

Consider the two force diagrams, one of the penguin alone and one of the combined system consisting of penguin plus sled. The normal force exerted on the penguin by the sled is n1 = w1 = m1 g

n2

n1

F f1

f2

w1 = 70.0 N

wtotal = 130 N

continued on next page

56157_04_ch04_p108-128.indd 124

10/15/10 3:34:40 PM

The Laws of Motion

125

and the normal force exerted on the combined system by the ground is n2 = wtotal = mtotal g = 130 N The penguin is accelerated forward by the static friction force exerted on it by the sled. When the penguin is on the verge of slipping, this acceleration is am ax =

(f )

=

1 m ax

m1

(

)

m s n1 m s m1 g = = m s g = ( 0.700 ) ( 9.80 m s 2 ) = 6.86 m s 2 m1 m1

Since the penguin does not slip on the sled, the combined system must have the same acceleration as the penguin. Applying Newton’s second law to this system gives ΣFx = F − f2 = mtotal am ax

⎛w ⎞ F = f2 + mtotal am ax = m k ( wtotal ) + ⎜ total ⎟ am ax ⎝ g ⎠

or

⎛ 130 N ⎞ which yields F = ( 0.100 ) (130 N ) + ⎜ (6.86 m s2 ) = 104 N ⎝ 9.80 m s 2 ⎟⎠ 4.79

First, we will compute the needed accelerations: (1)

Before it starts to move:

ay = 0

(2)

During the first 0.80 s:

ay =

(3)

While moving at constant velocity: a y = 0

v y − v0 y t v y − v0 y

=

1.2 m s − 0 = 1.5 m s 2 0.80 s

0 −1.2 m s = − 0.80 m s 2 1.5 s t The spring scale reads the normal force the scale exerts on the man. Applying Newton’s second law to the vertical motion of the man gives (4)

ΣFy = n − mg = ma y

4.80

ay =

During the last 1.5 s:

or

(

=

n = m g + ay

)

(a)

When a y = 0,

n = ( 72 kg ) ( 9.80 m s 2 + 0 ) = 7.1 × 10 2 N

(b)

When a y = 1.5 m s 2 ,

n = ( 72 kg ) ( 9.80 m s 2 + 1.5 m s 2 ) = 8.1 × 10 2 N

(c)

When a y = 0,

n = ( 72 kg ) ( 9.80 m s 2 + 0 ) = 7.1× 10 2 N

(d)

When a y = −0.80 m s 2 ,

n = ( 72 kg ) ( 9.80 m s 2 − 0.80 m s 2 ) = 6.5 × 10 2 N

The friction force exerted on the mug by the moving tablecloth is the only horizontal force the mug experiences during this process. Thus, the horizontal acceleration of the mug will be am ug =

fk 0.100 N = = 0.500 m s 2 mm ug 0.200 kg

The cloth and the mug both start from rest ( v0 x = 0 ) at time t = 0. Then, at time t > 0, the horizontal displacements of the mug and cloth are given by Δx = v0 x t + 12 ax t 2 as

(

) (

(

) (

)

Δx mug = 0 + 12 0.500 m s 2 t 2 = 0.250 m s 2 t 2 and

)

Δx cloth = 0 + 12 3.00 m s 2 t 2 = 1.50 m s 2 t 2 continued on next page

56157_04_ch04_p108-128.indd 125

10/15/10 3:34:42 PM

126

Chapter 4

In order for the edge of the cloth to slip under the mug, it is necessary that Δx cloth = Δx m ug + 0.300 m, or

(1.50

m s 2 ) t 2 = ( 0.250 m s 2 ) t 2 + 0.300 m t=

The elapsed time when this occurs is

0.300 m

(1.50 − 0.250 ) m s2

= 0.490 s

At this time, the mug has moved a distance of Δx m ug = ( 0.250 m s 2 ) ( 0.490 s ) = 6.00 × 10 −2 m = 6.00 cm 2

4.81

(a)

T 250 N

Consider the first free-body diagram in which the child and the chair are treated as a combined system. The weight T 250 N of this system is wtotal = 480 N, and its mass is mtotal

n

T 250 N a

w = total = 49.0 kg g

Taking upward as positive, the acceleration of this system is found from Newton’s second law as

a

wtotal 320 N 160 N

wchild 320 N

ΣFy = 2T − wtotal = mtotal a y ay =

Thus

2 ( 250 N ) − 480 N = + 0.408 m s 2 49.0 kg

or

0.408 m s 2 upward (b)

The downward force that the child exerts on the chair has the same magnitude as the upward normal force exerted on the child by the chair. This is found from the free-body diagram of the child alone as ΣFy = T + n − wchild = mchild a y Hence,

4.82

so

n = mchild a y + wchild − T

⎛ 320 N ⎞ n=⎜ ( 0.408 m s2 ) + 320 N − 250 N = 83.3 N ⎝ 9.80 m s 2 ⎟⎠

Let R represent the horizontal force of air resistance. Since the helicopter and bucket move at constant velocity, ax = a y = 0 . The second law then gives

Also,

mg cos 40.0°

ΣFy = T cos 40.0° − mg = 0

or

T=

ΣFx = T sin 40.0° − R = 0

or

R = T sin 40.0°

y

→

T

40.0° →

x

R

→

→

w mg

Thus,

56157_04_ch04_p108-128.indd 126

⎛ mg ⎞ R=⎜ sin 40.0° = ( 620 kg ) ( 9.80 m s 2 ) tan 40.0° = 5.10 × 10 3 N ⎝ cos 40.0° ⎟⎠

10/15/10 3:34:46 PM

The Laws of Motion

4.83

(a)

127

With the crate still in equilibrium, ax = a y = 0, and Newton’s second law gives ΣFy = ma y

⇒

or

n = Fg + P sinq

and

ΣFx = max

or

P cosq = fs

n − P sinq − Fg = 0 [1]

⇒ P cosq − fs = 0 [2]

We now assume that the crate is on the verge of sliding and make use of Equation [1] to find

(

fs = ( fs )max = m s n = m s Fg + P sinq

)

Equation [2] then becomes P cosq = m s Fg + m s P sinq , and yields P= (c)

m s Fg cosq − m s sinq

m s Fg (1 cosq )

1− m s ( sinq cosq )

=

m s Fg secq 1− m s tanq

Note that the answer to part (a) states that the magnitude of the applied force required to start the crate moving will approach infinity as m s tanq approaches the value of 1. Thus, it is impossible to start the crate sliding if the angle equals or exceeds a critical value, q c , for which m s tanq c = 1

4.84

=

q c = tan −1 (1 m s )

or

(a)

Observe that if block 2 moves downward one unit, two units of cord must pass over the pulley attached to the table. This is true because one unit of cord must be added to each of the vertical sections of cord above block 2. Thus, block 1 must always move at twice the speed of block 2. This means that v1 = 2v2 and a1 = 2a2 at all times.

(b)

Applying Newton’s second law to block 1 gives ΣFx = max

⇒

T = m1a1

a1

n

[1]

T

m1

Taking downward as the positive direction for block 2, Newton’s second law gives ΣFy = ma y

⇒

m2 g − 2T = m2 a2

m1g

[2] T

T

Substitute Equation [1] and the result of part (a) into Equation [2] to obtain m2 g − 2m1 ( 2a2 ) = m2 a2

or

a2 =

m2 g 4m1 + m2

If m2 = 1.30 kg, then m2 g = 12.7 N, and we have a2 =

(c)

m2

a2

m2 g

12.7 N 4m1 + 1.30 kg

If m1 Wbiceps , it is clear that additional muscles must be involved .

5.28

Applying Wnc = ( KE + PE ) f − ( KE + PE )i to the jump of the “original” ﬂea gives Fm d = ( 0 + mgy f ) − ( 0 + 0 )

Fm d mg where Fm is the force exerted by the muscle and d is the length of contraction. yf =

or

If we scale the ﬂea by a factor f, the muscle force increases by f 2 and the length of contraction increases by f. The mass, being proportional to the volume which increases by f 3, will also increase by f 3. Putting these factors into our expression for y f gives

(y ) f

( f F ) ( fd ) = F d = y ( f m ) g mg 2

super flea

=

m

3

m

f

≈ 0.5 m

so the “super ﬂea” cannot jump any higher! This analysis is used to argue that most animals should be able to jump to approximately the same height (~0.5 m). Data on mammals from elephants to shrews tend to support this. 5.29

(a)

(b)

Taking y = 0, and hence PEg = mgy = 0, at ground level, the initial total mechanical energy of the projectile is 1 ( Etotal )i = KEi + PEi = mvi2 + mgyi 2 2 1 = ( 50.0 kg ) (1.20 × 10 2 m s ) + ( 50.0 kg ) ( 9.80 m s 2 ) (142 m ) = 4.30 × 10 5 J 2 The work done on the projectile is equal to the change in its total mechanical energy. 1 (Wnc )rise = ( KE f + PE f ) − ( KEi + PEi ) = m ( v 2f − vi2 ) + mg ( y f − yi ) 2 1 2 2 = ( 50.0 kg ) ⎡⎣(85.0 m s ) − (120 m s ) ⎤⎦ + ( 50.0 kg ) ( 9.80 m s 2 ) ( 427 m − 142 m ) 2 = −3.97 × 10 4 J

continued on next page

56157_05_ch05_p141-168.indd 142

10/12/10 2:24:19 PM

Energy

(c)

143

If, during the descent from the maximum height to the ground, air resistance does one and a half times as much work on the projectile as it did while the projectile was rising to the top of the arc, the total work done for the entire trip will be

(W )

nc total

= (Wnc )rise + (Wnc )descent = (Wnc )rise + 1.50 (Wnc )rise = 2.50 ( −3.97 × 10 4 J ) = −9.93 × 10 4 J

Then, applying the work-energy theorem to the entire ﬂight of the projectile gives 1 2 1 mv f + mgy f − mvi2 − mgyi 2 2 and the speed of the projectile just before hitting the ground is

(W )

nc total

vf = = 5.30

(a)

= ( KE + PE ) just before

hitting ground

2 (Wnc )total m

(

− ( KE + PE )at launch =

+ vi2 + 2 g yi − y f

2 ( −9.93 × 10 4 J ) 50.0 kg

)

+ (120 m s ) + 2 ( 9.80 m s 2 ) (142 m − 0 ) = 115 m s 2

The work done by the gravitational force equals the decrease in the gravitational potential energy, or Wg = − ( PE f − PEi ) = PEi − PE f = mg ( yi − y f ) = mgh

(b)

The change in kinetic energy is equal to the net work done on the projectile, which, in the absence of air resistance, is just that done by the gravitational force. Thus, Wnet = Wg = ΔKE

5.31

⇒

ΔKE = mgh 1 m v02 + mgh 2

(c)

ΔKE = KE f − KEi = mgh

(d)

No. None of the calculations in parts (a), (b), or (c) involve the initial angle.

(a)

so

KE f = KEi + mgh =

The system will consist of the mass, the spring, and the Earth . The parts of this system interact via the spring force, the gravitational force, and a normal force .

(b)

The points of interest are where the mass is released from rest ( at x = 6.00 cm ) and the equilibrium point, x = 0 .

(c)

The energy stored in the spring is the elastic potential energy, PEs = 12 kx 2. 2 1 At x = 6.00 cm, PEs = (850 N m ) ( 6.00 × 10 −2 m ) = 1.53 J 2 1 2 and at the equilibrium position ( x = 0 ), PEs = k ( 0 ) = 0 2

(d)

The only force doing work on the mass is the conservative spring force (the normal force and the gravitational force are both perpendicular to the motion). Thus, the total mechanical energy of the mass will be constant. Because we may choose y = 0, and hence PEg = 0, at the level of the horizontal surface, the energy conservation equation becomes 1 1 1 1 KE f + ( PEs ) f = KEi + ( PEs )i or m v 2f + kx 2f = m vi2 + kxi2 2 2 2 2 and solving for the ﬁnal speed gives v f = vi2 +

k 2 xi − x 2f ) ( m continued on next page

56157_05_ch05_p141-168.indd 143

10/12/10 2:24:19 PM

144

Chapter 5

If the ﬁnal position is the equilibrium position ( x f = 0 ) and the object starts from rest

( vi = 0 ) at xi = 6.00 cm, the ﬁnal speed is vf = 0 + (e)

2 850 N m ⎡ 6.00 × 10 −2 m ) − 0 ⎤ = 3.06 m 2 s 2 = 1.75 m s ( ⎣ ⎦ 1.00 kg

When the object is halfway between the release point and the equilibrium position, we have vi = 0, xi = 6.00 cm, and x f = 3.00 cm, giving vf = 0 +

2 2 850 N m ⎡ (6.00 × 10−2 m ) − (3.00 × 10−2 m ) ⎤⎦ = 1.51 m s 1.00 kg ⎣

This is not half of the speed at equilibrium because the equation for ﬁnal speed is not a linear function of position . 5.32

Using conservation of mechanical energy, we have 1 1 m v 2f + mgy f = m vi2 + 0 2 2 or

5.33

yf =

vi2 − v 2f

=

2g

(10 m s )2 − (1.0 m s )2 2 ( 9.80 m s 2 )

= 5.1 m

Since no nonconservative forces do work, we use conservation of mechanical energy, with the zero of potential energy selected at the level of the base of the hill. Then, 1 1 m v 2f + mgy f = m vi2 + mgyi with y f = 0 yields 2 2 yi =

v 2f − vi2

=

2g

( 3.00 m s )2 − 0 2 ( 9.80 m s 2 )

= 0.459 m

Note that this result is independent of the mass of the child and sled. 5.34

(a)

The stretching force is the weight of the 7.50-kg object, and the elongation is Δx = Thus, the spring constant for this spring is k=

(b)

Fs Δx

(

)(

f

− i.

)

7.50 kg 9.80 m s 2 mg = = 1.13 × 10 3 N m = 1.13 kN m − 0 . 415 m − 0 . 350 m f i

=

The two men pulling on the ends of the spring create a tension Fs =190 N in the spring. The elongation this will produce is Δx =

Fs k

=

190 N = 1.68 × 10 −1 m = 16.8 cm 1.13 × 10 3 N m

The ﬁnal length of the spring is then f

5.35

(a)

=

i

+ Δx = 35.0 cm + 16.8 cm = 51.8 cm

On a frictionless track, no external forces do work on the system consisting of the block and the spring as the spring is being compressed. Thus, the total mechanical energy of the system is constant, or KE f + ( PEg ) f + ( PEs ) f = KEi + ( PEg )i + ( PEs )i . Because the track is horizontal, the gravitational potential energy when the mass comes to rest is the same as just before it made contact with the spring, or ( PEg ) f = ( PEg )i . This gives

continued on next page

56157_05_ch05_p141-168.indd 144

10/12/10 2:24:20 PM

145

Energy

1 1 1 1 m v 2f + kx 2f = m vi2 + kxi2 2 2 2 2 Since v f = 0 (the block comes to rest) and xi = 0 (the spring is initially undistorted), x f = vi

5.36

m 0.250 kg = (1.50 m s ) = 0.350 m k 4.60 N M

(b)

If the track was not frictionless, some of the original kinetic energy would be spent overcoming friction between the block and track. This would mean that less energy would be stored as elastic potential energy in the spring when the block came to rest. Therefore, the maximum compression of the spring would be less in this case.

(a)

From conservation of mechanical energy, 1 1 m vB2 + mgyB = m vA2 + mgyA , or 2 2

B 5.00 m

vB = vA2 + 2 g ( yA − yB ) = 0 + 2 ( 9.80 m s

2

)(1.80 m ) =

A

3.20 m

C 2.00 m

5.94 m s

Similarly, vC = vA2 + 2 g ( yA − yC ) = 0 + 2 ( 9.80 m s 2 )( 5.00 m − 2.00 m ) = 7.67 m s (b) 5.37

(a)

(W )

g A→C

= ( PEg ) A − ( PEg )C = mg ( yA − yC ) = ( 49.0 N )( 3.00 m ) = 147 J

We choose the zero of potential energy at the level of the bottom of the arc. The initial height of Tarzan above this level is yi = ( 30.0 m ) (1 − cos 37.0°) = 6.04 m

艎⫽30.0 m

q yi yf

Then, using conservation of mechanical energy, we ﬁnd 1 1 m v 2f + 0 = m vi2 + mgyi 2 2 or (b)

v f = vi2 + 2 gyi = 0 + 2 ( 9.80 m s 2 ) ( 6.04 m ) = 10.9 m s

In this case, conservation of mechanical energy yields v f = vi2 + 2 gyi =

5.38

(a)

( 4.00 m s )2 + 2 ( 9.80 m s2 ) ( 6.04 m ) = 11.6 m s

If the string does not stretch, the speeds of the two blocks must be equal at all times until m1 reaches the ﬂoor. Also, while m1 is falling, only conservative forces (the gravitational forces) do work on the system of two blocks. Thus, the total mechanical energy is constant, or KE1, f + KE2, f + ( PEg )1, f + ( PEg )2, f = KE1,i + KE2,i + ( PEg )1,i + ( PEg )2,i Choosing y = 0 at ﬂoor level, this becomes 1 1 m1 v 2f + m2 v 2f + 0 + m2 gh = 0 + 0 + m1 gh + 0 2 2 and yields

vf =

2 ( m1 − m2 ) gh m1 + m2

continued on next page

56157_05_ch05_p141-168.indd 145

10/12/10 2:24:20 PM

146

Chapter 5

(b)

Using the provided data values, the answer from part (a) gives vf =

(c)

2 ( 6.5 kg − 4.2 kg ) ( 9.80 m s 2 ) ( 3.2 m ) 6.5 kg + 4.2 kg

= 3.7 m s

From conservation of energy, 1 1 m1 v 2f + m2 v 2f + m1 gy1, f + m2 gy2, f = 0 + 0 + m1 gy1,i + m2 gy2,i 2 2 vf =

or

vf = 5.39

2 g ⎡⎣ m1 ( y1,i − y1, f ) + m2 ( y2,i − y2, f ) ⎤⎦ m1 + m2

2 ( 9.80 m s 2 ) ⎡⎣( 6.5 kg ) (1.6 m ) + ( 4.2 kg ) ( −1.6 kg ) ⎤⎦ = 2.6 m s 6.5 kg + 4.2 kg

(a)

Initially, all the energy is stored as elastic potential energy within the spring. When the gun is ﬁred, and as the projectile leaves the gun, most of the energy is in the form of kinetic energy along with a small amount of gravitational potential energy. When the projectile comes to rest momentarily at its maximum height, all of the energy is in the form of gravitational potential energy.

(b)

Use conservation of mechanical energy from when the projectile is at rest within the gun ( vi = 0, yi = 0, and xi = −0.120 m ) until it reaches maximum height where v f = 0, y f = ymax = 20.0 m, and x f = 0 (the spring is relaxed after the gun is ﬁred). Then, ( KE + PEg + PEs ) f = ( KE + PEg + PEs )i becomes 1 2 kxi 2 2 ( 20.0 × 10 −3 kg )( 9.80 m s 2 ) ( 20.0 m )

0 + mgymax + 0 = 0 + 0 + k=

or (c)

2 mgymax = xi2

( −0.120 m )2

= 544 N m

This time, we use conservation of mechanical energy from when the projectile is at rest within the gun ( vi = 0, yi = 0, and xi = −0.120 m) until it reaches the equilibrium position of the spring ( y f = + 0.120 m and x f = 0). This gives KE f = ( KE + PEg + PEs )i − ( PEg + PEs ) f

or

1 1 m v 2f = ⎛⎜ 0 + 0 + kxi2 ⎞⎟ − ( mgy f + 0 ) 2 2 ⎝ ⎠

⎛ k⎞ v 2f = ⎜ ⎟ xi2 − 2 gy f ⎝ m⎠ ⎛ 544 N m ⎞ =⎜ ( −0.120 m )2 − 2 ( 9.80 m s2 )( 0.120 m ) ⎝ 20.0 × 10 −3 kg ⎟⎠ yielding 5.40

(a)

v f = 19.7 m s

ΣFy = 0 ⇒ n + F sin θ − mg = 0 or

→

→

n

→

Δx

F

n = mg − F sin θ

The friction force is then fk = μ k n = μ k ( mg − F sin θ )

q

m →

fk →

mg

continued on next page

56157_05_ch05_p141-168.indd 146

10/12/10 2:24:21 PM

Energy

(b)

147

The work done by the applied force is WF = F Δx cos θ = Fx cos θ and the work done by the friction force is Wfk = fk Δx cos φ where φ is the angle between

the direction of fk and Δ x. Thus, W fk = fk x cos 180° = − μ k ( mg − F sin θ ) x . (c)

The forces that do no work are those perpendicular to the direction of the displacement Δx. These are n, m g, and the vertical component of F .

(d)

For part (a): n = mg − F sin θ = ( 2.00 kg ) ( 9.80 m s 2 ) − (15.0 N ) sin 37.0° = 10.6 N fk = μ k n = ( 0.400 )(10.6 N ) = 4.24 N For part (b): WF = Fx cos θ = (15.0 N )( 4.00 m ) cos 37.0° = 47.9 J Wfk = fk x cos φ = ( 4.24 N )( 4.00 m ) cos 180° = −17.0 J

5.41

(a)

When the child slides down a frictionless surface, the only nonconservative force acting on the child is the normal force. At each instant, this force is perpendicular to the motion and, hence, does no work. Thus, conservation of mechanical energy can be used in this case.

(b)

The equation for conservation of mechanical energy, ( KE + PE ) f = ( KE + PE )i , for this situation is 12 m v 2f + m gy f = 12 m vi2 + m gyi . Notice that the mass of the child cancels out of the equation, so the mass of the child is not a factor in the frictionless case.

(c)

Observe that solving the energy conservation equation from above for the ﬁnal speed gives v f = vi2 + 2 g ( yi − y f ). Since the child starts with the same initial speed (vi = 0) and has the same change in altitude in both cases, v f is the same in the two cases.

(d)

Work done by a nonconservative force must be acccounted for when friction is present. This is done by using the work-energy theorem rather than conservation of mechanical energy.

(e)

From part (b), conservation of mechanical energy gives the ﬁnal speed as v f = vi2 + 2 g ( yi − y f ) = 0 + 2 ( 9.80 m s 2 ) (12.0 m ) = 15.3 m s

5.42

(a)

No. The change in the kinetic energy of the plane is equal to the net work done by all forces doing work on it. In this case, there are two such forces, the thrust due to the engine and a resistive force due to the air. Since the work done by the air resistance force is negative, the net work done (and hence, the change in kinetic energy) is less than the positive work done by the engine thrust. Also, because the thrust from the engine and the air resistance force are nonconservative forces, mechanical energy is not conserved in this case .

(b)

Since the plane is in level ﬂight, PEg

(

) = ( PE ) , and the work-energy theorem reduces to g i

f

Wnc = Wthrust + Wresistance = KE f − KEi , or 1 2

1 2

( F cos 0°) s + ( f cos 180°) s = mv 2f − mvi2

continued on next page

56157_05_ch05_p141-168.indd 147

10/12/10 2:24:22 PM

148

Chapter 5

This gives v f = vi2 + 5.43

2(F − f ) s = m

2 ⎡( 7.5 − 4.0 ) × 10 4 N ⎤⎦ ( 500 m ) = 77 m s ( 60 m s )2 + ⎣ 4 1.5 × 10 kg

T3

Neglecting the mass of the rope and any friction in the pulleys, the tension T1 is uniform throughout the length of the rope and equal to the magnitude of the force F applied to the loose end of the rope. Also, since the load (and hence the moving pulley) have constant velocity, applying Newton’s second law to the three force diagrams at the right gives T2 − mg = 0

Load:

or

T2 = mg

[2] [3]

moving pulley:

2T1 − T2 = 0

or

T mg T1 = 2 = 2 2

ﬁxed pulley:

T3 − 2T1 = 0

or

T3 = 2T1 = mg

(a)

T1

T1

T1

T1 Moving pulley

[1] T2 T2 m

Load mg

If m = 76.0 kg, Equation [2] gives F = T1 =

(b)

Fixed pulley

( 76.0 kg ) ( 9.80 m s2 ) 2

= 372 N

From above, T1 = 372 N . From Equations [1] and [3], T2 = T3 = mg = ( 76.0 kg ) ( 9.80 m s 2 ) = 745 N

(c)

Observe that if the load is raised 1.80 m, this length of rope must be removed from each of the two vertical segments of rope supporting the moving pulley. Thus, the loose end of the rope must be pulled downward a distance d = 3.60 m. The work done by the applied force is then WF = F ⋅ d ⋅ cos θ = ( 372 N )( 3.60 m ) cos 0° = 1.34 × 10 3 J = 1.34 kJ

5.44

(a)

Choose PEg = 0 at the level of the bottom of the arc. The child’s initial vertical displacement from this level is yi = ( 2.00 m )(1 − cos 30.0° ) = 0.268 m In the absence of friction, we use conservation of mechanical energy as

( KE + PE ) = ( KE + PE ) , or g

g i

f

1 m v 2f + 0 = 0 + mgyi , which gives 2

v f = 2 gyi = 2 ( 9.80 m s 2 )( 0.268 m ) = 2.29 m s (b)

With a nonconservative force present, we use

(

Wnc = KE + PEg

) − ( KE + PE ) = ⎛⎜⎝ 12 mv f

g i

2 f

⎞ + 0 ⎟ − ( 0 + mgyi ), or ⎠

continued on next page

56157_05_ch05_p141-168.indd 148

10/15/10 9:58:16 AM

149

Energy

⎛ v 2f ⎞ Wnc = m ⎜ − gyi ⎟ ⎝ 2 ⎠ ⎡ ( 2.00 m s )2 ⎤ = ( 25.0 kg ) ⎢ − ( 9.80 m s 2 )( 0.268 m )⎥ = −15.7 J 2 ⎢⎣ ⎥⎦ Thus, 15.7 J of energy is spent overcoming friction. 5.45

Choose PEg = 0 at the level of the bottom of the driveway.

(

Then Wnc = KE + PEg

) − ( KE + PE )

g i

f

→

n

becomes

( f cos180°) s = ⎡⎢

1 m v 2f + 0 ⎤⎥ − ⎡⎣ 0 + mg ( s sin 20° ) ⎤⎦ ⎣2 ⎦

s →

Solving for the ﬁnal speed gives vf =

or 5.46

( 2gs ) sin 20° −

v f = 2 ( 9.80 m s

2

→

f

20°

y

mg

2f s m

)(5.0 m ) sin 20° −

2 ( 4.0 × 10 3 N )( 5.0 m ) 2.10 × 10 3 kg

= 3.8 m s

(a)

Yes. Two forces, a conservative gravitational force and a nonconservative normal force, act on the child as she goes down the slide. However, the normal force is perpendicular to the child’s motion at each point on the path and does no work. In the absence of work done by nonconservative forces, mechanical energy is conserved.

(b)

We choose the level of the pool to be the y = 0 and hence, PEg = 0 level. Then, when the child is at rest at the top of the slide, PEg = mgh and KE = 0 . Note that this gives the constant total mechanical energy of the child as Etotal = KE + PEg = mgh. At the launching point (where y = h 5 ), we have PEg = mgy = mgh 5 and KE = Etotal − PEg = 4 mgh 5 . At the pool level, PEg = 0 and KE = mgh .

(c)

At the launching point (i.e., where the child leaves the end of the slide),

(

KE =

)

1 4 mgh m v02 = 2 5

meaning that v0 = (d)

8 gh 5

After the child leaves the slide and becomes a projectile, energy conservation gives 1 KE + PEg = m v 2 + mgy = Etotal = mgh where v 2 = vx2 + vy2 . Here, vx = v0 x is constant, 2 but vy varies with time. At maximum height, y = ymax and vy = 0, yielding 1 m ( v02 x + 0 ) + mgymax = mgh 2

and

ymax = h −

v02 x 2g continued on next page

56157_05_ch05_p141-168.indd 149

10/12/10 2:24:23 PM

150

Chapter 5

(e)

If the child’s launch angle leaving the slide is θ, then v0 x = v0 cos θ . Substituting this into the result from part (d) and making use of the result from part (c) gives ymax = h −

(f)

5.47

v02 1 ⎛ 8 gh ⎞ 2 cos 2 θ = h − ⎜ ⎟ cos θ 2g 2g ⎝ 5 ⎠

or

4 ⎛ ⎞ ymax = h ⎜ 1 − cos 2 θ ⎟ ⎝ ⎠ 5

No. If friction is present, mechanical energy would not be conserved, so her kinetic energy at all points after leaving the top of the waterslide would be reduced when compared with the frictionless case. Consequently, her launch speed, maximum height reached, and ﬁnal speed would be reduced as well.

Choose PEg = 0 at the level of the base of the hill and let x represent the distance the skier moves along the horizontal portion before coming to rest. The normal force exerted on the skier by the snow while on the hill is n1 = mg cos 10.5° and, while on the horizontal portion, n2 = mg. Consider the entire trip, starting from rest at the top of the hill until the skier comes to rest on the horizontal portion. The work done by friction forces is Wnc = ⎡⎣( fk )1 cos 180° ⎤⎦ ( 200 m ) + ⎡⎣( fk )2 cos 180° ⎤⎦ x = − μ k ( mg cos 10.5° )( 200 m ) − μ k ( mg ) x

(

Applying Wnc = KE + PEg

) − ( KE + PE )

g i

f

to this complete trip gives

− μ k ( mg cos 10.5° )( 200 m ) − μ k ( mg ) x = [ 0 + 0 ] − [ 0 + mg ( 200 m ) sin 10.5° ] or 5.48

⎛ sin 10.5° ⎞ x=⎜ − cos 10.5°⎟ ( 200 m ). If μ k = 0.0750 , then x = 289 m . ⎝ μk ⎠

The normal force exerted on the sled by the track is n = mg cos θ and the friction force is fk = μ k n = μ k mg cos θ . If s is the distance measured along the incline that the sled travels, applying Wnc = KE + PEg − KE + PEg to the entire trip gives

(

) ( f

)

i

1 ⎡⎣( μ k mg cos θ ) cos 180° ⎤⎦ s = [ 0 + mg s ( sin θ )] − ⎢⎡ m vi2 + 0 ⎤⎥ ⎣2 ⎦ vi2 ( 4.0 m s ) = = 1.5 m 2 2 g ( sin θ + μ k cos θ ) 2 ( 9.80 m s ) ( sin 20° + 0.20 cos 20° ) 2

or 5.49

(a)

s=

(

Consider the entire trip and apply Wnc = KE + PEg

(f

1

or

) − ( KE + PE ) f

g i

to obtain

⎛1 ⎞ cos 180° ) d1 + ( f2 cos 180° ) d2 = ⎜ m v 2f + 0 ⎟ − ( 0 + mgyi ) ⎝2 ⎠ vf =

fd + f d ⎞ ⎛ 2 ⎜ g yi − 1 1 2 2 ⎟ ⎝ ⎠ m

⎛ ( 50.0 N )(800 m ) + ( 3600 N )( 200 m ) ⎞ = 2 ⎜ ( 9.80 m s 2 )(1000 m ) − ⎟⎠ 80.0 kg ⎝ which yields v f = 24.5 m s (b)

Yes , this is too fast for safety.

continued on next page

56157_05_ch05_p141-168.indd 150

10/12/10 2:24:24 PM

Energy

(c)

(

Again, apply Wnc = KE + PEg

) − ( KE + PE ) , now with d g i

f

2

151

considered to be a variable,

d1 = 1000 m − d2, and v f = 5.00 m s. This gives

(f

1

⎛1 ⎞ cos 180° ) (1000 m − d2 ) + ( f2 cos 180° ) d2 = ⎜ m v 2f + 0 ⎟ − ( 0 + mgyi ) ⎝2 ⎠

which reduces to − (1000 m ) f1 + f1 d2 − f2 d2 =

d2 =

=

5.50

( mg ) yi − (1000 m ) f1 − f2 − f1

1 m v 2f − mgyi. Therefore, 2

1 m v 2f 2

1 2

( 784 N )(1000 m ) − (1000 m )( 50.0 N ) − (80.0 kg )( 5.00 m s )2 3600 N − 50.0 N

= 206 m

(d)

In reality, the air drag will depend on the skydiver’s speed. It will be larger than her 784 N weight only after the chute is opened. It will be nearly equal to 784 N before she opens the chute and again before she touches down, whenever she moves near terminal speed.

(a)

Wnc = ΔKE + ΔPE , but ΔKE = 0 because the speed is constant. The skier rises a vertical distance of Δy = ( 60 m ) sin30° = 30 m. Thus, Wnc = ( 70 kg ) ( 9.80 m s 2 )( 30 m ) = 2.1 × 10 4 J = 21 kJ

(b)

5.51

The time to travel 60 m at a constant speed of 2.0 m s is 30 s. Thus, the required power input is 2 mg ( Δy ) ( 70 kg ) ( 9.80 m s )( 30 m ) W ⎛ 1 hp ⎞ P = nc = = = ( 686 W ) ⎜ = 0.92 hp ⎝ 746 W ⎟⎠ Δt Δt 30 s

As the piano is lifted at constant speed up to the apartment, the total work that must be done on it is

(

)

Wnc = ΔKE + ΔPEg = 0 + mg y f − yi = ( 3.50 × 10 3 N )( 25.0 m ) = 8.75 × 10 4 J The three workmen (using a pulley system with an efﬁciency of 0.750) do work on the piano at a rate of ⎛ ⎞ Pnet = 0.750 ⎜ 3Psingle ⎟ = 0.750 [ 3 (165 W )] = 371 W = 371 J s ⎝ worker ⎠ so the time required to do the necessary work on the piano is Δt = 5.52

Wnc 8.75 × 10 4 J ⎛ 1 min ⎞ = = 236 s = ( 236 s ) ⎜ = 3.93 min ⎝ 60 s ⎟⎠ Pnet 371 J s

Let ∆N be the number of steps taken in time ∆t. We determine the number of steps per unit time by Power =

work done ( work per step per unit mass )( mass )( # steps ) = Δt Δt

continued on next page

56157_05_ch05_p141-168.indd 151

10/12/10 2:24:25 PM

152

Chapter 5

or

⎛ J step ⎞ ΔN ΔN 70 W = ⎜ 0.60 = 1.9 steps s (60 kg) ⎛⎜⎝ ⎞⎟⎠ , giving ⎟ kg ⎠ Δt Δt ⎝

The running speed is then step ⎞ ⎛ m ⎞ Δx ⎛ ΔN ⎞ ⎛ = 2.9 m s =⎜ ⎟ ( distance traveled per step ) = ⎜⎝ 1.9 ⎟ 1.5 s ⎠ ⎜⎝ step ⎟⎠ Δt ⎝ Δt ⎠

vav = 5.53

Assuming a level track, PE f = PEi, and the work done on the train is Wnc = ( KE + PE ) f − ( KE + PE )i =

(

)

1 1 2 m v 2f − vi2 = ( 0.875 kg ) ⎡⎣( 0.620 m s ) − 0 ⎤⎦ = 0.168 J 2 2

The power delivered by the motor is then P= 5.54

Wnc 0.168 J = = 8.00 W Δt 21.0 × 10 -3 s

When the car moves at constant speed on a level roadway, the power used to overcome the total frictional force equals the power input from the engine, or Poutput = ftotal v = Pinput. This gives ftotal =

5.55

Pinput v

=

175 hp ⎛ 746 W ⎞ = 4.5 × 10 3 N 29 m s ⎜⎝ 1 hp ⎟⎠

(

The work done on the older car is (Wnet )old = KE f − KEi

)

old

=

The work done on the newer car is

(W )

net new

(

= KE f − KEi

)

new

=

1 1 mv 2 − 0 = mv 2 2 2

1 2 ⎛1 ⎞ m ( 2 v ) − 0 = 4 ⎜ m v 2 ⎟ = 4 (Wnet )old ⎝2 ⎠ 2

and the power input to this car is Pnew =

(W )

net new

Δt

=

4 (Wnet )old Δt

= 4 Pold

or the power of the newer car is 4 times that of thhe older car . 5.56

Neglecting any variation of gravity with altitude, the work required to lift a 3.20 × 10 7 kg load at constant speed to an altitude of Δy = 1.75 km is W = ΔPEg = mg ( Δy ) = ( 3.20 × 10 7 kg )( 9.80 m s 2 ) (1.75 × 10 3 m ) = 5.49 × 1011 J The time required to do this work using a P = 2.70 kW = 2.70 × 10 3 J s pump is Δt =

5.57

(a)

W 5.49 × 1011 J 1 yr ⎛ ⎞ = = 2.03 × 108 s = ( 2.03 × 108 s ) ⎜ = 6.43 yr ⎝ 3.156 × 10 7 s ⎟⎠ P 2.70 × 10 3 J s

The acceleration of the car is a=

v − v0 18.0 m s − 0 = = 1.50 m s 2 t 12 .0 s

continued on next page

56157_05_ch05_p141-168.indd 152

10/12/10 2:24:25 PM

153

Energy

Thus, the constant forward force due to the engine is found from ΣF = Fengine − Fair = ma as Fengine = Fair + ma = 400 N + (1.50 × 10 3 kg )(1.50 m s 2 ) = 2 .65 × 10 3 N The average velocity of the car during this interval is vav = ( v + v0 ) 2 = 9.00 m s , so the average power input from the engine during this time is ⎛ 1 hp ⎞ Pav = Fengine vav = ( 2.65 × 10 3 N ) ( 9.00 m s ) = ( 2.39 × 10 4 W ) ⎜ = 32.0 hp ⎝ 746 W ⎟⎠ (b)

At t = 12 .0 s, the instantaneous velocity of the car is v = 18.0 m s and the instantaneous power input from the engine is ⎛ 1 hp ⎞ P = Fengine v = ( 2.65 × 10 3 N ) (18.0 m s ) = ( 4.77 × 10 4 W ) ⎜ = 63.9 hp ⎝ 746 W ⎟⎠

5.58

(a)

The acceleration of the elevator during the ﬁrst 3.00 s is a=

v − v0 1.75 m s − 0 = = 0.583 m s 2 t 3.00 s

so Fnet = Fmotor − mg = ma gives the force exerted by the motor as Fmotor = m ( a + g ) = ( 650 kg ) ⎡⎣( 0.583 + 9.80 ) m s 2 ⎤⎦ = 6.75 × 10 3 N The average velocity during this interval is vav = ( v + v0 ) 2 = 0.875 m s so the average power input from the motor during this time is ⎛ 1 hp ⎞ Pav = Fmotor vav = ( 6.75 × 10 3 N ) ( 0.875 m s ) ⎜ = 7.92 hp ⎝ 746 W ⎟⎠ (b)

When the elevator moves upward with a constant speed of v = 1.75 m s, the upward force exerted by the motor is Fmotor = mg and the instantaneous power input from the motor is ⎛ 1 hp ⎞ P = ( mg ) v = ( 650 kg ) ( 9.80 m s 2 ) (1.75 m s ) ⎜ = 14.9 hp ⎝ 746 W ⎟⎠

5.59

The work done on the particle by the force F as the particle moves from x = xi to x = x f is the area under the curve from xi to x f.

Fx (N) B

6 4

(a)

For x = 0 to x = 8.00 m ,

2

1 W = area of triangle ABC = AC × altitude 2 W0→8 (b)

1 = (8.00 m )( 6.00 N ) = 24.0 J 2

=

56157_05_ch05_p141-168.indd 153

⫺2 ⫺4

A 2

4

6

C

E

8

10 12

x (m)

D

For x = 8.00 m to x = 10.0 m, W8→10 = area of triangle CDE =

(c)

0

1 CE × altitude 2

1 ( 2.00 m )( −3.00 N ) = − 3.00 J 2

W0→10 = W0→8 + W8→10 = 24.0 J + ( − 3.00 J ) = 21.0 J

10/12/10 2:24:26 PM

154

5.60

Chapter 5

The work done by a force equals the area under the Force versus Displacement curve. (a)

3

For the region 0 ≤ x ≤ 5.00 m , W0 to 5 =

(b)

Fx (N)

( 3.00 N )( 5.00 m ) 2

2

= 7.50 J

1 x (m)

0 0

For the region 5.00 m ≤ x ≤ 10.0 m ,

2

4

6

8

10 12 14 16

W5 to 10 = ( 3.00 N )( 5.00 m ) = 15.0 J

( 3.00 N )( 5.00 m )

(c)

For the region 10.0 m ≤ x ≤ 15.0 m , W10 to 15 =

(d)

KE x = x f − KE x = 0 = W0 to x f = area under F vs. x curve from x = 0 m to x = x f , or 1 1 m v 2f = m v02 + W0 to x f 2 2

2

= 7.50 J

⎛ 2⎞ v f = v02 + ⎜ ⎟ W0 to x f ⎝ m⎠

giving

For x f = 5.00 m: ⎛ 2⎞ v f = v02 + ⎜ ⎟ W0 to 5 = ⎝ m⎠

( 0.500

⎛ 2 ⎞ 2 m s) + ⎜ ( 7.50 J ) = 2.29 m s ⎝ 3.00 kg ⎟⎠

For x f = 15.0 m: ⎛ 2⎞ ⎛ 2⎞ v f = v02 + ⎜ ⎟ W0 to 15 = v02 + ⎜ ⎟ (W0 to 5 + W5 to 10 + W10 to 15 ) ⎝ m⎠ ⎝ m⎠ or vf = 5.61

(a) (b)

( 0.500

⎛ 2 ⎞ 2 m s) + ⎜ ( 7.50 J + 15.0 J + 7.50 J ) = 4.50 m s ⎝ 3.00 kg ⎟⎠

Fx = (8 x − 16 ) N; see the graph at the right. The net work done is the total area under the graph from x = 0 to x = 3.00 m . This consists of two triangular shapes, one below the axis (negative area) and one above the axis (positive). The net work is then Wnet

( 2 .00 m ) ( − 16.0 N ) + (1.00 m )(8.00 N ) = 2

2

Fx (N) 20 (3, 8)

10 0 ⫺10

1

2

3

4

x (m)

⫺20

= − 12.0 J 5.62

At the top of the arc, v y = 0, and vx = v0 x = v0 cos 30.0° Therefore v 2 = vx2 + vy2 = v02 x + 0 = ( v0 cos 30.0° ) , and 2

KE =

56157_05_ch05_p141-168.indd 154

1 1 2 m v 2 = ( 0.150 kg ) ⎡⎣( 40.0 m s ) cos 2 ( 30.0° )⎤⎦ = 90.0 J 2 2

10/12/10 2:24:26 PM

Energy

5.63

w 700 N = = 71.4 kg. The net upward force acting on the body g 9.80 m s 2 is Fnet = 2 ( 355 N ) − 700 N = 10.0 N. The ﬁnal upward velocity can then be calculated from the work-energy theorem as The person’s mass is m =

Wnet = KE f − KEi = or

(F

net

1 1 m v 2 − m vi2 2 2

cos θ ) s = [(10.0 N ) cos 0° ]( 0.250 m ) =

1 ( 71.4 kg) v 2 − 0 2

v = 0.265 m s upward

which gives 5.64

155

Taking y = 0 at ground level, and using conservation of energy from when the boy starts from rest ( vi = 0) at the top of the slide ( yi = H ) to the instant he leaves the lower end ( y f = h) of the frictionless slide with a horizontal velocity ( v0 x = v f , v0 y = 0), yields 1 m v 2f + mgh = 0 + mgH 2

v 2f = 2 g ( H − h )

or

[1]

Considering his ﬂight as a projectile after leaving the end of the slide, Δy = v0 y t + 12 ay t 2 gives the time to drop distance h to the ground as −h = 0 +

1 (−g)t 2 2

or

t=

2h g

The horizontal distance traveled (at constant horizontal velocity) during this time is d, so d = v0 x t = v f

2h g

vf = d

and

g = 2h

gd 2 2h

Substituting this result into Equation [1] above gives gd 2 = 2g ( H − h ) 2h 5.65

(a)

H = h+

or

d2 4h

If y = 0 at point B, then yA = ( 35.0 m ) sin 50.0° = 26.8 m and yB = 0. Thus, PEA = mgyA = (1.50 × 10 3 kg )( 9.80 m s 2 ) ( 26.8 m ) = 3.94 × 10 5 J PEB = mgyB = 0

(b)

and

ΔPEA→ B = PEB − PEA = 0 − 3.94 × 10 5 J = −3.94 × 10 5 J

If y = 0 at point C, then yA = ( 50.0 m ) sin 50.0° = 38.3 m and yB = (15.0 m ) sin 50.0° = 11.5 m . In this case, PEA = mgyA = (1.50 × 10 3 kg )( 9.80 m s 2 ) ( 38.3 m ) = 5.63 × 10 5 J PEB = mgyB = (1.50 × 10 3 kg )( 9.80 m s 2 ) (11.5 m ) = 1.69 × 10 5 J and

5.66

(a)

ΔPEA→ B = PEB − PEA = 1.69 × 10 5 J − 5.63 × 10 5 J = −3.94 × 10 5 J

Taking y = 0 at the initial level of the upper end of the spring, and applying conservation of energy from the instant the ball is released from rest to the instant just before it contacts the spring gives continued on next page

56157_05_ch05_p141-168.indd 155

10/12/10 2:24:27 PM

156

Chapter 5

(

) (

KE f = KEi + PEgi − PEgf + PEsi − PEsf 1 m v 2f = 0 + ( mgh − 0 ) + ( 0 − 0 ) 2

or

v f = 2 gh = 2 ( 9.80 m s 2 )( 0.650 m ) = 3.57 m s

and (b)

)

Since the spring is very light, we neglect any energy loss during the collision between the ball and spring, and apply conservation of energy from the instant the ball is released from rest at y = + h to the instant the ball comes to rest momentarily at y = − d. This yields

(

) (

KE f = KEi + PEgi − PEgf + PEsi − PEsf or

)

1 ⎛ ⎞ 0 = 0 + [ mgh − mg ( − d )] + ⎜ 0 − kd 2 ⎟ ⎝ ⎠ 2

Thus, the force constant of the spring is k=

or 5.67

(

2mg h + d d2

2

The equivalent spring constant of the bow is given by F = kx as

(a)

(b)

Ff xf

=

230 N = 575 N m 0.400 m

The work done pulling the bow is equal to the elastic potential energy stored in the bow in its ﬁnal conﬁguration, or W=

1 2 1 2 kx f = ( 575 N m )( 0.400 m ) = 46.0 J 2 2

Choose PEg = 0 at the level where the block comes to rest against the spring. Then, in the absence of work done by nonconservative forces, the conservation of mechanical energy gives

( KE + PE or

0+0+

Thus, 5.69

2

k = 3.22 × 10 3 N m = 3.22 kN m

k=

5.68

) = 2(1.80 kg)(9.80 m s )( 0.650 m + 0.0900 m ) ( 0.0900 m )

(a)

g

+ PEs

) = ( KE + PE f

g

+ PEs

)

i

1 2 kx f = 0 + mg L sin θ + 0 2

xf =

2 mg L sin θ = k

2 (12.0 kg ) ( 9.80 m s 2 )( 3.00 m ) sin 35.0° 3.00 × 10 4 N m

= 0.116 m

From v 2 = v02 + 2ay ( Δy ), we ﬁnd the speed just before touching the ground as v = 0 + 2 ( 9.80 m s 2 )(1.0 m ) = 4.4 m s

(b)

Choose PEg = 0 at the level where the feet come to rest. Then Wnc = KE + PEg − KE + PEg becomes

(

) ( f

)

i

continued on next page

56157_05_ch05_p141-168.indd 156

10/12/10 2:24:28 PM

Energy

(F

av

157

⎛1 ⎞ cos180° ) s = ( 0 + 0 ) − ⎜ m vi2 + mg s ⎟ ⎝2 ⎠ m vi2 ( 75 kg)( 4.4 m s ) + 75 kg 9.80 m s2 = 1.5 × 105 N + mg = ( )( ) 2s 2 ( 5.0 × 10 −3 m ) 2

Fav =

or 5.70

From the work-energy theorem,

(

Wnc = KE + PEg + PEs

) − ( KE + PE f

g

+ PEs

)

i

we have

(f or 5.71

(a)

k

1 ⎛1 ⎞ ⎛ ⎞ cos180° ) s = ⎜ m v 2f + 0 + 0 ⎟ − ⎜ 0 + 0 + kxi2 ⎟ ⎝2 ⎠ ⎝ ⎠ 2

(8.0

kxi2 − 2 fk s = m

vf =

N m ) ( 5.0 × 10 −2 m ) − 2 ( 0.032 N )( 0.15 m ) 2

5.3 × 10 −3 kg

= 1.4 m s

The two masses will pass when both are at y f = 2.00 m above the table. From conservation of energy, ( KE + PEg + PEs ) f = ( KE + PEg + PEs )i 1 ( m1 + m2 ) v 2f + ( m1 + m2 ) gy f + 0 = 0 + m1gy1i + 0, or 2 vf =

=

2 m1 g y1i − 2 g yf m1 + m2 2 ( 5.00 kg ) ( 9.80 m s 2 )( 4.00 m ) 8.00 kg

− 2 ( 9.80 m s 2 ) ( 2 .00 m )

This yields the passing speed as v f = 3.13 m s . (b)

When m1 = 5.00 kg reaches the table, m2 = 3.00 kg is y2 f = 4.00 m above the table. Thus, ( KE + PEg + PEs ) f = ( KE + PEg + PEs )i becomes 1 ( m1 + m2 ) v 2f + m2 g y2 f + 0 = 0 + m1gy1i + 0, or v f = 2

(

2 g m1 y1i − m2 y2 f m1 + m2

)

Thus, vf = (c)

When the 5.00-kg object hits the table, the string goes slack and the 3.00-kg object becomes a projectile launched straight upward with initial speed v0 y = 4.43 m s. At the top of its arc, vy = 0 and vy2 = v02 y + 2ay ( Δy) gives Δy =

56157_05_ch05_p141-168.indd 157

2 ( 9.80 m s 2 ) ⎡⎣( 5.00 kg )( 4.00 m ) − ( 3.00 kg )( 4.00 m )⎤⎦ = 4.43 m s 8.00 kg

vy2 − v02 y 2 ay

0 − ( 4.43 m s ) = 1.00 m 2 ( −9.80 m s 2 ) 2

=

10/12/10 2:24:29 PM

158

Chapter 5

5.72

(a)

The needle has maximum speed during the interval between when the spring returns to normal length and the needle tip ﬁrst contacts the skin. During this interval, the kinetic energy of the needle equals the original elastic potential energy of the spring, or 2 2 1 1 2 m vmax = 2 kxi . This gives k 375 N m = (8.10 × 10 −2 m ) = 21.0 m s m 5.60 × 10 −3 kg

vmax = xi (b)

If F1 is the force the needle must overcome as it penetrates a thickness x1 of skin and soft tissue while F2 is the force overcame while penetrating thickness x2 of organ material, application of the work-energy theorem from the instant before skin contact until the instant before hitting the stop gives 2 Wnet = − F1 x1 − F2 x2 = 12 m v 2f − 12 m vmax

2 v f = vmax −

or

vf = 5.73

( 21.0

2 ( F1 x1 + F2 x2 ) m

2 ⎡( 7.60 N ) ( 2.40 × 10 −2 m ) + ( 9.20 N )( 3.50 × 10 −2 m ) ⎤⎦ 2 = 16.2 m s m s) − ⎣ 5.60 × 10 −3 kg

Since the bowl is smooth (that is, frictionless), mechanical energy is conserved or ( KE + PE ) f = ( KE + PE )i . Also, if we choose y = 0 (and hence, PEg = 0) at the lowest point in the bowl, then y A = + R, yB = 0, and yC = 2 R 3. (a)

( PE ) = mgy = mgR or ( PE ) = ( 0.200 kg ) ( 9.80 m s )( 0.300 m ) = g A

2

g A

5.74

2R/3

A

0.588 J

(b)

KEB = KEA + PEA − PEB = 0 + mgyA − mgyB = 0.588 J − 0 = 0.588 J

(c)

KEB = 12 m vB2 ⇒ vB =

(d)

( PE )

(e)

KEC = KEB + PEB − PEC = 0.588 J + 0 − 0.392 J = 0.196 J

(a)

KEB =

(b)

The change in the altitude of the particle as it goes from A to B is yA − yB = R, where R = 0.300 m is the radius of the bowl. Therefore, the work-energy theorem gives

g C

2 KEB = m

2 ( 0.588 J ) = 2.42 m s 0.200 kg

⎡ 2 ( 0.300 m ) ⎤ = mgyC = ( 0.200 kg ) ( 9.80 m s 2 ) ⎢ ⎥ = 0.392 J 3 ⎣ ⎦

1 1 2 m vB2 = ( 0.200 kg )(1.50 m s ) = 0.225 J 2 2

Wnc = ( KEB − KEA ) + ( PEB − PEA )

= KEB − 0 + mg ( yB − yA ) = KEB + mg ( − R )

A m v ⫽0 A R⫽30.0 cm vB ⫽ 1.50 m/s m B

continued on next page

56157_05_ch05_p141-168.indd 158

10/12/10 2:24:29 PM

159

Energy

or

Wnc = 0.225 J + ( 0.200 kg ) ( 9.80 m s 2 )( −0.300 m ) = −0.363 J

The loss of mechanical energy as a result of friction is then 0.363 J .

5.75

(c)

No. Because the normal force, and hence the fricction force, varies with the position of the particle on its path, it is not possible to use the result from part (b) to determine the coefﬁcient of friction without using calculus.

(a)

Consider the sketch at the right. When the mass m = 1.50 kg is in equilibrium, the upward spring force exerted on it by the lower spring (i.e., the tension in this spring) must equal the weight of the object, or FS 2 = mg. Hooke’s law then gives the elongation of this spring as

k1 A k2

F mg x2 = S 2 = k2 k2 m

Now, consider point A where the two springs join. Because this point is in equilibrium, the upward spring force exerted on A by the upper spring must have the same magnitude as the downward spring force exerted on A by the lower spring (that is, the tensions in the two springs must be equal). The elongation of the upper spring must be x1 =

FS1 FS 2 mg = = k1 k1 k1

and the total elongation of the spring system is x = x1 + x2 =

⎛1 1⎞ mg mg + = mg ⎜ + ⎟ k1 k2 ⎝ k1 k2 ⎠

⎞ ⎛ 1 1 = 2.04 × 10 −2 m + = (1.50 kg ) ( 9.80 m s 2 ) ⎜ 3 3 ⎝ 1.20 × 10 N m 1.80 × 10 N m ⎟⎠ (b)

The spring system exerts an upward spring force of FS = mg on the suspended object and undergoes an elongation of x. The effective spring constant is then keffective =

5.76

2 FS mg (1.50 kg ) ( 9.80 m s ) = = = 7.20 × 10 2 N m x x 2.04 × 10 −2 m

Refer to the sketch given in the solution of Problem 5.75. (a)

Because the object of mass m is in equilibrium, the tension in the lower spring, FS 2, must equal the weight of the object. Therefore, from Hooke’s law, the elongation of the lower spring is x2 =

FS 2 mg = k2 k2

From the fact that the point where the springs join (A) is in equilibrium, we conclude that the tensions in the two springs must be equal, FS1 = FS 2 = mg. The elongation of the upper spring is then

continued on next page

56157_05_ch05_p141-168.indd 159

10/12/10 2:24:31 PM

160

Chapter 5

x1 =

FS1 mg = k1 k1

and the total elongation of the spring system is ⎛1 1⎞ x = x1 + x2 = mg ⎜ + ⎟ ⎝ k1 k2 ⎠ (b)

The two-spring system undergoes a total elongation of x and exerts an upward spring force FS = mg on the suspended mass. The effective spring constant of the two springs in series is then

or 5.77

(a)

keffective =

FS mg = = x x

keffective =

k1 k2 k1 + k2

⎛1 1⎞ mg =⎜ + ⎟ ⎛ 1 1 ⎞ ⎝ k1 k2 ⎠ mg ⎜ + ⎟ ⎝ k1 k2 ⎠

−1

The person walking uses Ew = (220 kcal)(4 186 J 1 kcal ) = 9.21 × 10 5 J of energy while going 3.00 miles. The quantity of gasoline which could furnish this much energy is V1 =

9.21 × 10 5 J = 7.08 × 10 −3 gal 1.30 × 108 J gal

This means that the walker’s fuel economy in equivalent miles per gallon is fuel economy = (b)

3.00 mi = 424 mi gal 7.08 × 10 −3 gal

In 1 hour, the bicyclist travels 10.0 miles and uses ⎛ 4 186 J ⎞ EB = ( 400 kcal ) ⎜ = 1.67 × 10 6 J ⎝ 1 kcal ⎟⎠ which is equal to the energy available in V2 =

1.67 × 10 6 J = 1.29 × 10 −2 gal 8 1.30 × 10 J gal

of gasoline. Thus, the equivalent fuel economy for the bicyclist is 10.0 mi = 775 mi gal 1.29 × 10 −2 gal 5.78

When 1 pound (454 grams) of fat is metabolized, the energy released is E = ( 454 g )( 9.00 kcal g ) = 4.09 × 10 3 kcal . Of this, 20.0% goes into mechanical energy (climbing stairs in this case). Thus, the mechanical energy generated by metabolizing 1 pound of fat is Em = ( 0.200 )( 4.09 × 10 3 kcal ) = 818 kcal Each time the student climbs the stairs, she raises her body a vertical distance of Δy = (80 steps)(0.150 m step) = 12.0 m. The mechanical energy required to do this is ΔPEg = mg( Δy), or

continued on next page

56157_05_ch05_p141-168.indd 160

10/12/10 2:24:32 PM

Energy

161

⎛ 1 kcal ⎞ ΔPEg = ( 50.0 kg ) ( 9.80 m s 2 )(12.0 m ) = ( 5.88 × 10 3 J ) ⎜ = 1.40 kcal ⎝ 4186 J ⎟⎠ (a)

The number of times the student must climb the stairs to metabolize 1 pound of fat is Em 818 kcal = = 584 trips ΔPEg 1.40 kcal trip

N=

It would be more practical for her to reduce food intake. (b)

The useful work done each time the student climbs the stairs is W = ΔPEg = 5.88 × 10 3 J Since this is accomplished in 65.0 s, the average power output is Pav =

5.79

(a)

Use conservation of mechanical energy, ( KE + PEg ) f = ( KE + PEg )i, from the start to the end of the track to ﬁnd the speed of the skier as she leaves the track. This gives 2 1 2 m v + mgy f = 0 + mgyi, or

(

v= (b)

W 5.88 × 10 3 J ⎛ 1 hp ⎞ = = 90.5 W = ( 90.5 W ) ⎜ = 0.121 hp ⎝ 746 W ⎟⎠ t 65.0 s

)

2 g yi − y f =

At the top of the parabolic arc the skier follows after leaving the track, vy = 0 and vx = (28.0 m s) cos 45.0° = 19.8 m s. Thus, vtop = vx2 + vy2 = 19.8 m s. Applying conservation of mechanical energy from the end of the track to the top of the arc 2 2 gives 12 m 19.8 m s + mgymax = 12 m 28.0 m s + mg 10.0 m , or

(

)

ymax = 10.0 m + (c)

2 ( 9.80 m s 2 )( 40.0 m ) = 28.0 m s

(

( 28.0

)

(

)

m s ) − (19.8 m s ) = 30.0 m 2 ( 9.80 m s 2 ) 2

2

Using Δy = v0 y t + 12 ay t 2 for the ﬂight from the end of the track to the ground gives −10.0 m = ⎡⎣( 28.0 m s ) sin 45.0° ⎤⎦ t +

1 ( −9.80 m s2 ) t 2 2

The positive solution of this equation gives the total time of ﬂight as t = 4.49 s. During this time, the skier has a horizontal displacement of Δx = v0 x t = ⎡⎣( 28.0 m s ) cos 45.0° ⎤⎦ ( 4.49 s ) = 88.9 m 5.80

First, determine the magnitude of the applied force by considering a free-body diagram of the block. Since the block moves with constant velocity, ΣFx = ΣFy = 0. From ΣFx = 0, we see that n = F cos30°.

→

n

5 kg →

30°

fk

→

F →

mg

Thus, fk = μ k n = μ k F cos30°, and ΣFy = 0 becomes F sin 30° = mg + μ k F cos 30°, or

continued on next page

56157_05_ch05_p141-168.indd 161

10/12/10 2:24:32 PM

162

Chapter 5

F= (a)

(5.0 kg) (9.80 m s2 ) mg = = 2 .0 × 10 2 N sin 30° − μ k cos 30° sin 30° − ( 0.30 ) cos 30°

The applied force makes a 60° angle with the displacement up the wall. Therefore, WF = ( F cos 60° ) s = ⎡⎣( 2.0 × 10 2 N ) cos 60° ⎤⎦ ( 3.0 m ) = 3.0 × 10 2 J

5.81

(b)

Wg = ( mg cos 180° ) s = ( 49 N )( −1.0 )( 3.0 m ) = − 1.5 × 10 2 J

(c)

Wn = ( n cos90° ) s = 0

(d)

PEg = mg ( Δy ) = ( 49 N )( 3.0 m ) = 1.5 × 10 2 J

We choose PEg = 0 at the level where the spring is relaxed ( x = 0), or at the level of position B. (a)

At position A, KE = 0 and the total energy of the system is given by

(

E = 0 + PEg + PEs or (b)

)

A

= mg x1 +

1 2 k x1 2

E = ( 25.0 kg ) ( 9.80 m s 2 )( −0.100 m ) +

1 ( 2.50 × 10 4 N m )( −0.100 m )2 = 101 J 2

In position C, KE = 0 and the spring is uncompressed, so PEs = 0. Hence, E = (0 + PEg + 0)C = mg x2 or

(c)

x2 =

E 101 J = 0.412 m = mg ( 25.0 kg ) ( 9.80 m s 2 )

At Position B, PEg = PEs = 0 and E = ( KE + 0 + 0 )B = Therefore, vB =

(d)

2 (101 J ) = 2.84 m s 25.0 kg

Where the velocity (and hence the kinetic energy) is a maximum, the acceleration (slope of the velocity versus time graph) is zero. Thus, ΣFy = − kx − mg = 0 and we ﬁnd x=−

(e)

2E = m

1 m vB2 2

mg 245 kg =− = −9.80 × 10 −3 m = − 9.80 mm k 2.50 × 10 4 N m

From the total energy, E = KE + PEg + PEs = v=

1 2 1 mv + mg x + k x 2, we ﬁnd 2 2

2E k − 2 g x − x2 m m

Where the speed, and hence kinetic energy, is a maximum (that is, at x = − 9.80 mm ), this gives vmax = or

56157_05_ch05_p141-168.indd 162

( 2.50 × 104 N m ) −9.80 × 10−3 m 2 2 (101 J ) − 2 ( 9.80 m s 2 ) ( −9.80 × 10 −3 m ) − ( ) 25.0 kg 25.0 kg

vmax = 2.86 m s

10/12/10 2:24:33 PM

Energy

5.82

163

When the hummingbird is hovering, the magnitude of the average upward force exerted by the air on the wings (and hence, the average downward force the wings exert on the air) must be Fav = mg, where mg is the weight of the bird. Thus, if the wings move downward distance d during a wing stroke, the work done each beat of the wings is Wbeat = Fav d = mgd = ( 3.0 × 10 −3 kg )( 9.80 m s 2 ) ( 3.5 × 10 −2 m ) = 1.0 × 10 −3 J In one minute, the number of beats of the wings that occur is N = (80 beats s )( 60 s min ) = 4.8 × 10 3 beats min so the total work preformed in one minute is beats ⎞ ⎛ J ⎞ ⎛ −3 Wtotal = NWbeat (1 min ) = ⎜ 4.8 × 10 3 ⎟⎠ ⎜⎝ 1.0 × 10 ⎟ (1 min ) = 4.8 J ⎝ min beat ⎠

5.83

Choose PEg = 0 at the level of the river. Then yi = 36.0 m, y f = 4.00, the jumper falls 32.0 m, and the cord stretches 7.00 m. Between the balloon and the level where the diver stops momentarily, ( KE + PEg + PEs ) f = ( KE + PEg + PEs )i gives 0 + ( 700 N )( 4.00 m ) + or

5.84

1 2 k ( 7.00 m ) = 0 + ( 700 N )( 36.0 m ) + 0 2

k = 914 N m

If a projectile is launched, in the absence of air resistance, with speed v0 at angle θ above the horizontal, the time required to return to the original level is found from Δy = v0 y t + 12 ay t 2 as 0 = ( v0 sin θ ) t − ( g 2)t 2 , which gives t = (2 v0 sin θ ) g. The range is the horizontal displacement occurring in this time. Thus, ⎛ 2 v sin θ ⎞ v02 ( 2 sin θ cos θ ) v02 sin ( 2θ ) R = v0 x t = ( v0 cos θ ) ⎜ 0 = ⎟⎠ = g g g ⎝ Maximum range occurs when θ = 45°, giving Rmax = v02 g or v02 = g Rmax. The minimum kinetic energy required to reach a given maximum range is then KE = (a)

(b)

1 1 m v02 = mg Rmax 2 2

The minimum kinetic energy needed in the record throw of each object is Javelin:

KE =

1 ( 0.80 kg) (9.80 m s2 )(98 m ) = 3.8 × 10 2 J 2

Discus:

KE =

1 ( 2.0 kg) (9.80 m s2 )( 74 m ) = 7.3 × 10 2 J 2

Shot:

KE =

1 ( 7.2 kg) (9.80 m s2 )( 23 m ) = 8.1 × 10 2 J 2

The average force exerted on an object during launch, when it starts from rest and is given KE − 0 . the kinetic energy found above, is computed from Wnet = Fav s = ΔKE as Fav = s Thus, the required force for each object is Javelin:

Fav =

3.8 × 10 2 J = 1.9 × 10 2 N 2.00 m

continued on next page

56157_05_ch05_p141-168.indd 163

10/12/10 2:24:34 PM

164

Chapter 5

(c)

Discus:

Fav =

7.3 × 10 2 J = 3.6 × 10 2 N 2.00 m

Shot:

Fav =

8.1 × 10 2 J = 4.1 × 10 2 N 2.00 m

Yes . If the muscles are capable of exerting 4.1 × 10 2 N on an object and giving that object a kinetic energy of 8.1 × 10 2 J, as in the case of the shot, those same muscles should be able to give the javelin a launch speed of

v0 =

2 KE = m

2 (8.1 × 10 2 J ) 0.80 kg

= 45 m s

with a resulting range of v02 ( 45 m s ) = 2.1 × 10 2 m = g 9.80 m s 2 2

Rmax =

Since this far exceeds the record range for the javelin, one must conclude that air resistance plays a very signiﬁcant role in these events. 5.85

(a)

From the work-energy theorem, Wnet = KE f − KEi . Since the package moves with constant velocity, KE f = KEi, giving Wnet = 0 . Note that the above result can also be obtained by the following reasoning: Since the object has zero acceleration, the net (or resultant) force acting on it must be zero. The net work done is Wnet = Fnet d = 0 .

(b)

The work done by the conservative gravitational force is

(

)

Wgravity = − ΔPEg = − mg y f − yi = − mg ( d sin θ ) or (c)

Wgravity = − ( 50 kg ) ( 9.80 m s 2 )( 340 m ) sin 7.0° = − 2.0 × 10 4 J

The normal force is perpendicular to the displacement. The work it does is Wnormal = n d cos90° = 0

(d)

Since the package moves up the incline at constant speed, the net force parallel to the incline is zero. Thus, ΣF = 0 ⇒ fs − mg sin θ = 0 or fs = mg sin θ . The work done by the friction force in moving the package distance d up the incline is Wfriction = fk d = ( mg sin θ ) d = ⎡⎣( 50 kg ) ( 9.80 m s 2 ) sin 7.0°° ⎤⎦ ( 340 m ) = 2.0 × 10 4 J

5.86

Each 5.00-m length of the cord will stretch 1.50 m when the tension in the cord equals the weight of the jumper (that is, when Fs = w = mg). Thus, the elongation in a cord of original length L when Fs = w = mg will be ⎛ L ⎞ x=⎜ (1.50 m ) = 0.300 L ⎝ 5.00 m ⎟⎠ and the force constant for the cord of length L is

k=

Fs mg = x 0.300 L

continued on next page

56157_05_ch05_p141-168.indd 164

10/12/10 2:24:35 PM

Energy

(a)

165

In the bungee-jump from the balloon, the daredevil drops yi − y f = 55.0 m. The stretch of the cord at the start of the jump is xi = 0, and that at the lowest point is x f = 55.0 m − L . Since KEi = KE f = 0 for the fall, conservation of mechanical energy gives

(

0 + PEg

) + ( PE ) f

s

f

(

)

= 0 + PEg + ( PEs )i i

⇒

(

)

(

1 k x 2f − xi2 = mg yi − y f 2

)

giving 1 ⎛ mg ⎞ ( 55.0 m − L )2 = mg ( 55.0 m ) 2 ⎜⎝ 0.300 L ⎟⎠

and

( 55.0 m − L )2 = ( 33.0 m ) L

which reduces to

( 55.0 m )2 − (110 m ) L + L2 = ( 33.0 m ) L or

L2 − (143 m ) L + ( 55.0 m ) = 0 2

and has solutions of L=

− ( −143 m ) ±

( −143 m )2 − 4 (1)( 55.0 m )2 2 (1)

This yields L=

143 m ± 91.4 m 2

and

L = 117 m or L = 25.8 m

Only the L = 25.8 m solution is physically acceptable!

(b)

During the jump, ΣFy = ma y

⇒ kx − mg = ma y

or

⎛ mg ⎞ ⎜ ⎟ x − m g = m ay ⎝ 0.300 L ⎠

⎛ x ⎞ ay = ⎜ −1 g ⎝ 0.300 L ⎟⎠

Thus,

which has maximum value at x = xmax = 55.0 m − L = 29.2 m

(a )

y max

5.87

(a)

⎤ ⎡ 29.2 m =⎢ − 1⎥ g = 2.77 g = 27.1 m s 2 ⎣ 0.300 ( 25.8 m ) ⎦

While the car moves at constant speed, the tension in the cable is F = mg sin θ , and the power input is P = F v = mgv sin θ or P = ( 950 kg ) ( 9.80 m s 2 ) ( 2.20 m s ) sin 30.0° = 1.02 × 10 4 W = 10.2 kW

(b)

While the car is accelerating, the tension in the cable is Δv ⎞ ⎛ Fa = mg sin θ + ma = m ⎜ g sin θ + ⎟ ⎝ Δt ⎠ 2.20 m s − 0 ⎤ 3 = ( 950 kg ) ⎡⎢( 9.80 m s 2 ) sin 30.0° + ⎥⎦ = 4.83 × 10 N 12.0 s ⎣

continued on next page

56157_05_ch05_p141-168.indd 165

10/12/10 2:24:35 PM

166

Chapter 5

Maximum power input occurs the last instant of the acceleration phase. Thus, Pmax = Fa vmax = ( 4.83 × 10 3 N ) ( 2.20 m s ) = 10.6 kW (c)

The work done by the motor in moving the car up the frictionless track is

(

Wnc = ( KE + PE ) f − ( KE + PE )i = KE f + PEg or 5.88

(a)

)

f

−0=

1 m v 2f + mg ( L sinn θ ) 2

1 2 Wnc = ( 950 kg ) ⎡⎢ ( 2.20 m s ) + ( 9.80 m s 2 ) (1 250 m ) sin 30.0° ⎤⎥ = 5.82 × 10 6 J 2 ⎣ ⎦

Since the tension in the string is always perpendicular to the motion of the object, the string does no work on the object. Then, mechanical energy is conserved:

( KE + PE ) = ( KE + PE ) g

g i

f

Choosing PEg = 0 at the level where the string attaches to the cart, this gives 0 + mg ( − L cos θ ) = (b)

2 g L (1 − cos θ )

If L = 1.20 m and θ = 35.0°, the result of part (a) gives v0 =

5.89

1 m v02 + mg ( − L ), or v0 = 2

2 ( 9.80 m s 2 )(1.20 m )(1 − cos 35.0° ) = 2.06 m s

Observe that when m3 moves downward distance d, m1 must move upward distance d and m2 must slide distance d to the right across the horizontal tabletop. Also, each block must always have the same speed as each of the other blocks. Therefore, if the system starts from rest, and f is the friction force the table exerts on m2, the work energy theorem (Wnc = ΔKE + ΔPEg ) gives − fd =

(

)

1 ( m1 + m2 + m3 ) v 2f − vi2 + ⎡⎣ m1g ( + d ) + m2 g ( 0 ) + m3 g ( − d )⎤⎦ 2

With vi = 0, m1 = 5.0 kg, m2 = 10 kg, m3 = 15 kg, f = 30 N, andd d = 4.0 m , this yields a ﬁnal speed of vf = 5.90

2d ⎡⎣( m3 − m1 ) g − f ⎤⎦ = m1 + m2 + m3

2 ( 4.0 m ) ⎡⎣(10 kg ) ( 9.8 m s 2 ) − 30 N ⎤⎦ = 4.3 m s 30 kg

Call the object on the left Block 1 and that on the right Block 2. Then, m1 = 0.250 kg and m2 = 0.500 kg. The spring force exerted on each block has magnitude F = kx = ( 3.85 N m ) (8.00 × 10 −2 m ) = 0.308 N

a1

a2

n1 m1 m1g

n2 F f1

F f2

m2 m2g

The magnitudes of normal forces are n1 = m1 g = 2.45 N and n2 = m2 g = 4.90 N . (a)

If the surface is frictionless ( μ = 0 ) , then f1 = f2 = 0 and the net horizontal force acting on each block is F = 0.308 N. Taking to the right as the positive x-direction, the acceleration of each block is

continued on next page

56157_05_ch05_p141-168.indd 166

10/12/10 2:24:36 PM

Energy

(b)

For m1:

a1 =

ΣF1,x − F − 0.308 N = = = −1.23 m s 2 m1 m1 0.250 kg

For m2:

a2 =

ΣF2,x + F + 0.308 N = = = + 0.616 m s 2 m2 m2 0.500 kg

167

If μ k = 0.100 for each block, the kinetic friction forces would be fk ,1 = μ k n1 = ( 0.100 )( 2.45 n ) = 0.245 N fk ,2 = μ k n2 = ( 0.100 )( 4.90 n ) = 0.490 N

and

Note that fk ,1 < F , so if the coefﬁcient of static friction is low enough to allow the spring force to start m1 moving, this block will have a net horizontal force acting on it and its acceleration will be a1 =

− F + f1,k − 0.308 N + 0.245 N = = − 0.252 m s 2 m1 0.250 kg

However, fk ,2 > F , and the maximum static friction force is even larger. Thus, the spring force will not be able to overcome static friction and start m2 moving. The acceleration of this block is a2 = 0 in this case. (c)

5.91

For μ k = 0.462, fk ,1 = ( 0.462 ) n1 = 1.13 N and fk ,2 = ( 0.462 ) n2 = 2.26 N, with even larger static friction forces possible. In this case, the spring force is unable to start either block, meaning that a1 = a2 = 0 .

When the cyclist travels at constant speed, the magnitude of the forward static friction force on the drive wheel equals that of the retarding air resistance force. Hence, the friction force is proportional to the square of the speed, and her power output may be written as P = fs v = ( k v 2 ) v = k v 3 where k is a proportionality constant. If the heart rate R is proportional to the power output, then R = k ′P = k ′( k v 3 ) = k ′k v 3, where k ′ is also a proportionality constant. The ratio of the heart rate R2 at speed v2 to the rate R1 at speed v1 is then R2 k ′k v23 ⎛ v2 ⎞ = = R1 k ′k v13 ⎜⎝ v1 ⎟⎠

3

⎛R ⎞ v2 = v1 ⎜ 2 ⎟ ⎝ R1 ⎠

giving

13

Thus, if R = 90.0 beats min at v = 22.0 km h , the speed at which the rate would be 136 beats min is ⎛ 136 beats min ⎞ v = ( 22.0 km h ) ⎜ ⎝ 90.0 beats min ⎟⎠

13

= 255.2 km h

and the speed at which the rate would be 166 beats min is ⎛ 166 beats min ⎞ v = ( 22.0 km h ) ⎜ ⎝ 90.0 beats min ⎟⎠

56157_05_ch05_p141-168.indd 167

13

= 277.0 km h

10/12/10 2:24:36 PM

168

5.92

Chapter 5

The normal force the incline exerts on block A is nA = ( mA g) cos37°, and the friction force is fk = μ k nA = μ k mA g cos37°. The vertical distance block A rises is ΔyA = (20 m)sin 37° = 12 m, while the vertical displacement of block B is ΔyB = −20 m. We ﬁnd the common ﬁnal speed of the two blocks by use of

(

Wnc = KE + PEg

) − ( KE + PE ) = ΔKE + ΔPE f

g i

g

1 This gives − ( μ k mA g cos37° ) s = ⎡⎢ ( mA + mB ) v 2f − 0 ⎤⎥ + ⎡⎣ mA g ( ΔyA ) + mB g ( ΔyB ) ⎤⎦ ⎣2 ⎦

or

v 2f =

=

2 g ⎡⎣ − mB ( ΔyB ) − mA ( ΔyA ) − ( μ k mA cos 37° ) s ⎤⎦ mA + mB

2 ( 9.80 m s 2 ) ⎡⎣ − (100 kg )( −20 m ) − ( 50 kg )(12 m ) − 0.25 ( 50 kg )( 20 m ) cos 37° ⎤⎦ 150 kg

which yields v 2f = 157 m 2 s 2 . The change in the kinetic energy of block A is then ΔKEA =

56157_05_ch05_p141-168.indd 168

1 1 mA v 2f − 0 = ( 50 kg ) (157 m 2 s 2 ) = 3.9 × 10 3 J = 3.9 kJ 2 2

10/12/10 2:24:37 PM

6 Momentum and Collisions QUICK QUIZZES 1.

Choice (b). The relation between the kinetic energy of an object and the magnitude of the 2 momentum of that object is KE = p 2m. Thus, when two objects having masses m1 < m2 have equal kinetic energies, we may write p2 p12 = 2 2m1 2m2

so

p1 = p2

m1 1, since mb > mg mg

Mechanical energy is gained because work is done by the skater's muscles as they push each other apart. The origin of this work is chemical energy within their bodies.

6.44

(a)

v2

y

m2

m148.0 kg v112.0 m/s m262.0 kg v215.0 m/s m3112 kg

v3 m3 105° q

(b)

(c)

m1

v1

x

x-direction:

Σpxf = Σpxi

⇒

m1 v1 cos 0° + m2 v2 cos105° + m3 v3 cosq = 0

y-direction:

Σpyf = Σpyi

⇒

m1 v1 sin 0° + m2 v2 sin105° + m3 v3 sinq = 0

p1x = m1 v1 cos 0° = ( 48.0 kg ) (12.0 m s ) (1) = 576 kg ⋅ m s p2 x = m2 v2 cos105° = ( 62.0 kg ) (15.0 m s ) ( −0.259 ) = −241 kg ⋅ m s

(d)

p1y = m1 v1 sin 0° = ( 48.0 kg ) (12.0 m s ) ( 0 ) = 0 p2 y = m2 v2 sin105° = ( 62.0 kg ) (15.0 m s ) ( +0.966 ) = 898 kg ⋅ m s

(e)

(f)

x-direction:

576 kg ⋅ m s − 241 kg ⋅ m s + (112 kg ) v3 cosq = 0

y-direction:

0 + 898 kg ⋅ m s + (112 kg ) v3 sinq = 0

x-direction:

v3 cosq =

−576 kg ⋅ m s + 241 kg ⋅ m s 112 kg

or

v3 cosq = −2.99 m s

y-direction:

v3 sinq =

−898 kg ⋅ m s 112 kg

or

v3 sinq = −8.02 m s

Then, squaring and adding these results, recognizing that cos 2 q + sin 2 q = 1, gives

(

)

v32 cos 2 q + sin 2 q = ( −2.99 m s ) + ( −8.02 m s ) and v3 = 73.3 m 2 s 2 = 8.56 m s (g)

v3 sinθ v3 cosθ

= tanq =

2

−8.02 m s = 2.68 −2.99 m s

2

so

q = tan −1 ( 2.68 ) + 180° = 250°

Note that the factor of 180° was included in the last calculation because it was recognized that both the sine and cosine of angle q were negative. This meant that q had to be a third quadrant angle. Use of the inverse tangent function alone yields only the principle angles ( −90° ≤ q ≤ +90°) that have the given value for the tangent function.

continued on next page

56157_06_ch06_p169-208.indd 189

10/12/10 2:26:01 PM

190

Chapter 6

(h)

6.45

Because the third fragment must have a momentum equal in magnitude and opposite direction to the resultant of the other two fragments momenta, all three pieces must travel in the same plane .

Conservation of momentum gives

( 25.0 g ) v

1f

or

+ (10.0 g ) v2 f = ( 25.0 g ) ( 20.0 cm s ) + (10.0 g ) (15.0 cm s )

2.50v1 f + v2 f = 65.0 cm s

[1]

(

)

For head-on, elastic collisions, we know that v1i − v2 i = − v1 f − v2 f . Thus, 20.0 cm s − 15.0 cm s = −v1 f + v2 f or v2 f = v1 f + 5.00 cm s

[2]

Substituting Equation [2] into [1] yields 3.50v1 f = 60.0 cm s, or v1 f = 17.1 cm s . Equation [2] then gives v2 f = 17.1 cm s + 5.00 cm s = 22.1 cm s . 6.46

First, consider conservation of momentum and write m1 v1i + m2 v2 i = m1 v1 f + m2 v2 f Since m1 = m2 , this becomes v1i + v2 i = v1 f + v2 f

[1]

For an elastic head-on collision, we also have v1i − v2 i = −(v1 f − v2 f ), which may be written as v1i − v2 i = −v1 f + v2 f

[2]

Adding Equations [1] and [2] yields v2 f = v1i

[3]

Subtracting Equation [2] from [1] gives v1 f = v2 i

[4]

Equations [3] and [4] show us that, under the conditions of equal mass objects striking one another in a head-on, elastic collision, the two objects simply exchange velocities. Thus, we may write the results of the various collisions as

6.47

(a)

v1 f = 0 , v2 f = 1.50 m s

(b)

v1 f = − 1.00 m s , v2 f = 1.50 m s

(c)

v1 f = 1.00 m s , v2 f = 1.50 m s

(a)

Over a the short time interval of the collision, external forces have no time to impart signiﬁcant impulse to the players. The two players move together after the tackle , so the collision is completely inelastic.

continued on next page

56157_06_ch06_p169-208.indd 190

10/12/10 2:26:05 PM

Momentum and Collisions

191

y (north)

(b)

vf m1 m2 v1i

m1

q

v2i

x (east)

m1 90.0 kg m2 95.0 kg

v1i 5.00 m/s v2i 3.00 m/s

m2

px f = Σpxi

⇒

or v f cosq =

( m1 + m2 ) v f cosq = m1v1i + 0

( 90.0 kg ) ( 5.00 m s ) m1 v1i = ( m1 + m2 ) 90.0 kg + 95.0 kg

v f cosq = 2.43 m s

and

( m1 + m2 ) v f sinq = 0 + m2 v2i ( 95.0 kg ) ( 3.00 m s ) m2 v2i v f sinq = = ( m1 + m2 ) 90.0 kg + 95.0 kg

py f = Σpyi giving

⇒

(

)

v f sinq = 1.54 m s

and

Therefore, v 2f sin 2 q + cos 2 q = v 2f = (1.54 m s ) + ( 2.43 m s ) , and 2

2

v f = 8.28 m 2 s 2 = 2.88 m s Also, tanq = Thus, (c)

6.48

v f cosq

=

1.54 m s = 0.634 2.43 m s

and

q = tan −1 ( 0.634 ) = 32.4°

v f = 2.88 m s at 32.4° north of east

KElost = KEi − KE f = =

(d)

v f sinq

1 1 1 m1 v1i2 + m2 v2i2 − ( m1 + m2 ) v 2f 2 2 2

1 1⎡ ( 90.0 kg ) ( 5.00 m s )2 + ( 95.0 kg ) ( 3.00 m s )2 ⎤⎦ − (185 kg ) ( 2.88 m s )2 = 785 J ⎣ 2 2

The lost kinetic energy is transformed into other forms of energy, such as thermal energy and sound.

Consider conservation of momentum in the ﬁrst event (twin A tossing the pack), taking the direction of the velocity given the backpack as positive. This yields

(

mA vAf + mpack vpack = mA + mpack

) (0) = 0

or vAf =

−mpack vpack mA

⎛ 12.0 kg ⎞ = −⎜ ( +3.00 m s ) = −0.655 m s and vAf = 0.655 m s ⎝ 55.0 kg ⎟⎠ continued on next page

56157_06_ch06_p169-208.indd 191

10/12/10 2:26:09 PM

192

Chapter 6

Conservation of momentum when twin B catches and holds onto the backpack yields

(m or 6.49

B

)

+ mpack vBf = mB ( 0 ) + mpack vpack

vBf =

mpack vpack mB + mpack

=

(12.0 kg ) ( +3.00

m s) = 0.537 m s 55.0 kg + 12.0 kg

Choose the +x-axis to be eastward and the +y-axis northward. If vi is the initial northward speed of the 3 000-kg car, conservation of momentum in the y-direction gives 0 + ( 3 000 kg ) vi = ( 3 000 kg + 2 000 kg ) ⎡⎣( 5.22 m s ) sin 40.0° ⎤⎦ or

vi = 5.59 m s

Observe that knowledge of the initial speed of the 2 000-kg car was unnecessary for this solution. 6.50

We use conservation of momentum for both eastward and northward components. For the eastward direction: M (13.0 m s ) = 2 M V f cos 55.0° For the northward direction: M v2 i = 2 M V f sin 55.0° Divide the northward equation by the eastward equation to ﬁnd 2MV f sin 55.0° M v2 i = M (13.0 m s ) 2MV f cos 55.0°

or

v2 i = (13.0 m s ) tan 55.0°

⎡ ⎛ 2.237 mi h ⎞ ⎤ yielding v2 i = ⎢(13.0 m s ) ⎜ ⎥ tan 55.0° = 41.5 mi h ⎝ 1 m s ⎟⎠ ⎥⎦ ⎢⎣ Thus, the driver of the northbound car was untruthful . 6.51

Choose the x-axis to be along the original line of motion. (a)

From conservation of momentum in the x-direction, m ( 5.00 m s ) + 0 = m ( 4.33 m s ) cos 30.0° + m v2 f cosq or

v2 f cosq = 1.25 m s

[1]

Conservation of momentum in the y-direction gives 0 = m ( 4.33 m s ) sin 30.0° + m v2 f sinq , or v2 f sinq = − 2 .16 m s Dividing Equation [2] by [1] gives tanq =

[2]

− 2 .16 = −1.73 and q = −60.0° 1.25

Then, either Equation [1] or [2] gives v2 f = 2 .50 m s, so the ﬁnal velocity of the second ball is v 2 f = 2.50 m s at − 60.0° .

continued on next page

56157_06_ch06_p169-208.indd 192

10/12/10 2:26:11 PM

Momentum and Collisions

(b)

(

KEi =

1 1 2 m v1i2 + 0 = m ( 5.00 m s ) = m 12 .5 m 2 s 2 2 2

KE f =

1 1 m v12 f + m v22 f 2 2

=

193

)

(

1 1 2 2 m ( 4.33 m s ) + m ( 2 .50 m s ) = m 12 .5 m 2 s 2 2 2

)

Since KE f = KEi, this is an elastic collision . 6.52

The recoil speed of the subject plus pallet after a heartbeat is V=

Δx 6.00 × 10 −5 m = = 3.75 × 10 −4 m s Δt 0.160 s

From conservation of momentum, mv − MV = 0 + 0, so the mass of blood leaving the heart is ⎛ 3.75 × 10 −4 m s ⎞ ⎛V⎞ m = M ⎜ ⎟ = ( 54.0 kg ) ⎜ = 4.05 × 10 −2 kg = 40.5 g ⎝ v⎠ ⎝ 0.500 m s ⎟⎠ 6.53

Choose the positive direction to be the direction of the truck’s initial velocity. Apply conservation of momentum to ﬁnd the velocity of the combined vehicles after collision:

( 4 000 kg + 800 kg ) v

f

= ( 4 000 kg ) ( +8.00 m s ) + ( 800 kg ) ( −8.00 m s )

which yields v f = + 5.33 m s.

(

)

Use the impulse-momentum theorem, I = Fav ( Δt ) = Δp = m v f − vi , to ﬁnd the magnitude of the average force exerted on each driver during the collision. Truck Driver: Fav =

m v f − vi

=

truck

Δt

( 80.0 kg ) 5.33

m s − 8.00 m s = 1.78 × 10 3 N 0.120 s

Car Driver: Fav = 6.54

m v f − vi Δt

car

=

( 80.0 kg ) 5.33

m s − ( −8.00 m s )

0.120 s

= 8.89 × 10 3 N

First, we use conservation of mechanical energy to ﬁnd the speed of m1 at B just before collision. 1 This gives m1 v12 + 0 = 0 + m1 ghi , 2 or

v12 =

2 g hi =

(

2 9.80 m s 2

) ( 5.00 m ) = 9.90 m s

Next, we apply conservation of momentum and knowledge of elastic collisions to ﬁnd the velocity of m1 at B just after collision.

continued on next page

56157_06_ch06_p169-208.indd 193

10/12/10 2:26:14 PM

194

Chapter 6

From conservation of momentum, with the second object initially at rest, we have m1 v1 f + m2 v2 f = m1 v1i + 0,

v2 f =

or

m1 v1i − v1 f m2

(

)

[1]

For head-on elastic collisions, v1i − v2 i = −(v1 f − v2 f ). Since v2 i = 0 in this case, this becomes v2 f = v1 f + v1i , and, combining this with Equation [1] above, we obtain v1 f + v1i =

so

m1 v1i − v1 f m2

(

)

(m

or

1

+ m2 ) v1 f = ( m1 − m2 ) v1i

⎛ m − m2 ⎞ ⎛ 5.00 − 10.0 ⎞ v1 f = ⎜ 1 v1i = ⎜ ( 9.90 m s ) = − 3.30 m s ⎟ ⎝ 5.00 + 10.0 ⎟⎠ m + m ⎝ 1 2 ⎠

Finally, use conservation of mechanical energy for m1 after the collision to ﬁnd the maximum rebound height. This gives (KE + PEg ) f = (KE + PEg )i or 6.55

1 = m1 v12 f + 0 2

0 + m1 ghm ax

hm ax =

and

v12 f 2g

( − 3.30 =

(

m s)

2

2 9.80 m s 2

)

= 0.556 m

Note that the initial velocity of the target particle is zero (that is, v2 i = 0). From conservation of momentum [1]

m1 v1 f + m2 v2 f = m1 v1i + 0

(

)

For head-on elastic collisions, v1i − v2 i = − v1 f − v2 f , and with v2 i = 0, this gives v2 f = v1i + v1 f

[2]

Substituting Equation [2] into [1] yields

(

)

m1 v1 f + m2 v1i + v1 f = m1 v1i or

( m1 + m2 ) v1 f = ( m1 − m2 ) v1i

⎛ m − m2 ⎞ v1 f = ⎜ 1 v1i ⎝ m1 + m2 ⎟⎠

and

[3]

Now, we substitute Equation [3] into [2] to obtain ⎛ m − m2 ⎞ v2 f = v1i + ⎜ 1 v1i ⎝ m1 + m2 ⎟⎠

or

⎛ 2m1 ⎞ v2 f = ⎜ v1i ⎝ m1 + m2 ⎟⎠

[4]

Equations [3] and [4] can now be used to answer both parts (a) and (b). (a)

If m1 = 2 .0 g, m2 = 1.0 g, and v1i = 8.0 m s, then v1 f =

8 ms 3

and v2 f =

32 ms 3

continued on next page

56157_06_ch06_p169-208.indd 194

10/12/10 2:26:17 PM

Momentum and Collisions

(b)

If m1 = 2 .0 g, m2 = 10 g, and v1i = 8.0 m s, we ﬁnd v1 f = −

(c)

195

16 8 m s and v2 f = ms 3 3

The ﬁnal kinetic energy of the 2.0 g particle in each case is: 2

Case (a): KE1 f =

1 1 ⎛8 ⎞ m1 v12 f = 2.0 × 10 −3 kg ⎜ m s⎟ = 7.1 × 10 −3 J ⎝3 ⎠ 2 2

Case (b): KE1 f =

1 1 ⎛ 16 ⎞ m1 v12 f = 2.0 × 10 −3 kg ⎜ − m s⎟ = 2 .8 × 10 −2 J ⎝ ⎠ 2 2 3

(

)

(

2

)

Since the incident kinetic energy is the same in cases (a) and (b), we observe that the incident particle loses more kinetic energy in case (a) . 6.56

If the pendulum bob barely swings through a complete circle, it arrives at the top of the arc (having risen a vertical distance of 2 ) with essentially zero velocity. From conservation of mechanical energy, we ﬁnd the minimum velocity of the bob at the bottom of the arc as (KE + PEg )bottom = (KE + PEg )top , or 12 M V 2 = 0 + M g (2 ) . This gives V = 2 g as the needed velocity of the bob just after the collision. Conserving momentum through the collision then gives the minimum initial velocity of the bullet as

(

⎛ v⎞ m⎜ ⎟ + M 2 g ⎝ 2⎠ 6.57

) = mv + 0

v=

or

4M m

g

Note: We consider the spring to have negligible mass and ignore any energy or momentum it may possess after being released. Also, we take toward the right as the positive direction. Conservation of momentum, p f = pi, gives m1 v1 + m2 v2 = 0

or

⎛m ⎞ v2 = − ⎜ 1 ⎟ v1 ⎝ m2 ⎠

[1]

Since the surface is frictionless, conservation of energy, KE1 f + KE2 f + PEs, f = KE1i + KE2i + PEs,i with KE1i = KE2i = PEs, f = 0 gives 1 1 1 m1 v12 + m2 v22 = kd 2 2 2 2

[2]

where d = 9.8 cm is the initial compression of the spring. Substituting Equation [1] into Equation [2] gives: 1 ⎛ m2 ⎞ 1 1 m1 v12 + m2 ⎜ 12 v12 ⎟ = kd 2 2 ⎝ m2 ⎠ 2 2

or

⎛ m ⎞ m1 ⎜ 1 + 1 ⎟ v12 = kd 2 m2 ⎠ ⎝

continued on next page

56157_06_ch06_p169-208.indd 195

10/12/10 2:26:21 PM

196

Chapter 6

Choosing the negative sign (since m1will recoil to the left), this result yields v1 = −d and

k = − 9.8 × 10 −2 m m1 (1 + m1 m2 )

(

v1 = −1.7 m s

or

)

280 N m ( 0.56 kg ) (1 + 0.56 0.88 )

1.7 m s to the left

Then, Equation [1] gives the velocity of the second object as ⎛ 0.56 ⎞ v2 = − ⎜ ( −1.7 m s ) = +1.1 m s ⎝ 0.88 ⎟⎠ 6.58

or

(

Use conservation of mechanical energy, KE + PEg

1.1 m s to the right

) = ( KE + PE ) , to ﬁnd the speed of the blue g

B

A

bead at point B just before it collides with the green bead. This gives v1i =

2 g yA =

(

1 m v1i2 + 0 = 0 + m g y A , or 2

)

2 9.80 m s 2 (1.50 m ) = 5.42 m s

Conservation of momentum during the collision gives + ( 0.600 kg ) v2 f = ( 0.400 kg ) ( 5.42 m s ) + 0

( 0.400 kg ) v

1f

or

v1 f + 1.50 v2 f = 5.42 m s

[1]

(

)

For a head-on elastic collision, we have v1i − v2 i = − v1 f − v2 f , and with v2i = 0, this becomes v1 f = v2 f − v1i

or

[2]

v1 f = v2 f − 5.42 m s

Substitute Equation [2] into [1] to ﬁnd the speed of the green bead just after collision as v2 f − 5.42 m s + 1.50 v2 f = 5.42 m s

v2 f =

or

2 ( 5.42 m s ) = 4.34 m s 2.50

Now, we use conservation of energy for the green bead after collision to ﬁnd the maximum height it will achieve. This gives 0 + m2 g ym ax = 6.59

1 m2 v22 f + 0 2

or

ym ax =

v22 f 2g

=

( 4.34

(

m s)

2

2 9.80 m s 2

)

= 0.961 m

We shall choose southward as the positive direction. The mass of the man is m=

w 730 N = = 74.5 kg g 9.80 m s 2

Then, from conservation of momentum, we ﬁnd

(m

m an

vm an + mbook vbook ) f = ( mm an vm an + mbook vbook )i

or

( 74.5 kg ) v

m an

+ (1.2 kg ) ( −5.0 m s ) = 0 + 0

and

vm an = 8.1 × 10 −2 m s

continued on next page

56157_06_ch06_p169-208.indd 196

10/12/10 2:26:24 PM

197

Momentum and Collisions

Therefore, the time required to travel the 5.0 m to shore is t= 6.60

Δx 5.0 m = = 62 s vm an 8.1 × 10 −2 m s

The mass of the third fragment must be m3 = mnucleus − m1 − m2 = (17 − 5.0 − 8.4 ) × 10 −27 kg = 3.6 × 10 −27 kg Conserving momentum in both the x- and y-directions gives y-direction:

m1 v1y + m2 v2 y + m3 v3y = 0

or

v3y = −

x-direction:

m1 v1x + m2 v2 x + m3 v3x = 0

or

v3x = −

and Also,

m1 v1y + m2 v2 y m3

=−

( 5.0 × 10

−27

)(

)

kg 6.0 × 10 6 m s + 0

3.6 × 10

(

−27

kg

)(

=−

30 × 10 6 m s 3.6

)

0 + 8.4 × 10 −27 kg 4.0 × 10 6 m s m1 v1x + m2 v2 x 34 =− × 10 6 m s =− m3 3.6 × 10 −27 kg 3.6

2 2 v3 = v3x + v3y =

( − ( 34 3.6 ) × 10

6

) (

m s + − ( 30 3.6 ) × 10 6 m s 2

)

2

= 1.3 × 10 7 m s

⎛ v3y ⎞ ⎛ 30 ⎞ q = tan −1 ⎜ + 180° = tan −1 ⎜ ⎟ + 180° = 2.2 × 10 2 degrees = 220° ⎟ ⎝ 34 ⎠ ⎝ v3x ⎠

Therefore, v 3 = 1.3 × 10 7 m s at 220° counterclockwise from the + x-axis . Note that the factor of 180° was included in the calculation for q because it was recognized that both v3x and v3y were negative. This meant that q had to be a third quadrant angle. Use of the inverse tangent function alone yields only the principal angles ( −90° ≤ θ ≤ +90° ) that have the given value for the tangent function. 6.61

The sketch at the right gives before and after views of the collision between these two objects. Since the collision is elastic, both kinetic energy and momentum must be conserved.

m1

v1i v0

v2i v0

m2

Before Impact v1f 0 m1

Conservation of Momentum:

m2

v2f v

After Impact

m1 v1 f + m2 v2 f = m1 v1i + m2 v2i m1 ( 0 ) + m2 v = m1 v0 + m2 ( −v0 ) or

⎛m ⎞ v = ⎜ 1 − 1⎟ v0 ⎝ m2 ⎠

[1]

(

)

Since this is a perfectly elastic collision, v1i − v2i = − v1 f − v2 f , and with the given velocities this becomes v0 − ( −v0 ) = − ( 0 − v )

or

v = 2v0

[2] continued on next page

56157_06_ch06_p169-208.indd 197

10/12/10 2:26:27 PM

198

6.62

Chapter 6

(a)

⎛m ⎞ Substituting Equation [2] into [1] gives 2 v0 = ⎜ 1 − 1⎟ v0 ⎝ m2 ⎠

(b)

From Equation [2] above, we have

(a)

Let v1i and v2 i be the velocities of m1 and m2 just before the collision. Then, using conservation of mechanical energy, KE + PEg = KE + PEg or 12 mvi2 + 0 = 0 + mgh0 ,

v v0 = 2

(

v1i = − v2 i =

gives

m1 m2 = 3

or

) (

)

i

(

2 g h0 =

2 9.80 m s 2

0

) ( 5.00 m ) = 9.90 m s

v1i = + 9.90 m s and v2 i = − 9.90 m s (b)

From conservation of momentum:

( 2.00 g ) v

1f

or

+ ( 4.00 g ) v2 f = ( 2.00 g ) ( 9.90 m s ) + ( 4.00 g ) ( −9.90 m s )

v1 f + ( 2.00 ) v2 f = −9.90 m s

[1]

(

)

For a perfectly elastic, head-on collision, v1i − v2 i = − v1 f − v2 f , giving + 9.90 m s − ( −9.90 m s ) = −v1 f + v2 f

v2 f = v1 f + 19.8 m s

or

(

[2]

)

Substituting Equation [2] into [1] gives v1 f + ( 2.00 ) v1 f + 19.8 m s = −9.90 m s or

v1 f =

−9.90 m s − 39.6 m s = −16.5 m s 3.00

Then, Equation [2] yields v2 f = −16.5 m s + 19.8 m s = +3.30 m s . (c)

Applying conservation of energy to each block after the collision gives 1 1 2 m ( 0 ) + mghmax = mv 2f + mg ( 0 ) 2 2 Thus,

and 6.63

(a)

h1 f =

h2 f =

v12 f 2g v22 f 2g

( −16.5 =

(

m s)

( 3.30 m s )

(

hmax =

v 2f 2g

2

2 9.80 m s 2

=

or

)

= 13.9 m

2

2 9.80 m s 2

)

= 0.556 m

Use conservation of mechanical energy to ﬁnd the speed of m1 just before collision. Taking y = 0 at the tabletop level, this gives 12 m1 v1i2 + mg ( 0 ) = 12 m1 ( 0 ) + m1 gh1, or v1i =

2 g h1 =

(

2 9.80 m s 2

) ( 2.50 m ) = 7.00 m s

Apply conservation of momentum from just before to just after the collision:

( 0.500 kg ) v

1f

or

+ (1.00 kg ) v2 f = ( 0.500 kg ) ( 7.00 m s ) + 0

v1 f + 2 v2 f = 7.00 m s

[1]

continued on next page

56157_06_ch06_p169-208.indd 198

10/12/10 2:26:31 PM

Momentum and Collisions

199

For a perfectly elastic head-on collision, v1i − v2 i = −( v1 f − v2 f ) . With v2i = 0, this becomes v2 f = v1 f + v1i

v2 f = v1 f + 7.00 m s

or

[2]

Substituting Equation [2] into [1] yields

(

)

v1 f + 2 v1 f + 7.00 m s = 7.00 m s

v1 f =

and

−7.00 m s = − 2.33 m s 3

Then, from Equation [2], v2 f = − 2.33 m s + 7.00 m s = 4.67 m s (b)

Apply conservation of mechanical energy to m1 after the collision to ﬁnd the rebound height of this object. 1 1 m1 ( 0 ) + m1 gh1′ = m1 v12f + mg ( 0 ) 2 2

(c)

h1′ =

or

v12 f 2g

=

( −2.33 m s )

(

2

2 9.80 m s 2

)

= 0.277 m

From Δy = v0 y t + 12 a y t 2, with v0 y = 0, the time for m2 to reach the ﬂoor after it ﬂies horizontally off the table is t=

2 ( Δy ) = ay

2 ( −2 .00 m ) = 0.639 s –9.80 m s 2

During this time it travels a horizontal distance Δx = v0 x t = ( 4.67 m s ) ( 0.639 s ) = 2.98 m (d)

After the 0.500 kg mass comes back down the incline, it ﬂies off the table with a horizontal velocity of 2.33 m/s. The time of the ﬂight to the ﬂoor is 0.639 s as found above and the horizontal distance traveled is Δx = v0 x t = ( 2.33 m s ) ( 0.639 s ) = 1.49 m

6.64

We label the two objects such that object 1 has mass m while object 2 has mass 3m. Conservation of the x-component of momentum gives

( 3m ) v

2x

+ 0 = −mv0 + ( 3m ) v0

or

v2 x =

2 v0 3

[1]

Likewise, conservation of the y-component of momentum gives −m v1 y + ( 3m ) v2 y = 0

and

v1 y = 3 v2 y

[2]

Since the collision is elastic, ( KE ) f = ( KE )i , or

(

)

1 1 1 1 m v12y + ( 3m ) v22 x + v22 y = m v02 + ( 3m ) v02 2 2 2 2

which reduces to

(

)

v12y + 3 v22 x + v22 y = 4 v02

[3] continued on next page

56157_06_ch06_p169-208.indd 199

10/12/10 2:26:35 PM

200

Chapter 6

Substituting Equations [1] and [2] into [3] yields ⎛4 ⎞ 9 v22 y + 3 ⎜ v02 + v22 y ⎟ = 4 v02 ⎠ ⎝9 (a)

6.65

(a)

2 3

From Equation [2], the particle of mass m has ﬁnal speed v1 y = 3 v2 y = v0 particle of mass 3m moves at v2 = v22 x + v22 y =

(b)

v2 y = v0

or

4 2 2 2 v0 + v0 = v0 9 9

2 , and the

2 3

⎛ v 2 3⎞ ⎛ v2 y ⎞ ⎛ 1 ⎞ = 35.3° = tan −1 ⎜ 0 q = tan −1 ⎜ = tan −1 ⎜ ⎟ ⎟ ⎝ 2 ⎟⎠ ⎝ v2 x ⎠ ⎝ 2v0 3 ⎠ The momentum of the system is initially zero and remains constant throughout the motion. Therefore, when m1 leaves the wedge, we must have m2 v2 + m1 v1 = 0, or ⎛m ⎞ 2.00 ⎛ 0.500 ⎞ v2 = − ⎜ 1 ⎟ v f = − ⎜ ms ⎟⎠ ( 4.00 m s ) = − ⎝ 3.00 3.00 ⎝ m2 ⎠ meaning

(b)

v 2 = 0.667 m s to the left

Using conservation of energy as the block slides down the wedge, we have

( KE + PE ) = ( KE + PE ) g

Thus, h =

=

6.66

i

g

f

0 + m1 gh =

or

1 1 m1 v12 + m2 v22 + 0 2 2

1 ⎡ 2 ⎛ m2 ⎞ 2 ⎤ v2 ⎥ ⎢ v1 + ⎜ 2 g ⎢⎣ ⎝ m1 ⎟⎠ ⎥⎦ 2 ⎡ 1 2 ⎛ 3.00 ⎞ ⎛ 2.00 ⎞ ⎤ 4.00 m s + − m s ( ) ⎢ ⎜⎝ ⎟⎜ ⎟⎠ ⎥ = 0.952 m 0.500 ⎠ ⎝ 3.00 19.6 m s 2 ⎣⎢ ⎦⎥

Choose the positive x-axis in the direction of the initial velocity of the cue ball. Let vci be the initial speed of the cue ball, vcf be the ﬁnal speed of the cue ball, vTf be the ﬁnal speed of the target, and q be the angle the target’s ﬁnal velocity makes with the x-axis. Conservation of momentum in the x-direction, recognizing that all billiard balls have the same mass, gives m vTf cosq + m vcf cos30.0° = 0 + mvci

or

vTf cosq = vci – vcf cos30.0°

[1]

The conservation equation for momentum in the y-direction is m vTf sinq + m vcf sin 30.0° = 0 + 0

or

vTf sinq = −vcf sin 30.0°

[2]

Since this is an elastic collision, kinetic energy is conserved, giving 1 1 1 m vTf2 + m vcf2 = m vci2 2 2 2

or

vTf2 = vci2 − vcf2

[3]

continued on next page

56157_06_ch06_p169-208.indd 200

10/12/10 2:26:38 PM

Momentum and Collisions

(b)

201

To solve, square Equations [1] and [2] and add the results to obtain vTf2 ( cos 2 q + sin 2 q ) = vci2 − 2 vci vcf cos30.0° + vcf2 ( cos 2 30.0° + sin 2 30.0° ) vTf2 = vci2 − 2 vci vcf cos30.0° + vcf2

or

Now, substitute this result into Equation [3] to get vci2 − 2 vci vcf cos30.0° + vcf2 = vci2 − vcf2

(

)

2vcf vcf − vci cos30.0° = 0

or

Since vcf ≠ 0, it is necessary that vcf = vci cos30.0° = ( 4.00 m s ) cos30.0° = 3.46 m s Then, Equation [3] yields vTf = vci2 − vcf2 , or vTf = (a)

( 4.00

m s ) − ( 3.46 m s ) = 2.00 m s 2

2

With the results found above, Equation [2] gives ⎛ vcf ⎞ ⎛ 3.46 m s ⎞ sinq = − ⎜ ⎟ sin 30.0° = − ⎜ 2.00 m s ⎟ sin 30.0° = − 0.866, or q = − 60.0° v ⎠ ⎝ ⎝ Tf ⎠ Thus, the angle between the velocity vectors after collision is f = 30.0° − ( − 60.0° ) = 90.0°

6.67

(a)

Use conservation of the horizontal component of momentum from just before to just after the cannon is ﬁred.

( Σp ) = ( Σp ) x

f

x

i

45.0°

gives

mshell ( vshell cos 45.0° ) + mcannon vrecoil = 0 , or ⎛ m ⎞ ⎛ 200 kg ⎞ vrecoil = − ⎜ shell ⎟ vshell cos 45.0° = − ⎜ (125 m s ) cos 45.0° = − 3.54 m s ⎝ 5 000 kg ⎟⎠ ⎝ mcannon ⎠ (b)

Use conservation of mechanical energy for the cannon-spring system from right after the cannon is ﬁred to the instant when the cannon comes to rest.

( KE + PE 0+0+

x m ax =

g

+ PEs

) = ( KE + PE f

g

+ PEs

)

i

1 2 1 2 kx m ax = mcannon vrecoil +0+0 2 2 2 mcannon vrecoil = k

(5 000 kg) ( −3.54

m s) = 1.77 m 2.00 × 10 4 N m 2

continued on next page

56157_06_ch06_p169-208.indd 201

10/12/10 2:26:42 PM

202

Chapter 6

(c) (d)

6.68

Fm ax = k x m ax = ( 2.00 × 10 4 N m )(1.77 m ) = 3.54 × 10 4 N No. The rail exerts a vertical external force (the normal force) on the cannon and prevents it from recoiling vertically. Momentum is not conserved in the vertical direction. The spring does not have time to stretch during the cannon ﬁring. Thus, no external horizontal force is exerted on the system (cannon plus shell) from just before to just after ﬁring. Momentum is conserved in the horizontal direction during this interval.

Observe from Figure P6.68, the platform exerts a 0.60-kN to support the weight of the standing athlete prior to t = 0.00 s. From this, we determine the mass of the athlete: m=

w 0.60 kN 600 N = = = 61 kg g g 9.8 m s 2

For the interval t = 0.00 s to t = 1.0 s, we subtract the 0.60 kN force used to counterbalance the weight to get the net upward force exerted on the athlete by the platform during the jump. The result is shown in the force-versus-time graph at the right. The net impulse imparted to the athlete is given by the area under this graph. Note that this area can be broken into two triangular areas plus a rectangular area.

0.40 0.30 0.20 0.10

0.00 0.50

The net upward impulse is then I=

Fnet (kN)

t(s) 0.00

0.50

1.0

1 1 ( 0.50 s )(100 N ) + ( 0.50 s )( 300 N ) + ( 0.50 N )(100 N ) = 150 N ⋅s 2 2

The upward velocity vi of the athlete as he lifts off of the platform (at t = 1.0 s) is found from I = Δp = mvi − mv0 = mvi − 0 ⇒ vi =

I 150 N ⋅s = = 2.5 m s m 61 kg

The height of the jump can then be found from v 2f = vi2 + 2a y Δy (with v f = 0 ) to be Δy = 6.69

v 2f − vi2 2a y

=

0 − ( 2.5 m s )

2

2 ( −9.8 m s 2 )

= 0.32 m

Let particle 1 be the neutron and particle 2 be the carbon nucleus. Then, we are given that m2 =12 m1. (a)

From conservation of momentum, m2 v2 f + m1 v1 f = m1 v1i + 0. Since m2 =12 m1, this reduces to

12v2 f + v1 f = v1i

For a head-on elastic collision,

v1i − v2 i = − v1 f − v2 f

Since v2 i = 0, this becomes

v2 f = v1i + v1 f

(

(

[1]

) [2]

)

Substitute Equation [2] into [1] to obtain 12 v1i + v1 f + v1 f = v1i or

13v1 f = −11v1i

and

v1 f = −

11 v1i 13

continued on next page

56157_06_ch06_p169-208.indd 202

10/12/10 2:26:45 PM

Momentum and Collisions

203

Then, Equation [2] yields v2 f = ( 2 13) v1i . The initial kinetic energy of the neutron is KE1i = 12 m1 v1i2 , and the ﬁnal kinetic energy of the carbon nucleus is 1 1 ⎛ 4 2 ⎞ 48 ⎛ 1 ⎞ 48 m v2 = KE m2 v22 f = (12 m1 ) ⎜ v = ⎝ 169 1i ⎟⎠ 169 ⎜⎝ 2 1 1i ⎟⎠ 169 1i 2 2

KE2 f =

The fraction of kinetic energy transferred is (b)

KE2 f KE1i

=

48 = 0.28 169

If KE1i = 1.6 × 10 −13 J, then KE2 f =

(

)

48 48 KE1i = 1.6 × 10 −13 J = 4.5 × 10 −14 J 169 169

The remaining energy, (1 − 48 169 ) KE1i = (121 169 ) (1.6 × 10 −13 J ) = 1.1 × 10 −13 J , stays with the neutron. 6.70

(a)

→

nB

→

nA

B

→

FAB

A

M

2M

→ FBA →

Mg

→

2Mg

(b)

From Newton’s third law, the force F BA exerted by B on A is at each instant equal in magnitude and opposite in direction to the force F AB exerted by A on B.

(c)

There are no horizontal external forces acting on System C, which consists of both blocks. The forces F BA and F AB are internal forces exerted on one part of System C by another part of System C. ΣF external =

Thus,

Δp

=0 ⇒

C

Δt

ΔpC = 0

This gives

(p ) = (p ) = (p ) + (p ) C

C

f

i

i

A

B

i

or

( M + 2M ) V = M ( +v ) + 0

so the velocity of the combined blocks after collision is

V = + v 3.

The change in momentum of A is then

( ) − ( p ) = MV − Mv = M ⎛⎜⎝ v3 − v⎞⎟⎠ =

Δp = p A

A

f

A

i

− 2Mv 3

and the change in momentum for B is

( ) − ( p ) = 2MV − 0 = 2M ⎛⎜⎝ +v3 ⎞⎟⎠ =

Δp = p B

B

f

B

i

+ 2Mv 3 continued on next page

56157_06_ch06_p169-208.indd 203

10/12/10 2:26:48 PM

204

Chapter 6

2

(d)

1 1 1 ⎛ v⎞ ΔKE = ( KEC ) f − ⎡⎣( KE A ) + ( KEB ) ⎤⎦i = ( 3M ) ⎜ ⎟ − ⎡⎢ Mv 2 + 0 ⎤⎥ = − Mv 2 ⎝ ⎠ 2 3 3 ⎣2 ⎦ Thus, kinetic energy is not conserved in this inelastic collision .

6.71

(a)

The owner’s claim should be denied . Immediately prior to impact, the total momentum of the two-car system had a northward component and an eastward component. Thus, after impact, the wreckage moved in a northeasterly direction and could not possibly have damaged the owner’s property on the southeast corner.

(b)

Choose east as the positive x-direction and north as the positive y-direction. From conservation of momentum:

(p ) x

or

vx =

(p ) y

or

after

(m

⇒

1

m1 ( v1i )x + m2 ( v2 i )x m1 + m2

after

vy =

= ( px )before

( )

= py

before

⇒

m1 + m2

(1 300 kg ) ( 30.0

km h ) + 0 = 16.3 km h 1 300 kg + 1100 kg

=

(m

m1 ( v1i ) y + m2 ( v2 i ) y

+ m2 ) vx = m1 ( v1i )x + m2 ( v2 i )x

1

=

+ m2 ) v y = m1 ( v1i ) y + m2 ( v2 i ) y 0 + (1100 kg ) ( 20.0 km h ) = 9.17 km h 1 300 kg + 1100 kg

Thus, the velocity of the wreckage immediately after impact is v = vx2 + v y2 = 18.7 km h or 6.72

(a)

and

⎛ vy ⎞ q = tan −1 ⎜ ⎟ = tan −1 ( 0.563) = 29.4° ⎝ vx ⎠

v = 18.7 km h at 29.4° north of east, consistent with part (a) .

Ignore any change in velocity due to the force of gravity during the brief collision time, and let vbf denote the velocity of the ball just after impact while v pf is that of the player. We use the conservation of momentum to obtain

( 0.45 kg ) v or

bf

+ ( 60 kg ) v pf = ( 0.45 kg ) ( −25 m s ) + ( 60 kg ) ( 4.0 m s )

(

)

v pf = 3.8 m s − 7.5 × 10 −3 vbf

[1]

(

)

Also, for an elastic collision ⇒ vbf − v pf = − vbi − v pi = − ( −25 m s − 4.0 m s ) or

vbf = 29 m s + v pf

[2]

Substituting Equation [1] into [2] yields vbf = (b)

The average acceleration of the ball during the collision is aav =

56157_06_ch06_p169-208.indd 204

29 m s + 3.8 m s = 33 m s 1 + 7.5 × 10 −3

vbf − vbi Δt

=

33 m s − ( −25 m s ) = 2.9 × 10 3 m s 2 20 × 10 −3 s

10/12/10 2:26:51 PM

Momentum and Collisions

6.73

(a)

205

The speed v of both balls just before the basketball reaches the ground may be found from v y2 = v02 y + 2a y Δy as

(

)

v = v02 y + 2a y Δy = 0 + 2 ( −g ) ( −h ) = 2gh = 2 9.80 m s 2 (1.20 m ) = 4.85 m s (b)

Immediately after the basketball rebounds from the ﬂoor, it and the tennis ball meet in an elastic collision. The velocities of the two balls just before collision are for the tennis ball: vti = −v

for the basketball: vbi = + v

and

We determine the velocity of the tennis ball immediately after this elastic collision as follows: Momentum conservation gives mt vt f + mb vb f = mt vti + mb vbi

mt v t f + mb v b f = ( mb − mt ) v

or

[1]

From the criteria for a perfectly elastic collision:

(

vti − vbi = − vt f − vb f

)

vb f = vt f + vti − vb i = vt f − 2v

or

[2]

Substituting Equation [2] into [1] gives

(

)

mt vt f + mb vt f − 2v = ( mb − mt ) v or the upward speed of the tennis ball immediately after the collision is ⎛ 3m − mt ⎞ ⎛ 3m − mt ⎞ v=⎜ b 2gh vt f = ⎜ b ⎟ ⎝ mt + mb ⎠ ⎝ mt + mb ⎟⎠ The vertical displacement of the tennis ball during its rebound following the collision is given by v y2 = v02 y + 2a y Δy as Δy =

v y2 − v02 y 2a y

=

0 − vt2f

2 ( −g )

=

1 ⎛ 3mb − mt ⎞ 2g ⎜⎝ mt + mb ⎟⎠

2

⎛

⎞

−m ( 2g h ) = ⎜⎝ 3m m + m ⎟⎠ b

t

t

2

h

b

⎡ 3 ( 590 g ) − ( 57.0 g ) ⎤ Δy = ⎢ ⎥ (1.20 m ) = 8.41 m ⎣ 57.0 g + 590 g ⎦ 2

or 6.74

The woman starts from rest (v0y = 0) and drops freely with a y = −g for 2.00 m before the impact with the toboggan. Then, v2i2 = v02 y + 2a y ( Δy ) gives her speed just before impact as

(

)

v2i = v02 y + 2a y ( Δy ) = 0 + 2 −9.80 m s 2 ( −2.00 m ) = 6.26 m s The sketches at the right show the situation just before and just after the woman’s impact with the toboggan. Since no external forces impart any signiﬁcant impulse directed parallel to the incline (+x-direction) to the system consisting of man, woman, and toboggan during the very brief duration of the impact,

56157_06_ch06_p169-208.indd 205

m1 mman mtoboggan90.0 kg m2 mwoman55.0 kg v1i 8.00 m/s v2i 6.26 m/s

m2 v2i 30.0° m1

m

1

v1

i

m

2

vf

x 30.0°

Just Before Impact

x 30.0°

Just After Impact continued on next page

10/12/10 2:26:54 PM

206

Chapter 6

we will consider the total momentum parallel to the incline to be conserved. That is,

( m1 + m2 ) v f

= m1 v1i + m2 ( v2i )x = m1 v1i + m2 v2i sin 30.0°

or the speed of the system immediately after impact is vf = 6.75

m1 v1i + m2 v2i sin 30.0° ( 90.0 kg ) ( 8.00 m s ) + ( 55.0 kg ) ( 6.26 m s ) sin 30.0° = = 6.15 m s 90.0 kg + 55.0 kg m1 + m2

First consider the motion of the block and embedded bullet from immediately after impact until the block comes to rest after sliding distance d across the horizontal table. During this time, a kinetic friction force fk = m k n = m k ( M + m ) g, directed opposite to the motion, acts on the block. The net work done on the block and embedded bullet during this time is Wnet = ( fk cos180° ) d = KE f − KEi = 0 −

1 ( M + m) V 2 2

so the speed, V, of the block and embedded bullet immediately after impact is

V=

−2 fk d = − ( M + m)

(

)

2m k M + m gd M+m

= 2m k gd

Now, make use of conservation of momentum from just before to just after impact to obtain pxi = px f

⇒

mv0 = ( M + m )V = ( M + m ) 2m k gd

and the initial velocity of the bullet was ⎛ M + m⎞ 2m k gd v0 = ⎜ ⎝ m ⎟⎠ 6.76

(a)

Apply conservation of momentum in the vertical direction to the squid-water system from the instant before to the instant after the water is ejected. This gives ⎛m ⎞ ⎛ 0.30 kg ⎞ ms vs + mw vw = ( ms + mw ) ( 0 ) or vs = − ⎜ w ⎟ vw = − ⎜ ( −20 m s ) = 7.1 m s ⎝ 0.85 kg ⎟⎠ ⎝ ms ⎠

(b)

Apply conservation of mechanical energy to the squid from the instant after the water is ejected until the squid reaches maximum height to ﬁnd 0 + ms gy f =

56157_06_ch06_p169-208.indd 206

1 ms vs2 + mgyi 2

( 7.1 m s ) = 2.6 m vs2 = 2g 2 9.8 m s 2 2

or

Δy = y f − yi =

(

)

10/12/10 2:26:56 PM

207

Momentum and Collisions

6.77

(a)

y

→

m1

v1i

x

m2

→

v1f

m1

x

53.0° q

At rest m2 →

v2f

Just Before Collision

Just After Collision

The situations just before and just after the collision are shown above. Conserving momentum in both the x- and y-directions gives

(p ) = (p ) y

(p ) x

y

f

f

i

⇒ m1 v1 f sin 53° − m2 v2 f sinf = 0

or

m2 v2 f sinf = m1 v1 f sin 53° [1]

= ( px )i ⇒ m1 v1 f cos 53° + m2 v2 f cosf = m1 v1i + 0 m2 v2 f cosf = m1 v1i - m1 v1 f cos 53°

or

[2]

Dividing Equation [1] by [2] yields tanf =

v1 f sin 53° v1i − v1 f cos 53°

Equation [1] then gives

(b)

=

(1.0 m s ) sin 53° = 0.57 2.0 m s ) − (1.0 m s ) cos 53° (

v2 f =

m1 v1 f sin 53° m2 sinf

=

or

( 0.20 kg ) (1.0 m s ) sin 53° = ( 0.30 kg ) sin 30°

f = 30°

1.1 m s

The fraction of the incident kinetic energy lost in this collision is KE f ΔKE KEi − KE f = = 1− = 1− KEi KEi KEi

1 2

( 0.20 kg ) (1.0 m s ) + ( 0.30 kg ) (1.1 m s ) ( 0.20 kg ) ( 2.0 m s ) 2

1 2

1 2

2

2

ΔKE = 0.30 or 30% KEi 6.78

(a)

Conservation of mechanical energy of the bullet-block-Earth system from just after impact until maximum height is reached may be used to relate the speed of the block and bullet just after collision to the maximum height. Then, conservation of momentum from just before to just after impact can be used to relate the initial speed of the bullet to the speed of the block and bullet just after collision.

(b)

Conservation of energy from just after impact until the block and embedded bullet come to rest momentarily at height h gives KEi + PEg,i = KE f + PEg, f

or

1 ( M + m ) V 2 + 0 = 0 + ( M + m ) gh 2 continued on next page

56157_06_ch06_p169-208.indd 207

10/12/10 2:26:58 PM

208

Chapter 6

and the speed of the block+bullet just after collision is V = 2gh . Now, using conservation of momentum during the collision gives mvi + 0 = ( M + m ) V

or

⎛ M + m⎞ V vi = ⎜ ⎝ m ⎟⎠

The initial speed of the bullet is then ⎛ M + m⎞ 2gh vi = ⎜ ⎝ m ⎟⎠ 6.79

(a)

Conservation of mechanical energy of the bullet-block-Earth system from just after impact until maximum height is reached may be used to relate the speed of the block and bullet just after collision to the maximum height. Then, conservation of momentum from just before to just after impact can be used to relate the initial speed of the bullet to the speed of the block and bullet just after collision.

(b)

Conservation of energy from just after impact until the block and embedded bullet come to rest momentarily at height h = 22.0 cm gives KEi + PEg,i = KE f + PEg, f

or

1 ( M + m ) V 2 + 0 = 0 + ( M + m ) gh 2

and the speed of the block+bullet just after collision is V = 2gh . Now, using conservation of momentum during the collision gives mvi + 0 = ( M + m ) V

or

⎛ M + m⎞ ⎛ M + m⎞ V =⎜ 2gh vi = ⎜ ⎝ m ⎟⎠ ⎝ m ⎟⎠

The initial speed of the bullet is then ⎛ 1.25 kg + 5.00 × 10 −3 kg ⎞ 2 vi = ⎜ ⎟⎠ 2 9.80 m s ( 0.220 m ) = 521 m s 5.00 × 10 −3 kg ⎝

(

)

and the initial velocity of the bullet is v i = 521 m s upward .

56157_06_ch06_p169-208.indd 208

10/12/10 2:27:01 PM

7 Rotational Motion and the Law of Gravity QUICK QUIZZES 1.

Choice (c). For a rotation of more than 180°, the angular displacement must be larger than p = 3.14 rad. The angular displacements in the three choices are (a) 6 rad − 3 rad = 3 rad, (b) 1 rad − ( −1) rad = 2 rad, (c) 5 rad − 1 rad = 4 rad.

2.

Choice (b). Because all angular displacements occurred in the same time interval, the displacement with the lowest value will be associated with the lowest average angular speed.

3.

Choice (b). From a =

w 2 − w 02 w 2 − 0 w2 = = 2 Δq 2 Δq 2 Δq

it is seen that the case with the smallest angular displacement involves the highest angular acceleration. 4.

Choice (b). All points in a rotating rigid body have the same angular speed.

5.

Choice (a). Andrea and Chuck have the same angular speed w , but Andrea moves in a circle with twice the radius of the circle followed by Chuck. Thus, from vt = rw , it is seen that Andrea’s tangential speed is twice Chuck’s.

6.

1. Choice (e). Since the tangential speed is constant, the tangential acceleration is zero. 2. Choice (a). The centripetal acceleration, ac = vt2 r, is inversely proportional to the radius when the tangential speed is constant. 3. Choice (b). The angular speed, w = vt r , is inversely proportional to the radius when the tangential speed is constant.

7.

Choice (c). Both the velocity and acceleration are changing in direction, so neither of these vector quantities is constant.

8.

Choices (b) and (c). According to Newton’s law of universal gravitation, the force between the ball and the Earth depends on the product of their masses, so both forces, that of the ball on the Earth, and that of the Earth on the ball, are equal in magnitude. This follows also, of course, from Newton’s third law. The ball has large motion compared to the Earth because, according to Newton’s second law, the force gives a much greater acceleration to the small mass of the ball.

9.

Choice (e). From F = G Mm r 2, the gravitational force is inversely proportional to the square of the radius of the orbit.

10.

Choice (d). The semimajor axis of the asteroid’s orbit is 4 times the size of Earth’s orbit. Thus, Kepler’s third law (T 2 r 3 = constant) indicates that its orbital period is 8 times that of Earth. 209

56157_07_ch07_p209-242.indd 209

10/12/10 2:27:26 PM

210

Chapter 7

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

Earth moves 2p radians around the Sun in 1 year. The average angular speed is then w av =

2p rad ⎛ 1y ⎞ −7 ⎜ ⎟ = 1.99 × 10 rad s 1 y ⎝ 3.156 × 10 7 s ⎠

which is choice (e). 2.

At the top of the circular path, both the tension in the string and the gravitational force act downward, toward the center of the circle, and together supply the needed centripetal force. Thus, Fc = T + mg = mrw 2 or

(

)

2 T = m rw 2 − g = ( 0.400 kg ) ⎡( 0.500 m ) ( 8.00 rad s ) − 9.80 m s2 ⎤ = 8.88 N ⎣ ⎦

making (a) the correct choice for this question. 3.

The wheel has a radius of 0.500 m and made 320 revolutions. The distance traveled is ⎛ 2p rad ⎞ = 1.0 × 10 3 m = 1.0 km s = rq = ( 0.500 m ) 3.2 × 10 2 rev ⎜ ⎝ 1 rev ⎟⎠

(

)

so choice (c) is the correct answer. 4.

The angular displacement will be ⎛ w f + wi ⎞ ⎛ 12.00 rad s + 4.00 rad s ⎞ Δt = ⎜ Δq = w av ⋅ Δt = ⎜ ⎟⎠ ( 4.00 s ) = 32.0 rad ⎟ ⎝ 2 2 ⎠ ⎝ which matches choice (d).

5.

According to Newton’s law of universal gravitation, the gravitational force one body exerts on the other decreases as the distance separating the two bodies increases. When on Earth’s surface, the astronaut’s distance from the center of the Earth is Earth’s radius r0 = RE . If h is the altitude at which the station orbits above the surface, her distance from Earth’s center when on the station is r ′ = RE + h > r0. Thus, she experiences a smaller force while on the space station and (c) is the correct choice.

6.

Any object moving in a circular path undergoes a constant change in the direction of its velocity. This change in the direction of velocity is an acceleration, always directed toward the center of the path, called the centripetal acceleration, ac = v 2 r = rw 2. The tangential speed of the object is vt = rw , where w is the angular velocity. If w is not constant, the object will have both an angular acceleration, a av = Δw Δt, and a tangential acceleration, at = ra . The only untrue statement among the listed choices is (b). Even when w is constant, the object still has centripetal acceleration.

7.

The required centripetal force is Fc = mac = mv 2 r = mrw 2. When m and w are both constant, the centripetal force is directly proportional to the radius of the circular path. Thus, as the rider moves toward the center of the merry-go-round, the centripetal force decreases and the correct choice is (c).

8.

The mass of a spherical body of radius R and density r is M = rV = r 4p R 3 3 . The escape velocity from the surface of this body may then be written in either of the following equivalent forms:

(

vesc =

56157_07_ch07_p209-242.indd 210

2GM R

and

vesc =

2G ⎛ 4p rR3 ⎞ = R ⎜⎝ 3 ⎟⎠

)

8p r GR 2 3

10/12/10 2:27:28 PM

Rotational Motion and the Law of Gravity

211

We see that the escape velocity depends on the three properties (mass, density, and radius) of the planet. Also, the weight of an object on the surface of the planet is Fg = mg = GMm R 2, giving g = GM R 2 =

G ⎡ ⎛ 4p R3 ⎞ ⎤ 4 ⎢r ⎥ = p r GR R 2 ⎣ ⎜⎝ 3 ⎟⎠ ⎦ 3

The acceleration of gravity at the planet surface then depends on the same properties as does the escape velocity. Changing the value of g would necessarily change the escape velocity. Of the listed quantities, the only one that does not affect the escape velocity is choice (e), the mass of the object. 9.

The satellite experiences a gravitational force, always directed toward the center of its orbit, and supplying the centripetal force required to hold it in its orbit. This force gives the satellite a centripetal acceleration, even if it is moving with constant angular speed. At each point on the circular orbit, the gravitational force is directed along a radius line of the path, and is perpendicular to the motion of the satellite, so this force does no work on the satellite. Therefore, the only true statement among the listed choices is (d).

10.

The total gravitational potential energy of this set of 4 particles is the sum of the gravitational energies of each distinct pair of particles in the set of four. There are six distinct pairs in a set of four particles, which are: 1 & 2, 1 & 3, 1 & 4, 2 & 3, 2 & 4, and 3 & 4. Therefore, the correct answer to this question is (b).

11.

The weight of an object of mass m at the surface of a spherical body of mass M and radius R is Fg = mg = GMm R 2. Thus, the acceleration of gravity at the surface is g = GM R 2. For Earth, gE = GM E RE2 and for the planet, gp =

GM p R

2 p

=

G ( 2M E )

( 2RE )

2

=

1 ⎛ GM E ⎞ 1 = gE = 0.5gE 2 ⎜⎝ RE2 ⎟⎠ 2

meaning that choice (b) is the correct response. 12.

In a circular orbit, the gravity force is always directed along a radius line of the circle, and hence, perpendicular to the object’s velocity which is tangential to the circle. In an elliptical orbit, the gravity force is always directed toward the center of the Earth, located at one of the foci of the orbit. This means that it is perpendicular to the velocity, which is always tangential to the orbit, only at the two points where the object crosses the major axis of the ellipse. These are the points where the object is nearest to and farthest from Earth. Since the gravity force is a conservative force, the total energy (kinetic plus gravitational potential energy) of the object is constant as it moves around the orbit. This means that it has maximum kinetic energy (and hence, greatest speed) when its potential energy is lowest (i.e., when it is closest to Earth. The only true statements among the listed choices are (a) and (b).

13.

We assume that the elliptical orbit is so elongated that the Sun, at one of the foci, is almost at one end of the major axis. If the period, T, is expressed in years and the semimajor axis, a, in astronomical units (AU), Kepler’s third law states that T 2 = a 3. Thus, for Halley’s comet, with a period of T = 76 y, the 2 semimajor axis of its orbit is a = 3 ( 76 ) = 18 AU. The length of the major axis, and the approximate maximum distance from the Sun, is 2a = 36 AU, making the correct answer for this question choice (e).

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

56157_07_ch07_p209-242.indd 211

The statement is wrong for two reasons. First, gravity is the centripetal force that keeps the astronauts and their spacecraft in orbit around the Earth, so the astronauts aren’t beyond gravity’s

10/12/10 2:27:31 PM

212

Chapter 7

influence. Second, weight is the magnitude of the gravitational force, so astronauts in Earth orbit have weight, although it’s lower than on Earth’s surface. Because they are freely falling around the Earth along with their spacecraft environment, gravity doesn’t press them against the cabin’s walls or floor, giving rise to the feeling of weightlessness. The same feeling of weightlessness would occur in a freely falling elevator. 4.

To a good first approximation, your bathroom scale reading is unaffected because you, the Earth, and the scale are all in free fall in the Sun’s gravitational field, in orbit around the Sun. To a precise second approximation, you weigh slightly less at noon and at midnight than you do at sunrise or sunset. The Sun’s gravitational field is a little weaker at the center of the Earth than at the surface sub-solar point, and a little weaker still on the far side of the planet. When the Sun is high in your sky, its gravity pulls up on you a little more strongly than on the Earth as a whole. At midnight the Sun pulls down on you a little less strongly than it does on the Earth below you. So you can have another doughnut with lunch, and your bedsprings will still last a little longer.

6.

Consider one end of a string connected to a spring scale and the other end connected to an object, of true weight w. The tension T in the string will be measured by the scale and construed as the apparent weight. We have w − T = mac. This gives T = w − m ac. Thus, the apparent weight is less than the actual weight by the term mac. At the poles the centripetal acceleration is zero, and T = w. However, at the equator the term containing the centripetal acceleration is non-zero, and the apparent weight is less than the true weight.

8.

(a)

If the acceleration is constant in magnitude and perpendicular to the velocity, the object is moving in a circular path at constant speed.

(b)

If the acceleration is parallel to the velocity, the object moves in a straight line, and is either speeding up (v and a in same direction) or slowing down (v and a in opposite directions).

10.

Kepler’s second law says that equal areas are swept out in equal times by a line drawn from the Sun to the planet. For this to be so, the planet must move fastest when it is closest to the Sun. This occurs during the winter season in the northern hemisphere.

12.

(a)

Velocity is north at A, west at B, and south at C.

(b)

The acceleration is west at A, nonexistent at B, east at C, to be radially inward.

ANSWERS TO EVEN NUMBERED PROBLEMS

4.

(a) 0.209 rad s 2

6.

−226 rad s 2

8.

(a) The tire is slowing down, so the second drop was released with a smaller initial velocity.

51 revolutions

12.

(a) 5.75 rad s

56157_07_ch07_p209-242.indd 212

(b) Yes.

−0.322 rad s 2

10.

(c)

(c)

7.7 × 10 2 m

(a)

(b)

2.1 m

(b) 1.2 × 10 2 m

2.

(b) vt = 1.29 m s , at = 0.563 m s2

76° counterclockwise from the +x-axis

10/18/10 6:45:46 PM

Rotational Motion and the Law of Gravity

5.25 s

(b)

27.6 rad

14.

(a)

16.

4.9 × 10 −2 rad s

18.

No. The required maximum tension in the vine is 1.38 × 10 3 N.

20.

(a)

the static friction force; slipping starts when mrw 2 > fs,max = m s mg.

(b)

6.55 s

22.

(a) 4.81 m s

24.

1.5 × 10 2 rev s

26.

(a)

a normal force exerted by the “floor” of the cabin

(b)

ΣFc = n = m vt2 r

28.

(b)

213

700 N

(c)

294 N

(d)

7.00 m s

(e) 0.700 rad s

(f)

8.98 s

(g)

slower; 5.74 m s

(a) T = m2 g

(b) T = m1 vt2 R

(c)

vt = m2 gR m1

(d) T = 9.8 m, vt = 6.3 m s 30.

(a)

the gravitational force and a contact force exerted by the pail

(b)

the contact force (the gravitational force alone would produce projectile motion)

(c) 3.13 m s; No, the motion would be identical to projectile motion. 32.

(a)

25 kN

(b) 12 m s

34.

(a)

− 4.76 × 10 9 J

(b)

36.

(a) 1.83 × 10 9 kg m 3

38.

2.59 × 10 8 m from the center of the Earth

40.

The two masses are 2.00 kg and 3.00 kg.

42.

(a) 1.23 × 10 6 m

(b)

6.89 m s2

44.

(a) 5.59 × 10 3 m s

(b)

3.98 h

(c) 1.47 × 10 3 N

46.

8.91 × 10 7 m

48.

1.63 × 10 4 rad s

50.

(a) 56.5 rad s

(b)

22.4 rad s

(c)

(d) 1.77 × 10 5 rad

(e)

5.81 km

52.

56157_07_ch07_p209-242.indd 213

(a)

vmin =

568 N

(b) 3.26 × 10 6 m s2

⎛ tanq − m ⎞ ⎛ tanq + m ⎞ Rg⎜ ; vmax = R g ⎜ ⎟ ⎝ 1 + m tanq ⎠ ⎝ 1 − m tanq ⎟⎠

(c) −2.08 × 1013 J

−7.62 × 10 −3 rad s2

(b) 8.6 m s ; 17 m s

10/12/10 3:28:04 PM

214

Chapter 7

54.

(a)

56.

See Solution.

58.

(a)

60.

(a)

8.42 N

(b)

64.8°

n A = 15 N ; nB = 30 N

(b)

m k = 0.17

See Solution.

(b)

See Solution.

(c)

1.67 N

(c) The seat exerts the greatest force on the passenger at the bottom of the loop. (d)

ntop = 546 N ; nbottom = 826 N

62.

12.8 N

64.

2.06 × 10 3 rev min

66.

(a)

68.

(a) 1.62 × 10 3 N

0

(b)

1.3 kN

(b)

196 N

(c)

2.1 kN

(c) With the right speed, the needed centripetal force at the top of the loop can be made exactly equal to the gravitational force. At this speed, the normal force exerted on the pilot by the seat (his apparent weight) will be zero, and the pilot will have the sensation of weightlessness. (d) 177 m s 70.

72.

(a)

a downward gravitation force and a tension force that is directed toward the center of the circular path

(b)

See Solution.

(c)

6.05 N

(d) 7.82 m s

(a)

4.1 m s

(b)

80°

(c)

(d)

No. Flying slower increases the minimum radius of achievable turns.

74.

(a) 2.1 m s

76.

0.75 m

(b)

54°

(c)

1.7 m

4.7 m s

PROBLEM SOLUTIONS 7.1

(a)

Earth rotates 2p radians (360°) on its axis in 1 day. Thus, w=

(b) 7.2

Δq 2p rad ⎛ 1 day ⎞ −5 = ⎜ ⎟ = 7.27 × 10 rad s Δt 1 day ⎝ 8.64 × 10 4 s ⎠

Because of its rotation about its axis, Earth bulges at the equator.

The distance traveled is s = rq , where q is in radians. (a)

For 30°,

⎡ ⎛ p rad ⎞ ⎤ = 2.1 m s = rq = ( 4.1 m ) ⎢ 30° ⎜ ⎝ 180° ⎟⎠ ⎥⎦ ⎣

continued on next page

56157_07_ch07_p209-242.indd 214

10/12/10 2:27:38 PM

Rotational Motion and the Law of Gravity

7.3

(b)

For 30 radians,

s = rq = ( 4.1 m ) ( 30 rad ) = 1.2 × 10 2 m

(c)

For 30 revolutions,

⎡ ⎛ 2 p rad ⎞ ⎤ = 7.7 × 10 2 m s = rq = ( 4.1 m ) ⎢ 30 rev ⎜ ⎝ 1 rev ⎟⎠ ⎥⎦ ⎣

(a)

q=

(b)

The car travels a distance equal to the circumference of the tire for every revolution the tire makes if there is no slipping of the tire on the roadway. Thus, the number of revolutions made during the warranty period is

s 60 000 mi ⎛ 5 280 ft ⎞ 8 = ⎜ ⎟ = 3.2 × 10 rad r 1.0 ft ⎝ 1 mi ⎠

n=

7.4

7.5

215

S 60 000 miles ⎛ 5 280 ft ⎞ 7 = ⎜ ⎟ = 5.0 × 10 rev 2p r 2p (1.0 ft ) ⎝ 1 mile ⎠

Δw 1.00 rev s − 0 ⎛ rev ⎞ ⎛ 2p rad ⎞ = = ⎜ 3.33 × 10 −2 2 ⎟ ⎜ = 0.209 rad s2 Δt 30.0 s s ⎠ ⎝ 1 rev ⎟⎠ ⎝

(a)

a =

(b)

Yes. When an object starts from rest, its angular speed is related to the angular acceleration and time by the equation w = a ( Δt ). Thus, the angular speed is directly proportional to both the angular acceleration and the time interval. If the time interval is held constant, doubling the angular acceleration will double the angular speed attained during the interval.

(a)

a =

(b)

⎡ 2.51 × 10 4 rev min ( 2p rad 1 rev ) (1 min 60.0 s ) + 0 ⎤ ⎛ w f + w0 ⎞ ⎢ ⎥ ( 3.20 s ) q =w t = ⎜ t = ⎝ 2 2 ⎟⎠ ⎥⎦ ⎢⎣

( 2.51 × 10

)

rev min − 0 ⎛ 2 p rad ⎞ ⎛ 1 min ⎞ 2 ⎟ = 821 rad s ⎜⎝ ⎟⎜ 3.20 s 1 rev ⎠ ⎝ 60.0 s ⎠ 4

(

)

= 4.21 × 10 3 rad 7.6

w i = 3 600

rev ⎛ 2 p rad ⎞ ⎛ 1 min ⎞ ⎜ ⎟⎜ ⎟ = 377 rad s min ⎝ 1 rev ⎠ ⎝ 60.0 s ⎠

⎛ 2p rad ⎞ = 314 rad Δq = 50.0 rev ⎜ ⎝ 1 rev ⎟⎠ w 2 − w i2 0 − ( 377 rad s ) = = − 226 rad s 2 2 Δq 2 ( 314 rad ) 2

Thus, a = 7.7

(a)

From w 2 = w 02 + 2a ( Δq ), the angular displacement is w 2 − w 02 ( 2.2 rad s ) − ( 0.06 rad s ) = = 3.5 rad 2a 2 0.70 rad s 2 2

Δq = (b)

56157_07_ch07_p209-242.indd 215

(

2

)

From the equation given above for Δq , observe that when the angular acceleration is constant, the displacement is proportional to the difference in the squares of the ﬁnal and initial angular speeds. Thus, the angular displacement would increase by a factor of 4 if both of these speeds were doubled.

10/12/10 2:27:40 PM

216

7.8

Chapter 7

(a)

(b)

The maximum height h depends on the drop’s vertical speed at the instant it leaves the tire and becomes a projectile. The vertical speed at this instant is the same as the tangential speed, vt = rw , of points on the tire. Since the second drop rose to a lesser height, the tangential speed decreased during the intervening rotation of the tire. From v 2 = v02 + 2a y ( Δy ), with v0 = vt , a y = −g, and v = 0 when Δy = h, the relation between the tangential speed of the tire and the maximum height h is found to be 0 = vt2 + 2 ( −g ) h

vt = 2gh

or

Thus, the angular speed of the tire when the ﬁrst drop left was w 1 = and when the second drop left, the angular speed was w 2 =

( v t )2

( vt )1 = r 2gh2

2gh1 r

,

= . From r r w = w + 2a ( Δq ), with Δq = 2p rad, the angular acceleration is found to be 2 0

2

a =

w 22 − w 12 2gh2 r 2 − 2gh1 r 2 g = = 2 ( h2 − h2 ) r ( Δq ) 2 ( Δq ) 2 ( Δq )

( 9.80 m s ) 2

a =

or

7.9

Main Rotor:

( 0.381 m ) ( 2p 2

rad )

( 0.510 m − 0.540 m ) =

− 0.322 rad s 2

rev ⎞ ⎛ 2 p rad ⎞ ⎛ 1 min ⎞ ⎛ v = rw = ( 3.80 m ) ⎜ 450 ⎟⎜ ⎟⎜ ⎟ = 179 m s ⎝ min ⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠

m ⎞ ⎛ vsound ⎞ ⎛ v = ⎜ 179 = 0.522 vsound ⎟ ⎝ s ⎠ ⎜⎝ 343 m s ⎟⎠ rev ⎞ ⎛ 2p rad ⎞ ⎛ 1 min ⎞ ⎛ v = rw = ( 0.510 m ) ⎜ 4138 ⎟⎜ ⎟⎜ ⎟ = 221 m s ⎝ min ⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠

Tail Rotor:

⎞ m⎞ ⎛ v ⎛ v = ⎜ 221 ⎟ ⎜ sound ⎟ = 0.644 vsound ⎝ s ⎠ ⎝ 343 m s ⎠ 7.10

We will break the motion into two stages: (1) an acceleration period and (2) a deceleration period. The angular displacement during the acceleration period is ⎡ ( 5.0 rev s ) ( 2 p rad 1 rev) + 0 ⎤ ⎛ w f + wi ⎞ 2 t=⎢ q1 = w av t = ⎜ ⎥ ( 8.0 s ) = 1.3 × 10 rad ⎟ 2 ⎠ 2 ⎝ ⎣ ⎦ and while decelerating, ⎡ 0 + ( 5.0 rev s ) ( 2 p rad 1 rev) ⎤ ⎛ w f + wi ⎞ 2 t=⎢ q2 = ⎜ ⎥ (12 s ) = 1.9 × 10 rad ⎟ 2 ⎠ 2 ⎝ ⎣ ⎦ ⎛ 1 rev ⎞ = 51 rev q = q1 +q 2 = ⎡⎣(1.3 + 1.9 ) × 10 2 rad ⎤⎦ ⎜ ⎝ 2p rad ⎟⎠

The total displacement is 7.11

(a)

The linear distance the car travels in coming to rest is given by v 2f = v02 + 2a ( Δx ) as Δx =

v 2f − v02 2a

=

0 − ( 29.0 m s )

(

2 −1.75 m s2

2

)

= 240 m

continued on next page

56157_07_ch07_p209-242.indd 216

10/12/10 2:27:43 PM

Rotational Motion and the Law of Gravity

217

Since the car does not skid, the linear displacement of the car and the angular displacement of the tires are related by Δx = r ( Δq ). Thus, the angular displacement of the tires is Δq = (b)

Δx 240 m ⎛ 1 rev ⎞ 116 rev = = ( 727 rad ) ⎜ = ⎝ 2p rad ⎟⎠ r 0.330 m

When the car has traveled 120 m (one half of the total distance), the linear speed of the car is v = v02 + 2a ( Δx ) =

( 29.0

(

)

m s ) + 2 −1.75 m s2 (120 m ) = 20.5 m s 2

and the angular speed of the tires is w= 7.12

v 20.5 m s = = 62.1 rad s r 0.330 m

(

)

(a)

The angular speed is w = w 0 + a t = 0 + 2.50 rad s 2 ( 2.30 s ) = 5.75 rad s

(b)

Since the disk has a diameter of 45.0 cm, its radius is r = ( 0.450 m ) 2 = 0.225 m. Thus, vt = rw = ( 0.225 m ) ( 5.75 rad s ) = 1.29 m s

(

(c)

)

at = ra = ( 0.225 m ) 2.50 rad s 2 = 0.563 m s2

and

The angular displacement of the disk is Δq = q f − q 0 =

w 2f − w 02 2a

( 5.75 =

(

rad s ) − 0 2

2 2.50 rad s

2

)

⎛ 360° ⎞ = ( 6.61 rad ) ⎜ = 379° ⎝ 2p rad ⎟⎠

and the ﬁnal angular position of the radius line through point P is q f = q 0 + Δq = 57.3° + 379° = 436° or it is at 76° counterclockwise from the + x-axis after turning 19° beyond one full revolution. 7.13

From Δq = w av t = ⎡⎣(w f + w i ) 2 ⎤⎦ t, we ﬁnd the initial angular speed to be wi =

2 ( 37.0 rev) ( 2p rad 1 rev) 2 Δq −w f = − 98.0 rad s = 57.0 rad s t 3.00 s

The angular acceleration is then a = 7.14

(a)

w f − wi t

=

98.0 rad s − 57.0 rad s = 13.7 rad s 2 3.00 s

The initial angular speed is w 0 = 1.00 × 10 2

rev ⎛ 2p rad ⎞ ⎛ 1 min ⎞ = 10.5 rad s min ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60.0 s ⎟⎠

The time to stop (i.e., reach a speed of w = 0) with a = −2.00 rad s 2 is

continued on next page

56157_07_ch07_p209-242.indd 217

10/12/10 2:27:46 PM

218

Chapter 7

t=

(b) 7.15

(a)

⎛ w + w0 ⎞ ⎛ 0 + 10.5 rad s ⎞ t=⎜ Δq = w av t = ⎜ ⎟⎠ ( 5.25 s ) = 27.6 rad ⎝ ⎝ 2 ⎟⎠ 2 The car travels 235 m at constant speed in an elapsed time of 36.0 s. Its constant speed is therefore v=

(b)

w − w 0 0 − 10.5 rad s = = 5.25 s −2.00 rad s 2 a

Δs 235 m = = 6.53 m s Δt 36.0 s

The angular displacement of the car during the 36.0 s time interval is one-fourth of a full circle or p 2 radians. Thus, the radius of the circular path is r=

Δs 235 m 470 = = m Δq p 2 rad p

During the 36.0 s interval, the car has zero tangential acceleration, but does have a centripetal acceleration of constant magnitude p ( 6.53 m s ) v 2 ( 6.53 m s ) ac = = = = 0.285 m s2 r 470 m ( 470 p ) m 2

2

This acceleration is always directed toward the center of the circle . Therefore, when the car is at point B, the vector expression for the car’s acceleration is a c = 0.285 m s2 at 35.0° north of west 7.16

⎛ 1609 m ⎞ = 4.0 × 10 3 m. Thus, from ac = r w 2, the The radius of the cylinder is r = 2.5 mi ⎜ ⎝ 1 mi ⎟⎠ required angular velocity is w=

7.17

(a)

ac = r

9.80 m s2 = 4.9 × 10 −2 rad s 4.0 × 10 3 m

The tangential acceleration of the bug as the disk speeds up is ⎛ w f − w0 ⎞ at = ra = r ⎜ ⎝ Δt ⎟⎠ ⎛ 78.0 rev min − 0 ⎞ ⎛ 1 m ⎞ ⎛ 1 min = 5.00 in ⎜ ⎟⎠ ⎜⎝ 39.37 in ⎟⎠ ⎜⎝ 60 s 3.00 s ⎝

(

(b)

)

⎞ ⎛ 2p rad ⎞ = 0.346 m s 2 ⎟⎠ ⎜ ⎝ 1 rev ⎟⎠

The ﬁnal tangential speed of the bug is ⎛ rev ⎞ ⎛ 1 m ⎞ ⎛ 1 min vt = rw f = 5.0 0 in ⎜ 78.0 ⎝ min ⎟⎠ ⎜⎝ 39.37 in ⎟⎠ ⎜⎝ 60 s

(

)

⎞ ⎛ 2p rad ⎞ ⎟⎠ ⎜⎝ 1 rev ⎟⎠ = 1.04 m s

(c)

Since the bug has constant angular acceleration, and hence constant tangential acceleration (at = ra ), the tangential acceleration at t = 1.00 is at = 0.346 m s 2 as above.

(d)

At t = 1.00 s, the tangential velocity of the bug is

(

)

vt = v0 + at t = 0 + 0.346 m s 2 (1.00 s ) = 0.346 m s continued on next page

56157_07_ch07_p209-242.indd 218

10/12/10 2:27:49 PM

Rotational Motion and the Law of Gravity

219

and the radial or centripetal acceleration is vt2 ( 0.346 m s ) = = 0.943 m s2 r ( 5.00 in )(1 m 39.37 in ) 2

ac = (e)

The total acceleration is a = ac2 + at2 = 1.00 m s 2 , and the angle this acceleration makes with the direction of a c is ⎛a ⎞ ⎛ 0.346 ⎞ q = tan −1 ⎜ t ⎟ = tan −1 ⎜ = 20.1° ⎝ 0.943 ⎟⎠ ⎝ ac ⎠

7.18

In order for the archeologist to make it safely across the river, the vine must be capable of giving 2 r upward as he passes through the lowest point on the swing him a net acceleration of ac = vmax with a speed of vmax = 8.00 m s. Thus, with T being the tension in the vine, the net force acting on the archeologist at the lowest point is ΣFy = T − mg = mac , giving the required minimum tensile strength of the vine as 2 ⎛ ⎛ v2 ⎞ m ( 8.00 m s ) ⎞ 3 T = mg + mac = m ⎜ g + max ⎟ = ( 85.0 kg ) ⎜ 9.80 2 + ⎟ = 1.38 × 10 N ⎜⎝ s r ⎠ 10.0 m ⎟⎠ ⎝

Since he chose a vine with breaking strength of 1 000 N, he does not make it across. 7.19

(a)

The tension in the string must counteract the radial component of the object’s weight, and also supply the needed centripetal acceleration. ΣFc = T − mg cosq = mac =

(

T = m v 2 r + g cosq

or

)

T

q

mv 2 r

q mg

⎡ ( 8.00 m s )2 ⎤ = ( 0.500 kg ) ⎢ + 9.80 m s2 cos 20.0° ⎥ ⎢⎣ 2.00 m ⎥⎦

(

)

= 20.6 N (b)

The net tangential force acting on the object is Ft = mg sinq , so the tangential acceleration has magnitude at =

(

)

Ft = g sinq = 9.80 m s2 sin 20.0° = 3.35 m s2 m

and is directed downward, tangential to the circular path. The radial component of the acceleration is v 2 ( 8.00 m s ) = = 32.0 m s2 toward the center of the path r 2.00 m 2

ac = (c)

→

The total acceleration has magnitude atotal = at2 + ac2 = or

atotal = 32.2 m s2

( 3.35 m s ) + ( 32.0 m s ) 2 2

2 2

→

atotal

q

ac

→

at

continued on next page

56157_07_ch07_p209-242.indd 219

10/12/10 2:27:51 PM

220

Chapter 7

⎛a ⎞ ⎛ 3.35 ⎞ q = tan −1 ⎜ t ⎟ = tan −1 ⎜ = 5.98° ⎝ 32.0 ⎟⎠ ⎝ ac ⎠

at

a total = 32.2 m s 2 at 5.98° to the cord and pointing below

Thus,

the center of the circular path

7.20

(d)

No change in answers if the object is swinging toward the equilibrium point instead of away from it.

(e)

If the object is swinging toward the equilibrium position, it is gaining speed, whereas it is losing speed if it is swinging away from the equilibrium position. In both cases, when the cord is 20.0° from the vertical, the tangential, centripetal, and total accelerations have the magnitudes and directions calculated in parts (a) through (c).

(a)

The natural tendency of the coin is to move in a straight line (tangent to the circular path of radius 15.0 cm), and hence, go farther from the center of the turntable. To prevent this, the force of static friction must act toward the center of the turntable and supply the needed centripetal force. When the necessary centripetal force exceeds the maximum value of the static friction force, ( fs )max = m s n = m s mg, the coin begins to slip.

(b)

When the turntable has angular speed w , the required centripetal force is Fc = mrw 2. Thus, if the coin is not to slip, it is necessary that mrw 2 ≤ m s mg w≤

or

( 0.350 ) ( 9.80

ms g = r

m s2

0.150 m

) = 4.78 rad s

With a constant angular acceleration of a = 0.730 rad s2, the time required to reach the critical angular speed is t= 7.21

(a)

w − w 0 4.78 rad s − 0 = = 6.55 s 0.730 rad s 2 a

From ΣFr = mac, we have ⎛ v 2 ⎞ ( 55.0 kg ) ( 4.00 m s ) T = m⎜ t ⎟ = = 1.10 × 10 3 N = 1.10 kN 0.800 m ⎝ r ⎠ 2

(b)

The tension is larger than her weight by a factor of 1.10 × 10 3 N T = = 2.04 times mg ( 55.0 kg ) 9.80 m s2

(

7.22

(a)

)

If T is the tension in each of the two support chains, the net force acting on the child at the lowest point on the circular path is ⎛ v2 ⎞ ΣFy = 2T − mg = mac = m ⎜ ⎟ ⎝ r ⎠ so the speed at this point is ⎛ 2T ⎞ v = r⎜ − g⎟ = ⎝ m ⎠

⎛ 2 ( 350 N ) ⎞ − 9.80 m s 2 ⎟ = 4.81 m s ⎝ 40.0 kg ⎠

( 3.00 m ) ⎜

continued on next page

56157_07_ch07_p209-242.indd 220

10/12/10 2:27:54 PM

Rotational Motion and the Law of Gravity

(b)

221

The upward force the seat exerts on the child at this lowest point is Fseat = 2T = 2 ( 350 N ) = 700 N

7.23

Friction between the tires and the roadway is capable of giving the truck a maximum centripetal acceleration of v2 ( 32.0 m s ) = 6.83 m s2 = t, max = r 150 m 2

ac, max

If the radius of the curve changes to 75.0 m, the maximum safe speed will be vt, max = r ac, max = 7.24

Since Fc = m

)

m s2 = 22.6 m s

vt2 = m r w 2 , the needed angular velocity is r Fc = mr

w =

( 75.0 m ) ( 6.83

(

4.0 × 10 −11 N 3.0 × 10 −16 kg ( 0.150 m )

)

⎛ 1 rev ⎞ = 1.5 × 10 2 rev s = 9.4 × 10 2 rad s ⎜ ⎝ 2p rad ⎟⎠

(

7.25

)

(a)

ac = rw 2 = ( 2.00 m ) ( 3.00 rad s ) = 18.0 m s 2

(b)

Fc = mac = ( 50.0 kg ) 18.0 m s2 = 900 N

(c)

We know the centripetal acceleration is produced by the force of friction. Therefore, the needed static friction force is fs = 900 N. Also, the normal force is n = mg = 490 N. Thus, the minimum coefﬁcient of friction required is

2

(

ms =

( fs )max n

=

)

900 N = 1.84 490 N

Such a large coefﬁcient of friction is unrealistic, and she will not be able to stay on the merry-go-round. 7.26

(a)

The only force acting on the astronaut is the normal force exerted on him by the “ﬂoor” of the cabin.

(b)

ΣF = n = mac

(c)

If n = 12 mgE , then

␻

n = mvt2 r

or

10.0 m

n=

(d)

(

)

1 ( 60.0 kg ) 9.80 m s2 = 294 N 2

From the equation in Part (b), vt =

nr = m

( 294 N ) (10.0 m ) 60.0 kg

= 7.00 m s

continued on next page

56157_07_ch07_p209-242.indd 221

10/12/10 2:27:57 PM

222

Chapter 7

(e)

Since vt = rw , we have vt 7.00 m s = = 0.700 rad s 10.0 m r

w= (f)

The period of rotation is T=

(g)

2p 2p = = 8.98 s w 0.700 rad s

Upon standing, the astronaut’s head is moving slower than his feet because his head is closer to the axis of rotation. When standing, the radius of the circular path followed by the head is rhead = 10.0 m − 1.80 m = 8.20 m, and the tangential speed of the head is

( vt )head = rheadw = ( 8.20 m ) ( 0.700 7.27

(a)

rad s ) = 5.74 m s

Since the 1.0-kg mass is in equilibrium, the tension in the string is

(

)

T = mg = (1.0 kg ) 9.8 m s 2 = 9.8 N (b)

The tension in the string must produce the centripetal acceleration of the puck. Hence, Fc = T = 9.8 N .

(c)

From Fc = m puck vt2 R , we ﬁnd

(

R Fc = mpuck

vt = 7.28

(a)

)

(1.0 m ) ( 9.8 N ) 0.25 kg

Since the mass m2 hangs in equilibrium on the end of the string, ΣFy = T − m2 g = 0

(b)

= 6.3 m s

T = m2 g

or

The puck moves in a circular path of radius R and must have an acceleration directed toward the center equal to ac = vt2 R. The only force acting on the puck and directed toward the center is the tension in the string. Newton’s second law requires ΣFtoward = m1ac

(

T = m1 vt2 R

giving

center

(c)

Combining the results from (a) and (b) gives m1

7.29

)

vt2 = m2 g R

vt = m2 gR m1

or

(d)

Substitution of the numeric data from Problem 7.27 into the results for (a) and (c) shown above will yield the answers given for that problem.

(a)

The force of static friction acting toward the road’s center of curvature must supply the briefcase’s required centripetal acceleration. The condition that it be able to meet this need is that Fc = m vt2 r ≤ ( fs )max = m s mg, or m s ≥ vt2 rg. When the tangential speed becomes large enough that m s = vt2 rg , the briefcase will begin to slide.

(b)

As discussed above, the briefcase starts to slide when m s = vt2 rg. If this occurs at the speed, vt = 15.0 m s, the coefﬁcient of static friction must be ms =

56157_07_ch07_p209-242.indd 222

(15.0

m s)

( 62.0 m ) ( 9.80

2

m s2

)

= 0.370

10/12/10 2:27:59 PM

Rotational Motion and the Law of Gravity

7.30

223

(a)

The external forces acting on the water are the gravitational force and the contact force exerted on the water by the pail .

(b)

The contact force exerted by the pail is the most important in causing the water to move in a circle. If the gravitational force acted alone, the water would follow the parabolic path of a projectile.

(c)

When the pail is inverted at the top of the circular path, it cannot hold the water up to prevent it from falling out. If the water is not to spill, the pail must be moving fast enough that the required centripetal force is at least as large as the gravitational force. That is, we must have m

v2 ≥ mg r

or

v ≥ rg =

(1.00 m ) ( 9.80 m s2 ) = 3.13 m s

If the pail were to suddenly disappear when is it at the top of the circle and moving at 3.13 m s, the water would follow the parabolic arc of a projectile launched with initial velocity components of v0 x = 3.13 m s , v0 y = 0. 7.31

(a)

The centripetal acceleration is ⎡⎛ rev ⎞ ⎛ 2 p rad ⎞ ⎛ 1 min ac = rw = ( 9.00 m ) ⎢⎜ 4.00 ⎟⎠ ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60 s min ⎢⎣⎝ 2

(b)

2

⎞⎤ 2 ⎟⎠ ⎥ = 1.58 m s ⎥⎦

At the bottom of the circular path, we take upward as positive and apply Newton’s second law. This yields ΣFy = n − mg = m(+ac ), or n = m ( g + ac ) = ( 40.0 kg ) ⎡⎣( 9.80 + 1.58 ) m s2 ⎤⎦ = 455 N upward

(c)

At the top of the path, we again take upward as positive and apply Newton’s second law to ﬁnd ΣFy = n − mg = m(−ac ), or n = m ( g − ac ) = ( 40.0 kg ) ⎡⎣( 9.80 − 1.58 ) m s2 ⎤⎦ = 329 N upward

(d)

At a point halfway up, the seat exerts an upward vertical component equal to the child’s weight (392 N) and a component toward the center having magnitude Fc = mac = ( 40.0 kg ) 1.58 m s2 = 63.2 N. The total force exerted by the seat is

(

FR =

)

( 392 N )2 + ( 63.2 N )2

= 397 N directed inward and at

⎛ 392 N ⎞ = 80.8° above the horizontal q = tan −1 ⎜ ⎝ 63.2 N ⎟⎠ 7.32

(a)

At A, taking upward as positive, Newton’s second law gives ΣFy = n − mg = m(+ac ). Thus, ⎡ ⎛ ( 20.0 m s )2 ⎤⎥ = 25 kN v2 ⎞ n = mg + mac = m ⎜ g + t ⎟ = ( 500 kg ) ⎢ 9.80 m s2 + r ⎠ 10 m ⎝ ⎢⎣ ⎥⎦

(b)

At B, still taking upward as positive, Newton’s second law yields ΣFy = n − mg = m ( −ac ) , or mg = n + mac = n + mvt2 r. If the car is on the verge of leaving the track, then n = 0 and mg = mvt2 r, giving vt = rg =

56157_07_ch07_p209-242.indd 223

(15 m ) ( 9.80

)

m s2 = 12 m s

10/12/10 2:28:03 PM

224

7.33

Chapter 7

At the half-way point the spaceship is 1.92 × 10 8 m from both bodies. The force exerted on the ship by the Earth is directed toward the Earth and has magnitude FE = =

GmE ms r2 6.67 × 10 −11 N ⋅ m 2 kg 2 5.98 × 10 24 kg 3.00 × 10 4 kg

(

)(

(1.92 × 10 m ) 8

)(

2

) = 325 N

The force exerted on the ship by the Moon is directed toward the Moon and has a magnitude of GmM ms r2 6.67 × 10 −11 N ⋅ m 2 kg 2 7.36 × 10 22 kg 3.00 × 10 4 kg

FM =

(

=

)(

(1.92 × 10 m ) 8

)(

2

) = 4.00 N

The resultant force is ( 325 N − 4.00 N ) = 321 N directed toward Earth . 7.34

The radius of the satellite’s orbit is r = RE + h = 6.38 × 10 6 m + 2.00 × 10 6 m = 8.38 × 10 6 m (a)

PEg = −

GME m r

(

)

24 ⎛ N ⋅ m 2 ⎞ 5.98 × 10 kg (100 kg ) = − ⎜ 6.67 × 10 −11 = − 4.76 × 10 9 J kg 2 ⎟⎠ 8.38 × 10 6 m ⎝

(b)

7.35

(

)

2 5.98 × 10 24 kg (100 kg ) GME m ⎛ −11 N ⋅ m ⎞ = 6.67 × 10 = 568 N 2 ⎜⎝ kg 2 ⎟⎠ r2 8.38 × 10 6 m

F=

(

)

The forces exerted on the 2.0-kg by the other bodies are Fx and Fy as shown in the diagram at the right. The magnitudes of these forces are Fx =

(6.67 × 10

−11

N ⋅ m kg 2

2

( 4.0 m )

) ( 2.0 kg) ( 4.0 kg)

Fy =

(

→

2.0 kg

)

6.67 × 10 −11 N ⋅ m 2 kg 2 ( 2.0 kg ) ( 3.0 kg )

( 2.0 m )

2

F

→

2.0 m Fy

2

= 3.3 × 10 −11 N and

3.0 kg

␪

4.0 kg

→ Fx

4.0 m

= 1.0 × 10 −10 N

The resultant force exerted on the 2.0-kg is F = Fx2 + Fy2 = 1.1 × 10 −10 N ⎛ Fy ⎞ directed at q = tan −1 ⎜ ⎟ = tan −1 ( 3.0 ) = 72° above the +x-axis ⎝ Fx ⎠ 7.36

(a)

The density of the white dwarf would be r=

MSun 3MSun M MSun = = = V VEarth 4p RE3 3 4p RE3

continued on next page

56157_07_ch07_p209-242.indd 224

10/12/10 2:28:06 PM

Rotational Motion and the Law of Gravity

225

Using data from Table 7.3,

(

(b)

(

kg m 3

)(

)

(

)

The general expression for the gravitational potential energy of an object of mass m at distance r from the center of a spherical mass M is PE = − GMm r. Thus, the potential energy of a 1.00 kg mass on the surface of the white dwarf would be

=− (a)

9

3

6.67 × 10 −11 N ⋅ m 2 kg 2 1.991 × 10 30 kg GMSun = = 3.26 × 10 6 m s2 2 RE2 6.38 × 10 6 m

PE = −

7.37

(

4p 6.38 × 10

6

Fg = mg = GMm r 2 , so the acceleration of gravity on the surface of the white dwarf would be g=

(c)

) = 1.83 × 10 m)

3 1.991 × 10 30 kg

r=

GMSun (1.00 kg ) RE

(6.67 × 10

−11

)(

)

N ⋅ m 2 kg2 1.991 × 10 30 kg (1.00 kg ) 6.38 × 10 m 6

= −2.08 × 1013 J

At the midpoint between the two masses, the forces exerted by the 200-kg and 500-kg masses are oppositely directed, so from F = GMm r 2 and r1 = r2 = r, we have ΣF =

or

ΣF =

GM1m GM 2 m Gm − = 2 ( M1 − M 2 ) r r12 r22

(6.67 × 10

−11

N ⋅ m 2 kg2 ) ( 50.0 kg ) ( 500 kg − 200 kg )

( 0.200 m )2

= 2.50 × 10 −5 N toward the 500-kg (b)

At a point between the two masses and distance d from the 500-kg mass, the net force will be zero when G ( 50.0 kg ) ( 200 kg )

( 0.400 m − d )

2

=

G ( 50.0 kg ) ( 500 kg ) d2

or

d = 0.245 m

Note that the above equation yields a second solution d = 1.09 m. At that point, the two gravitational forces do have equal magnitudes, but are in the same direction and cannot add to zero. 7.38

The equilibrium position lies between the Earth and the Sun on the line connecting their centers. At this point, the gravitational forces exerted on the object by the Earth and Sun have equal magnitudes and opposite directions. Let this point be located distance r from the center of the Earth. Then, its distance from the Sun is 1.496 × 1011 m − r , and we may determine the value of r by requiring that

(

)

GmS m GmE m = r2 1.496 × 1011 m − r

(

)

2

where mE and mS are the masses of the Earth and Sun, respectively. This reduces to

(1.496 × 10

11

r or

56157_07_ch07_p209-242.indd 225

m−r

)=

mS = 577 mE

1.496 × 1011 m = 577r, which yields r = 2.59 × 10 8 m from center of the Earth .

10/12/10 2:28:08 PM

226

Chapter 7

7.39

(a)

If air resistance is ignored, the only force acting on the projectile during its ﬂight is the gravitational force. Since the gravitational force is a conservative force, the total energy of the projectile remains constant. At r = RE , the projectile has speed v = vesc 3 = 2GM E RE 3, and its total energy is E = KE + PEg = E=−

or

1 2 ⎛ GM E m ⎞ 1 ⎛ 1 2GM E ⎞ GM E m mv + ⎜ − = m ⋅ − 2 RE ⎟⎠ RE ⎟⎠ 2 ⎜⎝ 9 RE ⎝

8 GM E m ⋅ 9 RE

When the projectile reaches maximum height at r = rmax, and is momentarily at rest, the kinetic energy is zero and we have E = KE + PEg = 0 − rmax =

or (b)

(

)

9 9 RE = 6.38 × 10 6 m = 7.18 × 10 6 m 8 8

The altitude of the projectile when at r = rmax is h = rmax − RE =

7.40

GM E m 8 GM E m =− ⋅ 9 rmax RE

9 R 6.38 × 10 6 m RE − RE = E = = 7.98 × 10 5 m 8 8 8

We know that m1 + m2 = 5.00 kg, or m2 = 5.00 kg − m1. F=

⎛ Gm1 m2 N ⋅ m 2 ⎞ m1 ( 5.00 kg − m1 ) ⇒ 1.00 × 10 −8 N = ⎜ 6.67 × 10 −11 2 r kg 2 ⎟⎠ ⎝ ( 0.200 m )2

( 5.00 kg ) m1 − m12 =

(1.00 × 10

−8

)

N ( 0.200 m )

6.67 × 10 −11 N ⋅ m 2 kg 2

2

= 6.00 kg 2

Thus, m12 − ( 5.00 kg ) m1 + 6.00 kg 2 = 0, or ( m1 − 3.00 kg ) ( m1 − 2.00 kg ) = 0. This yields m1 = 3.00 kg, so m2 = 2.00 kg . The answer m1 = 2 .00 kg and m2 = 3.00 kg is physically equivalent. 7.41

(a)

The radius of the satellite’s orbit is r = h + RE = 2.80 × 10 6 m + 6.38 × 10 6 m = 9.18 × 10 6 m Then, modifying Equation 7.23 (Kepler’s third law) for orbital motion about the Earth rather than the Sun, we have

(

)

3

4p 2 9.18 × 10 6 m ⎛ 4p 2 ⎞ 3 T =⎜ r = ⎝ GM E ⎟⎠ 6.67 × 10 −11 N ⋅ m 2 kg 2 5.98 × 10 24 kg 2

yielding

(

T 2 = 7.66 × 10 7 s2

and

)(

)

⎛ 1h ⎞ T = 8.75 × 10 3 s ⎜ = 2.43 h ⎝ 3 600 s ⎟⎠

continued on next page

56157_07_ch07_p209-242.indd 226

10/12/10 2:28:11 PM

Rotational Motion and the Law of Gravity

(b)

The constant tangential speed of the satellite is

(

6 circumference of orbit 2p r 2p 9.18 × 10 m vt = = = period T 8.75 × 10 3 s

or (c)

The satellite’s only acceleration is centripetal acceleration, so

(

(a)

)

vt = 6.59 × 10 3 m s = 6.59 km s

6.59 × 10 3 km s vt2 a = ac = = r 9.18 × 10 6 m 7.42

227

)

2

= 4.73 m s2 toward center of Earth

The satellite’s period is T = 110 min ( 60.0 s 1.00 min ) = 6.60 × 10 3 s. Using the form of Kepler’s third law (Equation 7.23) suitable for objects orbiting the Earth, we have 1

⎛ 4p 2 ⎞ 3 T =⎜ r ⎝ GM E ⎟⎠ 2

⎛ T 2 GM E ⎞ 3 r=⎜ 2 ⎝ 4p ⎟⎠

or

Thus, the radius of the orbit must be

(

⎛ 6.60 × 10 3 s r=⎜ ⎜⎝

) (6.67 × 10 2

−11

N ⋅ m kg

4p

2

2

) (5.98 × 10

24

2

)

1

kg ⎞ 3 ⎟ = 7.61 × 10 6 m ⎟⎠

and the altitude of the satellite is h = r − RE = 7.61 × 10 6 m − 6.38 × 10 6 m = 1.23 × 10 6 m (b)

The gravitational force is ⎛ GM ⎞ Fg = ⎜ 2 E ⎟ m = mg ⎝ r ⎠

g=

so

GM E r2

Thus, at the altitude of the satellite, the acceleration due to gravity is g= 7.43

( 6.67 × 10

−11

)(

N ⋅ m 2 kg 2 5.98 × 10 24 kg

( 7.61 × 10 m ) 6

2

) = 6.89 m s

From Kepler’s third law (Equation 7.23), written in the form suitable for bodies orbiting Mars, we have T 2 = 4p 2 GM Mars r 3, so the mass of Mars, computed from the given data, must be

(

)

⎛ 4p 2 ⎞ 3 ⎛ 4p 2 ⎜ M Mars = ⎜ = r 2 ⎝ GT ⎟⎠ ⎜⎝ 6.67 × 10 −11 N ⋅ m 2 kg 2 2.8 × 10 4 s

(

7.44

2

(a)

)(

)

⎞ ⎟ 9.4 × 10 6 m 2 ⎟⎠

(

)

3

= 6.3 × 10 23 kg

The satellite moves in an orbit of radius r = 2 RE and the gravitational force supplies the 2 required centripetal acceleration. Hence, m vt2 2 RE = GmE m ( 2 RE ) , or

(

vt =

)

(

) )

2 5.98 × 10 24 kg ⎛ GmE −11 N ⋅ m ⎞ = ⎜ 6.67 × 10 = 5.59 × 10 3 m s 2 RE kg 2 ⎟⎠ 2 6.38 × 10 6 m ⎝

(

continued on next page

56157_07_ch07_p209-242.indd 227

10/12/10 2:28:14 PM

228

Chapter 7

(b)

The period of the satellite’s motion is

(

)

6 2p r 2p ⎡⎣ 2 6.38 × 10 m ⎤⎦ ⎛ 1 h T= = ⎜⎝ 3 600 vt 5.59 × 10 3 m s

(c)

⎞ = 3.98 h s ⎟⎠

The gravitational force acting on the satellite is F = GmE m r 2 or

(

)

24 ⎛ N ⋅ m 2 ⎞ 5.98 × 10 kg ( 600 kg ) F = ⎜ 6.67 × 10 −11 = 1.47 × 10 3 N kg2 ⎟⎠ ⎡ 2 6.38 × 10 6 m ⎤ 2 ⎝ ⎣ ⎦

7.45

(

)

The radii of the orbits of the two satellites are rA = hA + RE = RE + RE = 2RE

rB = hB + RE = 2RE + RE = 3RE

and

From Kepler’s third law, the ratio of the squares of the periods of the two satellites is 3

27 TB2 ⎛ 4p 2 rB3 ⎞ ⎛ GME ⎞ rB3 ⎛ 3RE ⎞ = = = ⋅ = 8 TA2 ⎜⎝ GME ⎟⎠ ⎜⎝ 4p 2 rA3 ⎟⎠ rA3 ⎜⎝ 2RE ⎟⎠ Thus, the ratio of their periods is TB2 = TA2

TB = TA 7.46

27 = 1.84 8

A synchronous satellite will have an orbital period equal to Jupiter’s rotation period, so the satellite can have the red spot in sight at all times. Thus, the desired orbital period is ⎛ 3 600 T = 9.84 h ⎜ ⎝ 1h

s⎞ 4 ⎟⎠ = 3.54 × 10 s

(

)

Kepler’s third law gives the period of a satellite in orbit around Jupiter as T 2 = 4p 2 GM Jupiter r 3. The required radius of the circular orbit is therefore ⎛ GM Jupiter T 2 ⎞ r=⎜ ⎟ 4p 2 ⎝ ⎠ or

13

(

)(

)(

)

13

⎡ 6.67 × 10 −11 N ⋅ m 2 kg2 1.90 × 10 27 kg 3.54 × 10 4 s 2 ⎤ ⎥ =⎢ 4p 2 ⎢ ⎥ ⎣ ⎦

r = 1.59 × 10 8 m

The altitude of the satellite above Jupiter’s surface should be h = r − RJupiter = 1.59 × 10 8 m − 6.99 × 10 7 m = 8.91 × 10 7 m 7.47

From Kepler’s third law, the mass of Jupiter can be expressed in terms of one of its satellite’s orbital radius and period as M Jupiter = 4p 2 GT 2 r 3.

(

(a)

)

For Io, r = 4.22 × 10 8 m giving

M Jupiter =

T = 1.77 days ⎡⎣(8.64 × 10 4 s) 1 day ⎤⎦ = 1.53 × 10 5 s,

and

(

4p 2 4.22 × 108 m

(

)(

)

3

6.67 × 10 −11 N ⋅ m 2 kg2 1.53 × 10 5 s

)

2

= 1.90 × 10 27 kg

continued on next page

56157_07_ch07_p209-242.indd 228

10/12/10 2:28:17 PM

Rotational Motion and the Law of Gravity

(b)

For Ganymede, r = 1.07 × 10 9 m giving

(c)

7.48

M Jupiter =

T = 7.16 days ⎡⎣(8.64 × 10 4 s) 1 day ⎤⎦ = 6.19 × 10 5 s,

and

(

4p 2 1.07 × 10 9 m

(6.67 × 10

−11

)(

)

3

N ⋅ m 2 kg2 6.19 × 10 5 s

)

2

= 1.89 × 10 27 kg

Yes. The results of parts (a) and (b) are consistent. They predict the same mass within the limits of uncertainty of the data used to compute these results.

The gravitational force on a small parcel of material at the star’s equator supplies the centripetal acceleration, or GMs m Rs2 = m vt2 Rs = m Rs w 2 . Hence, w = GM s Rs3 .

(

( 6.67 × 10

w=

7.49

229

−11

)

(

)

) ( )

)

N ⋅ m 2 kg 2 ⎡⎣ 2 1.99 × 10 30 kg ⎤⎦ = 1.63 × 10 4 rad s 3 10.0 × 10 3 m

(

(a)

w=

vt ( 98.0 mi h ) ⎛ 0.447 m = r 0.742 m ⎜⎝ 1 mi h

(b)

a =

w 2 − w i2 ( 9.40 rev s ) − 0 = = 44.2 rev s 2 2 Δq 2 (1 rev)

s ⎞ ⎛ 1 rev ⎞ ⎟⎠ ⎜⎝ 2p rad ⎟⎠ = 9.40 rev s

2

2

⎡ ⎛ 0.447 m s ⎞ ⎤ 98.0 mi h ( ) ⎢ ⎜⎝ 1 mi h ⎟⎠ ⎥ v2 ⎢ ⎦⎥ = 2.59 × 10 3 m s2 ar = t = ⎣ r 0.742 m ⎡ rev ⎛ 2p rad ⎞ ⎤ 2 at = ra = ( 0.742 m ) ⎢ 44.2 2 ⎜ ⎟ = 206 m s s ⎝ 1 rev ⎠ ⎥⎦ ⎣ (c)

In the radial direction at the release point, the hand supports the weight of the ball and also supplies the centripetal acceleration. Thus, Fr = mg + mar = m ( g + ar ) or

(

)

Fr = ( 0.198 kg ) 9.80 m s2 + 2.59 × 10 3 m s2 = 515 N In the tangential direction, the hand supplies only the tangential acceleration, so Ft = mat = ( 0.198 kg ) ( 206 m s 2 ) = 40.8 N 7.50

(a)

wi =

vt 1.30 m s = = 56.5 rad s 2.30 × 10 −2 m ri

(b)

wf =

vt 1.30 m s = = 22.4 rad s 5.80 × 10 −2 m rf

(c)

The duration of the recording is Δt = ( 74 min ) ( 60 s min ) + 33 s = 4 473 s Thus, a av =

w f − wi Δt

=

( 22.4 − 56.5 ) 4 473 s

rad s

= − 7.62 × 10 −3 rad s2 continued on next page

56157_07_ch07_p209-242.indd 229

10/12/10 2:28:19 PM

230

Chapter 7

w 2f − w i2

rad s ) − ( 56.5 rad s ) 2

2

Δq =

(e)

The track moves past the lens at a constant speed of vt = 1.30 m s for 4 473 seconds. Therefore, the length of the spiral track is

2a

=

( 22.4

(d)

(

2 −7.62 × 10 −3 rad s 2

)

= 1.77 × 10 5 rad

Δs = vt ( Δt ) = (1.30 m s ) ( 4 473 s ) = 5.81 × 10 3 m = 5.81 km 7.51

The angular velocity of the ball is w = 0.500 rev s = p rad s. (a)

vt = rw = ( 0.800 m ) (p rad s ) = 2.51 m s

(b)

ac =

(c)

We imagine that the weight of the ball is supported by a frictionless platform. Then, the rope tension need only produce the centripetal acceleration. The force required to produce the needed centripetal acceleration is F = m ( vt2 r ). Thus, if the maximum force the rope can exert is 100 N, the maximum tangential speed of the ball is

vt2 2 = rw 2 = ( 0.800 m ) (p rad s ) = 7.90 m s2 r

( vt )max = 7.52

(a)

rFmax = m

( 0.800 m ) (100 N ) 5.00 kg

= 4.00 m s

When the car is about to slip down the incline, the friction force, f, is directed up the incline as shown and has the magnitude f = m n. Thus,

n=

mg cosq + m sinq

y n

f ␪

ΣFy = n cosq + m nsinq − mg = 0 or

␪

R mg

[1]

x

⎛ v2 ⎞ Also, ΣFx = n sinq − m n cosq = m ⎜ min ⎟ ⎝ R ⎠ or

vmin =

nR (sinq − m cosq ) m

[2]

Substituting Equation [1] into [2] gives vmin =

⎛ sinq − m cosq ⎞ Rg ⎜ = ⎝ cosq + m sinq ⎟⎠

⎛ tanq − m ⎞ Rg ⎜ ⎝ 1 + m tanq ⎟⎠

If the car is about to slip up the incline, f = mn is directed down the slope (opposite to what is shown in the sketch). Then, ΣFy = n cosq − mn sinq − mg = 0, or n =

mg cosq − m sinq

[3]

⎛ v2 ⎞ Also, ΣFx = n sinq + mn cosq = m ⎜ max ⎟ , or ⎝ R ⎠ vmax =

nR (sinq + m cosq ) m

[4]

continued on next page

56157_07_ch07_p209-242.indd 230

10/12/10 2:28:22 PM

Rotational Motion and the Law of Gravity

231

Combining Equations [3] and [4] gives ⎛ sinq + m cosq ⎞ vmax = Rg ⎜ = ⎝ cosq − m sinq ⎟⎠ (b)

If R = 100 m, q = 10°, and m = 0.10, the lower and upper limits of safe speeds are

and 7.53

⎛ tanq + m ⎞ Rg ⎜ ⎝ 1 − m tanq ⎟⎠

vmin =

(100 m ) ( 9.8

⎛ tan10° − 0.10 ⎞ m s2 ⎜ = 8.6 m s ⎝ 1 + 0.10 tan10° ⎟⎠

vmax =

(100 m ) ( 9.8

⎛ tan10° + 0.10 ⎞ m s2 ⎜ = 17 m s ⎝ 1 − 0.10 tan10° ⎟⎠

)

)

The radius of the satellite’s orbit is

(

)

r = RE + h = 6.38 × 10 6 m + 1.50 × 10 2 mi (1 609 m 1 mi ) = 6.62 × 10 6 m (a)

The required centripetal acceleration is produced by the gravitational force, so

(

)

m vt2 r = GM E m r 2, which gives vt = GM E r .

(

vt = (b)

)

2 5.98 × 10 24 kg ⎛ −11 N ⋅ m ⎞ = 7.76 × 10 3 m s ⎜⎝ 6.67 × 10 kg 2 ⎟⎠ 6.62 × 10 6 m

The time for one complete revolution is

(

)

6 2 p r 2 p 6.62 × 10 m T= = = 5.36 × 10 3 s = 89.3 min 3 vt 7.76 × 10 m s

7.54

(a)

At the lowest point on the path, the net upward force (i.e., the force directed toward the center of the path and supplying the centripetal acceleration) is ΣFup = T − mg = m vt2 r , so the tension in the cable is

(

)

⎛ ⎛ ( 3.00 m s )2 ⎞ = 8.42 N v2 ⎞ T = m ⎜ g + t ⎟ = ( 0.400 kg ) ⎜ 9.80 m s2 + ⎟ ⎜⎝ r ⎠ 0.800 m ⎟⎠ ⎝ (b)

(

) = ( KE + PE ) , as the bob goes

Using conservation of mechanical energy, KE + PEg from the lowest to the highest point on the path, gives

f

g i

v2 1 0 + mg ⎡⎣ L (1 − cosq max ) ⎤⎦ = mvi2 + 0, or cosq max = 1 − i 2gL 2 q max (c)

2 ⎛ ⎞ ⎛ 3.00 m s ) vi2 ⎞ ( −1 = cos = cos ⎜ 1 − 1 − ⎜ ⎟ = 64.8° 2 g L ⎟⎠ ⎝ 2 9.80 m s 2 ( 0.800 m ) ⎠ ⎝ −1

(

)

At the highest point on the path, the bob is at rest and the net radial force is ⎛ v2 ⎞ ΣFr = T − mg cosq max = m ⎜ t ⎟ = 0 ⎝ r ⎠ Therefore,

(

)

T = mg cosq max = ( 0.400 kg ) 9.80 m s2 cos ( 64.8° ) = 1.67 N

56157_07_ch07_p209-242.indd 231

10/12/10 2:28:24 PM

232

Chapter 7

7.55

(a)

When the car is at the top of the arc, the normal force is upward and the weight downward. The net force directed downward, toward the center of the circular path and hence supplying the centripetal acceleration, is ΣFdown = mg − n = m vt2 r . Thus, the normal force is n = m g − vt2 r .

(

(b)

)

)

If r = 30.0 m and n → 0, then g − vt2 r → 0. For this to be true, the speed of the car must be vt = rg =

7.56

(

( 30.0 m ) ( 9.80

)

m s 2 = 17.1 m s

The escape speed from the surface of a planet of radius R and mass M is given by ve = 2GM R. If the planet has uniform density, r, the mass is given by

(

)

M = r ( volume ) = r 4p R 3 3 = 4p r R 3 3 The expression for the escape speed then becomes 2 G ⎛ 4 p r R3 ⎞ ⎛ ⎟⎠ = ⎜⎝ R ⎜⎝ 3

ve =

8pr G ⎞ R= 3 ⎟⎠

(constant ) R

or the escape speed is directly proportional to the radius of the planet. 7.57

The speed the person has due to the rotation of the Earth is vt = rw , where r is the distance from the rotation axis and w is the angular velocity of rotation. The person’s apparent weight, Fg, apparent , equals the magnitude of the upward normal force exerted on him by the scales. The true weight, Fg, true = mg, is directed downward. The net downward force produces the needed centripetal acceleration, or ⎛ v2 ⎞ ΣFdown = − n + Fg,true = −Fg,apparent + Fg,true = m ⎜ t ⎟ = mrw 2 ⎝ r ⎠ (a)

At the equator, r = RE , so Fg,true = Fg,apparent + mRE w 2 > Fg,apparent

(b)

At the equator, it is given that ac = RE w 2 = 0.0340 m s 2, so the apparent weight is Fg,apparent = Fg,true − mRE w 2 = ( 75.0 kg ) ⎡⎣( 9.80 − 0.0340 ) m s2 ⎤⎦ = 732 N At either pole, r = 0 (the person is on the rotation axis) and

(

)

Fg,apparent = Fg,true = mg = ( 75.0 kg ) 9.80 m s2 = 735 N 7.58

Choosing y = 0 and PEg = 0 at the level of point B, applying the work-energy theorem Wnc = (KE + PEg ) f − (KE + PEg )i to the block’s motion gives Wnc = (a)

1 2 1 mv + mgy − mv02 − mg ( 2R ), 2 2

or

v 2 = v02 +

2Wnc + 2g ( 2 R − y ) m

[1]

At point A, y = R and Wnc = 0 (no non-conservative force has done work on the block yet). Thus, v 2A = v02 + 2 gR. The normal force exerted on the block by the track must supply the centripetal acceleration at point A, so ⎛ v2 ⎞ ⎛ v02 ⎛ v 2 + 2 gR ⎞ ⎞ nA = m ⎜ A ⎟ = m ⎜ 0 + 2 g⎟ = m ⎜ ⎟ R ⎝ R⎠ ⎝ R ⎝ ⎠ ⎠

continued on next page

56157_07_ch07_p209-242.indd 232

10/12/10 2:28:26 PM

Rotational Motion and the Law of Gravity

233

⎛ ( 4.0 m s )2 ⎞ n A = ( 0.50 kg ) ⎜ + 2 9.8 m s2 ⎟ = 15 N ⎜⎝ 1.5 m ⎟⎠

(

or

)

At point B, y = 0 and Wnc is still zero. Thus, v B2 = v02 + 4 gR. Here, the normal force must supply the centripetal acceleration and support the weight of the block. Therefore, ⎛ v2 ⎞ ⎛ v 2 + 4 gR ⎞ ⎛ v02 ⎞ nB = m ⎜ B ⎟ + mg = m ⎜ 0 + 5g ⎟ + mg = m ⎟ ⎜ R ⎝ R⎠ ⎝ ⎠ ⎝ R ⎠ ⎛ ( 4.0 m s )2 ⎞ nB = ( 0.50 kg ) ⎜ + 5 9.8 m s2 ⎟ = 30 N ⎜⎝ 1.5 m ⎟⎠

(

or (b)

)

When the block reaches point C, y = 2R and Wnc = − fk L = − m k ( mg ) L. At this point, the normal force is to be zero, so the weight alone must supply the centripetal acceleration. Thus, m vC2 R = mg, or the required speed at point C is vC2 = Rg. Substituting this into Equation [1] yields Rg = v02 − 2 m k gL + 0, or

(

)

(

)

2 v 2 − Rg ( 4.0 m s ) − (1.5 m ) 9.8 m s mk = 0 = = 0.17 2 gL 2 9.8 m s 2 ( 0.40 m ) 2

(

7.59

)

Deﬁne the following symbols: M M = mass of the Moon, M E = mass of the Earth, RM = radius of the Moon, RE = radius of the Earth, and r = radius of the Moon’s orbit around the Earth. We interpret “lunar escape speed” to be the escape speed from the surface of a stationary Moon alone in the universe. Then, vlaunch = 2 vescape = 2

2 GM M 8 GM M v2 = RM , or launch RM

Applying conservation of mechanical energy from launch to impact gives 2 2 2 2 1 1 , or vimpact = vlaunch + ⎡ PEg − PEg 2 m vimpact + PE g f = 2 m vlaunch + PE g i i m ⎢⎣

(

)

(

(

)

) (

)

f

⎤ ⎥⎦

The needed potential energies are

( PE )

g i

=−

(

GMM m GM E m − and PEg RM r

)

f

=−

GM E m GM M m − RE r

2 Using these potential energies and the expression for vlaunch from above, the equation for the impact speed reduces to

vimpact =

⎛ 3 MM ME ( ME − MM ) ⎞ 2G⎜ + − ⎟ RE r ⎝ RM ⎠

N ⋅ m2 , M E = 5.98 × 10 24 kg, M M = 7.36 × 10 22 kg, kg 2 RM = 1.74 × 10 6 m, RE = 6.38 × 10 6 m, and r = 3.84 × 10 8 m, we ﬁnd

With numeric values of G = 6.67 × 10 −11

vimpact = 1.18 × 10 4 m s = 11.8 km s

56157_07_ch07_p209-242.indd 233

10/12/10 2:28:29 PM

234

Chapter 7

7.60

(a)

When the passenger is at the top, the radial forces producing the centripetal acceleration are the upward force of the seat and the downward force of gravity. The downward force must exceed the upward force to yield a net force toward the center of the circular path.

→

n

→ Fg

(b)

→

At the lowest point on the path, the radial forces contributing to the centripetal acceleration are again the upward force of the seat and the downward force of gravity. However, the upward force must now exceed the downward force to yield a net force directed toward the center of the circular path.

n

→ Fg

(c)

The seat must exert the greatest force on the passenger at the lowest point on the circular path.

(d)

At the top of the loop, ΣFr = Fg − n = m v 2 r, or n = Fg − m

⎛ ⎛ ( 4.00 m s )2 ⎞ = 546 N v2 v2 ⎞ = m ⎜ g − ⎟ = ( 70.0 kg ) ⎜ 9.80 m s2 − ⎟ ⎜⎝ r r ⎠ 8.00 m ⎟⎠ ⎝

At the bottom of the loop, ΣFr = n − Fg = mv 2 r, or n = Fg + m 7.61

⎛ ⎛ ( 4.00 m s )2 ⎞ = 826 N v2 v2 ⎞ = m ⎜ g + ⎟ = ( 70.0 kg ) ⎜ 9.80 m s2 + ⎟ ⎜⎝ r r ⎠ 8.00 m ⎟⎠ ⎝

The item of clothing will fall away from the rotating drum when the component of its weight directed toward the center of the drum exceeds the needed centripetal force. That is, when

(

q mg

)

mg sinq ≥ m vt2 r = mrw 2

r q

Thus, if the clothing is to lose contact with the drum at q = 68.0°, we must have mrw 2 = mg sin 68.0°. The required angular velocity, expressed in revolutions per second, is therefore w= 7.62

g sin 68.0° ⎛ 1 rev ⎞ rad s = ⎜ ⎝ 2p rad ⎟⎠ r

( 9.80 m s ) sin 68.0° = 0.835 rev s 2

0.330 m

Since the airplane ﬂies in a horizontal circle, its vertical acceleration is zero, and ΣFy = ma y or

F=

⇒

F

Fcosq − T sinq − mg = 0

mg + T tanq cosq

y q x

q

[1]

T

mg

continued on next page

56157_07_ch07_p209-242.indd 234

10/12/10 2:28:33 PM

Rotational Motion and the Law of Gravity

235

Also, the components of the lift force and the tension in the wire directed toward the center of the circular path must supply the required centripetal acceleration. Hence, ⎛ v2 ⎞ ΣFradial = mac = m ⎜ t ⎟ ⎝ r ⎠

F sinq + T cosq =

or

mvt2 r

[2]

Substituting Equation [1] into [2] yields

( mg tanq + T sinq tanq ) + T cosq =

r q

mvt2 r

60.0 m

or the tension in the wire is T=

(

)

m vt2 r − mg tanq

where

sinq tanq + cosq

r = cosq

If q = 20.0°, m = 0.750 kg, vt = 35.0 m s, and

= 60.0 m, this gives

(

)

⎡ ( 35.0 m s )2 ⎡( 60.0 m ) cos 20.0° ⎤ − 9.80 m s 2 tan 20.0° ⎤ ⎣ ⎦ ⎥ T = ( 0.750 kg ) ⎢ sin 20.0° tan 20.0° + cos 20.0° ⎢ ⎥ ⎣ ⎦ T = 12.8 N

or 7.63

Choosing PEg = 0 at the top of the hill, the speed of the skier after dropping distance h is found using conservation of mechanical energy as 1 2

m v − mgh = 0 + 0 , or v = 2gh 2 t

2 t

The net force directed toward the center of the circular path, and providing the centripetal acceleration, is

(

ΣFr = mgcosq − n = m vt2 R

vi 艑 0 →

h

Rh

n

␪

R

␪ →

mg

)

Solving for the normal force, after making the substitutions vt2 = 2gh and cosq = ( R − h ) R = 1 − ( h R ), gives n = mg (1 − h R ) − m ( 2gh R ) = mg (1 − 3h R ) The skier leaves the hill when n → 0. This occurs when 1− 7.64

3h = 0 or h = R 3 R

The centripetal acceleration of a particle at distance r from the axis is ac = vt2 r = rw 2 . If we are to have ac = 100g, then it is necessary that rw 2 = 100g

or

w = 100g r

The required rotation rate increases as r decreases. In order to maintain the required acceleration for all particles in the casting, we use the minimum value of r and ﬁnd w=

56157_07_ch07_p209-242.indd 235

(

)

100 9.80 m s2 100g rad ⎛ 1 rev ⎞ ⎛ 60.0 s ⎞ 3 rev = = 216 ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = 2.06 × 10 −2 rmin min 2.10 × 10 m s 2p rad 1 min

10/12/10 2:28:35 PM

236

7.65

Chapter 7

→

The sketch at the right shows the car as it passes the highest point on the bump. Taking upward as positive, we have

(

ΣFy = ma y ⇒ n − mg = m − v 2 r

(

n

)

→ → Fg m g

)

or

n = m g − v2 r

(a)

If v = 8.94 m s, the normal force exerted by the road is

r

2 ⎡ m ( 8.94 m s ) ⎤ ⎥ = 1.06 × 10 4 N = 10.6 kN n = (1 800 kg ) ⎢ 9.80 2 − s 20.4 m ⎥ ⎢⎣ ⎦

When the car is on the verge of losing contact with the road, n = 0. This gives g = v 2 r and the speed must be v = rg =

When the rope makes angle q with the vertical, the net force directed toward the center of the circular path is ΣFr = T − mgcosq as shown in the sketch. This force supplies the needed centripetal acceleration, so

r →

T

⎛v ⎞ ⎛ v ⎞ T − mg cosq = m ⎜ ⎟ , or T = m ⎜ g cosq + ⎟ r ⎠ ⎝ r ⎠ ⎝ 2 t

␪

2 t

1.5 m

7.66

( 20.4 m ) ( 9.80 m s2 ) = 14.1 m s

2.5 m

(b)

␪ →

→

w m g Using conservation of mechanical energy, (KE + PEg ) f = (KE + PEg )i , with KE = 0 at q = 90° and PEg = 0 at the bottom of the arc, the speed when the rope is at angle q from the vertical is given by 12 mvt2 + mg ( r − r cosq ) = 0 + mgr, or vt2 = 2gr cosq . The expression for the tension in the rope at angle q then reduces to T = 3mg cosq .

(a)

At the beginning of the motion, q = 90° and T = 0 .

(b)

At 1.5 m from the bottom of the arc, cosq = 2 .5 m r = 2 .5 m 4.0 m = 0.63 and the tension is

(

)

T = 3 ( 70 kg ) 9.8 m s2 ( 0.63) = 1.3 × 10 3 N = 1.3 kN (c)

At the bottom of the arc, q = 0° and cosq = 1.0, so the tension is

(

)

T = 3 ( 70 kg ) 9.8 m s2 (1.0 ) = 2.1 × 10 3 N = 2.1 kN 7.67

(a)

The desired path is an elliptical trajectory with the Sun at one of the foci, the departure planet at the perihelion, rD and the target planet at the aphelion. The perihelion distance rD is the radius of the departure planet’s orbit, while Departure the aphelion distance rT is the radius Planet of the target planet’s orbit. The semimajor axis of the desired trajectory is then a = ( rD + rT ) 2.

2a

Sun

rT Target Planet

continued on next page

56157_07_ch07_p209-242.indd 236

10/12/10 2:28:38 PM

Rotational Motion and the Law of Gravity

If Earth is the departure planet,

rD = 1.496 × 1011 m = 1.00 AU.

With Mars as the target planet,

1 AU ⎛ ⎞ = 1.52 AU rT = 2.28 × 1011 m ⎜ 11 ⎝ 1.496 × 10 m ⎟⎠

237

Thus, the semimajor axis of the minimum energy trajectory is a=

rD + rT 1.00 AU + 1.52 AU = = 1.26 AU 2 2

Kepler’s third law, T 2 = a 3, then gives the time for a full trip around this path as T = a3 =

(1.26 AU )3

= 1.41 y

so the time for a one-way trip from Earth to Mars is Δt = T 2 = 1.41 y 2 = 0.71 y .

7.68

(b)

This trip cannot be taken at just any time. The departure must be timed so that the spacecraft arrives at the aphelion when the target planet is located there.

(a)

Consider the sketch at the right. At the bottom of the loop, the net force toward the center (i.e., the centripetal force) is Fc =

→

v

→ Fg

→

n

mv 2 = n − Fg R

R

so the pilot’s apparent weight (normal force) is n = Fg +

(

)

Fg g v 2 ⎛ mv 2 v2 ⎞ = Fg + = Fg ⎜ 1 + R R gR ⎟⎠ ⎝

→

n

→

or

(

)

2 ⎛ ⎞ 2.00 × 10 2 m s ⎟ n = ( 712 N ) ⎜ 1 + ⎜⎝ (9.80 m s2 ) 3.20 × 103 m s2 ⎟⎠

(

v

)

→ Fg

= 1.62 × 1013 N (b)

At the top of the loop, the centripetal force is Fc = mv 2 R = n + Fg, so the apparent weight is n=

(

)

Fg g v 2 ⎛ v2 ⎞ mv 2 − Fg = − Fg = Fg ⎜ − 1⎟ R R ⎝ gR ⎠

(

)

2 ⎛ ⎞ 2.00 × 10 2 m s ⎟ = 196 N − 1 = ( 712 N ) ⎜ ⎜⎝ 9.80 m s2 3.20 × 10 3 m s2 ⎟⎠

(

)

(c)

With the right speed, the needed centripetal force at the top of the loop can be made exactly equal to the gravitational force. At this speed, the normal force exerted on the pilot by the seat (his apparent weight) will be zero, and the pilot will have the sensation of weightlessness.

(d)

When n = 0 at the top of the loop, Fc = mv 2 R = mg = Fg, and the speed will be v=

56157_07_ch07_p209-242.indd 237

)(

mg = Rg = m R

( 3.20 × 10 m ) ( 9.80 m s ) = 177 m s 3

2

10/12/10 2:28:43 PM

238

Chapter 7

7.69

(a)

At the instant the mud leaves the tire and becomes a projectile, its velocity components are v0 x = 0, v0 y = vt = Rw . From Δy = v0 y t + 12 a y t 2 with a y = −g, the time required for the mud to return to its starting point ( with Δy = 0 ) is given by 0 = t ( Rw − gt 2 ), for which the nonzero solution is t = 2Rw g .

(b)

The angular displacement of the wheel (turning at constant angular speed w ) in time t is Δq = w t. If the displacement is Δq = 1 rev = 2p rad at t = 2Rw g, then 2p rad = w ( 2Rw g ), or w 2 = p g R and w = p g R .

(a)

At each point on the vertical circular path, two forces are acting on the ball. These are: (1) the downward gravitational force with constant magnitude Fg = mg, and

7.70

(2) the tension force in the string, always directed toward the center of the path . (b)

(c)

The sketch at the right shows the forces acting on the ball when it is at the bottom of the circular path, and when it is at the highest point on the path. Note that the gravitational force has the same magnitude and direction at each point on the circular path. The tension force varies in magnitude at different points and is always directed toward the center of the path.

T

At the top of the circle, Fc = T + Fg = mv 2 r , or

(

T = mv 2 r − Fg = mv 2 r − mg = m v 2 r − g

T

) Fg

⎡ ( 5.20 m s ) ⎤ = ( 0.275 kg ) ⎢ − 9.80 m s 2 ⎥ = 6.05 N ⎢⎣ 0.850 m ⎥⎦ 2

(d)

Fg

At the bottom of the circle, Fc = T − Fg = T − mg = mv 2 r, and solving for the speed gives v2 =

r T (T − mg ) = r ⎛⎜⎝ − g⎞⎟⎠ m m

and

⎛T ⎞ v = r ⎜ − g⎟ ⎝m ⎠

If the string is at the breaking point at the bottom of the circle, then T = 22.5 N, and the speed of the object at this point must be v= 7.71

⎞ ⎛ 22.5 N − 9.80 m s2 ⎟ = 7.82 m s ⎠ ⎝ 0.275 kg

( 0.850 m ) ⎜

From Figure (a) at the right, observe that the angle the strings make with the vertical is ⎛ 1.50 m ⎞ = 41.4° q = cos ⎜ ⎝ 2.00 m ⎟⎠ −1

r=

( 2.00 m )2 − (1.50 m )2

q x

3.00 m

Also, the radius of the circular path is

y

T1

2.00 m

q

2.00 m

q T2

Fg mg

= 1.32 m (a)

(b)

Figure (b) gives a force diagram of the object with the +y-axis vertical and the +x-axis directed toward the center of the circular path. continued on next page

56157_07_ch07_p209-242.indd 238

10/12/10 2:28:45 PM

Rotational Motion and the Law of Gravity

(a)

239

Since the object has zero vertical acceleration, Newton’s second law gives ΣFy = T1 cosq − T2 cosq − mg = 0

T1 − T2 =

or

mg cosq

[1]

In the horizontal direction, the object has the centripetal acceleration ac = v 2 r directed in the +x-direction (toward the center of the circular path). Thus, ΣFx = T1 sinq + T2 sinq =

mv 2 r

T1 + T2 =

or

mv 2 r sinq

[2]

⎛ g v2 ⎞ Adding Equations [1] and [2] gives 2T1 = m ⎜ + , so the tension in the ⎝ cosq r sinq ⎟⎠ upper string is T1 (b)

)

2 ⎤ 6.00 m s 2 m s2 ⎥ = 109 N + ⎢ cos 41.4° (1.32 m ) sin 41.4° ⎥ ⎣ ⎦

2

To compute the tension T2 in the lower string, subtract Equation [1] above from Equation [2] to obtain 2T2 = m v 2 r sinq − g cosq . Thus,

(

T2 7.72

(

( 4.00 kg ) ⎡⎢ 9.80 =

)

( 6.00 m s )

( 4.00 kg ) ⎡⎢ =

2 2

⎢ (1.32 m ) sin 41.4° ⎣

2

−

⎤ 9.80 m s2 ⎥ = 56.4 N cos 41.4° ⎥ ⎦

The maximum lift force is ( FL )max = Cv 2, where C = 0.018 N ⋅ s2 m 2 and v is the ﬂying speed. For the bat to stay aloft, the vertical component of the lift force must equal the weight, or FL cosq = mg, where q is the banking angle. The horizontal component of this force supplies the centripetal acceleration needed to make a turn, or FL sinq = m v 2 r , where r is the radius of the turn. (a)

To stay aloft while ﬂying at minimum speed, the bat must have q = 0 (to give cosq = ( cosq )max = 1) and also use the maximum lift force possible at that speed. That is, we need

( FL )max ( cosq )max = mg,

2 Cvmin (1) = mg

or

Thus, we see that minimum ﬂying speed is vmin = (b)

mg = C

m s2

0.018 N ⋅ s m 2

2

)=

4.1 m s

To maintain horizontal ﬂight while banking at the maximum possible angle, we must have m s, this yields

( FL )max cosq max = mg, or Cv 2 cosq max = mg. For v = 10 cosq max =

(c)

( 0.031 kg ) ( 9.8

(

)

( 0.031 kg) 9.8 m s2 mg = 2 = 0.17 Cv 2 0.018 N ⋅ s2 m 2 (10 m s )

(

)

or

q max = 80°

The horizontal component of the lift force supplies the centripetal acceleration in a turn, FL sinq = mv 2 r. Thus, the minimum radius turn possible is given by rmin =

mv 2 m v2 m = = 2 C sinq F sinq ( ) ( L )max C v sinq max max max continued on next page

56157_07_ch07_p209-242.indd 239

10/12/10 2:28:49 PM

240

Chapter 7

where we have recognized that sinq has its maximum value at the largest allowable value of q . For a ﬂying speed of v = 10 m s, the maximum allowable bank angle is q max = 80° as found in part (b). The minimum radius turn possible at this ﬂying speed is then rmin = (d)

(

0.031 kg = 1.7 m 0.018 N ⋅ s2 m 2 sin 80.0°

)

No . Flying slower actually increases the minimum radius of the achievable turns. As found in part (c), rmin = m C sinq max. To see how this depends on the ﬂying speed, recall that the vertical component of the lift force must equal the weight, or FL cosq = mg. At the maximum allowable bank angle, cosq will be a minimum. This occurs when FL = ( FL )max = Cv 2. Thus, cosq max = mg Cv 2 and ⎛ mg ⎞ sinq max = 1 − cos2 q max = 1 − ⎜ 2 ⎟ ⎝ Cv ⎠

2

This gives the minimum radius turn possible at ﬂying speed v as rm in =

m ⎛ mg ⎞ C 1− ⎜ 2 ⎟ ⎝ Cv ⎠

2

Decreasing the ﬂying speed v will decrease the denominator of this expression, yielding a larger value for the minimum radius of achievable turns. 7.73

The angular speed of the luggage is w = 2p T, where T is the time for one complete rotation of the carousel. The resultant force acting on the luggage must be directed toward the center of the horizontal circular path (that is, in the +x-direction) and supply the needed centripetal force Fc = mac = mrw 2 = 4p 2 mr T 2.

y → fs

q

ΣFy = ma y ⇒ fs sinq + n cosq − mg = 0

or

n=

n

x q →

mg

Thus, ΣFx = max ⇒ fs cosq − n sinq = 4p 2 mr T 2 and

→

mg − fs sinq cosq

[1]

[2]

Substituting Equation [2] into Equation [1] gives ⎛ ⎛ sin 2 q ⎞ 4p 2 r ⎞ fs ⎜ cosq + = m ⎜ g tanq + 2 ⎟ ⎟ cosq ⎠ T ⎠ ⎝ ⎝ or, multiplying by cosq and using the identity cos 2 q + sin 2 q = 1, we have ⎛ 4p 2 r cosq ⎞ fs = m ⎜ g sinq + ⎟⎠ T2 ⎝ (a)

[3]

With r = 7.46 m and T = 38.0 s, Equation [3] gives the required friction force as ⎡ 4p 2 ( 7.46 m ) cos 20.0° ⎤ fs = ( 30.0 kg ) ⎢ 9.80 m s2 sin 20.0° + ⎥ = 106 N ( 38.0 s )2 ⎢⎣ ⎥⎦ continued on next page

(

56157_07_ch07_p209-242.indd 240

)

10/12/10 2:28:53 PM

Rotational Motion and the Law of Gravity

(b)

If the bag is on the verge of slipping when r = 7.94 m and T = 34.0 s, then fs = ( fs )max = m s n and (using Equations [2] and [3]) the coefﬁcient of static friction is ms =

ms =

or

fs fs cosq cosq cosq = = = n mg − fs sinq ( mg fs ) − sinq ⎡ g g sinq + 4p 2 r cosq T 2 ⎤ − sinq ⎣ ⎦

(

giving

⎡ ⎤ 9.80 m s 2 ⎢ ⎥ − sin 20.0° 2 2 2 ⎢⎣ 9.80 m s sin 20.0° + 4p ( 7.94 m ) cos 20.0° ( 34.0 s ) ⎥⎦

)

m s = 0.396

The horizontal component of the tension in the cord is the only force directed toward the center of the circular path, so it must supply the centripetal acceleration. Thus, ⎛ vt2 ⎞ ⎛ v2 ⎞ T sinq = m ⎜ t ⎟ = m ⎜ ⎝ L sinq ⎟⎠ ⎝ r ⎠ or

)

cos 20.0°

(

7.74

241

T sin 2 q =

L

→

q

T

→

q

r

mg

mvt2 L

[1]

Also, the vertical component of the tension must support the weight of the ball, or T cosq = mg (a)

[2]

Dividing Equation [1] by [2] gives sin 2 q vt2 = cosq Lg

Lg cosq

vt = sinq

or

[3]

With L = 1.5 m s and q = 30°, vt = sin 30° (b)

(1.5 m ) ( 9.8

m s2

cos 30°

)=

2.1 m s

From Equation [3], with sin 2 q = 1 − cos 2 q , we ﬁnd 1 − cos2 q vt2 = cosq Lg

or

⎛ v2 ⎞ cos 2 q + ⎜ t ⎟ cosq − 1 = 0 ⎝ Lg ⎠

Solving this quadratic equation for cosq gives 2

⎛ v2 ⎞ ⎛ v2 ⎞ cosq = − ⎜ t ⎟ ± ⎜ t ⎟ + 1 ⎝ 2 Lg ⎠ ⎝ 2 Lg ⎠ If L = 1.5 m and vt = 4.0 m s, this yields solutions cosq = −1.7 (which is impossible), and cosq = + 0.59 (which is possible). Thus, q = cos−1 ( 0.59 ) = 54° . (c)

From Equation [2], when T = 9.8 N and the cord is about to break, the angle is

(

)

⎛ ( 0.50 kg ) 9.8 m s2 ⎞ ⎛ mg ⎞ −1 = cos q = cos −1 ⎜ ⎜ ⎟ = 60° ⎝ T ⎟⎠ ⎜⎝ ⎟⎠ 9.8 N continued on next page

56157_07_ch07_p209-242.indd 241

10/12/10 2:28:57 PM

242

Chapter 7

Then Equation [3] gives Lg = sin 60° cosq

vt = sinq 7.75

(1.5 m ) ( 9.8

m s2

cos 60°

)=

4.7 m s

The normal force exerted on the person by the cylindrical wall must provide the centripetal acceleration, so n = m r w 2 .

(

)

If the minimum acceptable coefﬁcient of friction is present, the person is on the verge of slipping and the maximum static friction force equals the person’s weight, or ( fs )max = ( m s )min n = mg. Thus, ( m s )min = 7.76

mg g 9.80 m s 2 = = = 0.131 n r w 2 ( 3.00 m ) ( 5.00 rad s )2

If the block will just make it through the top of the loop, the force required to produce the centripetal acceleration at point C must equal the block’s weight, or m vC2 R = mg.

(

)

This gives vC = Rg as the required speed of the block at point C. We apply the work-energy theorem in the form

(

Wnc = KE + PEg + PEs

) − ( KE + PE f

g

+ PEs

)

i

from when the block is ﬁrst released until it reaches point C to obtain

( )

fk AB cos180° =

1 2 1 mvC + mg ( 2R ) + 0 − 0 − 0 − kd 2 2 2

The friction force between points A and B is fk = uk ( mg ), and for minimum initial compression of the spring, vC2 = Rg as found above. Thus, the work-energy equation reduces to d min =

d min =

56157_07_ch07_p209-242.indd 242

( )

2 m k mg AB + mRg + 2 mg ( 2R ) k

( 0.50 kg ) ( 9.8

=

(

mg 2 m k AB + 5R

)

k

)

m s2 ⎡⎣ 2 ( 0.30 ) ( 2.5 m ) + 5 (1.5 m ) ⎤⎦ = 0.75 m 78.4 N m

10/12/10 2:29:01 PM

8 Rotational Equilibrium and Rotational Dynamics QUICK QUIZZES 1.

Choice (d). A larger torque is needed to turn the screw. Increasing the radius of the screwdriver handle provides a greater lever arm and hence an increased torque.

2.

Choice (b). Since the object has a constant net torque acting on it, it will experience a constant angular acceleration. Thus, the angular velocity will change at a constant rate.

3.

Choice (b). The hollow cylinder has the larger moment of inertia, so it will be given the smaller angular acceleration and take longer to stop.

4.

Choice (a). The hollow sphere has the larger moment of inertia, so it will have the higher rotational kinetic energy, KEr = 12 Iw 2 .

5.

Choice (c). Apply conservation of angular momentum to the system (the two disks) before and after the second disk is added to get the result: I1w 1 = ( I1 + I 2 ) w .

6.

Choice (a). Earth already bulges slightly at the Equator, and is slightly flat at the poles. If more mass moved towards the Equator, it would essentially move the mass to a greater distance from the axis of rotation, and increase the moment of inertia. Because conservation of angular momentum requires that I zw z = constant, an increase in the moment of inertia would decrease the angular velocity, and slow down the spinning of Earth. Thus, the length of each day would increase.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

Assuming a uniform, solid disk, its moment of inertia about a perpendicular axis through its center is I = MR 2 2, so t = Ia gives a =

2t 2 ( 40.0 N ⋅ m ) = = 5.00 rad s 2 MR 2 ( 25.0 kg ) ( 0.800 m )2

and the correct answer is (b). 2.

Using the left end of the plank as a pivot and requiring that Σt = 0 gives −mg ( 2.00 m ) + F2 ( 3.00 m ) = 0, or F2 =

(

F1

F2 3.00 m

)

2 2mg 2 ( 20.0 kg ) 9.80 m s = = 131 N 3 3

2.00 m

mg

and choice (d) is the correct response. 3.

t = rF sinq = ( 0.500 m )(80.0 N ) sin ( 60.0° ) = 34.6 N ⋅ m which is choice (a). 243

56157_08_ch08_p243-287.indd 243

10/12/10 2:29:53 PM

244

Chapter 8

4.

The reel will be given a greater angular acceleration when a constant tension of 50 N is maintained in the cord as in choice (a). When an object weighing 50 N is attached to the end of the cord and allowed to fall, the tension in the cord will be less than 50 N, with the difference between the 50 N weight of the object and the tension being the net downward force accelerating the object.

5.

In order for an object to be in equilibrium, it must be in both translational equilibrium and rotational equilibrium. Thus, it must meet two conditions of equilibrium, namely F net = 0 and s net = 0. The correct answer is therefore choice (d).

6.

When objects travel down ramps of the same length, the one with the greatest translational kinetic energy at the bottom will have the greatest final translational speed (and, hence, greatest average translational speed). This means that it will require less time to travel the length of the ramp. Of the objects listed, all will have the same total kinetic energy at the bottom, since they have the same decrease in gravitational potential energy (due to the ramps having the same vertical drop) and no energy has been spent overcoming friction. All of the block’s kinetic energy is in the form of translational kinetic energy. Of the rolling bodies, the fraction of their total kinetic energy that is in the translational form is f =

KEt = KEt + KEr

1 1 Mv 2 = 2 = 2 1 1 1 + I MR 2 1 + ( I M ) (w v ) 2 Mv + 2 Iw 1 2

(

2

)

Since the ratio I MR 2 equals 2 5 for a solid ball and 2 3 for a hollow sphere, the solid ball has the larger translational kinetic energy at the bottom and will arrive before the hollow sphere. The correct rankings of arrival times, from shortest to longest, is then block, solid ball, hollow sphere, and choice (e) is the correct response. 7.

The moment of inertia of a body is determined by its mass and the way that mass is distributed about the rotation axis. Also, the location of the body’s center of mass is determined by how its mass is distributed. As long as these qualities do not change, both the moment of inertia and the center of mass are constant. From t = Ia , we see that when a body experiences a constant, nonzero torque, it will have a constant, non-zero angular acceleration. However, since the angular acceleration is non-zero, the angular velocity w (and hence the angular momentum, L = Iw ) will vary in time. The correct responses to this question are then (b) and (e).

8.

In a rigid, rotating body, all points in that body rotate about the axis at the same rate (or have the same angular velocity). The centripetal acceleration, tangential acceleration, linear velocity, and total acceleration of a point in the body all vary with the distance that point is from the axis of rotation. Thus, the only correct choice is (b).

9.

Please read the answer to Question 6 above, since most of what is said there also applies to this question. The total kinetic energy of either the disk or the hoop at the bottom of the ramp will be KEtotal = Mgh, where M is the mass of the body in question and h is the vertical drop of the ramp. The translational kinetic energy of this body will then be KEt = f KEtotal = f Mgh, where f is the fraction discussed in Question 6. Hence, M v 2 2 = f M gh and the translational speed at the bottom is v = 2 f gh . Since f = 1 (1 + 1 2 ) = 2 3 for the disk and f = 1 (1 + 1) = 1 2 for the hoop, we see that the disk will have the greater translational speed at the bottom, and hence, will arrive first. Notice that both the mass and radius of the object have canceled in the calculation. Our conclusion is then independent of the object’s mass and/or radius. Therefore, the only correct response is choice (d).

10.

The ratio of rotational kinetic energy to the total kinetic energy for an object that rolls without slipping is KEr KEr = = KEtotal KEt + KEr

56157_08_ch08_p243-287.indd 244

1 1 Iw 2 = = 2 2 1 1 MR 2 M⎛ v⎞ 2 Mv + 2 Iw +1 ⎜⎝ ⎟⎠ + 1 I I w 1 2 2

10/12/10 2:29:56 PM

Rotational Equilibrium and Rotational Dynamics

245

For a solid cylinder, I = MR 2 2 and this ratio becomes 1 1 KEr = = KEtotal 2 + 1 3 and the correct answer is (c). 11.

Since the axle of the turntable is frictionless, no external agent exerts a torque about this vertical axis of the mouse-turntable system. This means that the total angular momentum of the mouse-turntable system will remain constant at its initial value of zero. Thus, as the mouse starts walking around the axis (and developing an angular momentum, Lmouse = I mw m , in the direction of its angular velocity), the turntable must start to turn in the opposite direction so it will possess an angular momentum, Ltable = I tw t , such that Ltotal = Lmouse + Ltable = I mw m + I tw t = 0. Thus, the angular velocity of the table will be w t = − ( I m I t ) w m . The negative sign means that if the mouse is walking around the axis in a clockwise direction, the turntable will be rotating in the opposite direction, or counterclockwise. The correct choice for this question is (d).

12.

Please review the answers given above for Questions 6 and 9. In the answer to Question 9, it is shown that the translational speed at the bottom of the hill of an object that rolls without slipping is v = 2 f gh , where h is the vertical drop of the hill and f is the ratio of the translational kinetic energy to the total kinetic energy of the rolling body. For a solid sphere, I = 2MR 2 5, so the ratio f is f =

1 1 1 = = 1 + 2 5 1.4 1 + I MR 2

(

)

and the translational speed at the bottom of the hill is v = 2gh 1.4 . Notice that this result is the same for all uniform, solid, spheres. Thus, the two spheres have the same translational speed at the bottom of the hill. This also means that they have the same average speed for the trip, and hence, both make the trip in the same time. The correct answer to this question is (d). 13.

If a car is to reach the bottom of the hill in the shortest time, it must have the greatest translational speed at the bottom (and hence, greatest average speed for the trip). To maximize its final translational speed, the car should be designed so as much as possible of the car’s total kinetic energy is in the form of translational kinetic energy. This means that the rotating parts of the car (i.e., the wheels) should have as little kinetic energy as possible. Therefore, the mass of these parts should be kept small, and the mass they do have should be concentrated near the axle in order to keep the moment of inertia as small as possible. The correct response to this question is (e).

14.

(i)

As the ponies walk inward toward the center of the turntable, the moment of inertia of the rotating system decreases. Thus, the angular speed of the system must increase to conserve angular momentum. Choice (a) is the correct answer.

(ii)

No. The mechanical energy of the system increases. The ponies must do work, converting chemical energy into additional mechanical energy, as they walk toward the center of the turntable.

(iii) Yes. The center of gravity of this system is located at the center of the turntable and is stationary. Thus, the linear momentum of the system has a constant value of zero. (iv) Yes. Since no external torque acts on the ponies-turntable system, the angular momentum of this system is constant.

56157_08_ch08_p243-287.indd 245

10/12/10 2:29:58 PM

246

Chapter 8

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The moment of inertia depends on the distribution of mass with respect to a given axis. If the axis is changed, then each bit of mass that makes up the object is at a different distance from the axis than before. Compare the moments of inertia of a uniform rigid rod about axes perpendicular to the rod, first passing through its center of mass, then passing through an end. For example, if you wiggle repeatedly a meter stick back and forth about an axis passing through its center of mass, you will find it does not take much effort to reverse the direction of rotation. However, if you move the axis to an end, you will find it more difficult to wiggle the stick back and forth. The moment of inertia about the end is much larger, because much of the mass of the stick is farther from the axis.

4.

(a)

The lever arm of a particular force is found with respect to some reference point or axis. Thus, an origin must be chosen to compute the torque of a force.

(b)

If the object is in translational equilibrium Fnet = ΣFi = 0 , the net torque acting on the system is independent of the origin or axis considered. However, if the resultant force acting on the object is not zero, the net torque has different values for different axis of rotation.

6.

(

)

The critical factor is the total torque being exerted about the line of the hinges. For simplicity, we assume that the paleontologist and the botanist exert equal magnitude forces. The free body diagram of the original situation is shown on the left and that for the modified situation is shown on the right in the sketches below: Pivot point →

8 cm F

Pivot point

→

→

F

d0 Original situation

F

d Modified situation

In order for the torque exerted on the door in the modified situation to equal that of the original situation, it is necessary that Fd = Fd 0 + F (8 cm ) or d = d 0 + 8 cm. Thus, the paleontologist would need to relocate about 8 cm farther from the hinge. 8.

Stable rotation axis

Unstable rotation axis Stable rotation axis

10.

After the head crosses the bar, the jumper should arch his back so the head and legs are lower than the midsection of the body. In this position, the center of gravity may pass under the bar while the midsection of the body is still above the bar. As the feet approach the bar, the legs should be straightened to avoid hitting the bar.

12.

(a)

56157_08_ch08_p243-287.indd 246

Consider two people, at the ends of a long table, pushing with equal magnitude forces directed in opposite directions perpendicular to the length of the table. The net force will be zero, yet the net torque is not zero.

10/12/10 2:30:00 PM

Rotational Equilibrium and Rotational Dynamics

(b) 14.

247

Consider a falling body. The net force acting on it is its weight, yet the net torque about the center of gravity is zero.

As the cat falls, angular momentum must be conserved. Thus, if the upper half of the body twists in one direction, something must get an equal angular momentum in the opposite direction. Rotating the lower half of the body in the opposite direction satisfies the law of conservation of angular momentum.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

3.55 N · m clockwise

4.

0.642 N · m counterclockwise

6.

ΣFx = Fx − Rx = 0, ΣFy = Fy + Ry − Fg = 0, Σt O = +Fy ( cosq ) − Fx ( sinq ) − Fg ( 2 ) cosq = 0

8.

(a)

226 N

(b)

117 N upward

10.

139 g

12.

(a)

See Solution.

(b) at x = 0

(c)

n1 = 0

(d)

n2 = 1.42 × 10 3 N

(e)

5.64 m

(f)

yes

(a)

n2 = ( m + M ) g

(b)

x = (1 + M m ) − ( M 2m ) L

14.

(c)

⎛ m + M 2⎞ =⎜ L ⎝ m + M ⎟⎠

16.

x cg = 0.459 m, ycg = 0.103 m

18.

T = 1.68 × 10 3 N, R = 2.33 × 10 3 N, q = 21.0°

20.

567 N, 333 N

22.

(a)

See Solution.

(b) T = 343 N, 172 N to the right, 683 N upward

24.

(c)

x max = 5.14 m

(a)

See Solution.

(b) To simplify equations by eliminating 2 unknowns. (c)

⎛L ⎞ Σt )hinge = 0 + 0 − mg ⎜ cosq ⎟ + T ( L sinq ) = 0 ⎝2 ⎠

(e)

ΣFx = Fx − T = 0, ΣFy = Fy − mg = 0

(f)

Fx = 136 N, Fy = 157 N

(d) T = 136 N

(g) Yes, this would reduce tension in the cable and stress on the hinge.

56157_08_ch08_p243-287.indd 247

10/12/10 2:30:00 PM

248

Chapter 8

26.

(a)

See Solution.

(c) T = m s mg

(b)

T = 12 mg cotq

(d)

m s = 12 cotq

(e) At smaller angles, the beam slips and the equation no longer applies. At larger angles, there is no longer impending motion and the equation is invalid. 28.

(a) T = 1.47 kN

30.

x min = 2.8 m

32.

t x = 149 N ⋅ m, t y = 66.0 N ⋅ m, t O = 215 N ⋅ m

34.

(a)

(b)

I = MR 2 + 12 mr 2

(b)

1.33 kN to the right, 2.58 kN upward

(c) t T > 0, a > 0, a < 0

0

(e) T − ( 2M + m ) g = ( 2M + m ) a

(d) a = − a r

− ( 2M + m ) g 2 2M + M ( R r ) + 3m 2

(f)

rT = Ia

(g) a =

(h)

a = −2.72 m s 2

(i) T = 35.4 N

(j)

36.

m k = 0.30

38.

(a)

40.

(a) To provide the net clockwise torque that accelerates the pulley. (b)

(b)

872 N

t = 0.857 s

1.40 kN

(c) T1 = 127 N, T2 = 138 N

2.88 m s 2

42.

(a) 1.37 × 108 J

44.

(a)

I hoop = 0.254 kg ⋅ m 2 , I cyl = 0.127 kg ⋅ m 2 , I sphere = 0.102 kg ⋅ m 2 , I shell = 0.169 kg ⋅ m 2

(b)

solid sphere; solid cylinder; thin spherical shell; and hoop

(c)

hoop; thin spherical shell; solid cylinder; and solid sphere

46.

36 rad s

48.

(a)

rotational KE,

(b)

static friction

50.

30.2 rev s

52.

10.9 rad s

54.

(a)

56.

1 2

5.11 h

Iw 2 ; translational KE, (c)

1 2

mv 2 ; gravitational PE, mgy

27

2.72 kg ⋅ m 2 s

(b) 1.36 kg ⋅ m 2 s

(c) 1.09 kg ⋅ m 2 s

(d) 1.81 kg ⋅ m 2 s

(a) Yes, the bullet effectively moves in a circle around the hinge just before impact. (b)

No, the bullet undergoes an inelastic collision with the door.

(c) 0.749 rad s

56157_08_ch08_p243-287.indd 248

(b)

(d) KE f = 1.68 J

KEi = 2.50 × 10 3 J

10/12/10 2:30:03 PM

Rotational Equilibrium and Rotational Dynamics

58.

0.91 km s

60.

(a)

62.

(a) 1.9 rad s (b)

0.360 rad s , counterclockwise

(b)

249

99.9 J

KEi = 2.5 J , KE f = 6.3 J

64.

12.3 m s 2

66.

(a)

68.

(a) 780 N

2 (b) 1.73 kg ⋅ m s

0.433 kg ⋅ m 2 s

(b)

R = 716 N at 70.4° above the horizontal to the right

70.

(a)

46.8 N

(b)

0.234 kg ⋅ m 2

(c)

72.

(a)

3.75 × 10 3 kg ⋅ m 2 s

(b)

1.88 kJ

(c) 3.75 × 10 3 kg ⋅ m 2 s

(e)

7.50 kJ

(f)

(d) 10.0 m s 74.

Tleft = 1.01 kN, Tright = 1.59 kN

76.

(a)

E = ( m1 − m2 ) gL

(b)

E=

1 2

40.0 rad s

5.62 kJ

( m1 + m2 ) L2w 2 + ( m1 − m2 ) gL sinq

(c) Equate the results of parts (a) and (b) and solve for w . (d) t net

vertical

= 0; t net

rotated

= ( m1 − m2 ) gL cosq ; Since ΔL Δt = t net , the angular momentum

changes at a non-uniform rate. (e) a = ( m1 − m2 ) g cosq 78.

22 N

80.

(a) vcg = 3gL 2

( m1 + m2 ) L; Yes, a → 0 as q → 90° and a is a maximum at q = 0°. (b)

vlower = 3gL = 2vcg end

82.

(a) A frictionless wall cannot exert a component of force parallel to its surface. (b)

L sinq

(c)

1 2

L sinq

84.

Answers are given in the problem statement.

86.

(a) T = 10.2 N

(b)

Rx = 6.56 N, Ry = 7.84 N

88.

(a)

See Solution.

(b)

218 N

(c)

72.4 N

(d)

2.41 m

(d)

2.5 m

(e) The analysis would need to include a horizontal friction force at the lower end of the ladder. The coefficient of static friction between the ladder and the floor would have to be known.

56157_08_ch08_p243-287.indd 249

10/12/10 2:30:08 PM

250

Chapter 8

PROBLEM SOLUTIONS 8.1

Resolve the 100-N force into components parallel to and perpendicular to the rod, as 2.00

F = F cos ( 20.0° + 37.0° ) = F cos57.0°

F 20.0 37.0

m

F = 100 N

F⬜

and

Pivot

F⊥ = F sin ( 20.0° + 37.0° ) = F sin 57.0° The lever arm of F⊥ about the indicated pivot is 2.00 m, while that of F is zero. The torque due to the 100-N force may be computed as the sum of the torques of its components, giving t = F ( 0 ) − F⊥ ( 2.00 m ) = 0 − [(100 N ) sin 57.0° ]( 2.00 m ) = −168 N ⋅ m or 8.2

t = 168 N ⋅ m clockwise

Note that each of the forces is perpendicular to the radius line of the wheel at the point where the force is tangent to the wheel. Thus, when considering torques about the center of the wheel, the radius line is the lever arm of the force. Taking counterclockwise torques as positive,

10.0 N b a

b O

12.0 N

Σt O = + (12.0 N ) a − (10.0 N ) b − ( 9.00 N ) b

9.00 N

= + (12.0 N )( 0.100 m ) − (19.0 N )( 0.250 m ) = −3.55 N ⋅ m 8.3

or

a = 10.0 cm b = 25.0 cm

3.55 N ⋅ m clockwise

First resolve all of the forces shown in Figure P8.3 into components parallel to and perpendicular to the beam as shown in the sketch below. (25 N) cos 30° (25 N) sin 30°

(30 N) cos 45° O (30 N) sin 45°

(10 N) cos 20°

C 2.0 m

(10 N) sin 20° 4.0 m

(a)

t O = + [( 25 N ) cos30° ]( 2.0 m ) − [(10 N ) sin 20° ]( 4.0 m ) = + 30 N ⋅ m or

(b)

t C = + [( 30 N ) sin 45° ]( 2.0 m ) − [(10 N ) sin 20° ]( 2.0 m ) = + 36 N ⋅ m or

8.4

t O = 30 N ⋅ m counterclockwise

t C = 36 N ⋅ m counterclockwise

The lever arm is d = (1.20 × 10 −2 m ) cos 48.0° = 8.03 × 10 −3 m, and the torque is t = Fd = (80.0 N )(8.03 × 10 −3 m ) = 0.642 N ⋅ m counterclockwise

56157_08_ch08_p243-287.indd 250

10/12/10 2:30:11 PM

Rotational Equilibrium and Rotational Dynamics

8.5

(a)

t = Fg ⋅ ( lever arm ) = ( mg ) ⋅ [ sinq ] = ( 3.0 kg ) ( 9.8 m s 2 ) ⋅ [( 2.0 m ) sin 5.0° ] = 5.1 N ⋅ m

(b)

251

q

ᐉ

The magnitude of the torque is proportional to sinq , where q is the angle between the direction of the force and the line from the pivot to the point where the force acts. Note from the sketch that this is the same as the angle the pendulum string makes with the vertical.

Pivot

Fg mg

q

Since sinq increases as q increases, the torque also increases with the angle. 8.6

The object is in both translational and rotational equilibrium. Thus, we may write ΣFx = 0 ⇒ Fx − Rx = 0 ΣFy = 0 ⇒ Σt O = 0 ⇒

and

Fy + Ry − Fg = 0 ⎛ ⎞ Fy ( cosq ) − Fx ( sinq ) − Fg ⎜ cosq ⎟ = 0 ⎝2 ⎠

8.7 0.080 m

Fsx

Fty Ft sin 12.0°

q

Fsx

Fsy

Ftx Ft cos 12.0° Fsy

Pivot

0.290 m

→ Fs

41.5 N

Requiring that Σt = 0 , using the shoulder joint at point O as a pivot, gives

Σt = ( Ft sin12.0°) ( 0.080 m ) − ( 41.5 N ) ( 0.290 m ) = 0, or Ft = 724 N Then ΣFy = 0 ⇒ −Fsy + ( 724 N ) sin12.0° − 41.5 N = 0 yielding Fsy = 109 N. ΣFx = 0 gives

Fs = Fsx2 + Fsy2 =

Therefore, 8.8

Fsx − ( 724 N ) cos12.0° = 0, or Fsx = 708 N

( 708 N )2 + (109 N )2

Since the beam is uniform, its center of gravity is at its geometric center.

= 716 N T d

d = 1.20 m cg O

Requiring that Σt = 0 about an axis through point O and perpendicular to the page gives F ( 0 ) + mg ( 2 ) − T ( − d ) = 0 (a)

/2

F

mg = 343 N = 5.00 m

The tension in the rope must then be T=

mg ( 2 ) ( 343 N ) ( 2.50 m ) = = 226 N −d 5.00 m − 1.20 m

continued on next page

56157_08_ch08_p243-287.indd 251

10/12/10 2:30:14 PM

252

Chapter 8

(b)

The force the column exerts is found from ΣFy = 0

F + T − mg = 0

F = mg − T = 343 N − 226 N = 717 N upward

or 8.9

⇒

Require that Σt = 0 about an axis through the elbow and perpendicular to the page. This gives

(

)

Σt = + ⎡⎣( 2.00 kg ) 9.80 m s 2 ⎤⎦ ( 25.0 cm + 8.00 cm ) − ( FB cos 75.0° ) (8.00 cm ) = 0 or 8.10

FB =

(19.6 N ) ( 33.0 cm ) = ( 8.00 cm ) cos 75.0°

312 N →

Since the bare meter stick balances at the 49.7 cm mark when placed on the fulcrum, the center of gravity of the meter stick is located 49.7 cm from the zero end. Thus, the entire weight of the meter stick may be considered to be concentrated at this point. The force diagram of the stick when it is balanced with the 50.0-g mass attached at the 10.0 cm mark is as given at the right.

R

39.2 cm

Point O

10.0 → cm (50.0 g)g →

49.7 cm

Mg

Requiring that the sum of the torques about point O be zero yields + ⎡⎣( 50.0 g ) g ⎤⎦ ( 39.2 cm − 10.0 cm ) − M g ( 49.7 cm − 39.2 cm ) = 0 or 8.11

⎛ 39.2 cm − 10.0 cm ⎞ = 139 g M = ( 50.0 g ) ⎜ ⎝ 49.7 cm − 39.2 cm ⎟⎠

Consider the remaining plywood to consist of two parts: A1 is a 4.00-ft by 4.00-ft section with center of gravity located at ( 2.00 ft, 2.00 ft ) , while A2 is a 2.00-ft by 4.00-ft section with center of gravity at ( 6.00 ft, 1.00 ft ) . Since the plywood is uniform, its mass per area s is constant and the mass of a section having area A is m = s A . The center of gravity of the remaining plywood has coordinates given by x cg =

and 8.12

(

y (ft) 4.00

A1

2.00 0

A2 2.00

x (ft)

4.00 6.00

) ( ) ( ) ( ) ) ( 2.00 ft ) + ( 8.00 ft ) (1.00 ft ) = (16.0 ft ) + ( 8.00 ft )

8.00

16.0 ft 2 ( 2.00 ft ) + 8.00 ft 2 ( 6.00 ft ) Σmi xi s A1 x1 + s A2 x 2 = = = 3.33 ft Σmi s A1 + s A2 16.0 ft 2 + 8.00 ft 2

(

16.0 ft 2 Σmi yi s A1 y1 + s A2 y2 ycg = = = Σmi s A1 + s A2

(a)

2

2

1.67 ft

mg

x

n1

2

3.00 m

Mg

n2

4.00 m

(

)

Mg = ( 90.0 kg ) 9.80 m s 2 = 882 N

(

)

mg = ( 55.0 kg ) 9.80 m s 2 = 539 N

continued on next page

56157_08_ch08_p243-287.indd 252

10/12/10 2:30:17 PM

Rotational Equilibrium and Rotational Dynamics

253

(b)

The woman is at x = 0 when n1 is greatest. With this location of the woman, she exerts her maximum possible counterclockwise torque about the center of the beam. Thus, n1 must be exerting its maximum clockwise torque about the center to hold the beam in rotational equilibrium.

(c)

n1 = 0 As the woman walks to the right along the beam, she will eventually reach a point where the beam will start to rotate clockwise about the rightmost pivot. At this point, the beam is starting to lift up off of the leftmost pivot and the normal force exerted by that pivot will have diminished to zero.

(d)

When the beam is about to tip, n1 = 0, and ΣFy = 0 gives 0 + n2 − Mg − mg = 0 n2 = Mg + mg = 882 N + 539 N = 1.42 × 10 3 N

or (e)

Requiring that Σt )rightmost = 0 when the beam is about to tip ( n1 = 0) gives pivot

− ( 4.00 m − x ) mg − ( 4.00 m − 3.00 m ) Mg = 0

(f)

or

( mg ) x = (1.00 m ) Mg + ( 4.00 m ) mg

and

x = (1.00 m )

M + 4.00 m m

Thus,

x = (1.00 m )

( 90.0 kg ) + 4.00 m = ( 55.0 kg )

5.64 m

When n1 = 0 and n2 = 1.42 × 10 3 N, requiring that Σt )left = 0 gives end

(

)

0 − ( 539 N ) x − ( 882 N ) ( 3.00 m ) + 1.42 × 10 N ( 4.00 m ) = 0 or

x=

3

−3.03 × 10 3 N ⋅ m = 5.62 N −539 N

which, within limits of rounding errors, is the same as the answer to part (e) . 8.13

Requiring that x cg =

Σmi xi = 0 gives Σmi

( 5.0 kg ) ( 0 ) + ( 3.0 kg ) ( 0 ) + ( 4.0 kg ) ( 3.0 m ) + ( 8.0 kg ) x = 0 ( 5.0 + 3.0 + 4.0 + 8.0 ) kg or 8.0 x + 12 m = 0 which yields x = −1.5 m Also, requiring that ycg = Σmi yi Σmi = 0 gives

( 5.0 kg ) ( 0 ) + ( 3.0 kg ) ( 4.0 m ) + ( 4.0 kg ) ( 0 ) + ( 8.0 kg ) y = 0 ( 5.0 + 3.0 + 4.0 + 8.0 ) kg or 8.0 y + 12 m = 0 yielding y = −1.5 m Thus, the 8.0-kg object should be placed at coordinates

56157_08_ch08_p243-287.indd 253

( −1.5 m,− 1.5 m ) .

10/12/10 2:30:20 PM

254

Chapter 8

8.14

(a)

As the woman walks to the right along the beam, she will eventually reach a point where the beam will start to rotate clockwise about the rightmost pivot. At this point, the beam is starting to lift up off of the leftmost pivot and the normal force, n1, exerted by that pivot will have diminished to zero. ΣFy = 0

Then, (b)

⇒

0 − mg − Mg + n2 = 0

mg

x

L/2

n1

Mg

n2

n2 = ( m + M ) g

or

When n1 = 0 and n2 = ( m + M ) g , requiring that Σt )left = 0 gives end

0 − ( mg ) x − ( Mg ) (c)

L + ( mg + Mg ) = 0 2

If the woman is to just reach the right end of the beam (x = L) when n1 = 0 and n2 = ( m + M ) g (i.e., the beam is ready to tip), then the result from part (b) requires that M⎞ ⎛ ⎛ M⎞ L L = ⎜1 + ⎟ − ⎜ ⎝ 2m ⎟⎠ ⎝ m⎠

8.15

M⎞ ⎛ ⎛ M⎞ x = ⎜ 1+ ⎟ − ⎜ L ⎝ 2m ⎟⎠ ⎝ m⎠

or

=

or

(1 + M 2m ) = (1 + M m )

⎛ m + M 2⎞ ⎜⎝ ⎟L m+M ⎠

In each case, the distance from the bar to the center of mass of the body is given by x cg =

Σmi xi marms x arms + mtorso x torso + mthighs x thighs + mlegs x legs = Σmi marms + mtorso + mthighs + mlegs

where the distance x for any body part is the distance from the bar to the center of gravity of that body part. In each case, we shall take the positive direction for distances to run from the bar toward the location of the head. Note that Σmi = ( 6.87 + 33.57 + 14.07 + 7.54 ) kg = 62.05 kg With the body positioned as shown in Figure P8.15b, the distances x for each body part is computed using the sketch given below: Positive direction Knee joint

Hip joint thighs

(rcg)legs

( )

xarms = + rcg x torso =

arms

x thighs = x legs =

arms

( )

+

(rcg)torso

(rcg)arms

= +0.239 m

+ rcg +

Bar arms

torso

(rcg)thighs

arms

arms

Shoulder joint

torso

torso

torso

= +0.548 m + 0.337 m = 0.885 m

( )

thighs

thighs

+ rcg

+ rcg

+

= ( +0.548 + 0.601+ 0.151 ) m = 1.30 m

( )

legs

= ( +0.548 + 0.601+ 0.374 + 0.227 ) m = 1.75 m

With these distances and the given masses we find x cg =

+ 62.8 kg ⋅ m = +1.01 m 62.05 kg

continued on next page

56157_08_ch08_p243-287.indd 254

10/12/10 2:30:24 PM

Rotational Equilibrium and Rotational Dynamics

255

With the body positioned as shown in Figure P8.15c, we use the following sketch to determine the distance x for each body part: Positive direction Hip joint

Knee joint

Shoulder Joint ᐉtorso (rcg)torso

ᐉthighs

(rcg)legs

(rcg)thighs

(rcg)arms ᐉarms Bar

( )

x arms = + rcg x torso =

arms

x thighs = x legs =

( )

− rcg

arms

arms

= +0.239 m

arms

−

−

torso

torso

torso

= +0.548 m − 0.337 m = +0.211 m

( )

thighs

thighs

− rcg

− rcg

−

= ( +0.548 − 0.601− 0.151 ) m = −0.204 m

( )

legs

= ( +0.548 − 0.601− 0.374 − 0.227 ) m = −0.654 m

With these distances, the location (relative to the bar) of the center of gravity of the body is x cg = 8.16

+ 0.924 kg ⋅ m = +0.015 m = 0.015 m towards the head 62.05 kg

With the coordinate system shown at the right, the coordinates of the center of gravity of each body part may be computed:

Bar

y

(rcg)legs

hig h

s

(rcg)arms

t

arms

(rcg)thighs 60.0°

Origin

x

(rcg)torso torso

x cg, arms = 0

ycg, arms =

( )

x cg, torso = rcg x cg, thighs = x cg, legs =

torso

torso

torso

= 0.337 m

+

( )

− rcg

arms

= 0.309 m

ycg, torso = 0

( )

thighs

thighs

cos 60.0° + rcg

+ rcg

arms

cos60.0° = 0.677 m

( )

legs

= 1.02 m

ycg, thighs = rcg

( )

thighs

ycg, legs =

thighs

sin 60.0° = 0.324 m

sin 60.0° = 0.131 m

With these coordinates for individual body parts and the masses given in Problem 8.15, the coordinates of the center of mass for the entire body are found to be continued on next page

56157_08_ch08_p243-287.indd 255

10/12/10 2:30:27 PM

256

Chapter 8

x cg =

marms x cg, arms + mtorso x cg, torso + mthighs x cg, thighs + mlegs x cg, legs marms + mtorso + mthighs + mlegs

=

28.5 kg ⋅ m = 0.459 m 62.05 kg

=

6.41 kg ⋅ m = 0.103 m 62.05 kg

and ycg = 8.17

marms ycg, arms + mtorso ycg, torso + mthighs ycg, thighs + mlegs ycg, legs marms + mtorso + mthighs + mlegs

The force diagram for the spine is shown below. Ty T sin12.0

Ry 2L/3

Tx T cos12.0

Rx Point O

L/2

200 N 350 N L

(a)

When the spine is in rotational equilibrium, the sum of the torques about the left end (point O) must be zero. Thus, ⎛ 2L ⎞ ⎛ L⎞ − ( 350 N ) ⎜ ⎟ − ( 200 N ) ( L ) = 0 +Ty ⎜ ⎝ 3 ⎟⎠ ⎝ 2⎠ yielding Ty = T sin12.0° = 563 N The tension in the back muscle is then T =

(b)

563 N = 2.71× 10 3 N = 2.71 kN . sin12.0°

The spine is also in translational equilibrium, so ΣFx = 0 ⇒ Rx − Tx = 0, and the compression force in the spine is Rx = Tx = T cos12.0° = ( 2.71 kN ) cos12.0° = 2.65 kN

8.18

In the force diagram of the foot given below, note that the force R (exerted on the foot by the tibia) has been replaced by its horizontal and vertical components. Employing both conditions of equilibrium (using point O as the pivot point) gives the following three equations: →

T

q

0 cm

25. 18.0

q

cm

Point O

Rx Rsin15.0

Ry Rcos15.0

n 700 N

continued on next page

56157_08_ch08_p243-287.indd 256

10/12/10 2:30:29 PM

Rotational Equilibrium and Rotational Dynamics

257

ΣFx = 0 ⇒ R sin15.0° − T sinq = 0 R=

or

T sinq sin15.0°

[1]

ΣFy = 0 ⇒ 700 N − R cos15.0° + T cosq = 0

[2]

Σt o = 0 ⇒ − ( 700 N ) [(18.0 cm ) cosq ] + T ( 25.0 cm − 18.0 cm ) = 0 T = (1 800 N ) cosq

or

[3]

Substituting Equation [3] into Equation [1] gives ⎛ 1 800 N ⎞ sinq cosq R=⎜ ⎝ sin15.0° ⎟⎠

[4]

Substituting Equations [3] and [4] into Equation [2] yields ⎛ 1 800 N ⎞ 2 ⎜⎝ ⎟ sinq cosq − (1 800 N ) cos q = 700 N tan15.0° ⎠ which reduces to: sinq cosq = ( tan15.0° ) cos 2 q + 0.104 Squaring this result and using the identity sin 2 q = 1 − cos 2 q gives ⎡⎣ tan 2 (15.0° ) + 1⎤⎦ cos 4 q + ⎡⎣( 2 tan15.0° ) ( 0.104 ) − 1⎤⎦ cos 2 + ( 0.104 ) = 0 2

In this last result, let u = cos 2 q and evaluate the constants to obtain the quadratic equation: (1.07 ) u 2 − ( 0.944 ) u + ( 0.010 8 ) = 0. The quadratic formula yields the solutions u = 0.871 and u = 0.011 6.

(

(

)

)

Thus q = cos −1 0.871 = 21.0° or q = cos −1 0.011 6 = 83.8°. We ignore the second solution since it is physically impossible for the human foot to stand with the sole inclined at 83.8° to the floor. We are the left with q = 21.0° . Equation [3] then yields:

T = (1 800 N ) cos 21.0° = 1.68 × 10 3 N

and Equation [1] gives:

R=

(1.68 × 10 N ) sin 21.0° = 3

8.19

sin15.0°

2.33 × 10 3 N

Consider the torques about an axis perpendicular to the page through the left end of the rod. Ty T cos30.0 6.00 m Ry Tx T sin30.0

Rx 3.00 m

4.00 m

100 N

500 N

continued on next page

56157_08_ch08_p243-287.indd 257

10/12/10 2:30:30 PM

258

Chapter 8

Σt = 0 ⇒ T =

(100 N ) ( 3.00 m ) + ( 500 N ) ( 4.00 m ) ( 6.00 m ) cos 30.0°

T = 443 N ΣFx = 0 ⇒ Rx = T sin 30.0° = ( 443 N ) sin 30.0° Rx = 222 N toward the right ΣFy = 0 ⇒ Ry + T cos 30.0° − 100 N − 500 N = 0 Ry = 600 N − ( 443 N ) cos 30.0° = 216 N upward 8.20

Consider the torques about an axis perpendicular to the page through the left end of the scaffold. Σt = 0 ⇒ T1 ( 0 ) − ( 700 N ) (1.00 m ) − ( 200 N ) (1.50 m ) + T2 ( 3.00 m ) = 0 T2 = 333 N

From which,

→ T1

→ T2

700 N

Then, from ΣFy = 0, we have

1.00 m

T1 + T2 − 700 N − 200 N = 0

2.00 m

1.50 m 200 N

or T1 = 900 N − T2 = 900 N − 333 N = 567 N 8.21

Consider the torques about an axis perpendicular to the page and through the left end of the plank. Σt = 0 gives − ( 700 N ) ( 0.500 m ) − ( 294 N ) (1.00 m ) + ( T1 sin 40.0° ) ( 2 .00 m ) = 0 or

→ T2

T1 = 501 N

0.500 m

Then, ΣFx = 0 gives − T3 + T1 cos 40.0° = 0, or

→ T3

T3 = ( 501 N ) cos 40.0° = 384 N

or

T2 = 994 N − ( 501 N ) sin 40.0° = 672 N

(a)

See the diagram below:

1.00 m

→

700 N

x

1.00 m

40.0°

mg 294 N

From ΣFy = 0, T2 − 994 N + T1 sin 40.0° = 0,

8.22

→ T1

700 N

T

60.0°

→

H

→

V

3.00 m 200 N

3.00 m 80.0 N

continued on next page

56157_08_ch08_p243-287.indd 258

10/12/10 2:30:33 PM

Rotational Equilibrium and Rotational Dynamics

(b)

259

If x = 1.00 m, then Σt )left end = 0 ⇒ − ( 700 N )(1.00 m ) − ( 200 N )( 3.00 m ) − (80.0 N )( 6.00 m ) + ( T sin 60.0° )( 6.00 m ) = 0

(c)

giving

T = 343 N

Then,

ΣFx = 0 ⇒ H − T cos 60.0° = 0, or

and

ΣFy = 0 ⇒ V − 980 N + ( 343 N ) sin 60.0° = 0, or V = 683 N upward

H = ( 343 N ) cos60.0° = 172 N to the right

When the wire is on the verge of breaking, T = 900 N and Σt )left end = − ( 700 N ) x max − ( 200 N ) ( 3.00 m ) −

(80.0 N )( 6.00 m ) + [( 900 N ) sin 60.0°]( 6.00 m ) = 0 which gives x max = 5.14 m (a)

Considering a pivot at the lower end of the beam, we get Σt )lower = 0

⇒

end

⎛ ⎞ + FS sinq − mg ⎜ cosq ⎟ = 0 ⎝2 ⎠

Fs

8.23

and the spring force is

2

mg ( cosq 2 ) mg = sinq 2 tanq

/

FS =

d= (b)

d=

or

q Rx

mg 2k tanq

From the first condition of equilibrium,

and 8.24

FS k

mg

Ry

From Hooke’s law, FS = kd , the distance the spring is stretched is then

ΣFx = 0

⇒

Rx − FS = 0

or

Rx = FS =

ΣFy = 0

⇒

Ry − mg = 0

or

Ry = mg

mg 2 tanq

T

(a)

q Fy q Fx

mg L/2

(b)

56157_08_ch08_p243-287.indd 259

L

The point of intersection of two unknown forces is always a good choice as the pivot point in a torque calculation. Doing this eliminates these two unknowns from the calculation (since they have zero lever arms about the chosen pivot) and makes it easier to solve the resulting equilibrium equation . continued on next page

10/12/10 2:30:36 PM

260

Chapter 8

⎛L ⎞ 0 + 0 − mg ⎜ cosq ⎟ + T ( L sinq ) = 0 ⎝2 ⎠

(c)

Σt )hinge = 0

(d)

Solving the above result for the tension in the cable gives T= T=

or

⇒

( mg 2 ) L cosq L sinq

(16.0 kg ) ( 9.80

mg 2 tanq

m s2

2 tan 30.0°

⇒

Fx − T = 0

) = 136 N

(e)

ΣFx = 0

(f)

Solving the above results for the components of the hinge force gives Fx = T = 136 N

(g) 8.25

=

and

ΣFy = 0

⇒

Fy − mg = 0

(

Attaching the cable higher up would allow the cable to bear some of the weight, thereby reducing the stress on the hinge. It would also reduce the tension in the cable.

When the refrigerator is on the verge of tipping (i.e., rotating counterclockwise about point O in the sketch at the right), the center of gravity of the refrigerator will be directly above point O and the line of action of the gravitational force will pass through point O. When this is true, increasing angle q by any amount will cause the line of action of Fg to pass to the left of point O, producing a counterclockwise torque with no force available to produce a counterbalancing clockwise torque about this point. At this critical value of angle q , we have tanq =

8.26

)

Fy = mg = (16.0 kg ) 9.80 m s 2 = 157 N

and

w2 h2

(a)

w

h w /2

h/2 q

Fg = mg

O

q

q = tan −1 ( w h )

or

T L/2 L/2 n

mg q

(b)

Σt )lower = 0 end

or (c)

fs (fs)max msn

T=

⇒

⎛L ⎞ 0 + 0 − mg ⎜ cosq ⎟ + T ( L sinq ) = 0 ⎝2 ⎠

mg ⎛ cosq ⎞ mg cotq ⎜ ⎟= 2 ⎝ sinq ⎠ 2

ΣFx = 0

⇒

− T + ms n = 0

or

T = ms n

[1]

ΣFy = 0

⇒

n − mg = 0

or

n = mg

[2]

Substitute Equation [2] into [1] to obtain T = m s mg continued on next page

56157_08_ch08_p243-287.indd 260

10/12/10 2:30:39 PM

Rotational Equilibrium and Rotational Dynamics

(d)

(e)

8.27

261

Equate the results of parts (b) and (c) to obtain m s = 12 cotq . This result is valid only at the critical angle q where the beam is on the verge of slipping (i.e., where fs = ( fs )max is valid). At angles below the critical angle (where fs = ( fs )max is valid), the beam will slip. At larger angles, the static friction force is reduced below the maximum value, and it is no longer appropriate to use m s in the calculation.

Consider the torques about an axis perpendicular to the page and through the point where the force T acts on the jawbone.

3.50 cm R

Fc 50.0 N

T

7.50 cm

Σt = 0 ⇒ ( 50.0 N ) ( 7.50 cm ) − R ( 3.50 cm ) = 0 which yields R = 107 N Then, ΣFy = 0 ⇒ − ( 50.0 N ) + T − 107 N = 0 , or T = 157 N 8.28

(a)

Observe that the cable is perpendicular to the boom. Then, using Σt = 0 for an axis perpendicular to the page and through the lower end of the boom gives ⎛L ⎞ ⎛3 ⎞ − (1.20 kN ) ⎜ cos 65.0°⎟ + T ⎜ L ⎟ − ( 2.00 kN ) ( L cos 65.0° ) = 0 ⎝4 ⎠ ⎝2 ⎠ or

T = 1.47 kN →

(b)

From ΣFx = 0,

L

w = 2.00 kN

3 4L

H = T cos 25.0° = 1.33 kN to the right

L 2

and ΣFy = 0 gives V = 3.20 kN − T sin 25.0° = 2.58 kN upward

T

25°

1.20 kN 65°

→

H →

V

8.29

Choose a reference frame with the x-axis parallel to the tibia and the y-axis perpendicular to it. Then, resolve all forces into their x- and y-components, as shown. Note that q = 40.0° and w y = ( 30.0 N ) sin 40.0° = 19.3 N

Ry

Rx

d/5 Ty

Tx

d

d/2

Fy = (12 .5 N ) sin 40.0° = 8.03 N and

Ty = T sin 25.0°

Using Σt = 0 for an axis perpendicular to the page and through the upper end of the tibia gives

(T sin 25.0°) d5 − (19.3 N ) d2 − ( 8.03 N ) d = 0, or

56157_08_ch08_p243-287.indd 261

wy

q

wx

Fy q

Fx

T = 209 N

10/12/10 2:30:42 PM

262

8.30

Chapter 8

When x = x min , the rod is on the verge of slipping, so

→

T

f = ( fs )max = m s n = 0.50 n

2.0 m

→

37°

n

From ΣFx = 0, n − T cos 37° = 0, or n = 0.80 T

→

Thus, f = 0.50 ( 0.80 T ) = 0.40 T.

2.0 m

x

f

→

w

→

w

From ΣFy = 0, f + T sin 37° − 2 w = 0, or 0.40 T + 0.60 T − 2 w = 0, giving T = 2 w. Using Σt = 0 for an axis perpendicular to the page and through the left end of the beam gives −w ⋅ x min − w ( 2.0 m ) + ⎡⎣( 2 w ) sin 37° ⎤⎦ ( 4.0 m ) = 0, which reduces to x min = 2.8 m . 8.31

The moment of inertia for rotations about an axis is I = Σmi ri2 , where ri is the distance mass mi is from that axis. (a)

For rotation about the x-axis, I x = ( 3.00 kg ) ( 3.00 m ) + ( 2 .00 kg ) ( 3.00 m ) + 2

2

( 2.00 kg ) ( 3.00 m )2 + ( 4.00 kg ) ( 3.00 m )2 = (b)

99.0 kg ⋅ m 2

When rotating about the y-axis, I y = ( 3.00 kg ) ( 2 .00 m ) + ( 2 .00 kg ) ( 2 .00 m ) + 2

2

( 2.00 kg ) ( 2.00 m )2 + ( 4.00 kg ) ( 2.00 m )2 = (c)

44.0 kg ⋅ m 2

For rotations about an axis perpendicular to the page through point O, the distance ri for each mass is ri =

( 2 .00 m )2 + ( 3.00 m )2

= 13.0 m. Thus,

(

)

IO = ⎡⎣( 3.00 + 2.00 + 2.00 + 4.00 ) kg ⎤⎦ 13.0 m 2 = 143 kg ⋅ m 2 8.32

The required torque in each case is t = I a . Thus,

( )( ) = I a = ( 44.0 kg ⋅ m ) (1.50 rad s ) = = I a = (143 kg ⋅ m ) (1.50 rad s ) =

t x = I x a = 99.0 kg ⋅ m 2 1.50 rad s 2 = 149 N ⋅ m ty and

8.33

tO

66.0 N ⋅ m

2

2

215 N ⋅ m

O

⇒

I=

t net rF sin 90° ( 0.330 m ) ( 250 N ) = = = 87.8 kg ⋅ m 2 a a 0.940 rad s 2

t net = Ia

(b)

For a solid cylinder, I = Mr 2 2, so

(

)

2 2I 2 87.8 kg ⋅ m = = 1.61 × 10 3 kg 2 r2 ( 0.330 m )

(

)

(c)

w = w 0 + a t = 0 + 0.940 rad s 2 ( 5.00 s ) = 4.70 rad s

(a)

I = 2I disk + I cylinder = 2 MR 2 2 + mr 2 2

continued on next page

56157_08_ch08_p243-287.indd 262

2

(a)

M=

8.34

2

y

(

)

or

I = MR 2 + mr 2 2

10/12/10 2:30:46 PM

263

Rotational Equilibrium and Rotational Dynamics

(b)

t g = 0 Since the line of action of the gravitational force passes through the rotation axis, it has zero lever arm about this axis and zero torque.

(c)

The torque due to the tension force is positive. Imagine gripping the cylinder with your right hand so your fingers on the front side of the cylinder point upward in the direction of the tension force. The thumb of your right hand then points toward the left (positive direction) along the rotation axis. Because t = Ia , the torque and angular acceleration have the same direction. Thus, a positive torque produces a positive angular acceleration. When released, the center of mass of the yoyo drops downward, in the negative direction. The translational acceleration is negative.

(d)

Since, with the chosen sign convention, the translational acceleration is negative when the angular acceleration is positive, we must include a negative sign in the proportionality between these two quantities. Thus, we write: a = −ra or a = − a r .

(e)

Translation:

ΣFy = mtotal a

(f)

Rotational:

Σt = Ia

(g)

Substitute the results of (d) and (a) into Equation [2] to obtain

⇒

⇒

T − ( 2M + m ) g = ( 2M + m ) a

rT sin 90° = Ia

[1]

rT = Ia

or

[2]

a 2 ⎛a ⎞ ⎛ −a r ⎞ T = I⎜ ⎟ = I⎜ = − MR 2 + mr 2 2 2 or T = − ⎡⎣ M ( R r ) + m 2 ⎤⎦ a ⎝r⎠ ⎝ r ⎟⎠ r

(

)

[3]

Substituting Equation [3] into [1] yields 2 − ⎡⎣ M ( R r ) + m 2 ⎤⎦ a − ( 2M + m ) g = ( 2M + m ) a

or

8.35

a=

− ( 2M + m ) g 2 2M + M ( R r ) + 3m 2

(

)

− ⎡⎣ 2 ( 2.00 kg ) + 1.00 kg ⎤⎦ 9.80 m s 2 = −2.72 m s 2 2 2 ( 2.00 kg ) + ( 2.00 kg ) (10.0 4.00 ) + 3 (1.00 kg ) 2

(h)

a=

(i)

From Equation [1], T = ( 2M + m ) ( g + a ) = ( 5.00 kg ) 9.80 m s 2 − 2.72 m s 2 = 35.4 N

(j)

Δy = ( 0 ) t + at 2 2

(a)

Consider the force diagrams of the cylinder and man given at the right. Note that we shall adopt a sign convention with clockwise and downward as the positive directions. Thus, both a and a are positive in the indicated directions and a = ra . We apply the appropriate form of Newton’s second law to each diagram to obtain:

(

⇒

t=

Rotation of Cylinder: t = Iα

2 ( Δy ) = a

⇒

)

2 ( −1.00 m ) = 0.857 s −2.72 m s 2

rT sin 90° = Ia

or

a M r T T

T = Ia r

a

m

so 1⎛1 ⎞ ⎛ a⎞ T = ⎜ Mr 2 ⎟ ⎜ ⎟ ⎠⎝r⎠ r⎝2

mg

giving

1 T = Ma 2

[1] continued on next page

56157_08_ch08_p243-287.indd 263

10/12/10 2:30:51 PM

264

Chapter 8

Translation of man: ΣFy = ma

⇒

mg − T = ma

Equating Equations [1] and [2] gives a=

8.36

T = m ( g − a)

or 1 2

[2]

Ma = m ( g − a ), or

(

)

( 75.0 kg) 9.80 m s2 mg = = 3.92 m s2 m+M 2 75.0 kg + ( 225 kg 2 ) a 3.92 m s 2 = = 9.80 rad s 2 r 0.400 m

(b)

From a = ra , we have

(c)

As the rope leaves the cylinder, the mass of the cylinder decreases, thereby decreasing the moment of inertia. At the same time, the weight of the rope leaving the cylinder would increase the downward force acting tangential to the cylinder, and hence increase the torque exerted on the cylinder. Both of these effects will cause the acceleration of the system to increase with time. (The increase would be slight in this case, given the large mass of the cylinder.)

a =

(

The angular acceleration is a = w f − w i

) Δt = − (w

i

Δt ) since w f = 0 .

The torque is t = I a = − ( I w i Δt ). But the torque is also t = − fr, so the magnitude of the required friction force is f=

(

)

12 kg ⋅ m 2 ( 50 rev min ) ⎛ 2p rad ⎞ ⎛ 1 min ⎞ I wi = ⎜⎝ ⎟⎜ ⎟ = 21 N r ( Δt ) 1 rev ⎠ ⎝ 60 s ⎠ ( 0.50 m )( 6.0 s )

Therefore, the coefficient of friction is mk =

8.37

8.38

f 21 N = = 0.30 n 70 N

(a)

t = F ⋅ r sinq = ( 0.800 N ) ( 30.0 m ) sin 90.0° = 24.0 N ⋅ m

(b)

a =

(c)

a t = r a = ( 30.0 m ) 0.035 6 rad s 2 = 1.07 m s 2

t t 24.0 N ⋅ m = = = 0.035 6 rad s 2 2 I mr ( 0.750 kg) (30.0 m )2

(

)

I = MR 2 = (1.80 kg )( 0.320 m ) = 0.184 kg ⋅ m 2 2

t net = t applied − t resistive = I a , yielding

F = ( I a + f ⋅ R) r

(a)

F=

( 0.184 kg ⋅ m ) ( 4.50 rad s ) + (120 N )( 0.320 m ) =

872 N

(b)

F=

( 0.184 kg ⋅ m ) ( 4.50 rad s ) + (120 N )( 0.320 m ) =

1.40 kN

2

2

4.50 × 10 −2 m

2

56157_08_ch08_p243-287.indd 264

F ⋅r − f ⋅ R = Ia

or

2

2.80 × 10 −2 m

10/12/10 2:30:54 PM

265

Rotational Equilibrium and Rotational Dynamics

8.39

I=

and

1 1 2 MR 2 = (150 kg ) (1.50 m ) = 169 kg ⋅ m 2 2 2 a =

w f − wi Δt

=

( 0.500

rev s − 0 ) ⎛ 2p rad ⎞ p rad s 2 ⎜⎝ ⎟= 2.00 s 1 rev ⎠ 2

Thus, t = F ⋅ r = I a gives

(169 kg ⋅ m ) ⎛⎜⎝ p2 2

Ia F= = r 8.40

(a)

⎞ rad s 2 ⎟ ⎠

1.50 m

= 177 N

It is necessary that the tensions T1 and T2 be different

a

in order to provide a net torque about the axis of

M r

the pulley and produce an angular acceleration of the pulley. Since intuition tells us that the system will accelerate in the directions shown in the diagrams at the right when m2 > m1, it is necessary that T2 > T1. (b)

a

We adopt a sign convention for each object with the positive direction being the indicated direction of the acceleration of that object in the diagrams at the right. Then, apply Newton’s second law to each object:

T1

T2

T1

T2

m1 m1g

m2

a

m2 g

For m1:

ΣFy = m1a

⇒ T1 − m1 g = m1a

or

T1 = m1 ( g + a )

[1]

For m2 :

ΣFy = m2 a

⇒ m2 g − T2 = m2 a

or

T2 = m2 ( g − a )

[2]

For M:

Σt = Ia

or

T2 − T1 = Ia r

[3]

⇒ rT2 − rT1 = Ia

Substitute Equations [1] and [2], along with the relations I = Mr 2 2 and a = a r, into Equation [3] to obtain m2 ( g − a ) − m1 ( g + a ) = a=

and (c)

8.41

( m2 − m1 ) g

m1 + m2 + M

Mr 2 2r

⎛ a ⎞ Ma ⎜⎝ ⎟⎠ = r 2

or

M⎞ ⎛ ⎜⎝ m1 + m2 + ⎟⎠ a = ( m2 − m1 ) g 2

( 20.0 kg − 10.0 kg) (9.80 m s2 ) = = 2 20.0 kg + 10.0 kg + (8.00 kg ) 2

(

)

(

)

2.88 m s 2

From Equation [1]:

T1 = (10.0 kg ) 9.80 m s 2 + 2.88 m s 2 = 127 N

From Equation [2]:

T2 = ( 20.0 kg ) 9.80 m s 2 − 2.88 m s 2 = 138 N

The initial angular velocity of the wheel is zero, and the final angular velocity is v 50.0 m s = = 40.0 rad s r 1.25 m Hence, the angular acceleration is wf =

a=

w f − wi Δt

=

40.0 rad s − 0 = 83.3 rad s 2 0.480 s continued on next page

56157_08_ch08_p243-287.indd 265

10/12/10 2:30:57 PM

266

Chapter 8

The torque acting on the wheel is t = fk ⋅r, so t = I a gives 2 2 I a (110 kg ⋅ m ) (83.3 rad s ) = = 7.33 × 10 3 N r 1.25 m

fk =

Thus, the coefficient of friction is fk 7.33 × 10 3 N = = 0.524 n 1.40 × 10 4 N

mk = 8.42

(a)

The moment of inertia of the flywheel is I=

1 1 2 MR 2 = ( 500 kg ) ( 2.00 m ) = 1.00 × 10 3 kg ⋅ m 2 2 2

and the angular velocity is rev ⎞ ⎛ 2p rad ⎞ ⎛ 1 min ⎞ ⎛ w = ⎜ 5000 ⎟⎜ ⎟⎜ ⎟ = 524 rad s ⎝ min ⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠ Therefore, the stored kinetic energy is KEstored = (b)

1 2 1 2 Iw = (1.00 × 10 3 kg ⋅ m 2 ) ( 524 rad s ) = 1.37 × 108 J 2 2

A 10.0-hp motor supplies energy at the rate of ⎛ 746 W ⎞ = 7.46 × 10 3 J s P = (10.0 hp )⎜ ⎝ 1 hp ⎟⎠ The time the flywheel could supply energy at this rate is t=

8.43

KEstored 1.37 × 108 J = = 1.84 × 10 4 P 7.46 × 10 3 J s

⎛ 1h ⎞ = 5.11 h s⎜ ⎝ 3 600 s ⎟⎠

The moment of inertia of the cylinder is I=

1 1 ⎛ w⎞ 1 ⎛ 800 N ⎞ MR 2 = ⎜ ⎟ R 2 = ⎜ (1.50 m )2 = 91.8 kg ⋅ m 2 2 2⎝ g⎠ 2 ⎝ 9.80 m s 2 ⎟⎠

The angular acceleration is given by a=

t F ⋅ R ( 50.0 N )(1.50 m ) = = = 0.817 rad s 2 I I 91.8 kg ⋅ m 2

At t = 3.00 s, the angular velocity is w = w i + a t = 0 + ( 0.817 rad s 2 )( 3.00 s ) = 2.45 rad s and the kinetic energy is KEr =

56157_08_ch08_p243-287.indd 266

1 2 1 2 Iw = ( 91.8 kg ⋅ m 2 ) ( 2.45 rad s ) = 276 J 2 2

10/12/10 2:31:01 PM

267

Rotational Equilibrium and Rotational Dynamics

8.44

(a)

(b)

Hoop:

I = MR 2 = ( 4.80 kg ) ( 0.230 m ) = 0.254 kg ⋅ m 2

Solid Cylinder:

I=

1 1 2 MR 2 = ( 4.80 kg ) ( 0.230 m ) = 0.127 kg ⋅ m 2 2 2

Solid Sphere:

I=

2 2 2 MR 2 = ( 4.80 kg ) ( 0.230 m ) = 0.102 kg ⋅ m 2 5 5

Thin Spherical Shell:

I=

2 2 2 MR 2 = ( 4.80 kg ) ( 0.230 m ) = 0.169 kg ⋅ m 2 3 3

2

When different objects of mass M and radius R roll without slipping ( ⇒ a = Ra ) down a ramp, the one with the largest translational acceleration a will have the highest translational speed at the bottom. To determine the translational acceleration for the various objects, consider the force diagram at the right: ΣFx = Ma t = Ia

n a

M f

R

q

a x

Mg

⇒ Mg sinq − f = Ma

q

[1]

⇒ f R = I ( a R ) or f = Ia R 2

[2]

Substitute Equation [2] into [1] to obtain Mg sinq − Ia R 2 = Ma

or

a=

Mg sinq M + I R2

Since M, R, g, and q are the same for all of the objects, we see that the translational acceleration (and hence, the translational speed) increases as the moment of inertia decreases. Thus, the proper rankings from highest to lowest by translational speed will be: solid sphere; solid cylinder; thin spherical shell; and hoop (c)

When an object rolls down the ramp without slipping, the friction force does no work and mechanical energy is conserved. Then, the total kinetic energy gained equals the gravitational potential energy given up: KEr + KEt = −ΔPEg = Mgh and KEr = Mgh − 12 Mv 2 , where h is the vertical drop of the ramp and v is the translational speed at the bottom. Since M, g, and h are the same for all of the objects, the rotational kinetic energy decreases as the translational speed increases. Using this fact, along with the result of part (b), we rank the object’s final rotational kinetic energies, from highest to lowest, as: hoop; thin spherical shell; solid cylinder; and solid sphere

8.45

(a)

Treating the particles on the ends of the rod as point masses, the total moment of inertia of 2 2 the rotating system is I = I rod + I1 + I 2 = mrod 2 12 + m1 ( 2 ) + m2 ( 2 ) . If the mass of 2 the rod can be ignored, this reduces to I = 0 + ( m1 + m2 ) ( 2 ) , and the rotational kinetic energy is KEr =

1 1 2 2 Iw 2 = ⎡⎣( 3.00 kg + 4.00 kg ) ( 0.500 m ) ⎤⎦ ( 2.50 rad s ) = 5.47 J 2 2 continued on next page

56157_08_ch08_p243-287.indd 267

10/12/10 2:31:03 PM

268

Chapter 8

(b)

If the rod has mass mrod = 2.00 kg, the rotational kinetic energy is KEr =

1 2 1⎡ 1 2 2 2 Iw = ⎢ ( 2.00 kg )(1.00 m ) + ( 3.00 kg + 4.00 kg )( 0.500 m ) ⎤⎥ ( 2.50 rad s ) 2 2 ⎣ 12 ⎦

or KEr = 5.99 J . 8.46

Using conservation of mechanical energy,

( KE + KE t

or

r

+ PE g

) = ( KE + KE t

f

r

+ PE g

)

i

1 1 M vt2 + Iw 2 + 0 = 0 + 0 + Mg( Lsinq ) 2 2

Since I = 25 MR 2 for a solid sphere and v t = Rw when rolling without slipping, this becomes 1 1 MR 2w 2 + MR 2w 2 = Mg( Lsinq ) 5 2 and reduces to

(

)

10 9.8 m s 2 ( 6.0 m ) sin 37° 10gLsinq w= = = 36 rad s 2 7R 2 7 ( 0.20 m ) 8.47

(a)

Assuming the disk rolls without slipping, the angular speed of the disk is w = v R, where v is the translational speed of the center of the disk. Also, if the disk does not slip, the friction force between disk and ramp does no work and total mechanical energy is conserved. Hence, (KEt + KEr + PE g ) f = (KEt + KEr + PE g )i , or 1 1 Mv 2 + Iw 2 + 0 = 0 + 0 + mghi 2 2 Since I = MR 2 2 , and hi = L sinq = ( 4.50 m ) sin15.0° , we have 1 ⎛ M R2 ⎞ ⎛ v2 ⎞ 1 Mv 2 + ⎜ = mgL sinq ⎟ 2 ⎝ 2 ⎠ ⎜⎝ R 2 ⎟⎠ 2

v=

and (b)

(a)

)

3

= 3.90 m s

The angular speed of the disk at the bottom is w=

8.48

(

4 9.80 m s 2 ( 4.50 m ) sin15.0°

4gL sinq = 3

v 3.90 m s = = 15.6 rad s R 0.250 m

Assuming the solid sphere starts from rest, and taking y = 0 at the level of the bottom of the incline, the total mechanical energy E = PE g = mgh will be split among i three distinct forms of energy as the sphere rolls down the incline. These are

(

rotational kinetic energy,

1 2

)

Iw 2 ; translational kinetic energy,

1 2

mv 2 ; and

gravitational potential energy, mgy , where y is the current height of the center of mass of the sphere above the level of the bottom of the incline. continued on next page

56157_08_ch08_p243-287.indd 268

10/12/10 2:31:06 PM

269

Rotational Equilibrium and Rotational Dynamics

(b)

The force of static friction, exerted on the sphere by the incline and directed up the incline, exerts a torque about the center of mass, giving the sphere an angular acceleration.

(c)

KEt = 12 Mv 2 and KEr = 12 Iw 2, where v = Rw (since the sphere rolls without slipping), and I = 25 MR 2 for a solid sphere. Therefore,

(

)

2MR 2 5 w 2 Iw 2 2 2 MR 2w 2 2 KEr = = = = 7 KEt + KEr Mv 2 2 + Iw 2 2 M ( Rw )2 + 2MR 2 5 w 2 5 MR 2w 2 + 2 MR 2w 2 8.49

)

Using Wnet = KE f − KEi = 12 Iw 2f − 0, we have 2Wnet = I

wf = 8.50

(

2 ( 5.57 N ) ( 0.800 m ) = 149 rad s 4.00 × 10 −4 kg ⋅ m 2

2F ⋅s = I

The work done on the grindstone is Wnet = F ⋅ s = F ⋅ ( r q ) = ( F ⋅ r )q = t ⋅q Thus, Wnet = ΔKE

becomes t ⋅q = 12 Iw 2f − 12 Iw i2 , or

( 25.0 N ⋅ m )(15.0 rev )⎛⎜⎝

2p rad ⎞ 1 2 2 ⎟⎠ = 0.130 kg ⋅ m w f − 0 1 rev 2

(

)

This yields rad ⎞ ⎛ 1 rev ⎞ ⎛ = 30.2 rev s w f = ⎜ 190 ⎟ ⎝ s ⎠ ⎜⎝ 2p rad ⎟⎠ 8.51

(a)

KEt =

1 2 1 2 mvt = (10.0 kg )(10.0 m s ) = 500 J 2 2

(b)

KEr =

1 2 1⎛ 1 Iw = ⎜ m R 2 2 2⎝ 2

= (c) 8.52

2 ⎞ ⎛ vt ⎞ ⎟⎠ ⎜ 2 ⎟ ⎝ R ⎠

1 1 2 mvt2 = (10.0 kg )(10.0 m s ) = 250 J 4 4

KE total = KEt + KEr = 750 J w

As the bucket drops, it loses gravitational potential energy. The spool gains rotational kinetic energy and the bucket gains translational kinetic energy. Since the string does not slip on the spool, v = rw where r is the radius of the spool. The moment of inertia of the spool is I = 12 Mr 2, where M is the mass of the spool. Conservation of energy gives

( KE + KE t

r

+ PE g

) = ( KE + KE f

t

r

+ PE g

)

i

1 2 1 2 mv + Iw + mgy f = 0 + 0 + mgyi 2 2 or

(

1⎛1 1 2 ⎞ m ( rw ) + ⎜ Mr 2 ⎟ w 2 = mg yi − y f ⎠ 2⎝2 2

I

→

V

m

) continued on next page

56157_08_ch08_p243-287.indd 269

10/12/10 2:31:09 PM

270

Chapter 8

This gives w= 8.53

(a)

(

2mg yi − y f

(m +

1 2

M)r

)=

2

(

)

2 ( 3.00 kg ) 9.80 m s 2 ( 4.00 m ) ⎡⎣3.00 kg +

1 2

(5.00 kg)⎤⎦ ( 0.600 m )2

= 10.9 rad s

The arm consists of a uniform rod of 10.0 m length and the mass of the seats at the lower end is negligible. The center of gravity of this system is then located at the geometric center of the arm, located 5.00 m from the upper end. From the sketch below, the height of the center of gravity above the zero level (chosen to be 10.0 m below the axis) is ycg = 10.0 m − ( 5.00 m ) cosq .

(b)

When q = 45.0°, ycg = 10.0 m − ( 5.00 m ) cos 45.0° and PE g = mgycg gives

(

PE g = ( 365 kg ) 9.80 m s

2

5.00 m

) [10.0 m − (5.00 m ) cos 45.0°]

q

= 2.31 × 10 J

cg

4

(c)

10.0 m

In the vertical orientation, q = 0° and cosq = 1, giving ycg = 10.0 m − 5.00 m = 5.00 m. Then,

(

mg

ycg

)

PE g = mgycg = ( 365 kg ) 9.80 m s2 ( 5.00 m ) = 1.79 × 10 4 J (d)

Using conservation of mechanical energy as the arm starts from rest in the 45° orientation and rotates about the upper end to the vertical orientation gives

( )

1 I endw 2f + mg ycg 2

f

( )

= 0 + mg ycg

or

i

wf =

( ) ( )

2mg ⎡ ycg − ycg i ⎣⎢ I end

f

⎤ ⎦⎥

[1]

For a long, thin rod, I end = mL2 3, and Equation [1] becomes wf =

( ) ( )

2 m g ⎡ ycg − ycg i ⎣⎢ m L2 3

f

⎤ ⎦⎥

=

( ) ( )

6g ⎡ ycg − ycg i ⎣⎢ L2

f

⎤ ⎦⎥

Then, from v = rw with r = L , the translational speed of the seats at the lower end of the rod is

( ) ( )

6g ⎡ ycg − ycg ⎢⎣ i v= L L2

(

= 6 9.80 m s 2 8.54

(

)

f

⎤ ⎥⎦

) [(10.0 − 5.00 cos 45.0°) m − 5.00 m ] =

9.28 m s

(a)

L = Iw = MR 2 w = ( 2.40 kg ) ( 0.180 m ) ( 35.0 rad s ) = 2.72 kg ⋅ m 2 s

(b)

1 2 ⎛1 ⎞ L = Iw = ⎜ MR 2 ⎟ w = ( 2.40 kg ) ( 0.180 m ) ( 35.0 rad s ) = 1.36 kg ⋅ m 2 s ⎝2 ⎠ 2

(c)

2 2 ⎛2 ⎞ L = Iw = ⎜ MR 2 ⎟ w = ( 2.40 kg ) ( 0.180 m ) ( 35.0 rad s ) = 1.09 kg ⋅ m 2 s ⎝5 ⎠ 5

2

continued on next page

56157_08_ch08_p243-287.indd 270

10/12/10 2:31:13 PM

271

Rotational Equilibrium and Rotational Dynamics

8.55

(d)

2 2 ⎛2 ⎞ L = Iw = ⎜ MR 2 ⎟ w = ( 2.40 kg ) ( 0.180 m ) ( 35.0 rad s ) = 1.81 kg ⋅ m 2 s ⎝3 ⎠ 3

(a)

The rotational speed of Earth is w E =

2p rad ⎛ 1d ⎜ 1 d ⎝ 8.64 × 10 4

⎞ −5 ⎟⎠ = 7.27 × 10 rad s s

⎛2 ⎞ Lspin = I spherew E = ⎜ M E RE2 ⎟ w E ⎝5 ⎠ 2 2 = ⎡⎢ 5.98 × 10 24 kg 6.38 × 10 6 m ⎤⎥ 7.27 × 10 −5 rad s = 7.08 × 10 33 J ⋅ s ⎣5 ⎦

(

(b)

)(

) (

(

(

(a)

)

2 w orbit . For Earth’s orbital motion, w orbit = 2p rad y and Lorbit = I pointw orbit = M E Rorbit Using data from Table 7.3, we find

)(

Lorbit = 5.98 × 10 24 kg 1.496 × 1011 m 8.56

)

) ( 2p 2

1y ⎛ rad y ) ⎜ ⎝ 3.156 × 10 7

⎞ 40 ⎟ = 2.66 × 10 J ⋅ s s⎠

Yes , the bullet has angular momentum about an axis through the hinges of the door before the collision. Consider the sketch at the right, showing the bullet the instant before it hits the door. The physical situation is identical to that of a point mass mB moving in a circular path of radius r with tangential speed vt = vi. For that situation the angular momentum is

Hinge r

⎛v ⎞ Li = I iw i = mB r 2 ⎜ i ⎟ = mB rvi ⎝ r⎠

(

)

and this is also the angular momentum of the bullet about the axis through the hinge at the instant just before impact.

mB

vi

(b)

No, mechanical energy is not conserved in the collision. The bullet embeds itself in the door with the two moving as a unit after impact. This is a perfectly inelastic collision in which a significant amount of mechanical energy is converted to other forms, notably thermal energy.

(c)

Apply conservation of angular momentum with Li = mB r vi as discussed in part (a). After impact, L f = I f w f = ( I door + I bullet ) w f = 13 M door L2 + mB r 2 w f where L = 1.00 m = the width of the door and r = L − 10.0 cm = 0.900 m. Then,

(

L f = Li

mB rvi

(

)

1 M L2 + mB r 2 3 door

=

( 0.005 00 kg)( 0.900 m )(1.00 × 103 m s ) 1 (18.0 kg)(1.00 m )2 + ( 0.005 kg)( 0.900 m )2 3

w f = 0.749 rad s

yielding (d)

⇒ wf =

)

The kinetic energy of the door-bullet system immediately after impact is KE f = or

1 1 1 2 2 2 I f w 2f = ⎡⎢ (18.0 kg )(1.00 m ) + ( 0.005 00 kg )( 0.900 m ) ⎤⎥ ( 0.749 rad s ) 2 2 ⎣3 ⎦

KE f = 1.68 J

The kinetic energy (of the bullet) just before impact was KEi =

56157_08_ch08_p243-287.indd 271

(

1 1 mB vi2 = ( 0.005 00 kg ) 1.00 × 10 3 m s 2 2

)

2

= 2.50 × 10 3 J ≈ 1 490 ⋅ KE f

10/12/10 2:31:16 PM

272

8.57

Chapter 8

Each mass moves in a circular path of radius r = 0.500 m s about the center of the connecting rod. Their angular speed is w=

v 5.00 m s = = 10.0 m s r 0.500 m

Neglecting the moment of inertia of the light connecting rod, the angular momentum of this rotating system is L = Iw = ⎡⎣ m1r 2 + m2 r 2 ⎤⎦ w = ( 4.00 kg + 3.00 kg ) ( 0.500 m ) (10.0 rad s ) = 17.5 J ⋅ s 2

8.58

( ) ( ) )v r = ( mr )v r , giving

Using conservation of angular momentum, Laphelion = Lperihelion. Thus, mra2 w a = mrp2 w p . Since

(

w = vt r at both aphelion and perihelion, this is equivalent to mr

2 a

a

a

2 p

p

p

⎛ rp ⎞ ⎛ 0.59 AU ⎞ va = ⎜ ⎟ v p = ⎜ (54 km s ) = 0.91 km s ⎝ 35 AU ⎟⎠ ⎝ ra ⎠ 8.59

(a)

I P = m ( d + d 3) + m ( d 3) + m ( 2d 3) 2

2

2

7 ⎛ 16 1 4 ⎞ 21 = md 2 ⎜ + + ⎟ = md 2 = md 2 ⎝ 9 9 9⎠ 9 3

1

m

2

m

d

(b)

(e)

m

d

Σt )P = mgd counterclockwise

From Σt )P = I Pa , we have a =

(d)

3

⎛4 ⎞ ⎛1 ⎞ ⎛2 ⎞ Σt )P = mg ⎜ d ⎟ + mg ⎜ d ⎟ − mg ⎜ d ⎟ = + mgd ⎝3 ⎠ ⎝3 ⎠ ⎝3 ⎠ or

(c)

2d 3

P

Σt )P +mgd 3g 3g = =+ = counterclockwise 2 7md 3 IP 7d 7d

2g ⎛ 2 ⎞ ⎛ 3g ⎞ at = ra = ⎜ d ⎟ ⎜ ⎟ = upward ⎝ 3 ⎠ ⎝ 7d ⎠ 7 Maximum kinetic energy occurs when the system’s gravitational potential energy is a minimum (i.e., when the center of gravity is at its lowest point). This occurs when the rod is vertical, where the y-coordinate (taking y = 0 at the level of point P) of the center of gravity is ycg =

Σ ( mi gyi ) Σmi g

d ⎛ 4 ⎞ ⎛ d⎞ ⎛ 2 ⎞ = m ⎜ − d⎟ + m ⎜ − ⎟ + m ⎜ + d⎟ = − ⎝ 3 ⎠ ⎝ 3⎠ ⎝ 3 ⎠ 3

Applying conservation of energy from when the rod is released from rest in the horizontal position until it reaches the vertical position gives

continued on next page

56157_08_ch08_p243-287.indd 272

10/12/10 2:31:19 PM

Rotational Equilibrium and Rotational Dynamics

(

)

(

KE f = KEi + PEi − PE f = 0 + ( mtotal g ) ycg,i − ycg, f

8.60

or

KE f = 3mg ⎡⎣ 0 − ( − d 3) ⎤⎦ = mgd

(f)

KE f =

1 2 I Pw max 2

(g)

⎛7 ⎞ Lmax = I Pw max = ⎜ md 2 ⎟ ⎝3 ⎠

(h)

( v2 )max = r2w max = d3

(a)

w max =

so

2 KE f IP

)

2 ( mgd ) = 7md 2 3

=

273

6g 7d

6g ⎛ 49 ⎞ 6g ⎛7 ⎞ = ⎜ m2d 4 ⎟ = m 2 ⎜ d 3 ⎟ 2g = m 14gd 3 3 ⎝ 9 ⎠ 7d ⎝3 ⎠ 7d

⎛ d 2 ⎞ 6g 6g ⎛ d ⎞ 2g = ⎜ ⎟ = ⎜ ⎟ = ⎝ 3⎠ 7 7d ⎝ 9 ⎠ 7d

2gd 21

The table turns counterclockwise, opposite to the way the woman walks. Its angular momentum cancels that of the woman so the total angular momentum maintains a constant value of Ltotal = Lwoman + Ltable = 0. Since the final angular momentum is Ltotal = I w w w + I t w t = 0, we have ⎛I ⎞ ⎛ m r2 ⎞ ⎛ v ⎞ ⎛ m r⎞ w t = − ⎜ w ⎟ w w = − ⎜ w ⎟ ⎜ w ⎟ = − ⎜ w ⎟ vw ⎝ It ⎠ ⎝ It ⎠ ⎝ r ⎠ ⎝ It ⎠ or (taking counterclockwise as positive), ⎡ ( 60.0 kg ) ( 2.00 m ) ⎤ wt = − ⎢ ⎥ ( −1.50 m s ) = + 0.360 rad s 500 kg ⋅ m 2 ⎣ ⎦ Hence,

(b)

1 2 1 mvw + I t w t2 2 2 1 1 2 2 = ( 60.0 kg )(1.50 m s ) + 500 kg ⋅ m 2 ( 0.360 rad s ) = 99.9 J 2 2

Wnet = ΔKE = KE f − 0 = Wnet

8.61

w table = 0.360 rad s counterclockwise

(

)

The moment of inertia of the cylinder before the putty arrives is Ii =

1 1 2 M R 2 = (10.0 kg )(1.00 m ) = 5.00 kg ⋅ m 2 2 2

After the putty sticks to the cylinder, the moment of inertia is I f = I i + mr 2 = 5.00 kg ⋅ m 2 + ( 0.250 kg )( 0.900 m ) = 5.20 kg ⋅ m 2 2

Conservation of angular momentum gives I f w f = I i w i , or 8.62

⎛I wf = ⎜ i ⎝ If

⎞ ⎛ 5.00 kg ⋅ m 2 ⎞ ( 7.00 rad s ) = 6.73 rad s ⎟ wi = ⎜ ⎝ 5.20 kg ⋅ m 2 ⎟⎠ ⎠

The total moment of inertia of the system is

(

)

I total = I masses + I student = 2 mr 2 + 3.0 kg ⋅ m 2 plus stool

continued on next page

56157_08_ch08_p243-287.indd 273

10/12/10 2:31:22 PM

274

Chapter 8

2 Initially, r = 1.0 m, and I i = 2 ⎡⎣( 3.0 kg )(1.0 m ) ⎤⎦ + 3.0 kg ⋅ m 2 = 9.0 kg ⋅ m 2

Afterward, r = 0.30 m, so 2 I f = 2 ⎡⎣( 3.0 kg )( 0.30 m ) ⎤⎦ + 3.0 kg ⋅ m 2 = 3.5 kg ⋅ m 2

(a)

From conservation of angular momentum, I f w f = I i w i , or ⎛I wf = ⎜ i ⎝ If

(b)

8.63

⎞ ⎛ 9.0 kg ⋅ m 2 ⎞ = w ( 0.75 rad s ) = 1.9 rad s ⎟ i ⎜ ⎝ 3.5 kg ⋅ m 2 ⎟⎠ ⎠

(

)

KEi =

1 1 2 I i w i2 = 9.0 kg ⋅ m 2 ( 0.75 rad s ) = 2.5 J 2 2

KE f =

1 1 2 I f w 2f = 3.5 kg ⋅ m 2 (1.9 rad s ) = 6.3 J 2 2

(

)

The initial angular velocity of the puck is wi =

( vt )i ri

=

0.800 m s rad = 2.00 0.400 m s

Since the tension in the string does not exert a torque about the axis of revolution, the angular momentum of the puck is conserved, or I f w f = I i w i . Thus, ⎛I wf = ⎜ i ⎝ If

2

2 ⎛ mri2 ⎞ ⎛ ri ⎞ ⎞ ⎛ 0.400 m ⎞ = = w = w w ⎟ ( 2.00 rad s ) = 5.12 rad s ⎜ ⎟ i ⎜⎝ ⎟ i ⎜ 2⎟ i 0.250 m ⎠ ⎝ mrf ⎠ ⎝ rf ⎠ ⎠

The net work done on the puck is Wnet = KE f − KEi = or

Wnet =

1 1 1 m I f w 2f − I i w i2 = ⎡⎣ mrf2 w 2f − mri2 w i2 ⎤⎦ = ⎡⎣ rf2w 2f − ri2w i2 ⎤⎦ 2 2 2 2

(

( 0.120 kg) ⎡( 0.250 m )2 2

⎣

(5.12

)

(

)

2 2 2 rad s ) − ( 0.400 m ) ( 2.00 rad s ) ⎤⎦

This yields Wnet = 5.99 × 10 −2 J . 8.64

With all crew members on the rim of the station, the apparent acceleration experienced is the centripetal acceleration, ac = rw 2 = g. Thus, the initial angular velocity of the station is w i = g r . The initial moment of inertia of the rotating system is I i = I crew + I station = 150mr 2 + I station After most of the crew move to the rotation axis, leaving only the managers on the rim, the moment of inertia is I f = I managers + I station = 50mr 2 + I station

(

)

Thus, conservation of angular momentum I f w f = I iw i gives the angular velocity during the union meeting as ⎛I ⎞ ⎛ 150 ( 65.0 kg ) (100 m )2 + 5.00 × 108 kg ⋅ m 2 ⎞ w f = ⎜ i ⎟ wi = ⎜ 2 8 2 ⎟ ⎝ 50 ( 65.0 kg ) (100 m ) + 5.00 × 10 kg ⋅ m ⎠ ⎝ If ⎠

g g = 1.12 r r

continued on next page

56157_08_ch08_p243-287.indd 274

10/12/10 2:31:25 PM

275

Rotational Equilibrium and Rotational Dynamics

The centripetal acceleration experienced by the managers still on the rim is ac = rw 2f = r (1.12 )

2

8.65

(a)

g 2 = (1.12 ) 9.80 m s 2 = 12.3 m s 2 r

(

)

From conservation of angular momentum, I f w f = I i w i , so

⎛I wf = ⎜ i ⎝ If

KE f =

⎛ I ⎞ ⎛ I ⎞ 1 ⎛ I ⎞ 1 1 I f w 2f = ( I1 + I 2 ) ⎜ 1 ⎟ w o2 = ⎜ 1 ⎟ ⎡⎢ I1 w o2 ⎤⎥ = ⎜ 1 ⎟ KEi 2 2 ⎦ ⎝ I1 + I 2 ⎠ ⎝ I1 + I 2 ⎠ ⎝ I1 + I 2 ⎠ ⎣ 2

⎞ ⎛ I1 ⎞ wo ⎟wi = ⎜ ⎝ I1 + I 2 ⎟⎠ ⎠ 2

(b)

or KE f KEi

=

I1 0, a y will be positive (upward), and the balloon rises . (d)

If the balloon and load are in equilibrium, ΣFy = ( B − wb − wHe ) − wload = 0 and wload = ( B − wb − wHe ) = 1.33 × 10 3 N. Thus, the mass of the load is mload =

9.35

wload 1.33 × 10 3 N = = 136 kg g 9.80 m s2

(e)

If mload < 136 kg, the net force acting on the balloon+load system is upward and the balloon and its load will accelerate upward .

(f)

As the balloon rises, decreasing atmospheric density decreases the buoyancy force. At some height the balloon will come to equilibrium and go no higher.

(a)

⎛ 4p r 3 ⎞ ⎛ 4p ⎞ B = rair gVballoon = rair g ⎜ = 1.29 kg m 3 9.80 m s 2 ⎜ ( 3.00 m )3 ⎟ ⎝ 3 ⎟⎠ ⎝ 3 ⎠

(

)(

)

= 1.43 × 10 3 N = 1.43 kN (b)

(

ΣFy = B − wtotal = 1.43 × 10 3 N − (15.0 kg ) 9.80 m s 2

)

= +1.28 × 10 3 N = 1.28 kN upward (c)

56157_09_ch09_p288-321.indd 303

The balloon expands as it rises because the external pressure (atmospheric pressure) decreases with increasing altitude.

10/12/10 2:40:08 PM

304

Chapter 9

9.36

(a)

Taking upward as positive, ΣFy = B − mg = ma y , or ma y = rw gV − mg .

(b)

Since m = rV, we have r V a y = rw g V − r V g ⎛r ⎞ a y = ⎜ w − 1⎟ g ⎝ r ⎠

or

(c)

⎛ 1.00 × 10 3 m kg 3 ⎞ ay = ⎜ − 1⎟ 9.80 m s 2 = −0.467 m s 2 = 0.467 m s 2 downward 3 ⎝ 1 050 m kg ⎠

(d)

From Δy = v0 y t + a y t 2 2, with v0 y = 0, we ﬁnd

(

t= 9.37

(a)

2 ( Δy ) = ay

)

2 ( −8.00 m ) = 5.85 s −0.467 m s2

⎡ ⎛ 4p 3 ⎞ ⎤ Btotal = 600 ⋅ Bsingle = 600 ( rair gVballoon ) = 600 ⎢ rair g ⎜ r ⎝ 3 ⎟⎠ ⎥⎦ balloon ⎣ 4p ⎡ = 600 ⎢ 1.29 kg m 3 9.80 m s 2 ( 0.50 m )3 ⎤⎥ = 4.0 × 10 3 N = 4.0 kN 3 ⎣ ⎦

(

(b) (c)

9.38

)(

)

(

)

ΣFy = Btotal − mtotal g = 4.0 × 10 3 N − 600 ( 0.30 kg ) 9.8 m s 2 = 2.2 × 10 3 N = 2.2 kN Atmospheric pressure at this high altitude is much lower than at Earth’s surface , so the balloons expanded and eventually burst.

The actual weight of the object is Fg, actual = mobject g = 5.00 N, and its mass is mobject = 5.00 N g. When fully submerged, the upward buoyant force (equal to the weight of the displaced water) and the upward force exerted on the object by the scale Fg, apparent = 3.50 N together support the actual weight of the object. That is,

(

ΣFy = 0

⇒

)

B + Fg, apparent − Fg, actual = 0

B = Fg, actual − Fg, apparent = 5.00 N − 3.50 N = 1.50 N

and

Thus, B = rwater gVobject gives Vobject = B ( rwater g ) and the density of the object is robject = 9.39

(a)

mobject Vobject

⎛ 5.00 N ⎞ ⎛ rwater g ⎞ 3 3 =⎜ ⎜ ⎟ = 3.33rwater = 3.33 × 10 kg m ⎝ g ⎟⎠ ⎝ 1.50 N ⎠

The wooden block sinks until the buoyant force (weight of the displaced water) equals the weight of the block. That is, when equilibrium is reached,

A ⫽ s2

B = rwater g ⎡⎣( s − h ) s 2 ⎤⎦ = rwood g ⋅ s 3, giving

s

s ⫽ 20.0 cm h

⎛r ⎞ s − h = ⎜ wood ⎟ ⋅ s ⎝ rwater ⎠ or

⎛ ⎛ r ⎞ 650 kg m 3 ⎞ h = s ⋅ ⎜ 1 − wood ⎟ = ( 20.0 cm ) ⎜ 1 − = 7.00 cm 3 rwater ⎠ ⎝ ⎝ 1000 kg m ⎟⎠

continued on next page

56157_09_ch09_p288-321.indd 304

10/12/10 2:40:11 PM

Solids and Fluids

(b)

305

When the upper surface of the block is level with the water surface, the buoyant force is B = rwater gVblock = rwater g ⋅ s 3 This must equal the weight of the block plus the weight of the added lead, or mPb g + mblock g = B, and B − mblock = rwater Vblock − rwoodVblock = ( rwater − rwood ) ⋅ s 3 g

mPb =

mPb = (1000 kg m 3 − 650 kg m 3 )( 0.200 m ) = 2.80 kg 3

giving 9.40

At equilibrium, ΣFy = B − Fspring − mg = 0, so the spring force is Fspring = B − mg = ⎡⎣( rwater Vblock ) − m ⎤⎦ g, where Vblock =

m rwood

=

5.00 kg = 7.69 × 10 −3 m 3 650 kg m 3

(

)(

)

(

)

Thus, Fspring = ⎡⎣ 10 3 kg m 3 7.69 × 10 −3 m 3 − 5.00 kg ⎤⎦ 9.80 m s 2 = 26.4 N. The elongation of the spring is then Δx = 9.41

(a)

Fspring k

26.4 N = 0.165 m = 16.5 cm 160 N m

The buoyant force is the difference between the weight in air and the apparent weight when immersed in the alcohol, or B = 300 N − 200 N = 100 N. But, from Archimedes’ principle, this is also the weight of the displaced alcohol, so B = ( ralcohol V ) g. Since the sample is fully submerged, the volume of the displaced alcohol is the same as the volume of the sample. This volume is V=

(b)

=

B ralcohol g

=

(

100 N = 1.46 × 10 −2 m 3 700 kg m 3 9.80 m s 2

)(

)

The mass of the sample is m=

weight in air 300 N = = 30.6 kg 9.80 m s 2 g

and its density is r= 9.42

m 30.6 kg = = 2.10 × 10 3 kg m 3 V 1.46 × 10 −2 m 3

The difference between the weight in air and the apparent weight when immersed is the buoyant force exerted on the object by the ﬂuid. (a)

The mass of the object is m=

weight in air 300 N = = 30.6 kg 9.80 m s 2 g continued on next page

56157_09_ch09_p288-321.indd 305

10/12/10 2:40:15 PM

306

Chapter 9

The buoyant force when immersed in water is the weight of a volume of water equal to the volume of the object, or Bwater = ( rwater V ) g. Thus, the volume of the object is V=

Bwater 300 N − 265 N = = 3.57 × 10 −3 m 3 3 3 2 rwater g 10 kg m 9.80 m s

(

)(

)

and its density is robject = (b)

The buoyant force when immersed in oil is equal to the weight of a volume V = 3.57 × 10 −3 m 3 of oil. Hence, Boil = ( roil V ) g, or the density of the oil is roil =

9.43

m 30.6 kg = = 8.57 × 10 3 kg m 3 V 3.57 × 10 −3 m 3

Boil 300 N − 275 N = = 715 kg m 3 Vg 3.57 × 10 −3 m 3 9.80 m s 2

(

)(

)

The volume of the iron block is V=

miron 2 .00 kg = = 2 .54 × 10 −4 m 3 7.86 × 10 3 kg m 3 riron

and the buoyant force exerted on the iron by the oil is

(

)(

)(

)

B = ( roil V ) g = 916 kg m 3 2 .54 × 10 −4 m 3 9.80 m s 2 = 2 .28 N Applying ΣFy = 0 to the iron block gives the support force exerted by the upper scale (and hence the reading on that scale) as Fupper = miron g − B = 19.6 N − 2.28 N = 17.3 N From Newton’s third law, the iron exerts force B downward on the oil (and hence the beaker). Applying ΣFy = 0 to the system consisting of the beaker and the oil gives Flower − B − ( moil + mbeaker ) g = 0 The support force exerted by the lower scale (and the lower scale reading) is then

(

)

Flower = B + ( moil + mbeaker ) g = 2.28 N + ⎡⎣( 2.00 + 1.00 ) kg ⎤⎦ 9.80 m s 2 = 31.7 N 9.44

(a)

The cross-sectional area of the hose is A = πr 2 = p d 2 4 = p ( 2.74 cm ) 4, and the volume ﬂow rate (volume per unit time) is Av = 25.0 L 1.50 min. Thus, 2

v=

25.0 L 1.50 min ⎛ 25.0 L =⎜ A ⎝ 1.50 min

⎤ ⎛ 1 min ⎞ ⎛ 10 3 cm 3 ⎞ ⎞⎡ 4 ⎢ ⎥ 2 ⎟⎠ ⎟⎜ ⎟ 2 ⎜ ⎢⎣ p ⋅ ( 2.74 ) cm ⎥⎦ ⎝ 60 s ⎠ ⎝ 1 L ⎠

⎛ 1m ⎞ = 0.471 m s = ( 47.1 cm s ) ⎜ 2 ⎝ 10 cm ⎟⎠ 2

(b)

1 A2 ⎛ p d 22 ⎞ ⎛ 4 ⎞ ⎛ d 2 ⎞ ⎛ 1⎞ =⎜ =⎜ ⎟ =⎜ ⎟ = 2 ⎟ ⎜ ⎟ ⎝ 3⎠ 9 A1 ⎝ 4 ⎠ ⎝ p d1 ⎠ ⎝ d1 ⎠ 2

or

A2 =

A1 9

continued on next page

56157_09_ch09_p288-321.indd 306

10/12/10 2:40:17 PM

Solids and Fluids

307

Then, from the equation of continuity, A2 v2 = A1 v1 and we ﬁnd ⎛A ⎞ v2 = ⎜ 1 ⎟ v1 = 9 ( 0.471 m s ) = 4.24 m s ⎝ A2 ⎠ 9.45

(a)

The volume ﬂow rate is Av, and the mass ﬂow rate is

(

)(

)

r Av = 1.0 g cm 3 2.0 cm 2 ( 40 cm s ) = 80 g s (b)

From the equation of continuity, the speed in the capillaries is ⎛ Aaorta ⎞ ⎛ 2 .0 cm 2 ⎞ vcapiliaries = ⎜ = 40 cm s ) v ⎟ aorta ⎜ 3 2 ( ⎝ 3.0 × 10 cm ⎟⎠ ⎝ Acapillaries ⎠ or

9.46

(a)

vcapiliaries = 2.7 × 10 −2 cm s = 0.27 mm s

From the equation of continuity, the ﬂow speed in the second section of the pipe is ⎛A ⎞ ⎛ 10.0 cm 2 ⎞ v2 = ⎜ 1 ⎟ v1 = ⎜ 2.75 m s ) = 11.0 m s 2 ( ⎝ A2 ⎠ ⎝ 2.50 cm ⎟⎠

(b)

Using Bernoulli’s equation and choosing y = 0 along the centerline of the pipe gives P2 = P1 +

(

1 r v12 − v22 2

= 1.20 × 10 5 Pa + or 9.47

) (

)

1 2 2 1.65 × 10 3 kg m 3 ⎡( 2.75 m s ) − (11.0 m s ) ⎤ ⎣ ⎦ 2

P2 = 2.64 × 10 4 Pa

From Bernoulli’s equation, choosing y = 0 at the level of the syringe and needle, P2 + 12 r v22 = P1 + 12 r v12 , so the ﬂow speed in the needle is v2 = v12 +

2 ( P1 − P2 ) r

In this situation, P1 − P2 = P1 − Patm = ( P1 )gauge =

F 2 .00 N = = 8.00 × 10 4 Pa A1 2 .50 × 10 −5 m 2

Thus, assuming v1 ≈ 0, v2 = 0 + 9.48

)

1.00 × 10 kg m 3 3

= 12.6 m s

We apply Bernoulli’s equation, ignoring the very small change in vertical position, to obtain 2 P1 − P2 = 12 r v22 − v12 = 12 r ⎡( 2 v1 ) − v12 ⎤ = 23 r v12 , or ⎣ ⎦

(

ΔP =

56157_09_ch09_p288-321.indd 307

(

2 8.00 × 10 4 Pa

(

)

)(

3 1.29 kg m 3 15 × 10 −2 m s 2

)

2

= 4.4 × 10 −2 Pa

10/12/10 2:40:20 PM

308

Chapter 9

9.49

(a)

Assuming the airplane is in level ﬂight, the net lift (the difference in the upward and downward forces exerted on the wings by the air ﬂowing over them) must equal the weight of the plane, or (Plower − Pupper )Awings = mg. This yields surface

surface

Plower − Pupper = surface

(b)

surface

)

1 1 2 2 r air vlower = Pupper + r air vupper 2 2

2 vupper = vlower +

so

9.50

)(

Neglecting the small difference in altitude between the upper and lower surfaces of the wings, and applying Bernoulli’s equation, yields Plower +

(c)

(

8.66 × 10 4 kg 9.80 m s 2 mg = = 9.43 × 10 3 Pa Awings 90.0 m 2

(

2 Plower − Pupper rair

) = ( 225 m s)

2

+

(

2 9.43 × 10 3 Pa 1.29 kg m

3

) = 255 m s

The density of air decreases with increasing height , resulting in a smaller pressure 2 2 difference, ΔP = 12 r air (vupper − vlower ). Beyond the maximum operational altitude, the pressure difference can no longer support the aircraft.

For level ﬂight, the net lift (difference between the upward and downward forces exerted on the wing surfaces by air ﬂowing over them) must equal the weight of the aircraft, or (Plower − Pupper )Awings = Mg. This gives the air pressure at the upper surface surface

surface

as Pupper = Plower − Mg Awings . surface

9.51

(a)

surface

Since the pistol is ﬁred horizontally, the emerging water stream has initial velocity components of (v0 x = vnozzle , v0 y = 0) . Then, Δy = v0 y t + a y t 2 2, with a y = −g, gives the time of ﬂight as t=

(b)

2 ( Δy ) = ay

With ax = 0 and v0 x = vnozzle , the horizontal range of the emergent stream is Δx = vnozzle t, where t is the time of ﬂight from above. Thus, the speed of the water emerging from the nozzle must be vnozzle =

(c)

2 ( −1.50 m ) = 0.553 s −9.80 m s2

Δx 8.00 m = = 14.5 m s t 0.553 s

From the equation of continuity, A1 v1 = A2 v2 , the speed of the water in the larger cylinder is v1 = ( A2 A1 ) v2 = ( A2 A1 ) vnozzle , or 2

2 ⎛ p r2 ⎞ ⎛r ⎞ ⎛ 1.00 mm ⎞ v1 = ⎜ 22 ⎟ vnozzle = ⎜ 2 ⎟ vnozzle = ⎜ (14.5 m s ) = 0.145 m s ⎟ ⎝ 10.0 mm ⎠ ⎝ p r1 ⎠ ⎝ r1 ⎠

(d)

The pressure at the nozzle is atmospheric pressure, or P2 = 1.013 × 10 5 Pa .

(e)

With the two cylinders horizontal, y1 ≈ y2 and gravity terms from Bernoulli’s equation can be neglected , leaving P1 + rwater v12 2 = P2 + r water v22 2, so the needed pressure in the larger cylinder is

continued on next page

56157_09_ch09_p288-321.indd 308

10/12/10 2:40:23 PM

Solids and Fluids

P1 = P2 +

r water 2 v2 − v12 2

(

= 1.013 × 10 5 Pa + or (f)

)

1.00 × 10 3 kg m 3 ⎡ 2 2 14.5 m s ) − ( 0.145 m s ) ⎤ ⎣( ⎦ 2

P1 = 2.06 × 10 5 Pa

To create an overpressure of ΔP = 2.06 × 10 5 Pa − 1.013 × 10 5 Pa = 1.05 × 10 5 Pa in the larger cylinder, the force that must be exerted on the piston is

(

) (

) (

F1 = ( ΔP ) A1 = ( ΔP ) p r12 = 1.05 × 10 5 Pa p 1.00 × 10 −2 m 9.52

(a)

309

)

2

= 33.0 N P2

From Bernoulli’s equation, P1 +

or

v2 r water v12 r + r water gy1 = P2 + water 2 + r water gy2 2 2

P1

⌬y ⫽ y2 ⫺ y1

⎡ P − P2 ⎤ v22 − v12 = 2 ⎢ 1 − g ( y2 − y1 ) ⎥ ⎣ r water ⎦

and using the given data values, we obtain ⎡ 1.75 × 10 5 Pa − 1.20 × 10 5 Pa ⎤ v22 − v12 = 2 ⎢ − 9.80 m s 2 ( 2.50 m ) ⎥ 3 3 1.00 × 10 kg m ⎣ ⎦

(

and

)

v22 − v12 = 61.0 m 2 s2

[1]

From the equation of continuity, 2

⎛A ⎞ ⎛ p r2 ⎞ ⎛r ⎞ ⎛ 3.00 cm ⎞ v2 = ⎜ 1 ⎟ v1 = ⎜ 12 ⎟ v1 = ⎜ 1 ⎟ v1 = ⎜ v ⎝ 1.50 cm ⎟⎠ 1 ⎝ A2 ⎠ ⎝ p r2 ⎠ ⎝ r2 ⎠ or

2

v2 = 4v1

[2]

Substituting Equation [2] into [1] gives (16 − 1) v12 = 61.0 m 2 s2, or v1 =

61.0 m 2 s2 = 2.02 m s 15

(b)

Equation [2] above now yields v2 = 4 ( 2.02 m s ) = 8.08 m s .

(c)

The volume ﬂow rate through the pipe is flow rate = A1 v1 = A2 v2 Looking at the lower point:

(

)

(

flow rate = p r12 v1 = p 3.00 × 10 −2 m 9.53

) ( 2.02 m s) = 5.71× 10 2

−3

m3 s

First, consider the path from the viewpoint of projectile motion to ﬁnd the speed at which the water emerges from the tank. From Δy = v0 y t + 12 a y t 2 with v0 y = 0, Δy = −1.00 m, and a y = −g, we ﬁnd the time of ﬂight as t=

2 ( Δy ) = ay

2.00 m g continued on next page

56157_09_ch09_p288-321.indd 309

10/12/10 2:40:26 PM

310

Chapter 9

From the horizontal motion, the speed of the water coming out of the hole is v2 = v0 x

( 0.600 m )2 g

Δx g = = ( 0.600 m ) = t 2.00 m

=

2.00 m

(1.80 × 10

−1

)

m g

We now use Bernoulli’s equation, with point 1 at the top of the tank and point 2 at the level of the hole. With P1 = P2 = Patm and v1 ≈ 0, this gives r g y1 = 12 r v22 + r g y2 , or

(

)

1.80 × 10 −1 m g v22 = 9.00 × 10 −2 m = 9.00 cm h = y1 − y2 = = 2g 2g 9.54

(a)

Apply Bernoulli’s equation with point 1 at the open top of the tank and point 2 at the opening of the hole. Then, P1 = P2 = Patm and we assume v1 ≈ 0. This gives 2 1 2 r v2 + r g y2 = r g y1 , or 2 g ( y1 − y2 ) =

v2 = (b)

(

)

2 9.80 m s 2 (16.0 m ) = 17.7 m s

The area of the hole is found from A2 =

flow rate 2 .50 × 10 −3 m 3 min ⎛ 1 min ⎞ −6 2 = ⎜⎝ ⎟ = 2 .35 × 10 m v2 17.7 m s 60 s ⎠

But, A2 = p d 22 4 and the diameter of the hole must be d2 = 9.55

(

4 2.35 × 10 −6 m 2

4 A2 = p

p

) = 1.73 × 10

−3

m = 1.73 mm

First, determine the ﬂow speed inside the larger section from v1 =

flow rate 1.80 × 10 −4 m 3 s = = 0.367 m s 2 A1 p 2.50 × 10 −2 m 4

(

)

The absolute pressure inside the large section on the left is P1 = Patm + r gh1, where h1 is the height of the water in the leftmost standpipe. The absolute pressure in the constriction is P2 = Patm + r gh2 , so P1 − P2 = r g( h1 − h2 ) = r g ( 5.00 cm ) The ﬂow speed inside the constriction is found from Bernoulli’s equation with y1 = y2 (since the pipe is horizontal). This gives v22 = v12 + or

v2 =

2 ( P1 − P2 ) = v12 + 2 g ( h1 − h2 ) r

( 0.367

(

)

m s ) + 2 ( 9.80 m s ) 5.00 × 10 −2 m = 1.06 m s 2

The cross-sectional area of the constriction is then A2 =

flow rate 1.80 × 10 −4 m 3 s = = 1.70 × 10 −4 m 2 v2 1.06 m s

and the diameter is d2 =

56157_09_ch09_p288-321.indd 310

4 A2 = p

(

4 1.70 × 10 −4 m 2 p

) = 1.47 × 10

−2

m = 1.47 cm

10/12/10 2:40:30 PM

Solids and Fluids

9.56

(a)

For minimum input pressure so the water will just reach the level of the rim, the gauge pressure at the upper end is zero (i.e., the absolute pressure inside the upper end of the pipe is atmospheric pressure), and the ﬂow rate is zero. Thus, Bernoulli’s equation, P + 12 r v 2 + r gy = P + 12 r v 2 + r gy , becomes

(

)

river

(

)

rim

( Priver )min + 0 = 1 atm + 0 + r g( yrim − yriver ) = 1 atm + r g( yrim − yriver ) or,

( Priver)min = 1.013 × 105 Pa + ⎛⎜⎝ 103

[1]

kg ⎞ ⎛ m⎞ ⎟ ⎜ 9.80 2 ⎟⎠ ( 2096 m − 564 m ) m3 ⎠ ⎝ s

( Priver)min = (1.013 × 105 + 1.50 × 10 7 ) Pa = 1.51× 107 Pa = (b)

311

15.1 MPa

When volume ﬂow rate is ⎛ p d2 ⎞ flow rate = Av= ⎜ v = 4 500 m 3 d ⎟ 4 ⎠ ⎝ the velocity in the pipe is v=

(c)

(

)

3 4 ( flow rate ) 4 4 500 m d ⎛ 1 d ⎞ = 2.95 m s = 2 p d2 p ( 0.150 m ) ⎜⎝ 86 400 s ⎟⎠

We imagine the pressure being applied to stationary water at river level, so Bernoulli’s equation becomes Priver + 0 = ⎡⎣1 atm + r g( yrim − yriver ) ⎤⎦ +

1 2 r vrim 2

or, using Equation [1] from above, Priver = ( Priver )min +

1 1⎛ kg ⎞ ⎛ m⎞ 2 r vrim = ( Priver )min + ⎜ 10 3 ⎟⎠ ⎜⎝ 2.95 ⎟ ⎝ 2 2 m3 s⎠

2

= ( Priver )min + 4.35 kPa The additional pressure required to achieve the desired ﬂow rate is ΔP = 4.35 kPa 9.57

(a)

For the upward ﬂight of a water-drop projectile from geyser vent to fountain-top, v 2y = v02 y + 2 a y ( Δy ), with v y = 0 when Δy = Δymax , gives

(

v0 y = 0 − 2 a y ( Δy )max = −2 −9.80 m s2 (b)

(

)

2 g ytop − yvent =

(

2 9.80 m s 2

) ( 40.0 m ) =

28.0 m s

Between the chamber and the geyser vent, Bernoulli’s equation with vchamber ≈ 0 yields 2 + r g yvent , or ( P + 0 + r g y )chamber = Patm + 12 r vvent

56157_09_ch09_p288-321.indd 311

28.0 m s

Because of the low density of air and the small change in altitude, atmospheric pressure at the fountain top will be considered equal to that at the geyser vent. Bernoulli’s equation, 2 with vtop = 0, then gives 12 r vvent = 0 + r g(ytop − yvent ), or vvent =

(c)

) ( 40.0 m ) =

continued on next page

10/12/10 2:40:33 PM

312

Chapter 9

⎡1 2 ⎤ P − Patm = r ⎢ vvent + g ( yvent − ychamber ) ⎥ ⎣2 ⎦ 2 ⎤ kg ⎞ ⎡ ( 28.0 m s ) ⎛ m⎞ ⎛ ⎢ = ⎜ 10 3 + ⎜ 9.80 2 ⎟ (175 m ) ⎥ = 2.11 × 10 6 Pa 3⎟ ⎝ ⎝ m ⎠⎢ 2 s ⎠ ⎥⎦ ⎣

or 9.58

(a)

Pgauge = P − Patm = 2.11 MPa = 20.8 atmospheres

Since the tube is horizontal, y1 = y2 and the gravity terms in Bernoulli’s equation cancel, leaving P1 +

P1

2

ã

v1

1

1 1 r v12 = P2 + r v22 2 2

P2

ã

v2

A2 A1

or

(

)

2 1.20 × 10 3 Pa 2 ( P1 − P2 ) v −v = = r 7.00 × 10 2 kg m 3 2 2

2 1

and v22 − v12 = 3.43 m 2 s2

[1]

From the continuity equation, A1 v1 = A2 v2 , we ﬁnd 2

⎛A ⎞ ⎛r ⎞ ⎛ 2.40 cm ⎞ v2 = ⎜ 1 ⎟ v1 = ⎜ 1 ⎟ v1 = ⎜ v ⎝ 1.20 cm ⎟⎠ 1 ⎝ A2 ⎠ ⎝ r2 ⎠ 2

or v2 = 4v1

[2]

Substituting Equation [2] into [1] yields 15v12 = 3.43 m 2 s2, and v1 = 0.478 m s. Then, Equation [2] gives v2 = 4 ( 0.478 m s ) = 1.91 m s . (b)

The volume ﬂow rate is

(

)

(

A1 v1 = A2 v2 = p r22 v2 = p 1.20 × 10 −2 m 9.59

) (1.91 m s) = 8.64 × 10 2

−4

m3 s

From ΣFy = T − mg − Fy = 0, the balance reading is found to be T = mg + Fy , where Fy is the vertical component of the surface tension force. Since this is a two-sided surface, the surface tension force is F = g ( 2 L ) and its vertical component is Fy = g ( 2 L )cosf , where f is the contact angle. Thus, T = mg + 2g Lcosf . T = 0.40 N when f = 0° ⇒

mg + 2g L= 0.40 N

[1]

T = 0.39 N when f = 180° ⇒

mg − 2g L= 0.39 N

[2]

Subtracting Equation [2] from [1] gives g =

56157_09_ch09_p288-321.indd 312

0.40 N − 0.39 N 0.40 N − 0.39 N = = 8.3 × 10 −2 N m 4L 4 3.0 × 10 −2 m

(

)

10/12/10 2:40:36 PM

Solids and Fluids

9.60

Because there are two edges (the inside and outside of the ring), we have g =

= 9.61

F F = Ltotal 2 (circumference) F 1.61 × 10 –2 N = = 7.32 × 10 −2 N m 4p r 4p 1.75 × 10 –2 m

(

)

The total vertical component of the surface tension force must equal the weight of the column of ﬂuid, or F cosf = g (2p r) ⋅ cosf = r(p r 2 )h ⋅ g. Thus, g =

9.62

313

(

)(

)(

)(

)

2.1× 10 −2 m 1 080 kg m 3 9.80 m s 2 5.0 × 10 −4 m hr gr = = 5.6 × 10 −2 N m 2 cosf 2 cos 0°

The blood will rise in the capillary until the weight of the ﬂuid column equals the total vertical component of the surface tension force, or until r(p r 2 )h ⋅ g = F cosf = g (2p r) ⋅ cosf This gives h=

9.63

)(

)

(

)

1.79 × 10 −3 N ⋅ s m 2 ⎡⎣( 0.800 m ) (1.20 m ) ⎤⎦ ( 0.50 m s ) h Av = = 8.6 N d 0.10 × 10 −3 m

(

)

1 500 × 10 −3 N ⋅ s m 2 ⎡⎣( 0.010 m ) ( 0.040 m ) ⎤⎦ ( 0.30 m s ) h Av = = 0.12 N d 1.5 × 10 −3 m

Poiseuille’s law gives flow rate = ( P1 − P2 ) p R 4 8h L , and P2 = Patm in this case. Thus, the desired gauge pressure is P1 − Patm = or

9.66

)(

From the deﬁnition of the coefﬁcient of viscosity, h = F d A v, the required force is F=

9.65

(

From the deﬁnition of the coefﬁcient of viscosity, h = F d A v, the required force is F=

9.64

2 ( 0.058 N m ) cos 0° 2g cosf = = 5.6 m r gr 1 050 kg m 2 9.80 m s 2 2.0 × 10 −6 m

(

)

(

2 −5 3 8h L ( flow rate ) 8 0.12 N ⋅ s m ( 50 m ) 8.6 × 10 m s = 4 p R4 p 0.50 × 10 −2 m

(

)

)

P1 − Patm = 2.1 × 10 6 Pa = 2.1 MPa

From Poiseuille’s law, the ﬂow rate in the artery is flow rate =

( ΔP )p R 4 8h L

=

( 400 Pa )p ( 2.6 × 10 −3 m )

4

8 ( 2.7 × 10 −3 N ⋅s m 2 )(8.4 × 10 −2 m )

= 3.2 × 10 −5 m 3 s

Thus, the ﬂow speed is v=

56157_09_ch09_p288-321.indd 313

flow rate 3.2 × 10 −5 m 3 s = 2 = 1.5 m s A p 2.6 × 10 –3 m

(

)

10/12/10 2:40:40 PM

314

9.67

Chapter 9

If a particle is still in suspension after 1 hour, its terminal velocity must be less than

( vt )max = ⎛⎜⎝ 5.0

cm ⎞ ⎛ 1 h ⎞ ⎛ 1 m ⎞ −5 ⎟ ⎜ ⎟ = 1.4 × 10 m s h ⎠ ⎜⎝ 3 600 s ⎟⎠ ⎝ 100 cm ⎠

Thus, from vt = 2 r 2 g(r − rf ) 9h , we ﬁnd the maximum radius of the particle: 9h water ( vt )max

rmax =

(

2 g rprotein − rwater

( )( 2 ( 9.80 m s ) ⎡⎣(1 800 − 1 000 )

9 1.00 × 10 −3 N ⋅s m 2 1.4 × 10 −5 m s

= 9.68

)

2

) = 2.8 × 10

−6

kg m ⎤⎦ 3

m = 2.8 m m

From Poiseuille’s law, the pressure difference required to produce a given volume ﬂow rate of ﬂuid with viscosity h through a tube of radius R and length L is ΔP =

8h L ( ΔV Δt ) p R4

If the mass ﬂow rate is ( Δm Δt ) = 1.0 × 10 –3 kg s, the volume ﬂow rate of the water is ΔV Δm Δt 1.0 × 10 –3 kg s = = = 1.0 × 10 −6 m 3 s r Δt 1.0 × 10 3 kg m 3 and the required pressure difference is ΔP = 9.69

(

)( )( p ( 0.15 × 10 m )

8 1.0 × 10 −3 Pa ⋅ s 3.0 × 10 −2 m 1.0 × 10 −6 m 3 s −3

4

)=

1.5 × 10 5 Pa

With the IV bag elevated 1.0 m above the needle and atmospheric pressure in the vein, the pressure difference between the input and output points of the needle is

(

)(

)

ΔP = ( Patm + r gh ) − Patm = r gh = 1.0 × 10 3 kg m 3 9.8 m s 2 (1.0 m ) = 9.8 × 10 3 Pa The desired ﬂow rate is

(

)

3 3 6 3 ΔV 500 cm 1 m 10 cm = = 2.8 × 10 −7 m 3 s Δt 30 min ( 60 s 1 min )

Poiseuille’s law then gives the required radius of the needle as ⎡ 8hL ( ΔV Δt ) ⎤ R=⎢ ⎥ ⎣ p ( ΔP ) ⎦

1 4

or 9.70

⎡ 8 (1.0 × 10 −3 Pa ⋅s ) ( 2.5 × 10 −2 m ) ( 2.8 × 10 −7 m 3 s ) ⎤ =⎢ ⎥ p ( 9.8 × 10 3 Pa ) ⎢⎣ ⎥⎦

1 4

ã

R = 2.1 × 10 −4 m = 0.21 mm

vout

We write Bernoulli’s equation as 0.500 m 2 Pout + 12 r vout + r gyout = Pin + 12 r vin2 + r gyin

or

(

)

2 Pgauge = Pin − Pout = r ⎡⎣ 12 vout − vin2 + g ( yout − yin ) ⎤⎦

continued on next page

56157_09_ch09_p288-321.indd 314

10/12/10 2:40:43 PM

Solids and Fluids

315

Approximating the speed of the ﬂuid inside the tank as vin ≈ 0, we ﬁnd 2 ⎡1 ⎤ Pgauge = 1.00 × 10 3 kg m 3 ⎢ ( 30.0 m s ) − 0 + 9.80 m s 2 ( 0.500 m ) ⎥ ⎣2 ⎦

(

or 9.71

)

(

)

Pgauge = 4.55 × 10 5 Pa = 455 kPa

The Reynolds number is

(

)

(

)

1 050 kg m 3 ( 0.55 m s ) 2 .0 × 10 −2 m r vd 3 RN = = = 4.3 × 10 −3 2 h 2 .7 × 10 N ⋅ s m In this region (RN > 3 000), the ﬂow is turbulent . 9.72

From the deﬁnition of the Reynolds number, the maximum ﬂow speed for streamlined (or laminar) ﬂow in this pipe is vmax =

9.73

h ⋅ ( RN )max rd

=

D=

8.0 cm s

( Diffusion rate) L DA

(5.3 × 10

(5.0 × 10

−10

)

kg s ( 0.10 m )

−15

)(

m s 6.0 × 10 2

−4

m

2

)

= 1.8 × 10 −3 kg m 3

(

5.7 × 10 −15 kg s = 9.5 × 10 −10 m 2 s 2 −2 4 2.0 × 10 m ⋅ 3.0 × 10 kg m −4

)(

)

Fr 3.0 × 10 −13 N = = 1.4 × 10 −5 N ⋅ s m 2 6p r v 6p 2 .5 × 10 −6 m 4.5 × 10 −4 m s

(

(

)(

)

)

Using vt = 2 r 2 g r − rf 9h , the density of the sphere is found to be rsphere = r water + 9h water vt 2r 2 g. Thus, if r = d 2 = 0.500 × 10 −3 m and vt = 1.10 × 10 −2 m s when falling through 20°C water, r sphere = 1.00 × 10 3

9.77

−2

3

Stokes’s law gives the viscosity of the air as h=

9.76

2

Fick’s law gives the diffusion coefﬁcient as D = (Diffusion rate) [A ⋅(ΔC L)], where ΔC L is the concentration gradient. Thus,

9.75

−3

The observed diffusion rate is (8.0 × 10 −14 kg) (15 s) = 5.3 × 10 −15 kg s. Then, from Fick’s law, the difference in concentration levels is found to be C2 − C1 =

9.74

(1.0 × 10 N ⋅ s m ) ( 2 000 ) = 0.080 m s = = (1 000 kg m ) ( 2.5 × 10 m )

(a)

(

)( )(

)

−3 2 −2 kg 9 1.00 × 10 N ⋅ s m 1.10 × 10 m s + = 1.02 × 10 3 kg m 3 2 −4 2 m3 2 5.00 × 10 m 9.80 m s

(

)

Both iron and aluminum are denser than water, so both blocks will be fully submerged. Since the two blocks have the same volume, they displace equal amounts of water and the buoyant forces acting on the two blocks are equal. continued on next page

56157_09_ch09_p288-321.indd 315

10/12/10 2:40:45 PM

316

Chapter 9

(b)

Since the block is held in equilibrium, the force diagram at the right shows that

B

T

ΣFy = 0 ⇒ T = mg − B The buoyant force B is the same for the two blocks, so the spring scale reading T is largest for the iron block , which has a higher density, and hence weight, than the aluminum block. (c)

Fg ⫽ mg

The buoyant force in each case is

(

)(

)(

)

B = ( r water V ) g = 1.00 × 10 3 kg m 3 0.20 m 3 9.80 m s 2 = 2.0 × 10 3 N For the iron block:

(

)(

)(

)

Tiron = ( r ironV ) g − B = 7.86 × 10 3 kg m 3 0.20 m 3 9.8 m s 2 − B Tiron = 1.5 × 10 4 N − 2.0 × 10 3 N = 13 × 10 3 N

or

For the aluminum block:

(

)(

)(

)

Taluminum = ( r aluminum V ) g − B = 2.70 × 10 3 kg m 3 0.20 m 3 9.8 m s 2 − B Taluminum = 5.3 × 10 3 N − 2.0 × 10 3 N = 3.3 × 10 3 N

or 9.78

The object has volume V and true weight Fg . It also experiences a buoyant force B = (rair V )g exerted on it by the surrounding air. The counterweight, of density r , on the opposite end of the balance has a true weight Fg′ and experiences a buoyant force

艎

艎

weight

object

⎛ m′ ⎞ ⎛r ⎞ ⎛r ⎞ B ′ = ( rair V ′ ) g = rair ⎜ ⎟ g = ⎜ air ⎟ m ′g = ⎜ air ⎟ Fg′ ⎝ r⎠ ⎝ r ⎠ ⎝ r ⎠

B

B⬘

Fg

F⬘g

The balance is in equilibrium when the net downward forces acting on its two ends are equal, that is, when Fg − B = Fg′ − B′. Thus, the true weight of the object may be expressed as ⎛r ⎞ Fg = Fg′ + B − B′ = Fg′ + ( rair V ) g − ⎜ air ⎟ Fg′ ⎝ r ⎠ Fg′ ⎞ ⎛ Fg = Fg′ + ⎜ V − rair g r g ⎟⎠ ⎝

or 9.79

(a)

From Archimedes’ principle, the granite continent will sink down into the peridotite layer until the weight of the displaced peridotite equals the weight of the continent. Thus, at equilibrium, ⎡⎣ rg ( At ) ⎤⎦ g = ⎡⎣ rp ( Ad ) ⎤⎦ g or

Area A Continent (granite)

␳g

t

d Peridotite

␳p

rg t = rp d

continued on next page

56157_09_ch09_p288-321.indd 316

10/12/10 2:40:48 PM

Solids and Fluids

(b)

317

If the continent sinks 5.0 km below the surface of the peridotite, then d = 5.0 km, and the result of part (a) gives the ﬁrst approximation of the thickness of the continent as ⎛ rp ⎞ ⎛ 3.3 × 10 3 kg m 3 ⎞ t=⎜ ⎟d=⎜ 5.0 km ) = 5.9 km 3 3 ( ⎝ 2.8 × 10 kg m ⎟⎠ ⎝ rg ⎠

9.80

(a)

Starting with P = P0 + r gh, we choose the reference level at the level of the heart, so P0 = PH . The pressure at the feet, a depth hH below the reference level in the pool of blood in the body, is PF = PH + r ghH . The pressure difference between feet and heart is then PF − PH = r ghH .

(b)

Using the result of part (a),

(

)(

)

PF − PH = 1.05 × 10 3 kg m 3 9.80 m s 2 (1.20 m ) = 1.23 × 10 4 Pa 9.81

The cross-sectional area of the aorta is A1 = p d12 4 and that of a single capillary is Ac = p d 22 4. If the circulatory system has N such capillaries, the total cross-sectional area carrying blood from the aorta is A2 = NAc = Np d 22 4. From the equation of continuity, A2 = (v1 v2 )A1, or Np d 22 ⎛ v1 ⎞ p d12 =⎜ ⎟ 4 ⎝ v2 ⎠ 4 yielding 2

2

⎛ v ⎞⎛ d ⎞ ⎛ 1.0 m s ⎞ ⎛ 0.50 × 10 −2 m ⎞ N =⎜ 1⎟⎜ 1⎟ =⎜ = 2.5 × 10 7 −2 −6 ⎝ v2 ⎠ ⎝ d 2 ⎠ ⎝ 1.0 × 10 m s ⎟⎠ ⎜⎝ 10 × 10 m ⎟⎠ 9.82

(a)

We imagine that a superhero is capable of producing a perfect vacuum above the water in the straw. Then P = P0 + r gh, with the reference level at the water surface inside the straw and P being atmospheric pressure on the water in the cup outside the straw, gives the maximum height of the water in the straw as hmax =

9.83

Patm − 0 P 1.013 × 10 5 N m 2 = atm = = 10.3 m rwater g rwater g 1.00 × 10 3 kg m 3 9.80 m s2

(

)(

)

(b)

The Moon has no atmosphere so Patm = 0, which yields hmax = 0 .

(a)

P = 160 mm of H 2 O = r H O g (160 mm ) 2

m⎞ kg ⎞ ⎛ ⎛ = ⎜ 10 3 ⎟ ⎜ 9.80 2 ⎟⎠ ( 0.160 m ) = 1.57 kPa ⎝ s m3 ⎠ ⎝ 1 atm ⎛ ⎞ = 1.55 × 10 −2 atm P = 1.57 × 10 3 Pa ⎜ ⎝ 1.013 × 10 5 Pa ⎟⎠

(

)

The pressure is P = r H O ghH 2

2O

= r Hg ghHg , so

⎛ rHO⎞ ⎛ 10 3 kg m 3 ⎞ hHg = ⎜ = h 160 mm ) = 11.8 mm of Hg ⎟ HO ⎜ 3 3⎟( ⎝ 13.6 × 10 kg m ⎠ ⎝ r Hg ⎠ 2

2

continued on next page

56157_09_ch09_p288-321.indd 317

10/12/10 2:40:51 PM

318

9.84

Chapter 9

(b)

The ﬂuid level in the tap should rise.

(c)

Blockage of ﬂow of the cerebrospinal ﬂuid

When the rod ﬂoats, the weight of the displaced ﬂuid equals the weight of the rod, or (r f Vdisplaced )g = (r 0Vrod )g. Assuming a cylindrical rod, Vrod = p r 2 L . The volume of ﬂuid displaced is the same as the volume of the rod that is submerged or

Vdisplaced = p r 2 ( L − h ).

Thus, rf ⎡⎣p r 2 ( L − h )⎤⎦ g = r0 ⎡⎣p r 2 L ⎤⎦ g which reduces to rf = r 0 L ( L − h ) . 9.85

Consider the diagram and apply Bernoulli’s equation to points A and B, taking y = 0 at the level of point B, and recognizing that v A ≈ 0. This gives PA + 0 + rw g( h − L sinq ) = PB +

A

h

1 rw v B2 + 0 2

Recognize that PA = PB = Patm since both points are open to the atmosphere. Thus, we obtain v B = 2 g( h − L sinq ) .

L Valve

B ␪

Now the problem reduces to one of projectile motion with v0 y = v B sinq . At the top of the arc v y = 0, y = ymax , and v 2y = v02 y + 2 a y (Δy) gives ymax − 0 =

v 2y − v02 y 2a y

=

2g ( h − L sinq ) sin 2 q 0 − v B2 sin 2 q = 2(−g) 2g

or ymax = ⎡⎣10.0 m − ( 2.00 m ) sin 30.0° ⎤⎦ sin 2 30.0° = 2.25 m above the level of point B 9.86

When the balloon comes into equilibrium, the weight of the displaced air equals the weight of the ﬁlled balloon plus the weight of string that is above ground level. If ms and L are the total mass and length of the string, the mass of string that is above ground level is (h L) ms . Thus, ( rairVballoon )g = mballoon g + ( rheliumVballoon )g + ( h L ) ms g, which reduces to ⎡ ( r − rhelium )Vballoon − mballoon ⎤ h = ⎢ air ⎥L ms ⎣ ⎦ This yields

(

)

⎡ 1.29 kg m 3 − 0.179 kg m 3 ⎡ 4p ( 0.40 m ) 3 3⎤ − 0.25 kg ⎤ ⎣ ⎦ ⎥ ( 2.0 m ) = 1.9 m h=⎢ ⎢ ⎥ 0.050 kg ⎢⎣ ⎥⎦ 9.87

Four forces are acting on the balloon: an upward buoyant force exerted by the surrounding air, B = (rair Vballoon )g; the downward weight of the balloon envelope, Fg, balloon = mg; the downward weight of the helium ﬁlling the balloon, Fg, He = (r HeVballoon )g; and the downward spring force, Fs = k Δx . At equilibrium, Δx = L, and we have ΣFy = 0

⇒

B − Fs − Fg, balloon − Fg, He = 0

continued on next page

56157_09_ch09_p288-321.indd 318

10/12/10 2:40:55 PM

Solids and Fluids

319

Fs = kL = B − Fg, balloon − Fg, He = ( rair Vballoon ) g − mg − ( r HeVballoon ) g

or and

⎡( rair − r He ) Vballoon − m ⎤⎦ g L= ⎣ k

This yields

L=

or

L = 0.605 m

{⎡⎣(1.29 − 0.179 ) kg m ⎤⎦ ( 5.00 m ) − 2.00 × 10 3

3

−3

}(

kg 9.80 m s 2

)

90.0 N m

9.88

Shield

→

v

A

B H

h L L A

B C

(b)

(a)

D

(c)

Consider the pressure at points A and B in part (b) of the ﬁgure by applying P = P0 + rf gh. Looking at the left tube gives PA = Patm + rwater g( L − h ), and looking at the tube on the right, PB = Patm + roil gL. Pascal’s principle says that PB = PA. Therefore, Patm + roil gL = Patm + rwater g( L − h ), giving ⎛ ⎛ r ⎞ 750 kg m 3 ⎞ h = ⎜ 1 − oil ⎟ L = ⎜ 1 − 5.00 cm ) = 1.25 cm 3 ( rwater ⎠ ⎝ ⎝ 1 000 kg m ⎟⎠

(b)

Consider part (c) of the diagram showing the situation when the air ﬂow over the left tube equalizes the ﬂuid levels in the two tubes. First, apply Bernoulli’s equation to points A and B. This gives PA + 12 rair v 2A + rair g y A = PB + 12 rair v B2 + rair g yB . Since y A = yB , v A = v, and v B = 0, this reduces to PB − PA =

1 rair v 2 2

[1]

Now use P = P0 + rf gh to ﬁnd the pressure at points C and D, both at the level of the oil–water interface in the right tube. From the left tube, PC = PA + rwater gL, and from the right tube, PD = PB + roil gL . Pascal’s principle says that PD = PC , and equating these two gives PB + roil gL = PA + rwater gL, or PB − PA = ( rwater − roil ) gL

[2] continued on next page

56157_09_ch09_p288-321.indd 319

10/12/10 2:40:59 PM

320

Chapter 9

Combining Equations [1] and [2] yields v= 9.89

2 ( rwater − roil ) gL = rair

The pressure on the surface of the two hemispheres is constant at all points, and the force on each element of surface area is directed along the radius of the hemispheres. The applied force along the horizontal axis must balance the net force on the “effective” area, which is the cross-sectional area of the sphere, Aeffective = p R

)(

1.29

)=

13.8 m s

R

F

F P P0

2

F = Pgauge Aeffective = ( P0 − P ) p R 2

and 9.90

(

2 (1 000 − 750 ) 9.80 m s 2 5.00 × 10 −2 m

Since the block is ﬂoating, the total buoyant force must equal the weight of the block. Thus, roil ⎡⎣ A( 4.00 cm − x ) ⎤⎦ g + rwater [ A ⋅ x ]g

Oil 4.00 cm ⫺ x 4.00 cm

= rwood ⎡⎣ A( 4.00 cm ) ⎤⎦ g

x

where A is the surface area of the top or bottom of the rectangular block.

Water

Solving for the distance x gives ⎛r ⎛ 960 − 930 ⎞ − roil ⎞ x = ⎜ wood ( 4.00 cm ) = ⎜ ( 4.00 cm ) = 1.71 cm ⎟ r − r ⎝ water ⎝ 1 000 − 930 ⎟⎠ oil ⎠ 9.91

A water droplet emerging from one of the holes becomes a projectile with v0 y = 0 and v0 x = v. The time for this droplet to fall distance h to the ﬂoor is found from Δy = v0 y t + 12 a y t 2 to be t = 2 h g. The horizontal range is R = vt = v 2 h g. If the two streams hit the ﬂoor at the same spot, it is necessary that R1 = R2 , or v1

2 h1 2 h2 = v2 g g

With h1 = 5.00 cm and h2 = 12.0 cm, this reduces to v1 = v2 h2 h1 = v2 12.0 cm 5.00 cm , or v1 = v2 2 .40

3

2 h3 1

h2 h1

R1 ⫽ R2

[1]

Apply Bernoulli’s equation to points 1 (the lower hole) and 3 (the surface of the water). The pressure is atmospheric pressure at both points and, if the tank is large in comparison to the size of the holes, v3 ≈ 0. Thus, Patm + 12 r v12 + r gh1 = Patm + 0 + r gh3, or v12 = 2 g( h3 − h1 )

[2]

continued on next page

56157_09_ch09_p288-321.indd 320

10/12/10 2:41:02 PM

Solids and Fluids

321

Similarly, applying Bernoulli’s equation to point 2 (the upper hole) and point 3 gives Patm + 12 r v22 + r gh2 = r atm + 0 + r gh3, or v22 = 2 g( h3 − h2 )

[3]

Square Equation [1] and substitute from Equations [2] and [3] to obtain 2 g( h3 − h1 ) = 2 .40 ⎡⎣ 2 g( h3 − h2 ) ⎤⎦ Solving for h3 yields h3 =

2 .40 h2 − h1 2 .40 (12 .0 cm ) − 5.00 cm = = 17.0 cm 1.40 1.40

so the surface of the water in the tank is 17.0 cm above floor level .

56157_09_ch09_p288-321.indd 321

10/12/10 2:41:05 PM

10 Thermal Physics QUICK QUIZZES 1.

Choice (c). When two objects having different temperatures are in thermal contact, energy is transferred from the higher temperature object to the lower temperature object. As a result, the temperature of the hotter object decreases and that of the cooler object increases until thermal equilibrium is reached at some intermediate temperature.

2.

Choice (b). The glass surrounding the mercury expands before the mercury does, causing the level of the mercury to drop slightly. The mercury rises after it begins to get warmer and approaches the temperature of the hot water, because its coefﬁcient of expansion is greater than that for glass.

3.

Choice (c). Gasoline has the highest coefﬁcient of expansion so it undergoes the greatest change in volume per degree change in temperature.

4.

Choice (c). A cavity in a material expands in exactly the same way as if the cavity were ﬁlled with the surrounding material. Thus, both spheres will expand by the same amount.

5.

Unlike land-based ice, ice ﬂoating in the ocean already displaces a quantity of liquid water whose weight equals the weight of the ice. This is the same situation as will exist after the ice melts, so the melting ocean-based ice will not change ocean levels much.

6.

Choice (b). Since the two containers are at the same temperature, the average kinetic energy per molecule is the same for the argon and helium gases. However, helium has a lower molar mass than does argon, so the rms speed of the helium atoms must be higher than that of the argon atoms.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 9 ( −25° ) + 32° = 5

1.

TF = 95 TC + 32 =

2.

The correct choice is (b). When an object, containing a cavity, is heated, the cavity expands in the same way as it would if ﬁlled with the material making up the rest of the object.

3.

4.

5.

−13° F, and the correct response is choice (e).

−1 ΔL = a C u L0 ( ΔT ) = ⎡⎣17 × 10 −6 (°C ) ⎤⎦ ( 93 m )( 5°C ) = 8 × 10 −3 m ∼ 10 −2 m = 1 cm and choice (c) is the correct order of magnitude.

TC = 95 ( TF − 32 ) = 95 (162 − 32 ) = 72.2°C, then TK = TC + 273 = 72.2 + 273 = 345 K, so choice (c) is the correct answer. Remember that one must use absolute temperatures and pressures in the ideal gas law. Thus, the original temperature is TK = TC + 273.13 = 25.0 + 273.15 = 298.2 K, and with the mass of the gas constant, the ideal gas law gives ⎛ P ⎞⎛V ⎞ ⎛ 1.07 × 10 6 Pa ⎞ T2 = ⎜ 2 ⎟ ⎜ 2 ⎟ T1 = ⎜ ( 3.00 )( 298.2 K ) = 191 K ⎝ 5.00 × 10 6 Pa ⎟⎠ ⎝ P1 ⎠ ⎝ V1 ⎠ and (d) is the best choice. 322

56157_10_ch10_p322-343.indd 322

10/12/10 2:43:30 PM

Thermal Physics

6.

323

From the ideal gas law, with the mass of the gas constant, P2V2 T2 = P1V1 T1. Thus, ⎛V ⎞⎛T ⎞ ⎛ 1⎞ P2 = ⎜ 1 ⎟ ⎜ 2 ⎟ P1 = ⎜ ⎟ ( 4 ) P1 = 2P1 ⎝ 2⎠ ⎝ V2 ⎠ ⎝ T1 ⎠ and (d) is the correct choice.

7.

From the ideal gas law, with the mass of the gas constant, P2V2 T2 = P1V1 T1. Thus, ⎛ P ⎞⎛T ⎞ V2 = ⎜ 1 ⎟ ⎜ 2 ⎟ V1 = ( 4 ) (1) 0.50 m 3 = 2.0 m 3 ⎝ P2 ⎠ ⎝ T1 ⎠ and (c) is the correct choice.

(

8.

The internal energy of n moles of a monatomic ideal gas is U = 32 nRT , where R is the universal gas constant and T is the absolute temperature of the gas. For the given neon sample, T = TC + 273.15 = (152 + 273.15) K = 425 K, and n=

m 26.0 g = = 1.29 mol molar mass 20.18 g mol

Thus, U = 9.

)

3 (1.29 2

mol ) (8.31 J mol ⋅ K ) ( 425 K ) = 6.83 × 10 3 J and (b) is correct answer.

If two objects at different temperatures are in thermal contact, energy ﬂows from the warmer object to the cooler object. Thermal equilibrium occurs only when the two objects are at the same temperature. Thus, since the contents of the bowl have reached thermal equilibrium, choice (b) is the correct answer.

10.

The kinetic theory of gases does assume that the molecules in a pure substance obey Newton’s laws and undergo elastic collisions, and the average distance between molecules is very large in comparison to molecular sizes. However, it also assumes that the number of molecules in the sample is large so that statistical averages are meaningful. The untrue statement included in the list of choices is (a).

11.

In a head-on, elastic collision with a wall, the change in momentum of a gas molecule is Δp = m v f − v0 = m ( −v0 − v0 ) = −2mv0 . If the molecule should stick to the wall instead of rebounding, the change in the molecule’s momentum would be Δp = m ( 0 − v0 ) = −mv0 , which is half that in the elastic collision. Since a gas exerts a pressure on its container by molecules imparting impulses to the walls during collisions, and the impulse imparted equals the magnitude of the change in the molecular momentum, decreasing the change in momentum during the collisions by a factor of 2 would reduce the pressure by a factor of 2. Thus, the correct response is choice (b).

12.

The rms speed of molecules in the gas is vrms = 3RT M . Thus, the ratio of the ﬁnal speed to the original speed would be

(

( vrms ) f ( vrms )0

)

=

3RT f M 3RT0 M

=

Tf T0

=

600 K = 3 200 K

Therefore, the correct answer to this question is choice (d). 13.

56157_10_ch10_p322-343.indd 323

Since, when the balloon is fully inﬂated, the inside and outside pressures are nearly equal at one atmosphere, the walls of the balloon must be very easily stretched. As the air cools, the walls will contract, just maintaining equality of the pressure inside and outside. Thus, this is an isobaric cooling process. Since, even at 100 K, the temperature is far above the liquefaction point, we treat

10/12/10 2:43:32 PM

324

Chapter 10

the air as an ideal gas under constant pressure and ﬁnd:

(

)

(i) V f = T f Ti Vi = (100 K 300 K )(1 L ) = 13 L, and choice (b) is the correct answer. (ii) As noted above, the pressure will remain constant at nearly one atmosphere, so the correct answer is choice (d). 14.

Consider the ideal gas law, PV = nRT , recognizing that the two cylinders contain equal quantities of gas (nA = nB) at equal temperatures (TA = TB). Thus, PAVA = PBVB and we have PA = (VB VA ) PB = (1 3) PB, meaning that the correct answer is choice (d).

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The lower temperature will make the power line decrease in length. This increases the tension in the line so it is closer to the breaking point.

4.

The pressure inside the balloon is greater than the ambient atmospheric pressure because the pressure inside must not only resist the external pressure, but also the force exerted by the elastic material of the balloon.

6.

At high temperature and pressure, the steam inside exerts large forces on the pot and cover. Strong latches hold them together, but they would explode apart if you tried to open the hot cooker.

8.

The measurements are too short. At 22°C the tape would read the width of the object accurately, but an increase in temperature causes the divisions ruled on the tape to be farther apart than they should be. This “too long” ruler will, then, measure objects to be shorter than they really are.

10.

The existence of an atmosphere on a planet is due to the gravitational force holding the gas of the atmosphere to the planet. On a small planet, the gravitational force is very small, and the escape speed is correspondingly small. If a small planet starts its existence with atmosphere, the molecules of the gas will have a distribution of speeds, according to kinetic theory. Some of these molecules will have speeds higher than the escape speed of the planet and will leave the atmosphere. As the remaining atmosphere is warmed by radiation from the Sun, more molecules will attain speeds high enough to escape. As a result, the atmosphere bleeds off into space.

12.

Doubling the volume while reducing the pressure by half results in no change in the quantity PV that appears in the ideal gas law equation. Consequently, the temperature and hence the internal energy remain the same.

14.

(a)

The sphere, which initially could just barely ﬁt through the ring, expands when it is heated and thus becomes too large to ﬁt through the unheated ring.

(b)

If the ring is heated, every linear dimension of the ring (including the inner circumference, and hence the diameter of the hole), increases. This occurs because, when a material is heated, the atoms in that material push against each other and get farther apart. The only way atoms on the inner circumference of the ring can get farther apart is for the circumference—and corresponding diameter—to increase. Thus, if the ring is heated while the sphere remains cool, the ring ﬁts over the sphere with additional room to spare.

56157_10_ch10_p322-343.indd 324

10/12/10 2:43:34 PM

Thermal Physics

325

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a)

−251°C

(b)

1.36 atm

4.

(a)

56.7°C, − 62.1°C

(b)

330 K, 211 K

6.

(a)

−270°C

(b)

1.27 atm

8.

(a) 31.7°C

(b)

31.7 K TK = 95 TR

10.

(a)

TR = TF + 459.67

(b)

12.

(a)

L = 1.3 m − 0.49 mm

(b) The clock will run fast.

14.

(a)

2.542 cm

(b)

3.0 × 10 2 °C

16.

See Solution.

18.

(a) 396 N

(b)

−101°C

(c)

1.74 atm

(c) The initial length of the wire cancels during the calculation, so the results are not affected by doubling the length of the bridge. 20.

See Solution.

22.

18.702 m

24.

(a) 1.5 km (b) Accordion-like expansion joints are placed in the pipeline at periodic intervals.

26.

27.7 kg

(b)

1.02 m 3

(d) 27.2 kg

(e)

0.5 kg

(a)

(c)

716 kg m 3

28.

1.2 cm

30.

(a) 292 K

(b)

7.83 mol

(c)

44.0 g mol

(d) 345 g

(e)

516 K, 5.98 mol

(f)

Pf = n f T f ni Ti Pi

(b)

4.43 atm = 449 kPa

(b)

1.81 × 10 24 molecules

(g)

1.28 × 10 6 Pa

32.

(a)

3.95 atm = 400 kPa

34.

(a) 3.00 mol

36.

0.131 kg m 3

56157_10_ch10_p322-343.indd 325

(

)

10/12/10 2:43:35 PM

326

38.

Chapter 10

(a)

r = nM V

(b)

(c)

See Solution.

69 kg m 3

(d) The shell would rise because rshell = 47.7 kg m 3 < ratmosphere . 40.

36.5 kN

42.

(a)

44.

(a) 385 K

vrms vrms

( (

35 37

) = 1.03 Cl )

Cl

(b)

The less massive atom,

(b)

7.97 × 10 −21 J

35

Cl, moves faster.

(c) The molar mass, from which the mass of a molecule can be computed. 46.

18 kPa

48.

(a) 0.176 mm

(b)

50.

(a) 0.34%

(b)

8.78 m m

(c)

9.30 × 10 −8 m 3

0.48%

(c) The moment of inertia for each of the shapes has the same mathematical form: the product of a constant, the mass, and the square of a length. 52.

28 m

54.

(a) See Solution.

(b)

greater than atmospheric pressure

(c) The equilibrium value of h increases as the temperature T increases. 56.

See Solution.

58.

(a) 343 K

(b)

12.5%

60.

(a) 16.9 cm

(b)

1.35 × 10 5 Pa

62.

(a) 6.0 cm

(b)

See Solution.

(c) The bridge does not crumble. The stress developed is 4.8 × 10 6 Pa < 2.0 × 10 7 Pa. 64.

1.15 atm

PROBLEM SOLUTIONS 10.1

56157_10_ch10_p322-343.indd 326

(a)

TF =

9 9 TC + 32 = ( −273.15) + 32 = − 460°F 5 5

(b)

TC =

5 (TF − 32) = 95 (98.6 − 32) = 37°C 9

(c)

TF =

9 9 9 TC + 32 = ( TK − 273.15) + 32 = ( −173.15) + 32 = − 280°F 5 5 5

10/12/10 2:43:37 PM

Thermal Physics

10.2

327

When the volume of a low-density gas is held constant, pressure and temperature are related by a linear equation P = AT + B, where A and B are constants to be determined. For the given constant-volume gas thermometer, P = 0.700 atm when T = 100°C ⇒ 0.700 atm = A (100°C ) + B

[1]

P = 0.512 atm when T = 0°C ⇒ 0.512 atm = A ( 0 ) + B

[2]

From Equation [2], B = 0.512 atm. Substituting this result into Equation [1] yields A=

0.700 atm − 0.512 atm = 1.88 × 10 −3 atm °C 100°C

(

)

so, the linear equation for this thermometer is: P = 1.88 × 10 −3 atm °C T + 0.512 atm

10.3

10.4

P − B 0.0400 atm − 0.512 atm = − 251°C = A 1.88 × 10 −3 atm °C

(a)

If P = 0.0400 atm, then

T=

(b)

If T = 450°C, then

P = 1.88 × 10 −3 atm °C ( 450°C ) + 0.512 atm = 1.36 atm

(a)

TC = TK − 273.15 = 20.3 − 273.15 = −253°C

(b)

TF =

9 9 TC + 32 = ( −253) + 32 = − 423°F 5 5

(a)

TC =

5 (TF − 32) = 95 (134 − 32) = 56.7°C , and 9

TC =

5 ( −79.8 − 32 ) = − 62.1°C 9

(b)

(

)

T = TC + 273.15 = 56.7 + 273.15 = 330 K , and TK = TC + 273.15 = −62.1+ 273.15 = 211 K

10.5

Start with TF = − 40°F and convert to Celsius. TC =

5 5 TF − 32 ) = ( − 40 − 32 ) = − 40°C ( 9 9

Since Celsius and Fahrenheit degrees of temperature change are different sizes (1 Celsius degree = 1.8 Fahrenheit degrees), this is the only temperature with the same numeric value on both scales. 10.6

Since we have a linear graph, we know that the pressure is related to the temperature as P = A + BTC , where A and B are constants. To ﬁnd A and B, we use the given data: 0.900 atm = A + B ( −78.5°C )

[1]

1.635 atm = A + B ( 78.0°C )

[2]

and

Solving Equations [1] and [2] simultaneously, we ﬁnd: A = 1.27 atm, and

B = 4.70 × 10 −3 atm °C

continued on next page

56157_10_ch10_p322-343.indd 327

10/12/10 2:43:39 PM

328

Chapter 10

(

(a)

At absolute zero the gas exerts zero pressure ( P = 0 ) , so TC =

(b)

)

P = 1.27 atm + 4.70 × 10 −3 atm °C TC

Therefore,

−1.27 atm = − 270°C 4.70 × 10 −3 atm °C

At the freezing point of water, TC = 0°C and P = 1.27 atm + 0 = 1.27 atm

(c)

At the boiling point of water, TC = 100°C , so

(

)

P = 1.27 atm + 4.70 × 10 −3 atm °C (100°C ) = 1.74 atm 10.7

Apply TF = 95 TC + 32 to two different Celsius temperatures, ( TC )1 and ( TC )2 ,

(TF )1 = 95 (TC )1 + 32

to obtain

[1]

(TF )2 = 95 (TC )2 + 32

and

[2]

Subtracting Equation [1] from [2] yields

(TF )2 − (TF )1 = 95 ⎡⎣(TC )2 − (TC )1 ⎤⎦ or 10.8

10.9

10.10

ΔTF = ( 9 5) ΔTC

( ΔTF ) = 95 (57.0 ) °C =

(a)

Using the result of Problem 10.7 above, ΔTC =

(b)

ΔTK = TC , out + 273.15 − TC , in + 273.15 = TC , out − TC , in = ΔTC = 31.7 K

(a)

TF = 95 TC + 32 =

(b)

Yes. The normal body temperature is 98.6°F, so this patient has a high fever and needs immediate attention.

(a)

Since temperature differences on the Rankine and Fahrenheit scales are identical, the temperature readings on the two thermometers must differ by no more than an additive constant (i.e., TR = TF + constant). To evaluate this constant, consider the temperature readings on the two scales at absolute zero. We have TR = 0°R at absolute zero, and

(

) (

9 ( 41.5) + 32 = 5

TF = 95 TC + 32.00 =

) (

5 9

31.7°C

)

107°F

9 ( −273.15) + 32.00 = 5

− 459.67°F

Substituting these temperatures in our Fahrenheit to Rankine conversion gives 0° = − 459.67° + constant giving

or

constant = 459.67°

TR = TF + 459.67

continued on next page

56157_10_ch10_p322-343.indd 328

10/12/10 2:43:43 PM

Thermal Physics

(b)

We start with the Kelvin temperature and convert to the Rankine temperature in several stages, using the Fahrenheit to Rankine conversion from part (a) above. TK = TC + 273.15 = =

or 10.11

329

5 (TF − 32.00 ) + 273.15 = 95 ⎡⎣(TR − 459.67) − 32.00 ⎤⎦ + 273.15 9

5 (TR − 491.67) + 273.15 = 95 TR − 95 ( 491.67) + 273.15 = 95 TR − 273.15 + 273.15 9

TK =

5 TR 9

The increase in temperature is ΔT = 35°C − ( −20°C ) = 55°C. −1 Thus, ΔL = a L0 ( ΔT ) = ⎡⎣11 × 10 −6 ( °C ) ⎤⎦ ( 518 m ) ( 55°C ) = 0.31 m = 31 cm

10.12

(a)

As the temperature drops by 20°C, the length of the pendulum changes by −1 ΔL = a L0 ( ΔT ) = ⎡⎣19 × 10 −6 ( °C ) ⎤⎦ (1.3 m ) ( −20°C )

or ΔL = − 4.9 × 10 −4 m = − 0.49 mm Thus, the ﬁnal length of the rod is L = 1.3 m − 0.49 mm . (b)

10.13

From the expression for the period, T = 2p L g , we see that as the length decreases the period decreases. Thus, the pendulum will swing too rapidly and the clock will run fast .

We choose the radius as our linear dimension. Then, from ΔL = a L0 ( ΔT ) , ΔT = TC − 20.0°C =

10.14

L − L0 2.21 cm − 2.20 cm = = 35.0°C a L0 ⎡1.30 × 10 − 4 ( °C )−1 ⎤( 2.20 cm ) ⎣ ⎦

or

TC = 55.0°C

(a)

The diameter is a linear dimension, so we consider the linear expansion of steel:

(

d = d 0 [1 + a ( ΔT )] = ( 2.540 cm ) ⎡1 + 11 × 10 −6 (°C ) ⎣ (b)

) (100.0°C − 25.00°C)⎤⎦ = 2.542 cm

If the volume increases by 1.000%, then ΔV = (1.000 × 10 −2 ) V0 . Then, using ΔV = b V0 ( ΔT ) , where b = 3a is the volume expansion coefﬁcient, we ﬁnd ΔT =

10.15

−1

ΔV V0 1.000 × 10 −2 = = 3.0 × 10 2 °C −1 b 3 ⎡⎣11 × 10 −6 (°C ) ⎤⎦

From ΔL = L − L0 = a L0 ( ΔT ) , the ﬁnal value of the linear dimension is L = L0 + a L0 ( ΔT ). To remove the ring from the rod, the diameter of the ring must be at least as large as the diameter of the rod. Thus, we require that LBrass = LAl , or ( L0 )Brass + a Brass ( L0 )Brass ( ΔT ) = ( L0 )Al + a Al ( L0 )Al ( ΔT )

continued on next page

56157_10_ch10_p322-343.indd 329

10/12/10 2:43:46 PM

330

Chapter 10

ΔT =

This gives

(a)

( L0 )Al − ( L0 )Brass a Brass ( L0 )Brass − a Al ( L0 )Al

If ( L0 )Al = 10.01 cm, ΔT =

⎡19 × 10 ⎣

−6

( °C )

−1

10.01− 10.00 = −199°C ⎤(10.00 ) − ⎡ 24 × 10 −6 ( °C )−1 ⎤(10.01) ⎦ ⎣ ⎦

so T = T0 + ΔT = 20.0°C − 199°C = − 179°C which is attainable (b)

If ( L0 )Al = 10.02 cm ΔT =

10.02 − 10.00 = −396°C ⎡19 × 10 −6 ( °C )−1 ⎤(10.00 ) − ⎡ 24 × 10 −6 ( °C )−1 ⎤(10.02 ) ⎣ ⎦ ⎣ ⎦

and T = T0 + ΔT = − 376°C which is below absolute zero and unattainable 10.16

r=

10.17

(a)

m V0 r0 m m m = = = = V V0 + ΔV V0 + b V0 ( ΔT ) 1 + b ( ΔT ) 1 + b ( ΔT ) Using the result of Problem 10.16, with b = 3a , gives r=

10.18

r0 11.3 × 10 3 kg m 3 = = 11.2 × 10 3 kg m 3 1+ b ( ΔT ) 1+ 3 ⎡ 29 × 10 −6 (°C )−1 ⎤ ( 90°C − 0°C ) ⎣ ⎦

(b)

No . Although the density of gold would be less on a warm day, the mass of the bar would be the same, regardless of its temperature, and that is what you are paying for. (Note that the volume of the bar increases with increasing temperature, whereas its density decreases. Its mass, however, remains constant.)

(a)

When a wire undergoes a decrease in temperature of magnitude ΔT , it will attempt to contract by an amount ΔL = a L0 ΔT . If the ends of the wire are held ﬁxed, not allowing it to contract, the wire will develop a tension sufﬁcient to stretch it by the amount of its normal contraction, ΔL . This tension is ⎛ a L0 ΔT ⎞ ⎛ ΔL ⎞ F = YA ⎜ = YA ⎜ ⎟ = YAa ΔT L0 ⎝ L0 ⎟⎠ ⎝ ⎠ where Y is Young’s modulus for the wire material and A is the cross-sectional area of the wire. For the given steel wire, with ΔT = −10.0 − 35.0 °C, the tension that develops in the wire is

(

)(

)

−1 F = 20.0 × 1010 N m 2 4.00 × 10 −6 m 2 ⎡⎣11.0 × 10 −6 ( °C ) ⎤⎦ ( 45.0°C ) = 396 N

(b)

As the temperature decreases while the wire is prevented from contracting, the stress that develops in the wire is stress = (F A) = Ya ΔT . The decrease in temperature required to reach the elastic limit is ΔT = −

( stress )limit Ya

=−

(

3.00 × 108 N m 2 = −136°C −1 20.0 × 1010 N m 2 ⎡⎣11× 10 −6 ( °C ) ⎤⎦

)

continued on next page

56157_10_ch10_p322-343.indd 330

10/12/10 2:43:49 PM

Thermal Physics

331

Thus, the temperature of the wire when it reaches its elastic limit is T = T0 + ΔT = 35.0°C + ( −136°C ) = −101°C (c)

10.19

Observe that the initial length of the wire cancels out in the calculations of parts (a) and (b). Thus, the results obtained would not be changed if the initial length of the wire were doubled.

The difference in Celsius temperature in the underground tank and the tanker truck is ΔTC =

5 ( ΔTF ) = 95 (95.0 − 52.0 ) = 23.9°C 9

If V52°F is the volume of gasoline that ﬁlls the tank at 52.0°F, the volume this quantity of gas would occupy on the tanker truck at 95.0°F is V95°F = V52°F + ΔV = V52°F + b V52°F ( ΔT ) = V52°F [1 + b ( ΔT )]

(

)

(

)

−1 = 1.00 × 10 3 gal ⎡1 + 9.6 × 10 −4 (°C ) ( 23.9°C )⎤ = 1.02 × 10 3 gal ⎣ ⎦

10.20

Consider a regular solid with initial volume given by V0 = A0 L0 at temperature T0 . Here, A is the cross-sectional area and L is the length of the regular solid. When temperature undergoes a change ΔT = T − T0 , the change in the cross-sectional area is ΔA = A − A0 = g A0 ( ΔT ) = 2a A0 ( ΔT ), giving A = A0 + 2a A0 ( ΔT ) . Similarly, the new length will be L = L0 + a L0 ( ΔT ), so the new volume is V = ⎡⎣ A0 + 2a A0 ( ΔT )⎤⎦ ⎡⎣ L0 + a L0 ( ΔT )⎤⎦ = A0 L0 + 3a A0 L0 ( ΔT ) + 2a 2 A0 L0 ( ΔT )

2

The term involving a 2 is negligibly small in comparison to the other terms, so V ≈ A0 L0 + 3a A0 L0 ( ΔT ) = V0 + 3a V0 ( ΔT ) This is of the form ΔV = V − V0 = b V0 ( ΔT ) where 10.21

(a)

b = 3a .

The volume of turpentine that overﬂows as the temperature rises equals the difference in the increase in the volume of the turpentine and the increase in the volume of the aluminum container. Voverflow = ΔVT − ΔVAl = b TV0 ( ΔT ) − b Al V0 ( ΔT ) = ( b T − b Al ) V0 ( ΔT ) = ( b T − 3a Al ) V0 ( ΔT )

(

)

= ⎡⎣9.0 × 10 − 4 °C−1 − 3 24 × 10 −6 °C−1 ⎤⎦ ( 2.000 L )(80.0°C − 20.0°C ) or (b)

Voverflow = 9.9 × 10 −2 L = 99 mL

The volume of turpentine remaining in the cylinder at 80.0°C is the same as the volume of the aluminum cylinder at 80.0°C. This is

(

)

V1 = V0 [1+ 3a Al ( ΔT )] = ( 2.000 L ) ⎡⎣1+ 3 24 × 10 −6 °C−1 (80.0 − 20.0 ) °C ⎤⎦ = 2.000 L + 0.008 6 L = 2.009 L or

V1 = Vturpentine = 2.009 L remaining

continued on next page

56157_10_ch10_p322-343.indd 331

10/12/10 2:43:53 PM

332

Chapter 10

(c)

The volume of the aluminum cylinder when cooled back to 20.0°C will be V0 = 2.000 L. The volume of the remaining turpentine when cooled to 20.0°C will be

(

)

V2 = V1 [1+ b T ( ΔT )] = ( 2.009 L ) ⎡⎣1+ 9.0 × 10 −4 °C−1 ( 20.0 − 80.0 ) °C ⎤⎦ = 2.009 L – 0.11 L = 1.90 L The fraction of the cylinder’s volume that is now empty will be fraction empty =

V0 − V2 2.000 L − 1.90 L = = 5.00 × 10 −2 V0 2.000 L

so, the empty height above the remaining turpentine at 20.0°C is

(

)

hempty = h ( fraction empty ) = ( 20.0 cm ) 5.00 × 10 −2 = 1.00 cm 10.22

[Note that some rules concerning signiﬁcant ﬁgures are deliberately violated in this solution to better illustrate the method of solution.] Let L be the ﬁnal length of the aluminum column. This will also be the ﬁnal length of the quantity of tape now stretching from one end of the column to the other. In order to determine what the scale reading now is, we need to ﬁnd the initial length this quantity of tape had at 21.2°C (when the scale markings were presumably put on the tape). Thus, we let this initial length of tape be ( L0 )tape and require that L = ( L0 )tape ⎡⎣1 + a steel ( ΔT )⎤⎦ = ( L0 )column [1 + a Al ( ΔT )] , which gives

( L0 )tape = or

( L0 )tape =

( L0 )column [1 + a Al ( ΔT )] 1 + a steel ( ΔT )

(18.700 m ) ⎡⎣1 + ( 24 × 10 −6 ( °C )−1 ) ( 29.4°C − 21.2°C )⎤⎦

(

1 + 11 × 10 −6 ( °C )

−1

) ( 29.4°C − 21.2°C)

= 18.702 m

The measured length of the column, according to the markings on the tape, at 29.4°C is therefore 18.702 m . 10.23

If allowed to do so, the amount the band (with initial length L0 ) would contract as it cools to 37°C is ΔL = a L0 ΔT . Since the band is not allowed to contract, it will develop a tensile stress given by ⎛ a L0 ΔT ⎞ ⎛ ΔL ⎞ Stress = Y ⎜ =Y⎜ ⎟ = Ya ΔT ⎟ L0 ⎝ L0 ⎠ ⎝ ⎠ If A = (height ⋅ thickness) = (4.0 mm)(0.50 mm) = 2.0 × 10 −6 m 2 is the cross-sectional area of the band, the tension in the band will be N⎞ ⎛ −6 −1 2 F = A ⋅ ( Stress ) = 2.0 × 10 −6 m 2 ⎜ 18 × 1010 ⎟ 17.3 × 10 °C (80°C − 37°C ) = 2.7 × 10 N ⎝ m2 ⎠

(

10.24

(a)

)

(

)

The expansion of the pipeline will be ΔL = a L0 ( ΔT ), or −1 ΔL = ⎡⎣11 × 10 −6 ( °C ) ⎤⎦ (1300 km ) [ 35°C − ( −73°C )] = 1.5 km

(b)

56157_10_ch10_p322-343.indd 332

This is accommodated by accordion-like expansion joints placed in the pipeline at periodic intervals.

10/12/10 2:43:56 PM

Thermal Physics

10.25

The drum and the carbon tetrachloride, both having an initial volume of V0 = 50.0 gal, expand at different rates as the temperature rises by ΔT = 20.0°C. From ΔV = b V0 ( ΔT ) , with b = 3a as the coefﬁcient of volume expansion for the steel drum, we obtain Vspillage = ΔVcarbon tetrachloride

or

10.26

333

(a) (b)

(

Vspillage = ⎡5.81 × 10 −4 (°C ) − 3 11 × 10 −6 (°C ) ⎣ −1

−1

)⎤⎦ (50.0 gal ) ( 20.0°C) =

0.548 gal

⎛ 3.80 × 10 −3 m 3 ⎞ m0 = r0V = 7.30 × 10 2 kg m 3 (10.0 gal ) ⎜ ⎟⎠ = 27.7 kg 1 gal ⎝

(

)

V = V0 + ΔV = V0 + b V0 ( ΔT ) = V0 [1 + b ( ΔT )] or

(c)

⎛ ⎞ − ΔVsteel = ⎜ b carbon − 3a steel ⎟ V0 ( ΔT ) ⎝ tetrachloride ⎠ drum

(

(

)

)

−1 V = 1.000 m 3 ⎡1 + 9.60 × 10 −4 (°C ) ( 20.0°C )⎤ = 1.02 m 3 ⎣ ⎦

Gasoline having a mass of m = 7.30 × 10 2 kg occupies a volume of V0 = 1.000 m 3 at 0°C and a volume of V = 1.02 m 3 at 20.0°C. The density of gasoline at 20.0°C is then m 7.30 × 10 2 kg = = 716 kg m 3 1.02 m 3 V ⎛ 3.80 × 10 −3 m 3 ⎞ m20 = r20V = 7.16 × 10 2 kg m 3 (10.0 gal ) ⎜ ⎟⎠ = 27.2 kg 1 gal ⎝ r20 =

(d)

10.27

(

)

(e)

Δm = m0 − m20 = 27.7 kg − 27.2 kg = 0.5 kg

(a)

The gap width is a linear dimension, so it increases in “thermal enlargement” as the temperature goes up. The gap expands in the same way the material removed to create the gap would have expanded.

(b)

At 190°C, the length of the piece of steel that is missing, or has been removed to create the gap, is L = L0 + ΔL = L0 [1 + a ( ΔT )] . This gives

(

)

L = (1.600 cm ) ⎡⎣1 + 11 × 10 −6 °C−1 (190°C − 30.0°C )⎤⎦ = 1.603 cm 10.28

Each slab will undergo an increase in length of ΔL = a L0 ( ΔT ) , and the gap between successive slabs must be at least this wide to accommodate this expansion. Thus, the minimum gap size should be

(

)

ΔL = a concrete L0 ( T − T0 ) = 12 × 10 −6 °C−1 ( 25.0 m ) ( 50.0°C − 10.0°C ) or 10.29

(a)

ΔL = 1.2 × 10 −2 m = 1.2 cm From the ideal gas law, PV = nRT , we ﬁnd P T = nR V . Thus, if both n and V are constant as the gas is heated, the ratio P T is constant, giving Pf T f = Pi Ti , or ⎛ Pf ⎞ ⎛ 3P ⎞ T f = Ti ⎜ ⎟ = ( 300 K ) ⎜ i ⎟ = 900 K = 627°C ⎝ Pi ⎠ ⎝ Pi ⎠

(b)

If both pressure and volume double as n is held constant, the ideal gas law gives ⎛ ( 2Pi ) ( 2Vi ) ⎞ ⎛ Pf V f ⎞ = Ti ⎜ T f = Ti ⎜ ⎟ = 4Ti = 4 ( 300 K ) = 1 200 K = 927°C ⎟ PiVi ⎝ PiVi ⎠ ⎝ ⎠

56157_10_ch10_p322-343.indd 333

10/12/10 2:43:58 PM

334

10.30

Chapter 10

(a)

Ti = TC + 273.15 = (19.0 + 273.15) K = 292 K

(b)

9.50 × 10 5 Pa ⎡⎣( 20.0 L ) 10 3 cm 3 1 L 1 m 3 10 6 cm 3 ⎤⎦ PiVi ni = = = 7.83 mol RTi (8.31 J mol ⋅ K ) ( 292 K )

(c)

M CO = [12.0 + 2 (16.0 )]

(d)

mi = ni M CO = ( 7.83 mol ) ( 44.0 g mol ) = 345 g

(e)

T f = Ti + ΔT = 292 K + 224 K = 516 K

(

)(

)

2

mf M CO

= 2

mi − Δm 345 g − 82.0 g = = 5.98 mol 44.0 g mol M CO 2

Neglecting any change in volume of the tank, V f ≈ Vi , and we have Pf V f

=

Pi Vi

10.31

(

g = 44.0 g mol mol

2

nf = (f)

)

n f RT f ni RTi

(g)

⎛ n f ⎞ ⎛ Tf Pf = ⎜ ⎟ ⎜ ⎝ ni ⎠ ⎝ Ti

(a)

n=

⇒

⎛ n f Tf Pf = ⎜ ⎝ ni Ti

⎞ ⎟⎠ Pi

⎞ ⎛ 5.98 mol ⎞ ⎛ 516 K ⎞ 5 6 ⎟⎠ Pi = ⎜⎝ 7.83 mol ⎟⎠ ⎜⎝ 292 K ⎟⎠ 9.50 × 10 Pa = 1.28 × 10 Pa

(

(

)(

)

)

1.013 × 10 5 Pa atm 1.0 × 10 −6 m 3 PV = = 4.2 × 10 −5 mol RT (8.31 J mol ⋅ K )( 293 K )

molecules ⎞ ⎛ 19 Thus, N = n ⋅ N A = 4.2 × 10 −5 mol ⎜ 6.02 × 10 23 ⎟ = 2.5 × 10 molecules . ⎝ mol ⎠

(

(b)

)

(

Since both V and T are constant, n2 n1 = P2 V2 RT2

) (P V 1

1

)

RT1 = P2 P1, or

⎛P ⎞ ⎛ 1.0 × 10 −11 Pa ⎞ 4.2 × 10 −5 mol = 4.1× 10 −21 mol n2 = ⎜ 2 ⎟ n1 = ⎜ ⎝ 1.013 × 10 5 Pa ⎟⎠ ⎝ P1 ⎠

(

10.32

(a)

)

With P0 = 1.00 atm, T0 = 10.0°C = 283 K, T1 = 40.0°C = 313 K, and V1 = 0.280V0 , we ﬁnd n1 RT1 P1V1 = P0V0 n0 RT0

⇒

(

⎛V ⎞⎛ T ⎞ ⎛ 1 ⎞ ⎛ 313 K ⎞ P1 = ⎜ 0 ⎟ ⎜ 1 ⎟ P0 = ⎜ (1.00 atm ) = 3.95 atm ⎝ 0.280 ⎟⎠ ⎜⎝ 283 K ⎟⎠ ⎝ V1 ⎠ ⎝ T0 ⎠

)

and P1 = ( 3.95 atm ) 1.013 × 10 5 Pa 1 atm = 4.00 × 10 5 Pa = 400 kPa . (b)

If now, conditions inside the tire change so that V f = 1.02V1 and T f = 85.0°C = 358 K, we form a new ratio to ﬁnd Pf V f P1V1

=

n f RT f n1 RT1

⇒

⎛ V ⎞ ⎛ Tf ⎞ ⎛ 1 ⎞ ⎛ 358 K ⎞ Pf = ⎜ 1 ⎟ ⎜ ⎟ P1 = ⎜ ( 3.95 atm ) = 4.43 atm ⎝ 1.02 ⎟⎠ ⎜⎝ 313 K ⎟⎠ V T ⎝ ⎠ ⎝ f⎠ 1

⎛ 1.013 × 10 5 Pa ⎞ 5 and Pf = ( 4.43 atm ) ⎜ ⎟⎠ = 4.49 × 10 Pa = 449 kPa 1 atm ⎝

56157_10_ch10_p322-343.indd 334

10/12/10 2:44:02 PM

Thermal Physics

10.33

335

The initial and ﬁnal absolute temperatures are Ti = TC ,i + 273 = ( 25.0 + 273) K = 298 K

T f = TC, f + 273 = ( 75.0 + 273) K = 348 K

and

The volume of the tank is assumed to be unchanged, or V f = Vi . Also, two-thirds of the gas is withdrawn, so n f = ni 3 . Thus, from the ideal gas law, Pf V f Pi Vi 10.34

=

n f RT f

⇒

ni RTi

⎛ n f ⎞ ⎛ Tf Pf = ⎜ ⎟ ⎜ ⎝ ni ⎠ ⎝ Ti

⎞ ⎛ 1 ⎞ ⎛ 348 K ⎞ ⎟⎠ Pi = ⎜⎝ 3 ⎟⎠ ⎜⎝ 298 K ⎟⎠ (11.0 atm ) = 4.28 atm

The volume of the gas is V = 8.00 L = 8.00 × 10 3 cm 3 = 8.00 × 10 −3 m 3 , and the absolute temperature is T = ( 20.0 + 273) K = 293 K. The ideal gas law then gives the number of moles present as

(a)

(

)(

)

5 −3 3 PV ⎡⎣9.00 atm 1.013 × 10 Pa 1 atm ⎤⎦ 8.00 × 10 m n= = = 3.00 mol RT (8.31 J mol ⋅ K ) ( 293 K )

(b)

The number of molecules present in the container is

(

)

N = n ⋅ N A = ( 3.00 mol ) 6.02 × 10 23 molecules mol = 1.81 × 10 24 molecules 10.35

With n held constant, the ideal gas law gives V1 ⎛ P2 ⎞ ⎛ T1 ⎞ ⎛ 0.030 atm ⎞ ⎛ 300 K ⎞ −2 = =⎜ ⎟⎜ ⎟ = 4.5 × 10 V2 ⎜⎝ P1 ⎟⎠ ⎜⎝ T2 ⎟⎠ ⎝ 1.0 atm ⎠ ⎝ 200 K ⎠ Since the volume of a sphere is V = ( 4p 3)r 3 , V1 V2 = ( r1 r2 ) . 3

⎛V ⎞ r1 = ⎜ 1 ⎟ ⎝ V2 ⎠

Thus, 10.36

13

(

r2 = 4.5 × 10 −2

)

13

( 20 m ) = 7.1 m

The mass of the gas in the balloon does not change as the temperature increases. Thus, rf ri

=

(m V ) = V f

( m Vi )

i

⎛V ⎞ rf = ri ⎜ i ⎟ ⎝ Vf ⎠

or

Vf

From the ideal gas law with both n and P constant, we ﬁnd Vi V f = Ti T f , and now have ⎛T ⎞ ⎛ 273 K ⎞ rf = ri ⎜ i ⎟ = 0.179 kg m 3 ⎜ = 0.131 kg m 3 ⎝ 373 K ⎟⎠ ⎝ Tf ⎠

(

10.37

)

The pressure 100 m below the surface is found, using P1 = Patm + r gh, to be

(

)(

)

P1 = 1.013 × 10 5 Pa + 10 3 kg m 3 9.80 m s2 (100 m ) = 1.08 × 10 6 Pa The ideal gas law, with both n and T constant, gives the volume at the surface as ⎛ P ⎞ ⎛P⎞ ⎛ 1.08 × 10 6 Pa ⎞ 1.50 cm 3 = 16.0 cm 3 V2 = ⎜ 1 ⎟ V1 = ⎜ 1 ⎟ V = ⎜ 5 ⎟ P P 1.013 × 10 Pa ⎝ ⎠ ⎝ 2⎠ ⎝ atm ⎠

(

10.38

(a)

)

We assume the density r can be written as r = n aV b M c, where n is the number of moles, V is the volume, and M is the molar mass in kilograms per mole, while a, b, and c are constants to be determined by dimensional analysis. In terms of mass (M), length (L), time (T), and number of moles (N), the fundamental units of density are [ r ] = [ mass / volume ] = ML−3, those of n are [ n ] = N, for volume [V ] = L3, and molar mass [ M ] = [ kg mol ] = MN −1 . In terms of basic units, our assumed equation for density becomes

[ r ] = [n ]a [V ]b [ M ]c

or

( ) ( MN )

M1L−3 = N a L3

b

−1 c

= N a−c L3b Mc continued on next page

56157_10_ch10_p322-343.indd 335

10/12/10 2:44:06 PM

336

Chapter 10

and equating the powers of each of the basic units on the two sides of the equation gives: 1 = c ⇒ c = 1;

− 3 = 3b ⇒ b = −1;

0 = a−c ⇒ a = c =1

so our expression for density, derived by dimensional analysis is r = n1V −1 M 1 , or r = nM V where M is expressed in kilograms per mole. (b)

From the ideal gas law, PV = nRT , or P = ( n V ) RT . From the result of part (a), we may write n V = r M, so the ideal gas law may be written in terms of the density of the gas as P = (r M)RT , where M is again expressed in kilograms per mole.

(c)

For carbon dioxide, M = 44 g mol = 44 × 10 −3 kg mol. Then, if the pressure is P = ( 90.0 atm ) 1.013 × 10 5 Pa 1 atm = 9.12 × 10 6 Pa, and T = 7.00 × 10 2 K, the density of the atmosphere on Venus is

(

r= (d)

)

(

)(

)

9.12 × 10 6 Pa 44 × 10 −3 kg mol PM = = 69 kg m 3 RT (8.31 J mol ⋅ K ) 7.00 × 10 2 K

(

)

The density of the evacuated steel shell would be rshell =

mshell 2.00 × 10 2 kg = = 47.7 kg m 3 3 Vshell 4p (1.00 m ) 3

Since rshell < ratmosphere , this shell would rise in the atmosphere on Venus. 10.39

The average kinetic energy of the molecules of any ideal gas at 300 K is KE =

10.40

1 3 3⎛ J⎞ −21 mv 2 = k BT = ⎜ 1.38 × 10 −23 ⎟ ( 300 K ) = 6.21 × 10 J 2 2 2⎝ K⎠

Since the sample contains three times Avogadro’s number of molecules, there must be 3 moles of gas present. The ideal gas law then gives P=

nRT ( 3 mol ) (8.31 J mol ⋅ K ) ( 293 K ) = = 9.13 × 10 5 Pa V ( 0.200 m )3

The force this gas will exert on one face of the cubical container is

(

)

F = PA = 9.13 × 10 5 Pa ( 0.200 m ) = 3.65 × 10 4 N = 36.5 kN 10.41

2

One mole of any substance contains Avogadro’s number of molecules and has a mass equal to the molar mass, M. Thus, the mass of a single molecule is m = M N A . For helium, M = 4.00 g mol = 4.00 × 10 −3 kg mol, and the mass of a helium molecule is m=

4.00 × 10 −3 kg mol = 6.64 × 10 −27 kg molecule 6.02 × 10 23 molecule mol

Since a helium molecule contains a single helium atom, the mass of a helium atom is matom = 6.64 × 10 −27 kg

56157_10_ch10_p322-343.indd 336

10/12/10 2:44:10 PM

Thermal Physics

10.42

(a)

337

The rms speed of molecules in a gas of molar mass M and absolute temperature T is vrms = 3RT M . Since the molar masses of 35 Cl and 37 Cl are 35.0 × 10 −3 kg mol and 37.0 × 10 −3 kg mol, respectively, the desired ratio is vrms ( 35 Cl) = vrms ( 37 Cl)

(b) 10.43

3RT 35.0 × 10 −3 kg mol 3RT 37.0 × 10

−3

kg mol

37.0 = 1.03 35.0

Since the above ratio is larger than 1, the less massive atom,

35

Cl, moves faster .

If vrms = vesc , we must have vrms = 3k BT m = vesc , where k B = 1.38 × 10 −23 J K is Boltzmann’s constant and m is the mass of a molecule (for helium, m = 6.64 × 10 −27 kg). Thus, the required 2 absolute temperature is T = mvesc 3k B. (a)

To have vrms = vesc on Earth where vesc = 1.12 × 10 4 m s, the required temperature for the helium gas is

(6.64 × 10 kg) (1.12 × 10 T= 3 (1.38 × 10 J K) −27

4

ms

−23

(b)

−27

3

ms

−23

(a)

)

2

= 2.01 × 10 4 K

If vrms = vesc on the Moon where vesc = 2.37 × 10 3 m s, the temperature must be

(6.64 × 10 kg) ( 2.37 × 10 T= 3 (1.38 × 10 J K) 10.44

=

)

2

= 901 K

The volume occupied by this gas is V = 7.00 L = 7.00 × 10 3 cm 3 = 7.00 × 10 −3 m 3 Then, the ideal gas law gives

(

(b)

KE molecule = (c)

10.45

)

(

)

3 3 k B T = 1.38 × 10 −23 J K ( 385 K ) = 7.97 × 10 −21 J 2 2

You would need to know the mass of the gas molecule to ﬁnd its rms speed, which in turn requires knowledge of the molar mass of the gas.

Consider a time interval of 1.0 min = 60 s, during which 150 bullets bounce off Superman’s chest. From the impulse-momentum theorem, the magnitude of the average force exerted on Superman is Fav = =

10.46

)(

1.60 × 10 6 Pa 7.00 × 10 −3 m 3 PV = = 385 K nR ( 3.50 mol ) (8.31 J mol ⋅ K ) The average kinetic energy per molecule in this gas is T=

150 Δp bullet 150 m ( v − v0 ) I = = Δt Δt Δt

(

)

150 8.0 × 10 −3 kg ⎡⎣( 400 m s ) − ( −400 m s ) ⎤⎦ = 16 N 60 s

From the impulse-momentum theorem, the average force exerted on the wall is Fav =

N Δp molecule N m ( v − v0 ) I = = Δt Δt Δt

(5.0 × 10 ) ( 4.68 × 10 23

or

Fav =

−26

)

kg ⎡⎣( 300 m s ) − ( −300 m s ) ⎤⎦ = 14 N 1.0 s continued on next page

56157_10_ch10_p322-343.indd 337

10/12/10 2:44:13 PM

338

Chapter 10

The pressure on the wall is then P= 10.47

Fav 14 N ⎛ 10 4 cm 2 ⎞ = 1.8 × 10 4 N m 2 = 18 kPa = A 8.0 cm 2 ⎜⎝ 1 m 2 ⎟⎠

As the pipe undergoes a temperature change ΔT = 46.5°C − 18.0°C = 28.5°C, the expansion toward the right of the horizontal segment is ΔLx = a L0 x ( ΔT ) −1 = ⎡⎣17 × 10 −6 ( °C ) ⎤⎦ ( 28.0 cm )( 28.5°C ) = 1.4 × 10 −2 cm = 0.14 mm

The downward expansion of the vertical section is −1 ΔL y = a L0 y ( ΔT ) = ⎡⎣17 × 10 −6 ( °C ) ⎤⎦ (134 cm )( 28.5°C ) = 6.5 × 10 −2 cm = 0.65 mm

The total displacement of the pipe elbow is ΔL = ΔL2x + ΔL2y =

10.48

( 0.14 mm )2 + ( 0.65 mm )2 = 0.66 mm

at

⎛ ΔL y ⎞ ⎛ 0.65 mm ⎞ q = tan −1 ⎜ = 78° = tan −1 ⎜ ⎟ ⎝ 0.14 mm ⎟⎠ ⎝ ΔLx ⎠

or

ΔL = 0.66 mm toward the right at 78° below the horizontal

(a)

−1 ΔL = a L0 ( ΔT ) = ⎡⎣9.00 × 10 −6 ( °C ) ⎤⎦ ( 30.0 cm )( 65.0°C ) =1.76 × 10 −2 cm = 0.176 mm

(b)

−1 ΔD = a D0 ( ΔT ) = ⎡⎣9.00 × 10 −6 ( °C ) ⎤⎦ (1.50 cm )( 65.0°C ) = 8.78 × 10 −4 cm = 8.78 m m

(c)

⎛ p D02 ⎞ p The initial volume is V0 = ⎜ 1.50 × 10 −2 m L0 = ⎟ 4 ⎝ 4 ⎠

(

)

2

( 0.300 m ) = 5.30 × 10 −5 m 3

ΔV = b V0 ( ΔT )

(

)

−1 = 3a V0 ( ΔT ) = 3 ⎡⎣9.00 × 10 −6 ( °C ) ⎤⎦ 5.30 × 10 −5 m 3 ( 65.0°C ) = 9.30 × 10 −8 m 3

10.49

The number of moles of CO 2 present is n = m M, where m = 6.50 g and M = 44.0 g mol. Thus, at the given temperature ( 20.0°C = 293 K ) and pressure (1.00 atm = 1.013 × 10 5 Pa), the volume will be V=

10.50

nRT mRT ( 6.50 g ) (8.31 J mol ⋅ K ) ( 293 K ) = = = 3.55 × 10 −3 m 3 = 3.55 L 5 P MP 44.0 g mol 1.013 × 10 Pa ( )

(

)

Note that the moment of inertia of each object in Table 8.1 is of the form I = CM 2 , where C is a constant, M is the mass of the object, and is some length. As the temperature increases, the factors C and M are unchanged, while the value of the length measurement at temperature T is = 0 (1 + a ⋅ ΔT ). Therefore, the percentage increase in the moment of inertia of the object is ⎧⎪ CM ⎡ I − I0 ⎣ % change = × 100% = ⎨ I0 ⎩⎪ or

0

2 (1+ a ⋅ ΔT )⎤⎦ − CM 20 ⎫⎪

CM

2 0

⎬ × 100% ⎭⎪

2 % change = ⎡⎣(1 + a ⋅ ΔT ) − 1⎤⎦ × 100%

continued on next page

56157_10_ch10_p322-343.indd 338

10/12/10 2:44:17 PM

Thermal Physics

(a)

339

For copper, with a = 17 × 10 −6 °C−1 and ΔT = 100°C,

{

(

)

}

2

% change = ⎡⎣1 + 17 × 10 −6 °C−1 (100°C )⎤⎦ − 1 × 100% = 0.34% (b)

For aluminum, with a = 24 × 10

{

−6

°C

(

−1

and ΔT = 100°C ,

)

2

}

% change = ⎡⎣1 + 24 × 10 −6 °C−1 (100°C )⎤⎦ − 1 × 100% = 0.48% (c)

10.51

The moment of inertia for each of the shapes has the same mathematical form: the product of a constant, the mass, and the square of a length.

For a temperature change ΔTF = TF − TF ,0 on the Fahrenheit scale, the corresponding temperature change on the Celsius scale is ΔTC = TC − TC ,0 =

(

)

(

)

5 (TF − 32) − 95 TF ,0 − 32 = 95 TF − TF ,0 = 95 ( ΔTF ) 9

Therefore, if L0 = 35.000 m and ΔTF = 90.000°F − 15.000°F = 75.000°F , the ﬁnal length of the beam is 5 L = L0 [1 + a ( ΔT )] = ( 35.000 m ) ⎡⎢1 + 11 × 10 −6 °C−1 ( 75.000 ) °C ⎤⎥ = 35.016 m 9 ⎣ ⎦

(

10.52

)

When air trapped in the tube is compressed, at constant temperature, into a cylindrical volume 0.40 m long, the ideal gas law gives its pressure as ⎛V ⎞ ⎛L ⎞ ⎛ 1.5 m ⎞ P2 = ⎜ 1 ⎟ P1 = ⎜ 1 ⎟ P1 = ⎜ 1.013 × 10 5 Pa = 3.8 × 10 5 Pa ⎝ 0.40 m ⎟⎠ ⎝ V2 ⎠ ⎝ L2 ⎠

(

)

This is also the water pressure at the bottom of the lake. Thus, P = Patm + r gh gives the depth of the lake as h= 10.53

P2 − Patm ( 3.8 − 1.013) × 105 Pa = = 28 m rg 10 3 kg m 3 9.80 m s2

(

)(

)

The mass of CO 2 produced by three astronauts in 7.00 days is m = 3 (1.09 kg d ) ( 7.00 d ) = 22.9 kg, and the number of moles of CO 2 available is n=

m 22.9 kg = = 520 mol M 44.0 × 10 −3 kg mol

The recycling process will generate 520 moles of methane to be stored. In a volume of V = 150 L = 0.150 m 3 and at temperature T = − 45.0°C = 228 K, the pressure of the stored methane is P= 10.54

(a)

nRT ( 520 mol ) (8.31 J mol ⋅ K ) ( 228 K ) = = 6.57 × 10 6 Pa = 6.57 MPa V 0.150 m 3

The piston in this vertical cylinder has three forces acting on it. These are: (1) a downward gravitational force, mg, the piston’s own weight; (2) a downward pressure force, Fd = P0 A, due to the atmospheric pressure above the piston; and (3) an upward pressure force, Fu = PA, due to the absolute pressure of the gas trapped inside the cylinder. Since the piston is in equilibrium, Newton’s second law requires ΣFy = 0

⇒

Fu − mg − Fd = 0

or

PA = mg + P0 A

[1] continued on next page

56157_10_ch10_p322-343.indd 339

10/12/10 2:44:21 PM

340

Chapter 10

From the ideal gas law, the absolute pressure of the trapped gas is P=

nRT nRT = V Ah

[2]

Substituting Equation [2] into [1] yields ⎛ nRT ⎞ ⎜⎝ A h ⎟⎠ A = mg + P0 A (b)

10.55

or

h=

nRT mg + P0 A

mg + P0 A where P0 is atmospheric pressure. This is greater than atmospheric pressure because mg A > 0 . From Equation [1] above, the absolute pressure inside the cylinder is P =

(c)

Observe from the result of part (a) above, if the absolute temperature T increases, the equilibrium value of h also increases.

(a)

As the acetone undergoes a change in temperature ΔT = ( 20.0 − 35.0 ) °C = −15.0 °C , the ﬁnal volume will be V f = V0 + ΔV = V0 + b V0 ( ΔT ) = V0 [1 + b ( ΔT )]

(

= (100 mL ) ⎡1 + 1.50 × 10 −4 (°C ) ⎣ (b)

−1

)( −15.0 °C)⎤⎦ = 99.8 mL

When acetone at 35°C is poured into the Pyrex ﬂask that was calibrated at 20°C, the volume of the ﬂask temporarily expands to be larger than its calibration markings indicate. −1 However, the coefﬁcient of volume expansion for Pyrex ⎡⎣ b = 3a = 9.6 × 10 −6 (°C ) ⎤⎦

−1 is much smaller than that of acetone ⎡⎣ b = 1.5 × 10 −4 (°C ) ⎤⎦ . Hence, the temporary increase in the volume of the ﬂask will be much smaller than the change in volume of the acetone as the materials cool back to 20°C, and this change in volume of the ﬂask has negligible effect on the answer.

10.56

If Pi is the initial gauge pressure of the gas in the cylinder, the initial absolute pressure is Pi, abs = Pi + P0 , where P0 is atmospheric pressure. Likewise, the ﬁnal absolute pressure in the cylinder is Pf , abs = Pf + P0 , where Pf is the ﬁnal gauge pressure. The initial and ﬁnal masses of gas in the cylinder are mi = ni M and m f = n f M , where n is the number of moles of gas present and M is the molar mass of this gas. Thus, m f mi = n f ni . We assume the cylinder is a rigid container whose volume does not vary with internal pressure. Also, since the temperature of the cylinder is constant, its volume does not expand nor contract. Then, the ideal gas law (using absolute pressures) with both temperature and volume constant gives Pf , abs V Pi, abs V

=

n f RT ni RT

=

mf mi

or

⎛ Pf , abs ⎞ mf = ⎜ ⎟ mi ⎝ Pi, abs ⎠

and in terms of gauge pressures, ⎛ Pf + P0 ⎞ mf = ⎜ mi ⎝ Pi + P0 ⎟⎠

56157_10_ch10_p322-343.indd 340

10/12/10 2:44:24 PM

Thermal Physics

10.57

(a)

The volume of the liquid expands by ΔVliquid = bV0 ( ΔT ) and the volume of the glass ﬂask expands by ΔVflask = ( 3a ) V0 ( ΔT ). The amount of liquid that must overﬂow into the capillary is Voverflow = ΔVliquid − ΔVflask = V0 ( b − 3a ) ( ΔT ). The distance the liquid will rise into the capillary is then Δh =

(b)

341

Voverflow ⎛V ⎞ = ⎜ 0 ⎟ ( b − 3a ) ( ΔT ) ⎝ A⎠ A

For a mercury thermometer, b Hg = 1.82 × 10 −4 ( °C ) and (assuming Pyrex glass), −1 −1 3a glass = 3 3.2 × 10 −6 ( °C ) = 9.6 × 10 −6 ( °C ) . Thus, the expansion of the mercury −1

(

)

is almost 20 times the expansion of the flask , making it a rather good approximation to neglect the expansion of the ﬂask. 10.58

(a)

The initial absolute pressure in the tire is P1 = Patm + P1, gauge = 2.80 atm, and the ﬁnal absolute pressure is P2 = Patm + P2, gauge = 3.20 atm. The ideal gas law, with both n and V constant, gives ⎛P ⎞ ⎛ 3.20 atm ⎞ T2 = ⎜ 2 ⎟ T1 = ⎜ ( 300 K ) = 343 K ⎝ 2.80 atm ⎟⎠ ⎝ P1 ⎠

(b)

When the quantity of gas varies, while volume and temperature are constant, the ideal gas law gives n3 = P3 . Thus, when air is released to lower the absolute n2 P2 pressure back to 2.80 atm, we have n3 2.80 atm = = 0.875 n2 3.20 atm At the end, we have 87.5% of the original mass of air remaining, or 12.5% of the original mass was released.

10.59

L

After expansion, the increase in the length of one span is

y

L0 ⫽ 125 m

ΔL = a L.0 ( ΔT ) −1 = ⎡⎣12 × 10 −6 ( °C ) ⎤⎦ (125 m ) ( 20.0°C ) = 3.0 × 10 −2 m

giving a ﬁnal length of L = L0 + ΔL = 125 m + 3.0 × 10 −2 m From the Pythagorean theorem, y = L2 − L20 = 10.60

(a)

[(125 + 0.030 ) m ] − (125 m ) 2

From the ideal gas law,

2

= 2.7 m

⎛ P ⎞⎛ V ⎞ ⎛ T ⎞ P2V2 P1V1 = , or ⎜ 2 ⎟ ⎜ 2 ⎟ = ⎜ 2 ⎟ T2 T1 ⎝ P1 ⎠ ⎝ V1 ⎠ ⎝ T1 ⎠

The initial conditions are: P1 = 1 atm, V1 = 5.00 L = 5.00 × 10 −3 m 3 , and T1 = 20.0°C = 293 K The ﬁnal conditions are: P2 = 1 atm +

F k ⋅h = 1 atm + , V2 = V1 + A ⋅ h, and T2 = 250 °C = 523 K A A continued on next page

56157_10_ch10_p322-343.indd 341

10/12/10 2:44:27 PM

342

Chapter 10

Thus,

⎛ k ⋅h ⎞⎛ A ⋅ h ⎞ ⎛ 523 K ⎞ ⎜⎝ 1 + A (1 atm ) ⎟⎠ ⎜⎝ 1 + V ⎟⎠ = ⎜⎝ 293 K ⎟⎠ 1

(

)

( (

)

⎛ ⎞⎛ 2.00 × 10 3 N m ⋅ h 0.010 0 m 2 ⋅ h ⎞ ⎛ 523 ⎞ 1+ ⎜ 1+ ⎟ ⎜ ⎟ =⎜ ⎟ 0.010 0 m 2 1.013 × 10 5 N m 2 ⎠ ⎝ 5.00 × 10 −3 m 3 ⎠ ⎝ 293 ⎠ ⎝

or

(

)(

)

)

Simplifying and using the quadratic formula yields h = 0.169 m = 16.9 cm (b) 10.61

(a)

P2 = 1 atm +

(

)

2.00 × 10 3 N m ( 0.169 m ) k ⋅h = 1.013 × 10 5 Pa + = 1.35 × 10 5 Pa A 0.0100 m 2

The two metallic strips have the same length L0 at the initial temperature T0 . After the temperature has changed by ΔT = T − T0 , the lengths of the two strips are L1 = L0 [1 + a 1 ( ΔT )]

L2 = L0 [1 + a 2 ( ΔT )]

and

The lengths of the circular arcs are related to their radii by L1 = r1q and L2 = r2q , where q is measured in radians. Thus,

10.62

Δr = r2 − r1 =

(a − a 1 ) L0 ( ΔT ) L2 L1 (a 2 − a 1 ) L0 ( ΔT ) − = , or q = 2 q q q Δr

(b)

As seen in the above result, q = 0 if either ΔT = 0 or a 1 = a 2 .

(c)

If ΔT < 0, then q is negative so the bar bends in the opposite direction .

(a)

If the bridge were free to expand as the temperature increased by ΔT = 20°C, the increase in length would be

(

)

ΔL = a L0 ( ΔT ) = 12 × 10 −6 °C−1 ( 250 m )( 20°C ) = 6.0 × 10 −2 m = 6.0 cm (b)

When the bridge is not allowed to expand naturally, stress builds up in the bridge, effectively compressing it the distance ΔL that it would normally have expanded. Combining the deﬁning equation for Young’s modulus, Y = Stress Strain = Stress ( ΔL L ) with the expression, ΔL = a L ( ΔT ) , for the linear expansion when the temperature changes by ΔT yields ⎛ a L ( ΔT ) ⎞ ⎛ ΔL ⎞ Stress = Y ⎜ =Y⎜ ⎟⎠ = a Y ( ΔT ) ⎝ L ⎟⎠ ⎝ L

(c)

When ΔT = 20°C, the stress in the speciﬁed bridge would be

(

)(

)

Stress = a Y ( ΔT ) = 12 × 10 −6 °C−1 2.0 × 1010 Pa ( 20°C ) = 4.8 × 10 6 Pa Since this is considerably less than the maximum stress, 2.0 × 10 7 Pa , that concrete can withstand, the bridge will not crumble .

56157_10_ch10_p322-343.indd 342

10/12/10 2:44:31 PM

Thermal Physics

10.63

(a)

343

Yes, the angular speed will increase as the disk cools. Since no external torque acts on the disk, the angular momentum of the disk will be conserved. As the disk cools, its radius, and hence its moment of inertia will decrease. Then, in order to keep angular momentum ( L = Iw ) constant, the angular speed must increase. Since angular momentum is conserved, Iw = I 0w 0 or w = ( I 0 I ) w 0 . Thus,

(b)

2

2 ⎛ ⎞ R0 ⎛ 1 MR02 ⎞ ⎛ R0 ⎞ w = ⎜ 12 w = ⎟ w0 ⎜⎝ ⎟⎠ w 0 = ⎜ 0 2⎟ R ⎝ 2 MR ⎠ ⎝ R0 [1 + a ( ΔT )] ⎠

or w= 10.64

w0

[1 + a ( ΔT )]

2

=

25.0 rad s

(

⎡1 + 17 × 10 ⎣

−6

)

°C−1 ( 20.0 − 850 ) °C ⎤⎦

2

= 25.7 rad s

Let container 1 be maintained at T1 = T0 = 0°C = 273 K, while the temperature of container 2 is raised to T2 = 100°C = 373 K. Both containers have the same constant volume, V, and the same initial pressures, ( P0 )2 = ( P0 )1 = P0 . As the temperature of container 2 is raised, gas ﬂows from one container to the other until the ﬁnal pressures are again equal, P2 = P1 = P. The total mass of gas is constant, so n2 + n1 = ( n0 )2 + ( n0 )1 From the ideal gas law, n =

[1] PV , so Equation [1] becomes RT

PV PV P0V P0V , + = + RT2 RT1 RT0 RT0

or

⎛ 1 1 ⎞ 2P P⎜ + ⎟ = 0 T0 ⎝ T2 T1 ⎠

Thus, P=

56157_10_ch10_p322-343.indd 343

2P0 T0

⎛ T1T2 ⎞ 2 (1.00 atm ) ⎛ 273 ⋅ 373 ⎞ ⎜⎝ 273 + 373 ⎟⎠ = 1.15 atm ⎜⎝ T + T ⎟⎠ = 273 1 2

10/12/10 2:44:35 PM

11 Energy in Thermal Processes QUICK QUIZZES 1.

(a) Water, glass, iron. Because it has the highest speciﬁc heat ( 4186 J kg ⋅ °C ) , water has the smallest change in temperature. Glass is next (837 J kg ⋅ °C ), and iron ( 448 J kg ⋅ °C ) is last. (b) Iron, glass, water. For a given temperature increase, the energy transfer by heat is proportional to the speciﬁc heat.

2.

Choice (b). The slopes are proportional to the reciprocal of the speciﬁc heat, so larger speciﬁc heat results in a smaller slope, meaning more energy to achieve a given change in temperature.

3.

Choice (c). The blanket acts as a thermal insulator, slowing the transfer of energy by heat from the air into the cube.

4.

Choice (b). The rate of energy transfer by conduction through a rod is proportional to the difference in the temperatures of the ends of the rod. When the rods are in parallel, each rod experiences the full difference in the temperatures of the two regions. If the rods are connected in series, neither rod will experience the full temperature difference between the two regions, and hence neither will conduct energy as rapidly as it did in the parallel connection.

5.

(a) PA PB = 4. From Stefan’s law, the power radiated from an object at absolute temperature T is proportional to the surface area of that object. Star A has twice the radius and four times the surface area of star B. (b) PA PB = 16. From Stefan’s law, the power radiated from an object having surface area A is proportional to the fourth power of the absolute temperature. Thus, 4 PA = s Ae ( 2TB ) = 24 s AeTB4 = 16PB. (c) PA PB = 64. When star A has both twice the radius and twice the absolute temperature of star B, the ratio of the radiated powers is

(

)

( (

) )

2 4 PA s AA eTA4 s 4p RA (1) TA ( 2RB ) ( 2TB ) = = = = 22 24 = 64 PB s ABeTB4 s 4p RB2 (1) TB4 RB2TB4 2

4

( )( )

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

From the mechanical equivalent of heat, 1 cal = 4.186 J. Therefore, ⎛ 4.186 J ⎞ 3.50 × 10 3 cal = 3.50 × 10 3 cal ⎜ = 1.47 × 10 4 J ⎝ 1 cal ⎟⎠

(

)

and (b) is the correct choice for this question. 2.

From Q = mc(ΔT ), we see that when equal-mass samples have equal amounts of energy transferred to them by heat, the expected temperature rise is inversely proportional to the speciﬁc heat: ΔT =

(Q m) c

344

56157_11_ch11_p344-370.indd 344

10/12/10 2:45:16 PM

Energy in Thermal Processes

345

Thus, the alcohol sample, with a speciﬁc heat that is about one-half that of water, should experience a temperature rise that is approximately twice that of the water sample. The correct choice is (c). 3.

The rate of energy transfer by conduction through a wall of area A and thickness L is P = k A ( Th − Tc ) L, where k is the thermal conductivity of the material making up the wall, while Th and Tc are the temperatures on the hotter and cooler sides of the wall, respectively. For the case given, the transfer rate will be J ⎛ ⎞ 2 P = ⎜ 0.08 ⎟ 48.0 m ⎝ s ⋅ m ⋅ °C ⎠

(

25°C − 14°C ) = 1.1 × 10 ) (4.00 ( × 10 m ) −2

3

J s = 1.1 × 10 3 W

and (d) is the correct answer. 4.

The energy which must be added to the 0°C ice to melt it, leaving liquid at 0°C, is

(

)

Q1 = mL f = ( 2.00 kg ) 3.33 × 10 5 J kg = 6.66 × 10 5 J Once this is done, there is Q2 = Qtotal − Q1 = 9.30 × 10 5 J − 6.66 × 10 5 J = 2.64 × 10 5 J of energy still available to raise the temperature of the liquid. The change in temperature this produces is ΔT = T f − 0°C =

Q2 2.64 × 10 5 J = = 31.5°C mcwater ( 2.00 kg ) ( 4 186 J kg ⋅ °C )

so the ﬁnal temperature is T f = 0°C + 31.5°C = 31.5°C and the correct choice is (c). 5.

The required energy input is Q = mc ( ΔT ) = ( 5.00 kg ) (128 J kg ⋅ °C ) ( 327°C − 20.0°C ) = 1.96 × 10 5 J and the correct response is (e).

6.

The power radiated by an object with emissivity e, surface area A, and absolute temperature T, in a location with absolute ambient temperature T0, is given by P = s Ae T 4 − T04 , where s = 5.669 6 × 10 −8 W m 2 ⋅ K 4 is a constant. Thus, for the given spherical object A = 4p r 2 with T = 273 + 135 = 408 K and T0 = 273 + 25 = 298 K, we have

(

(

(

)

)

)

2 4 4 P = 5.669 6 × 10 −8 W m 2 ⋅ K 4 4p ( 2.00 m ) ( 0.450 ) ⎡⎣( 408 K ) − ( 298 K ) ⎤⎦

yielding 7.

P = 2.54 × 10 4 W,

so (e) is the correct choice.

The temperature of the ice must be raised to the melting point, ΔT = +20.0°C, before it will start to melt. The total energy input required to melt the 2.00-kg of ice is Q = mc ( ΔT ) + mL f = ( 2.00 kg ) ⎡⎣( 2 090 J kg ⋅ °C ) ( 20.0°C ) + 3.33 × 10 5 J kg ⎤⎦ = 7.50 × 10 5 J The time the heating element will need to supply this quantity of energy is Δt =

Q 7.50 × 10 5 J = = 750 P 1.00 × 10 3 J s

⎛ 1 min ⎞ s⎜ = 12.5 min ⎝ 60 s ⎟⎠

making (d) the correct choice.

56157_11_ch11_p344-370.indd 345

10/12/10 2:45:19 PM

346

Chapter 11

8.

If energy transfer between the environment and the contents of the calorimeter cannot be avoided, one would like the initial temperature to be such that the contents of the calorimeter would gain as much energy from the environment in one part of the process as it loses to the environment in another part of the process. Thus, we would like (after a few trial runs) to choose an initial temperature such that room temperature will be about halfway between the initial and ﬁnal temperatures of the calorimeter contents. The best response is therefore choice (a).

9.

Since less energy was required to produce a 5°C rise in the temperature of the ice than was required to produce a 5°C rise in temperature of an equal mass of water, we conclude that the ⎛ Q ⎞ speciﬁc heat of ice ⎜ c = is less than that of water. Thus, choice (c) is correct. m ( ΔT ) ⎟⎠ ⎝

10.

One would like the poker to be capable of absorbing a large amount of energy, but undergo a small rise in temperature. This means it should be made of a material with a high speciﬁc heat capacity. Also, it is desirable that energy absorbed by the end of the poker in the ﬁre be conducted to the person holding the other end very slowly. Thus, the material should have a low thermal conductivity. The correct choice is (d).

11.

With e A = eB, rA = 2rB, and TA = 2TB, the ratio of the power output of A to that of B is 2

4

4 PA s AA e A TA 4p rA2 TA4 ⎛ rA ⎞ ⎛ TA ⎞ 2 4 6 = = = = ( 2 ) ( 2 ) = ( 2 ) = 64 PB s AB eB TB4 4p rB2 TB4 ⎜⎝ rB ⎟⎠ ⎜⎝ TB ⎟⎠

making (e) the correct choice. 12.

By agitating the coffee inside this sealed, insulated container, the person is raising the internal energy of the coffee, which will result is a rise in the temperature of the coffee. However, doing this for only a few minutes, the temperature rise will be quite small. The correct response to this question is (d).

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The high thermal capacity of the barrel of water and its high heat of fusion mean that a large amount of energy would have to leak out of the cellar before the water and produce froze solid. Also, evaporation of the water keeps the relative humidity high to protect foodstuffs from drying out.

4.

(a)

Yes, wrap the blanket around the ice chest. The environment is warmer than the ice, so the low thermal conductivity of the blanket slows energy transfer by heat from the environment to the ice.

(b)

Explain to your little sister that her body is warmer than the environment and requires energy transfer by heat into the air to remain at a ﬁxed temperature. The blanket will slow this conduction and cause her to feel warmer, not cool like the ice.

6.

Yes, if you know the speciﬁc heat of zinc and copper, you can determine the relative fraction of each by heating a known weight of pennies to a speciﬁc initial temperature, say 100°C, then dump them into a known quantity of water, at say 20°C. The equation for conservation of energy will be mpennies ⎡⎣ x ⋅ cCu + (1 − x ) c Zn ⎤⎦ (100°C − T ) = mwater cwater ( T − 20°C )

56157_11_ch11_p344-370.indd 346

10/12/10 2:45:22 PM

Energy in Thermal Processes

347

The equilibrium temperature, T, and the masses will be measured. The speciﬁc heats are known, so the fraction of metal that is copper, x, can be computed. 8.

Write mwater cwater (1°C ) = ( rair V ) cair (1°C ), to ﬁnd V=

(

)

1.0 × 10 3 kg ( 4 186 J kg ⋅ °C ) mwater cwater = = 3.2 × 10 3 m 3 rair cair 1.3 kg m 3 1.0 × 10 3 J kg ⋅ °C

(

)(

)

10.

The black car absorbs more of the incoming energy from the Sun than does the white car, making it more likely to cook the egg.

12.

Keep them dry. The air pockets in the pad conduct energy slowly. Wet pads absorb some energy in warming up themselves, but the pot would still be hot and the water would quickly conduct a lot of energy to your hand.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

0.234 kJ kg ⋅ °C

4.

0.15 mm

6.

(a)

2.3 × 10 6 J

(b)

2.8 × 10 4 stairs

(c)

7.0 × 10 3 stairs

8.

(a)

P = Fv

(b)

P = ma 2 t

(c)

2.20 m s2

(d)

P = ( 363 W s ) ⋅ t

(e) 1.74 Cal s

10.

0.105°C

12.

(a)

9.9 × 10 −3 °C

14.

(a)

Stress =

F = Y [a ( ΔT )] A

(d) 6.7 × 10 6 J

(b)

It is absorbed by the rough horizontal surface.

(b)

Q=

(e)

79°C

mc ⎛ F ⎞ ⎜ ⎟ Yα ⎝ A ⎠

(c)

96.0 kg

(f)

∼4 h

16.

467

18.

The copper bullet wins: Tcopper = 89.7°C, Tsilver = 89.8°C

20.

29.6°C

22.

47°C

24.

1.18 × 10 3 J kg ⋅ °C

26.

49 kJ

28.

0.12 MJ

30.

(a) ice at −10.0°C to ice at 0°C; ice at 0°C to liquid water at 0°C; water at 0°C to water at T; aluminum at 20.0°C to aluminum at T; ethyl alcohol at 30.0°C to ethyl alcohol at T . (b)

56157_11_ch11_p344-370.indd 347

See Solution.

10/12/10 2:45:23 PM

348

Chapter 11

(c) mice cice (10.0°C ) + mice L f + mice cwater ( T − 0 ) + mAl cAl [T − 20.0°C ] + malc calc [T − 30.0°C ] = 0 (d) 4.81°C 32.

0.33 kg, 0.067 or 6.7%

34.

403 cm 3

36.

11.1 W

38.

(a) 6 × 10 3 W

40.

(a)

(b)

5 × 108 J

Rskin = 5.0 × 10 −2 m 2 ⋅ K W, Rfat = 2.5 × 10 −2 m 2 ⋅ K W, Rtissue = 6.4 × 10 −2 m 2 ⋅ K W, Rtotal = 0.14 m 2 ⋅ K W

(b) 5.3 × 10 2 W 42.

39 m 3

44.

(a) 52 W

46.

7.3 × 10 −2 W m ⋅ °C

48.

330 K

50.

1.1 × 10 −5 m 2

52.

1.83 h

54.

(a) 1.1 × 10 2 W

(b)

2 kW

(b) The positive sign indicates that the body is radiating energy away faster than it absorbs energy from the environment. 56.

1.8 kg

58.

(a) 1.6 × 10 2 W

(b)

2.7 × 10 2 W

(b)

4.5°C

(c)

11 W

(d) 1.2 × 10 2 W 60.

45°C

62.

(a) 2.0 kW

64.

28°C

66.

(a)

68.

(a) 0.457 kg

2.03 × 10 3 J s

(b) 7.84 ft 2 ⋅ °F ⋅ h Btu

(b) If the samples and inner surface of the insulation are preheated, nothing undergoes a temperature change during the test. Therefore, only the mass of the wax, which undergoes a change of phase, needs to be known.

56157_11_ch11_p344-370.indd 348

10/12/10 2:45:26 PM

Energy in Thermal Processes

349

70.

0.9 kg

72.

(a) The processes involved are the removal of energy to: (1) cool liquid water from 20.0 °C to 0°C, (2) convert liquid water at 0°C to solid water (ice) at 0°C, and (3) cool ice from 0°C to −8.00°C. (b) 32.5 kJ

PROBLEM SOLUTIONS 11.1

As mass m of water drops from the top to the bottom of the falls, the gravitational potential energy given up (and hence, the kinetic energy gained) is Q = mgh. If all of this goes into raising the temperature, the rise in temperature will be ΔT =

(

)

9.80 m s2 (807 m ) Q m gh = = = 1.89°C mcwater m cwater 4 186 J kg ⋅ °C T f = Ti + ΔT = 15.0°C +1.89°C = 16.9°C

and the ﬁnal temperature is

Q 1.23 × 10 3 J = = 234 J kg ⋅ °C = 0.234 kJ kg ⋅ °C m ( ΔT ) ( 0.525 kg ) (10.0°C )

11.2

c=

11.3

The mass of water involved is kg ⎞ ⎛ m = rV = ⎜ 10 3 3 ⎟ 4.00 × 1011 m 3 = 4.00 × 1014 kg ⎝ m ⎠

(

(

)

(a)

Q = mc ( ΔT ) = 4.00 × 1014 kg ( 4 186 J kg ⋅ °C ) (1.00°C ) = 1.67 × 1018 J

(b)

The power input is P = 1 000 MW = 1.00 × 10 9 J s, so

11.4

)

t=

Q 1.67 × 1018 J ⎛ 1 yr = ⎜⎝ 9 P 1.00 × 10 J s 3.156 × 10 7

⎞ ⎟ = 52.9 yr s⎠

The change in temperature of the rod is ΔT =

Q 1.00 × 10 4 J = = 31.7°C mc ( 0.350 kg ) ( 900 J kg°C )

and the change in the length is ΔL = a L0 ( ΔT ) −1 = ⎡⎣ 24 × 10 −6 ( °C ) ⎤⎦ ( 20.0 cm )( 31.7°C ) = 1.5 × 10 −2 cm = 0.15 mm

11.5

(a)

(

)

Q = 0.600 ΔPEg = 0.600 ( mg h ) = 0.600 ⋅ m 9.80 m s2 ( 50.0 m ) or

(

)

Q = 294 m 2 s2 ⋅ m continued on next page

56157_11_ch11_p344-370.indd 349

10/12/10 2:45:28 PM

350

Chapter 11

From Q = mc(ΔT ) = mc(T f − Ti ), we ﬁnd the ﬁnal temperature as T f = Ti + (b)

11.6

(a) (b)

⎛ 10 3 cal ⎞ ⎛ 4.186 J ⎞ Q = 540 Cal ⎜ = 2.3 × 10 6 J ⎝ 1 Cal ⎟⎠ ⎜⎝ 1 cal ⎟⎠ The work done lifting her weight mg up one stair of height h is W1 = mgh. Thus, the total work done in climbing N stairs is W = Nmgh, and we have W = Nmgh = Q or Q 2.3 × 10 6 J = = 2.8 × 10 4 stairs mgh ( 55 kg ) 9.80 m s2 ( 0.15 m )

(

)

If only 25% of the energy from the donut goes into mechanical energy, we have N=

11.7

)

Observe that the mass of the coin cancels out in the calculation of part (a). Hence, the result is independent of the mass of the coin .

N= (c)

(

294 m 2 s2 ⋅ m Q = 25.0°C + = 25.8°C mc m ( 387 J kg ⋅ °C )

⎛ Q ⎞ 0.25Q = 0.25 ⎜ = 0.25 2.8 × 10 4 stairs = 7.0 × 10 3 stairs mgh ⎝ mgh ⎟⎠

(

(

)

)

1 1 2 m v 2f − v02 = ( 75 kg ) ⎡⎣(11.0 m s ) − 0 ⎤⎦ = 4.54 × 10 3 J → 4.5 × 10 3 J 2 2

(a)

Wnet = ΔKE =

(b)

P=

(c)

If the mechanical energy is 25% of the energy gained from converting food energy, then Wnet = 0.25 ( ΔQ ) and P = 0.25 ( ΔQ ) Δt , so the food energy conversion rate is

Wnet 4.54 × 10 3 J = = 9.1 × 10 2 J s = 910 W Δt 5.0 s

P ΔQ ⎛ 910 J s ⎞ ⎛ 1 Cal ⎞ = =⎜ = 0.87 Cal s ⎟ 0.25 ⎝ 0.25 ⎠ ⎜⎝ 4 186 J ⎟⎠ Δt

11.8

(d)

The excess thermal energy is transported by conduction and convection to the surface of the skin and disposed of through the evaporation of perspiration.

(a)

The instantaneous power is P = Fv , where F is the applied force and v is the instantaneous velocity.

(b)

From Newton’s second law, Fnet = ma, and the kinematics equation v = v0 + at with v0 = 0, the instantaneous power expression given above may be written as P = Fv = ( ma ) ( 0 + at )

(c)

a=

or

P = ma 2 t

Δv v − 0 11.0 m s = = = 2.20 m s2 Δt t−0 5.00 s

continued on next page

56157_11_ch11_p344-370.indd 350

10/12/10 2:45:30 PM

Energy in Thermal Processes

(

)

(

)

(d)

P = ma 2 t = ( 75.0 kg ) 2.20 m s2

(e)

Maximum instantaneous power occurs when t = tmax = 5.00 s, so

(

2

351

t = 363 kg ⋅ m 2 s4 ⋅ t = ( 363 W s ) ⋅ t

)

Pmax = 363 J s2 ( 5.00 s ) = 1.82 × 10 3 J s If this corresponds to 25.0% of the rate of using food energy, that rate must be P ΔQ 1.82 × 10 3 J s ⎛ 1 Cal ⎞ = max = ⎜⎝ 4 186 J ⎟⎠ = 1.74 Cal s Δt 0.250 0.250 11.9

The mechanical energy transformed into internal energy of the bullet is 1 1⎛1 ⎞ 1 Q = ( KEi ) = ⎜ mvi2 ⎟ = mvi2. Thus, the change in temperature of the bullet is ⎠ 4 2 2⎝2 1 m v2 Q (300 m s ) = 176 °C i = 4 = m clead 4 (128 J kg ⋅ °C ) mc 2

ΔT = 11.10

The internal energy added to the system equals the gravitational potential energy given up by the 2 falling blocks, or Q = ΔPEg = 2mb gh. Thus, ΔT =

11.11

(

)

2 2mb gh 2 (1.50 kg ) 9.80 m s ( 3.00 m ) Q = = = 0.105°C mw cw mw cw ( 0.200 kg) ( 4 186 J kg ⋅ °C)

The quantity of energy transferred from the water-cup combination in a time interval of 1 minute is Q = ⎡⎣( mc )water + ( mc )cup ⎤⎦ ( ΔT ) ⎡ ⎛ ⎛ J ⎞ J ⎞⎤ 3 = ⎢( 0.800 kg ) ⎜ 4 186 + ( 0.200 kg ) ⎜ 900 ⎟ ⎟⎠ ⎥ (1.5°C ) = 5.3 × 10 J kg ⋅ °C kg ⋅ °C ⎝ ⎠ ⎝ ⎣ ⎦ The rate of energy transfer is P=

11.12

(a)

Q 5.3 × 10 3 J J = = 88 = 88 W Δt 60 s s

The mechanical energy converted into internal energy of the block is ⎛1 ⎞ Q = 0.85 ( KEi ) = 0.85 ⎜ mvi2 ⎟ . The change in temperature of the block will be ⎝2 ⎠ ΔT =

(b)

11.13

(

)

2 0.85 12 m vi2 0.85 ( 3.0 m s ) Q = = = 9.9 × 10 −3 °C mcCu m cCu 2 ( 387 J kg ⋅ °C )

The remaining energy is absorbed by the horizontal surface on which the block slides.

From ΔL = a L0 ( ΔT ), the required increase in temperature is found, using Table 10.1, as ΔT =

⎛ 1 yd ΔL 3.0 × 10 −3 m = ⎜ −1 −6 a steel L0 11 × 10 (°C ) 13 yd ⎝ 3.0 ft

(

)(

)

⎞ ⎛ 3.281 ft ⎞ ⎟ ⎜⎝ 1 m ⎟⎠ = 23°C ⎠ continued on next page

56157_11_ch11_p344-370.indd 351

10/12/10 2:45:33 PM

352

Chapter 11

The mass of the rail is m=

(

)

70 lb yd (13 yd ) ⎛ 4.448 N ⎞ w 2 = ⎜⎝ 1 lb ⎟⎠ = 4.1 × 10 kg g 9.80 m s2

so the required thermal energy (assuming that csteel = ciron ) is

(

)

Q = mcsteel ( ΔT ) = 4.1 × 10 2 kg ( 448 J kg ⋅ °C ) ( 23°C ) = 4.2 × 10 6 J 11.14

(a)

From the relation between compressive stress and strain, F A = Y ( ΔL L0 ), where Y is Young’s modulus of the material. From the discussion on linear expansion, the strain due to thermal expansion can be written as ( ΔL L0 ) = a ( ΔT ), where a is the coefﬁcient of linear expansion. Thus, the stress becomes F A = Y [a ( ΔT )] .

(b)

If the concrete slab has mass m, the thermal energy required to produce a change in temperature ΔT is Q = mc ( ΔT ), where c is the speciﬁc heat of concrete. Using the result from part (a), the absorbed thermal energy required to produce compressive stress F A is ⎛ F A⎞ Q = mc ⎜ ⎝ Ya ⎟⎠

(c)

Q=

or

mc ⎛ F ⎞ ⎜ ⎟ Ya ⎝ A ⎠

The mass of the given concrete slab is

(

)(

)

m = r V = 2.40 × 10 3 kg m 3 ⎡⎣ 4.00 × 10 −2 m (1.00 m ) (1.00 m )⎤⎦ = 96.0 kg (d)

If the maximum compressive stress concrete can withstand is F A = 2.00 × 10 7 Pa, the maximum thermal energy this slab can absorb before starting to break up is found, using Table 10.1, to be Qmax =

(e)

)(

(

)

( 2.00 × 10

7

)

Pa = 6.7 × 10 6 J

The change in temperature of the slab as it absorbs the thermal energy computed above is ΔT =

(f)

mc ⎛ F ⎞ (96.0 kg) (880 J kg ⋅ °C) = ⎜⎝ ⎟⎠ −1 Ya A max 2.1 × 1010 Pa 12 × 10 −6 (°C )

Q 6.7 × 10 6 J = = 79°C mc ( 96.0 kg ) (880 J kg ⋅ °C )

The rate the slab absorbs solar energy is

(

)

Pabsorbed = 0.5Psolar = 0.5 1.00 × 10 3 W = 5 × 10 2 J s so the time required to absorb the thermal energy computed in (d) above is t= 11.15

Q Pabsorbed

=

6.7 × 10 6 J ⎛ 1 h ⎞ ∼ 4h 5 × 10 2 J s ⎜⎝ 3 600 s ⎟⎠

When thermal equilibrium is reached, the water and aluminum will have a common temperature of T f = 65.0°C. Assuming that the water-aluminum system is thermally isolated from the environment, Qcold = −Qhot, so mw cw T f − Ti,w = −mAl cAl T f − Ti,Al , or

(

mw =

56157_11_ch11_p344-370.indd 352

(

−mAl cAl T f − Ti,Al

(

cw T f − Ti,w

)

)

(

)

) = − (1.85 kg) (900 J kg ⋅ °C) (65.0°C − 150°C) = ( 4 186

J kg ⋅ °C ) ( 65.0°C − 25.0°C )

0.845 kg

10/12/10 2:45:36 PM

Energy in Thermal Processes

11.16

353

If N pellets are used, the mass of the lead is Nmpellet . Since the energy lost by the lead must equal the energy absorbed by the water, Nmpellet c ( ΔT )

lead

= [ mc ( ΔT )]water

or the number of pellets required is N=

= 11.17

mw cw ( ΔT )w mpellet clead ΔT lead

( 0.500 kg) ( 4 186 J kg ⋅ °C) ( 25.0°C − 20.0°C) = (1.00 × 10−3 kg) (128 J kg ⋅ °C) ( 200°C − 25.0°C)

467

The total energy absorbed by the cup, stirrer, and water equals the energy given up by the silver sample. Thus, ⎡⎣ mc cAl + ms cCu + mw cw ⎤⎦ ( ΔT )w = [ mc ΔT

]

Ag

Solving for the mass of the cup gives

or 11.18

mc =

ΔT Ag ⎤ 1 ⎡ − ms cCu − mw cw ⎥ , ⎢ mAg cAg cAl ⎣ ΔT ( )w ⎦

mc =

⎤ 1 ⎡ (87 − 32 ) − ( 40 g ) ( 387 ) − ( 225 g ) ( 4 186 ) ⎥ = 80 g ⎢( 400 g ) ( 234 ) 900 ⎣ ( 32 − 27 ) ⎦

(

)

The mass of water is

(

)(

)

mw = rwVw = 1.00 g cm 3 100 cm 3 = 100 g = 0.100 kg For each bullet, the energy absorbed by the bullet equals the energy given up by the water, so mb cb ( T − 20.0°C ) = mw cw ( 90.0°C − T ). Solving for the ﬁnal temperature gives T=

mw cw ( 90.0°C ) + mb cb ( 20.0°C ) mw cw + mb cb

For the silver bullet, mb = 5.00 × 10 −3 kg and cb = 234 J kg ⋅ °C, giving Tsilver =

( 0.100 ) ( 4 186 ) ( 90.0°C ) + ( 5.00 × 10 −3 ) ( 234 )( 20.0°C ) = 89.8°C ( 0.100 ) ( 4 186 ) + ( 5.00 × 10 −3 ) ( 234 )

For the copper bullet, mb = 5.00 × 10 −3 kg and cb = 387 J kg ⋅ °C, which yields Tcopper =

( 0.100 ) ( 4 186 ) ( 90.0°C ) + ( 5.00 × 10 −3 ) ( 387 )( 20.0°C ) = 89.7°C ( 0.100 ) ( 4 186 ) + ( 5.00 × 10 −3 ) ( 387 )

Thus, the copper bullet wins the showdown of the water cups. 11.19

(a)

The total energy given up by the copper and the unknown sample equals the total energy absorbed by the calorimeter and water. Hence, mCu cCu ΔT Cu + munk cunk ΔT unk = ⎡⎣ mc cAl + mw cw ⎤⎦ ( ΔT )w continued on next page

56157_11_ch11_p344-370.indd 353

10/12/10 2:45:39 PM

354

Chapter 11

Solving for the speciﬁc heat of the unknown material gives ⎡ mc cAl + mw cw ⎤⎦ ( ΔT )w − mCu cCu ΔT Cu , or cunk = ⎣ munk ΔT unk cunk =

( 70.0 g) (80.0°C) { ⎣ 1

⎡(100 g ) ( 900 J kg ⋅ °C ) + ( 250 g ) ( 4 186 J kg ⋅ °C ) ⎤⎦ (10.0°C )

− ( 50.0 g ) ( 387 J kg ⋅ °C ) ( 60.0°C ) } = 1.82 × 10 3 J kg ⋅ °C

11.20

(b)

The unknown could be beryllium, but other possibilities also exist.

(c)

The material could be an unknown alloy or a material not listed in the table.

Qcold = −Qhot

⇒

Tf =

or

=

)

(

)

mw cw Ti ,w + mFe cFe Ti ,Fe mw cw + mFe cFe

( 20.0 kg) ( 4 186 J kg ⋅ °C) ( 25.0°C) + (1.50 kg) ( 448 J kg ⋅ °C) (600°C) ( 20.0 kg) ( 4 186 J kg ⋅ °C) + (1.50 kg) ( 448 J kg ⋅ °C)

T f = 29.6°C

and 11.21

(

mw cw T f − Ti ,w = −mFe cFe T f − Ti ,Fe

Since the temperature of the water and the steel container is unchanged, and neither substance undergoes a phase change, the internal energy of these materials is constant. Thus, all the energy given up by the copper is absorbed by the aluminum, giving mAl cAl ( ΔT )Al = mC u cC u ΔT C u , or ⎛ c ⎞ ⎡ ΔT C u mAl = ⎜ C u ⎟ ⎢ ⎝ cAl ⎠ ⎣ ( ΔT )Al

⎤ ⎥ mC u ⎦

⎛ 387 ⎞ ⎛ 85°C − 25°C ⎞ =⎜ ( 200 g) = 2.6 × 10 2 g = 0.26 kg ⎝ 900 ⎟⎠ ⎜⎝ 25°C − 5.0°C ⎟⎠ 11.22

The kinetic energy given up by the car is absorbed as internal energy by the four brake drums (a total mass of 32 kg of iron). Thus, ΔKE = Q = mdrum s cFe ( ΔT ), or ΔT =

11.23

(a)

1 mcar vi2 (1 500 kg) (30 m s ) = 47°C = 2 mdrum s cFe ( 32 kg ) ( 448 J kg ⋅ °C ) 2

1 2

Assuming that the tin-lead-water mixture is thermally isolated from the environment, we have Qcold = −Qhot

or

(

)

(

)

(

mw cw T f − Ti ,w = −mSn cSn T f − Ti ,Sn − mPb cPb T f − Ti ,Pb

)

and since mSn = mPb = mm etal = 0.400 kg and Ti ,Sn = Ti ,Pb = Thot = 60.0°C, this yields Tf = =

mw cw Ti ,w + mm etal ( cSn + cPb ) Thot mw cw + mm etal ( cSn + cPb )

(1.00 kg) ( 4 186 J kg ⋅ °C) ( 20.0°C) + ( 0.400 kg) ( 227 (1.00 kg) ( 4 186 J kg ⋅ °C) + ( 0.400 kg) ( 227

yielding

J kg ⋅ °C + 128 J kg ⋅ °C ) ( 60.0°C ) J kg ⋅ °C + 128 J kg ⋅ °C )

T f = 21.3°C

continued on next page

56157_11_ch11_p344-370.indd 354

10/12/10 2:45:42 PM

Energy in Thermal Processes

355

If an alloy containing a mass mSn of tin and a mass mPb of lead undergoes a rise in temperature ΔT, the thermal energy absorbed would be Q = QSn + QPb , or

(b)

(m

Sn

(

)

(

)

(

)

+ mPb ) calloy ΔT = mSn cSn ΔT + mPb cPb ΔT , giving calloy =

If the alloy is a half-and-half mixture, mSn = mPb , then this reduces to calloy =

and yields (c)

mSn cSn + mPb cPb mSn + mPb cSn + cPb 2

227 J kg ⋅ °C + 128 J kg ⋅ °C = 178 J kg ⋅ °C 2

calloy =

For a substance forming monatomic molecules, the number of atoms in a mass equal to the molecular weight of that material is Avogadro’s number, N A. Thus, the number of tin atoms in mSn = 0.400 kg = 400 g of tin with a molecular weight of MSn = 118.7 g mol is ⎛m ⎞ ⎛ ⎞ 400 g N Sn = ⎜ Sn ⎟ N A = ⎜ (6.02 × 1023 mol −1 ) = 2.03 × 1024 ⎝ 118.7 g mol ⎟⎠ ⎝ MSn ⎠ ⎛m ⎞ ⎛ ⎞ 400 g and, for the lead, N Pb = ⎜ Pb ⎟ N A = ⎜ (6.02 × 1023 mol −1 ) = 1.16 × 1024 ⎝ 207.2 g mol ⎟⎠ ⎝ M Pb ⎠

(d)

N Sn 2.03 × 10 24 = = 1.75 N Pb 1.16 × 10 24

We have

cSn 227 J kg ⋅ °C = = 1.77 cPb 128 J kg ⋅ °C

and observe that

from which we conclude that the specific heat of an element is proportional to the number of atoms per unit mass of that element. 11.24

Assuming that the unknown-water-calorimeter system is thermally isolated from the environment, −Qhot = Qcold , or −mx cx T f − Ti ,x = mw cw T f − Ti ,w + mAl cAl T f − Ti ,Al and, since Ti ,w = Ti ,Al = Tcold = 25.0°C, we have

(

(

cx = ( mw cw + mAl cAl ) T f − Tcold

(

(

mx Ti ,x − T f

)

(

)

)

⎡( 0.285 kg ) ( 4 186 J kg ⋅ °C ) + ( 0.150 kg ) ( 900 J kg ⋅ °C ) ⎤⎦ ( 32.0 − 25.0 ) °C cx = ⎣ ( 0.125 kg) (95.0°C − 32.0°C)

or

yielding 11.25

)

)

cx = 1.18 × 10 3 J kg ⋅ °C

Remember that energy must be supplied to melt the ice before its temperature will begin to rise. Then, assuming a thermally isolated system, Qcold = −Qhot, or

(

)

(

mice L f + mice cwater T f − 0°C = −mwater cwater T f − 25°C

)

and Tf = yielding

56157_11_ch11_p344-370.indd 355

mwater cwater ( 25°C ) − mice L f

(m

ice

+ mwater ) cwater

=

(825 g ) ( 4 186 J kg ⋅ °C) ( 25°C) − ( 75 g ) (3.33 × 10 ( 75 g + 825 g ) ( 4 186 J kg ⋅ °C)

5

J kg )

T f = 16°C

10/12/10 2:45:44 PM

356

11.26

Chapter 11

The total energy input required is Q = ( energy to melt 50 g of ice )

+ ( energy to warm 50 g of water to 100°C )

+ ( energy to vaporize 5.0 g water )

= ( 50 g ) L f + ( 50 g ) cwater (100°C − 0°C ) + ( 5.0 g ) Lv ⎛ J ⎞ Thus, Q = ( 0.050 kg ) ⎜ 3.33 × 10 5 kg ⎟⎠ ⎝

⎛ J ⎞ + ( 0.050 kg ) ⎜ 4 186 (100°C − 0°C ) kg ⋅ °C ⎟⎠ ⎝ ⎛ J ⎞ + ( 5.0 × 10 −3 kg ) ⎜ 2 .26 × 10 6 kg ⎟⎠ ⎝ which gives Q = 4.9 × 10 4 J = 49 kJ 11.27

The conservation of energy equation for this process is

( energy to melt ice ) + ( energy to warm melted ice to T ) = ( energy to cool water to T ) or

mice L f + mice cw ( T − 0°C ) = mw cw (80°C − T )

This yields T =

mw cw (80°C ) − mice L f

(m

ice

+ mw ) cw

so T= 11.28

(1.0 kg) ( 4 186

J kg ⋅ °C ) (80°C ) − ( 0.100 kg ) ( 3.33 × 10 5 J kg )

(1.1 kg) ( 4 186

J kg ⋅ °C )

= 65°C

The energy required is the following sum of terms: Q = ( energy to reach melting point )

+ ( energy to melt ) + ( energy to reach boiling point )

+ ( energy to vaporize ) + ( energy to reach 110°C )

Mathematically, Q = m ⎡cice [ 0°C − ( −10°C )] + L f + cwater (100°C − 0°C ) + Lv + csteam (110°C − 100°C )⎤ ⎣ ⎦ This yields ⎡⎛ ⎛ J ⎞ J ⎞ Q = ( 40 × 10 −3 kg ) ⎢⎜ 2 090 (10°C ) + ⎜ 3.33 × 105 ⎟ kg ⋅ °C ⎠ kg ⎟⎠ ⎝ ⎣⎝ ⎤ ⎛ ⎛ J ⎞ J ⎞ ⎛ J ⎞ + ⎜ 4 186 + ⎜ 2 010 (100°C ) + ⎜ 2.26 × 106 (10°C )⎥ ⎟ ⎟ ⎟ kg ⋅ °C ⎠ kg ⎠ ⎝ kg ⋅ °C ⎠ ⎝ ⎝ ⎦ or

56157_11_ch11_p344-370.indd 356

Q = 1.2 × 10 5 J = 0.12 MJ

10/12/10 2:45:48 PM

Energy in Thermal Processes

11.29

Assuming all work done against friction is used to melt snow, the energy balance equation is f ⋅ s = msnow L f . Since f = m k ( mskier g ), the distance traveled is s=

11.30

357

(a)

msnow L f

m k ( mskier

(1.0 kg) (3.33 × 10 g ) 0.20 ( 75 kg ) ( 9.80 =

5

J kg )

m s2 )

= 2.3 × 10 3 m = 2.3 km

Observe that the equilibrium temperature will lie between the two extreme temperatures ( −10.0°C and +30.0°C ) of the mixed materials. Also, observe that a water-ice change of phase can be expected in this temperature range, but that neither aluminum nor ethyl alcohol undergoes a change of phase in this temperature range. The thermal energy transfers we can anticipate as the system comes to an equilibrium temperature are: ice at − 10.0°C to ice at 0°C; ice at 0°C to liquid water at 0°C; water at 0°C to water at T; aluminum at 20.0°C to aluminum at T; ethyl alcohol at 30.0°C to ethyl alcohol at T.

(b)

Q

m (kg)

c (j/kg·°C) L (j/kg)

Tf (°C)

Ti (°C)

Expression

Qice

1.00

2 090

0

−10.0

mice cice [ 0 − ( −10.0°C )]

Qm elt

1.00

0

0

mice L f

Qwater

1.00

4 186

T

0

mice cwater ( T − 0 )

QAl

0.500

900

T

20.0

mAl cAl [T − 20.0°C ]

Qalc

6.00

2 430

T

30.0

malc calc [T − 30.0°C ]

3.33 × 10 5

(c)

mice cice (10.0°C ) + mice L f + mice cwater ( T − 0 ) + mAl cAl [T − 20.0°C ] + malc calc [T − 30.0°C ] = 0

(d)

T=

mAl cAl ( 20.0°C ) + malc calc ( 30.0°C ) − mice ⎡⎣cice (10.0°C ) + L f ⎤⎦ mice cwater + mAl cAl + malc calc

Substituting in numeric values from the table in (b) above gives T=

( 0.500 )( 900 )( 20.0 ) + ( 6.00 )( 2 430 )( 30.0 ) − (1.00 ) ⎡⎣( 2 090 )(10.0 ) + 3.33 × 105 ⎤⎦ (1.00 )( 4 186 ) + ( 0.500 )( 900 ) + ( 6.00 )( 2 430 )

and yields 11.31

T = 4.81°C

Assume that all the ice melts. If this yields a result T > 0, the assumption is valid, otherwise the problem must be solved again based on a different premise. If all ice melts, energy conservation (Qcold = −Qhot ) yields mice ⎡⎣cice [ 0°C − ( −78°C )] + L f + cw ( T − 0°C )⎤⎦ = − ( mw cw + mcal cC u )( T − 25°C ) or

T=

(m

w

cw + mcal cC u )( 25°C ) − mice ⎡⎣cice ( 78°C ) + L f ⎤⎦ ( mw + mice ) cw + mcal cC u continued on next page

56157_11_ch11_p344-370.indd 357

10/12/10 2:45:50 PM

358

Chapter 11

mw = 0.560 kg, mcal = 0.080 g, mice = 0.040 g, cw = 4 186 J kg ⋅°C,

With

cC u = 387 J kg ⋅°C, cice = 2 090 J kg ⋅°C, and L f = 3.33 × 10 5 J kg, this gives ⎡( 0.560 )( 4 186 ) + ( 0.080 )( 387 )⎤⎦ ( 25°C ) − ( 0.040 ) ⎡⎣( 2 090 )( 78°C ) + 3.33 × 10 5 ⎤⎦ T=⎣ ( 0.560 + 0.040 )( 4 186 ) + 0.080 ( 387 ) or T = 16°C and the assumption that all ice melts is seen to be valid. 11.32

At a rate of 400 kcal h, the excess internal energy that must be eliminated in a half-hour run is cal ⎞ ⎛ 4.186 J ⎞ ⎛ 5 Q = ⎜ 400 × 10 3 ⎟⎜ ⎟ ( 0.500 h ) = 8.37 × 10 J ⎝ h ⎠ ⎝ 1 cal ⎠ The mass of water that will be evaporated by this amount of excess energy is Q 8.37 × 10 5 J = 0.33 kg = Lv 2.5 × 10 6 J kg

mevaporated =

The mass of fat burned (and thus, the mass of water produced at a rate of 1 gram of water per gram of fat burned) is mproduced =

( 400

kcal h )( 0.500 h ) = 22 g = 22 × 10 −3 kg 9.0 kcal gram of fat

so the fraction of water needs provided by burning fat is f= 11.33

(a)

mproduced mevaporated

=

22 × 10 −3 kg = 0.067 or 6.7% 0.33 kg

The mass of 2.0 liters of water is mw = rV = (10 3 kg m 3 ) ( 2.0 × 10 −3 m 3 ) = 2.0 kg. The energy required to raise the temperature of the water (and pot) up to the boiling point of water is Qboil = ( mw cw + mAl cAl )( ΔT ) ⎡ ⎛ ⎛ J ⎞ J ⎞⎤ 5 + ( 0.25 kg ) ⎜ 900 Qboil = ⎢( 2.0 kg ) ⎜ 4 186 ⎟ ⎟⎠ ⎥ (100°C − 20°C ) = 6.9 × 10 J kg kg ⎝ ⎠ ⎝ ⎣ ⎦ The time required for the 14 000 Btu h burner to produce this much energy is or

t boil = (b)

Qboil 6.9 × 10 5 J ⎛ 1 Btu ⎞ −2 = ⎜ ⎟ = 4.7 × 10 h = 2.8 min 14 000 Btu h 14 000 Btu h ⎝ 1.054 × 10 3 J ⎠

Once the boiling temperature is reached, the additional energy required to evaporate all of the water is Qevaporate = mw Lv = ( 2.0 kg ) ( 2.26 × 10 6 J kg ) = 4.5 × 10 6 J and the time required for the burner to produce this energy is tevaporate =

56157_11_ch11_p344-370.indd 358

Qevaporate 14 000 Btu h

=

4.5 × 10 6 J ⎛ 1 Btu ⎞ ⎜ ⎟ = 0.30 h = 18 min 14 000 Btu h ⎝ 1.054 × 10 3 J ⎠

10/12/10 2:45:54 PM

Energy in Thermal Processes

11.34

359

In one hour, the energy dissipated by the runner is ΔE = P ⋅ t = ( 300 J s ) ( 3 600 s ) = 1.08 × 10 6 J Ninety percent, or Q = 0.900 (1.08 × 10 6 J ) = 9.72 × 10 5 J, of this is used to evaporate bodily ﬂuids. The mass of ﬂuid evaporated is m=

Q 9.72 × 10 5 J = 0.403 kg = Lv 2.41× 10 6 J kg

Assuming the ﬂuid is primarily water, the volume of ﬂuid evaporated in one hour is V= 11.35

⎛ 10 6 cm 3 ⎞ m 0.403 kg = = ( 4.03 × 10 −4 m 3 ) ⎜ = 403 cm 3 3 r 1 000 kg m ⎝ 1 m 3 ⎟⎠

The energy required to melt 50 g of ice is Q1 = mice L f = ( 0.050 kg ) ( 333 kJ kg ) = 17 kJ The energy needed to warm 50 g of melted ice from 0°C to 100°C is Q2 = mice cw ( ΔT ) = ( 0.050 kg ) ( 4.186 kJ kg ⋅ °C ) (100°C ) = 21 kJ (a)

If 10 g of steam is used, the energy it will give up as it condenses is Q3 = msteam Lv = ( 0.010 kg ) ( 2 260 kJ kg ) = 23 kJ Since Q3 > Q1, all of the ice will melt. However, Q3 < Q1 + Q2, so the ﬁnal temperature is less than 100°C. From conservation of energy, we ﬁnd Qcold = −Qhot mice ⎡⎣ L f + cw ( T − 0°C )⎤⎦ = −msteam ⎡⎣ −Lv + cw ( T − 100°C )⎤⎦ T=

or

giving (b)

msteam ⎡⎣ Lv + cw (100°C )⎤⎦ − mice L f , ( mice + msteam ) cw T=

(10 g) ⎡⎣ 2.26 × 106 + ( 4 186) (100 )⎤⎦ − (50 g) (3.33 × 105 ) = (50 g + 10 g) ( 4 186)

40 °C

If only 1.0 g of steam is used, then Q3′ = msteam Lv = 2.26 kJ. The energy 1.0 g of condensed steam can give up as it cools from 100°C to 0°C is

(

)

Q4 = msteam cw ( ΔT ) = 1.0 × 10 −3 kg ( 4.186 kJ kg ⋅ °C ) (100°C ) = 0.42 kJ Since Q3′ + Q4 is less than Q1, not all of the 50 g of ice will melt, so the ﬁnal temperature will be 0 °C . The mass of ice which melts as the steam condenses and the condensate cools to 0°C is m=

56157_11_ch11_p344-370.indd 359

Q3′ + Q4 ( 2.26 + 0.42 ) kJ = = 8.0 × 10 −3 kg = 8.0 g Lf 333 kJ kg

10/12/10 2:45:57 PM

360

11.36

Chapter 11

First, we use the ideal gas law (with V = 0.600 L = 0.600 × 10 −3 m 3 and T = 37.0°C = 310 K ) to determine the quantity of water vapor in each exhaled breath: PV = nRT ⇒ n =

(

)(

)

3.20 × 10 3 Pa 0.600 × 10 −3 m 3 PV = = 7.45 × 10 −4 mol RT (8.31 J mol ⋅ K ) (310 K )

⎛ g ⎞ ⎛ 1 kg ⎞ −5 m = nM water = 7.45 × 10 −4 mol ⎜ 18.0 ⎜ ⎟ = 1.34 × 10 kg mol ⎟⎠ ⎝ 10 3 g ⎠ ⎝ The energy required to vaporize this much water, and hence the energy carried from the body with each breath, is

(

or

)

(

)(

)

Q = mLv = 1.34 × 10 −5 kg 2.26 × 10 6 J kg = 30.3 J The rate of losing energy by exhaling humid air is then breaths ⎞ ⎛ 1 min ⎞ J ⎞⎛ ⎛ P = Q ⋅ ( respiration rate ) = ⎜ 30.3 ⎟ ⎜ 22.0 ⎟⎜ ⎟ = 11.1 W ⎝ min ⎠ ⎝ 60 s ⎠ breath ⎠ ⎝ 11.37

(a)

The bullet loses all of its kinetic energy as it is stopped by the ice. Also, thermal energy is transferred from the bullet to the ice as the bullet cools from 30.0°C to the ﬁnal temperature. The sum of these two quantities of energy equals the energy required to melt part of the ice. The final temperature is 0°C because not all of the ice melts.

(b)

The total energy transferred from the bullet to the ice is Q = KEi + mbullet clead 0°C − 30.0°C =

(

= 3.00 × 10

−3

(

1 mbullet vi2 + mbullet clead ( 30.0°C ) 2

⎡ 2.40 × 10 2 m s kg ⎢ 2 ⎢ ⎣

)

) + (128 J kg ⋅ °C) (30.0°C)⎤⎥ = 97.9 J ⎥ 2

⎦

The mass of ice that melts when this quantity of thermal energy is absorbed is m=

Q

=

(L )

f water

11.38

(a)

⎛ 10 3 g ⎞ 97.9 J = 2.94 × 10 −4 kg ⎜ = 0.294 g 5 3.33 × 10 J kg ⎝ 1 kg ⎟⎠

The rate of energy transfer by conduction through a material of area A, thickness L, with thermal conductivity k, and temperatures Th > Tc on opposite sides is P = k A ( Th − Tc ) L. For the given windowpane, this is J ( 25°C − 0°C ) ⎛ ⎞ = 6 × 10 3 J s = 6 × 10 3 W P = ⎜ 0.8 ⎟ [(1.0 m ) ( 2.0 m )] ⎝ 0.62 × 10 −2 m s ⋅ m ⋅ °C ⎠

(b)

The total energy lost per day is

(

)(

)

E = P ⋅ Δt = 6 × 10 3 J s 8.64 × 10 4 s = 5 × 108 J 11.39

The rate of energy transfer by conduction through a material having thermal conductivity k , cross-sectional area A, thickness L and a temperature change of Th − Tc across it is P = k A ( Th − Tc ) L. Hence, with k = 0.6 J s ⋅ m ⋅ °C for water, the rate of energy transfer by conduction to the bottom of the pond is P=

56157_11_ch11_p344-370.indd 360

( 0.6

(

)

J s ⋅ m ⋅ °C ) 820 m 2 ( 25°C − 12°C ) 2.0 m

= 3 × 10 3 J s = 3 × 10 3 W

10/12/10 2:46:01 PM

Energy in Thermal Processes

11.40

361

The R value of a material is R = L k , where L is its thickness and k is the thermal conductivity. The R values of the three layers covering the core tissues in this body are:

(a)

1.0 × 10 −3 m = 5.0 × 10 −2 m 2 ⋅ K W 0.020 W m ⋅ K

Rskin = Rfat =

0.50 × 10 −2 m = 2.5 × 10 −2 m 2 ⋅ K W 0.020 W m ⋅ K

Rtissue =

and

3.2 × 10 −2 m = 6.4 × 10 −2 m 2 ⋅ K W 0.50 W m ⋅ K

so the total R value of the three layers taken together is Rtotal = ∑ Ri = ( 5.0 + 2.5 + 6.4 ) × 10 −2 3

i =1

(b)

The rate of energy transfer by conduction through these three layers with a surface area of A = 2.0 m 2 and temperature difference of ΔT = ( 37 − 0 ) °C = 37°C = 37 K is P=

11.41

m 2 ⋅K m 2 ⋅K m 2 ⋅K = 14 × 10 −2 = 0.14 W W W

(

)

2.0 m 2 ( 37 K ) A ( ΔT ) = = 5.3 × 10 2 W Rtotal 0.14 m 2 ⋅ K W

⎛ 10 2 cm ⎞ ⎛ 4.186 J ⎞ cal J ⎛ ΔT ⎞ P = k A⎜ , with k = 0.200 ⎟ ⎜⎝ ⎟⎠ = 83.7 ⎜ ⎟ ⎝ L ⎠ cm ⋅ °C ⋅ s ⎝ 1 m ⎠ 1 cal s ⋅ m ⋅ °C Thus, the energy transfer rate is J ⎛ ⎞ ⎛ 200°C − 20.0°C ⎞ P = ⎜ 83.7 ⎟ [(8.00 m ) ( 50.0 m )] ⎜⎝ ⎟ ⎝ s ⋅ m ⋅ °C ⎠ 1.50 × 10 −2 m ⎠ = 4.02 × 108 J s = 4.02 × 108 W = 402 MW

11.42

The total surface area of the house is A = Aside walls + Aend walls + Agables + Aroof where

Aside walls = 2 [( 5.00 m ) × (10.0 m )] = 100 m 2 Aend walls = 2 [( 5.00 m ) × (8.00 m )] = 80.0 m 2 Agables = 2 ⎡⎣ 12 ( base ) × ( altitude ) ⎤⎦ = 2 [ 12 (8.00 m ) × ( 4.00 m ) tan 37.0° ] = 24.1 m 2 Aroof = 2 ⎡⎣(10.0 m ) × ( 4.00 m cos 37.0° ) ⎤⎦ = 100 m 2

Thus,

A = 100 m 2 + 80.0 m 2 + 24.1 m 2 + 100 m 2 = 304 m 2

With an average thickness of 0.210 m, average thermal conductivity of 4.8 × 10 −4 kW m ⋅°C, and a 25.0°C difference between inside and outside temperatures, the energy transfer from the house to the outside air each day is

(

)(

)

⎡ 4.8 × 10 −4 kW m ⋅ °C 304 m 2 ( 25.0°C ) ⎤ ⎡ k A ( ΔT ) ⎤ ⎢ ⎥ (86 400 s ) Δt E = P ( Δt ) = ⎢ = ( ) ⎥ L 0.210 m ⎣ ⎦ ⎢⎣ ⎥⎦ continued on next page

56157_11_ch11_p344-370.indd 361

10/12/10 2:46:04 PM

362

Chapter 11

E = 1.5 × 10 6 kJ = 1.5 × 10 9 J

or

The volume of gas that must be burned to replace this energy is V= 11.43

E 1.5 × 10 9 J = = 39 m 3 heat of combustion 9 300 kcal m 3 ( 4 186 J kcal )

(

)

Because the two pots hold the same quantity of water at the same initial temperature, the same amount, Q, of thermal energy is required to boil away the water in each pot. The time required to do this for one of the pots is t = Q P, where P is the rate of energy conduction from the stove to the water through the bottom of the pot. The ratio of the times required for the two pots is

( (

t Al ⎛ Q ⎞ ⎛ PCu ⎞ k Cu A ΔT = = tCu ⎜⎝ PAl ⎟⎠ ⎜⎝ Q ⎟⎠ k Al A ΔT

) )

L L

=

k Cu k Al

Note that in the above calculation everything except the thermal conductivities canceled because the two pots are identical except for the material making up the bottoms. Thus, the time required to boil away the water in the aluminum bottomed pot is ⎛k ⎞ ⎛ 397 W m ⋅ °C ⎞ t Al = ⎜ Cu ⎟ tCu = ⎜ ( 425 s ) = 709 s ⎝ 238 W m ⋅ °C ⎟⎠ ⎝ k Al ⎠ 11.44

The rate of energy transfer through a compound slab is P= (a)

A ( ΔT ) , where R = Σ Li k i R

For the thermopane, R = Rpane + Rtrapped air + Rpane = 2Rpane + Rtrapped air ⎛ 0.50 × 10 −2 m ⎞ m 2 ⋅ °C 1.0 × 10 −2 m m 2 ⋅ °C Thus, R = 2 ⎜ + = 0.01 + 0.43 = 0.44 ( ) W W ⎝ 0.8 W m ⋅ °C ⎟⎠ 0.023 4 W m ⋅ °C

(1.0 m ) ( 23°C) = P= 2

and (b)

0.44 m 2 ⋅ °C W

52 W

For the 1.0 cm thick pane of glass, R=

1.0 × 10 −2 m m 2 ⋅ °C = 1 × 10 −2 0.8 W m ⋅ °C W

(1.0 m ) ( 23°C) 2

so 11.45

P=

1 × 10 −2 m 2 ⋅ °C W

= 2 × 10 3 W = 2 kW , about 38 times greater

When the temperature of the junction stabilizes, the energy transfer rate must be the same for each of the rods, or PCu = PAl . The cross-sectional areas of the rods are equal, and if the temperature of the junction is 50°C, the temperature difference is ΔT = 50°C for each rod. Thus,

⎛ ΔT ⎞ ⎛ ΔT ⎞ PCu = k Cu A ⎜ = k Al A ⎜ = PAl , which gives ⎟ ⎝ LAl ⎟⎠ ⎝ LCu ⎠ ⎛k ⎞ ⎛ 238 W m ⋅ °C ⎞ LAl = ⎜ Al ⎟ LCu = ⎜ (15 cm ) = 9.0 cm ⎝ 397 W m ⋅ °C ⎟⎠ ⎝ k Cu ⎠

56157_11_ch11_p344-370.indd 362

10/12/10 2:46:08 PM

Energy in Thermal Processes

11.46

The energy transfer rate is P =

(

363

)

5 ΔQ mice L f ( 5.0 kg ) 3.33 × 10 J kg = = 58 W = Δt Δt (8.0 h ) ( 3 600 s 1 h )

⎛ ΔT ⎞ Thus, P = k A ⎜ gives the thermal conductivity as ⎝ L ⎟⎠ k = 11.47

( )

)

( 58 W ) 2.0 × 10 −2 m P⋅L = = 7.3 × 10 −2 W m ⋅ °C A ( ΔT ) 0.80 m 2 ( 25°C − 5.0°C )

(

The window will consist of the glass pane and a stagnant air layer on each side (see Example 11.10 in text). From Tables 11.3 and 11.4, the R-values for these layers are Rpane =

L k glass

Rair = 0.17

and

layer

=

0.40 × 10 −2 m m 2 ⋅ °C = 5.0 × 10 −3 0.80 W m ⋅ °C W

2 ft 2 ⋅ °F ⎛ 1 Btu h ⎞ ⎛ 1 m ⎞ ⎛ 1 °C ⎞ m 2 ⋅ °C = ⋅⎜ 0 . 030 ⎟⎜ ⎟ Btu h ⎝ 0.293 W ⎠ ⎝ 3.281 ft ⎠ ⎜⎝ 9 5 °F ⎟⎠ W

Rtotal = ΣRi = ( 0.030 + 5.0 × 10 −3 + 0.030 )

Thus,

m 2 ⋅ °C m 2 ⋅ °C = 0.065 W W

The energy loss through the window in a 12 hour interval is then Q = P⋅t = 11.48

A ( Th − Tc ) ΣRi

( 2.0 m )( 20°C) ⋅ (12 h ) ⎛ 3 600 s ⎞ = 2

⋅t =

0.065 m 2 ⋅ °C W

⎜⎝

⎟ 1h ⎠

2.7 × 10 7 J

Since 97% of the incident energy is reﬂected, the rate of energy absorption from the sunlight is Pabsorbed = 3.0% × ( I ⋅ A ) = 0.030 ( I ⋅ A ), where I is the intensity of the solar radiation.

(

)(

Pabsorbed = 0.030 1.40 × 10 3 W m 2 1.00 × 10 3 m

)

2

= 4.2 × 10 7 W

Assuming the sail radiates equally from both sides (so A = 2 (1.00 km ) = 2.00 × 10 6 m 2 ), the rate at which it will radiate energy to a 0 K environment when it has absolute temperature T is 2

(

Pradiated = s Ae T 4 − 0

)

W ⎞ W⎞ ⎛ ⎛ 2.00 × 10 6 m 2 ( 0.03) ⋅ T 4 = ⎜ 3.4 × 10 −3 4 ⎟ ⋅ T 4 = ⎜ 5.669 6 × 10 −8 2 4 ⎟ ⎝ ⎠ ⎝ m ⋅K K ⎠

(

)

At the equilibrium temperature, where Pradiated = Pabsorbed , we then have 14

⎛ −3 W ⎞ 4 7 ⎜⎝ 3.4 × 10 ⎟ ⋅ T = 4.2 × 10 W K4 ⎠ 11.49

or

⎡ 4.2 × 10 7 W ⎤ T=⎢ −3 4 ⎥ ⎣ 3.4 × 10 W K ⎦

= 330 K

The absolute temperatures of the two stars are TX = 5 727 + 273 = 6 000 K and TY = 11 727 + 273 = 12 000 K. Thus, the ratio of their radiated powers is 4

PY s AeTY4 ⎛ TY ⎞ 4 = = ⎜ ⎟ = ( 2 ) = 16 4 PX s AeTX ⎝ TX ⎠ 11.50

From Stefan’s law, the power radiated by an object at absolute temperature T and surface area A is P = s AeT 4 , where s = 5.669 6 × 10 −8 W m 2 ⋅ K and e is the emissivity. Thus, the surface area of the ﬁlament must be A=

56157_11_ch11_p344-370.indd 363

P 75 W −5 = m2 4 = 1.1 × 10 4 −8 s eT 5.669 6 × 10 W m 2 ⋅ K 4 (1.0 ) ( 3 300 K )

(

)

10/12/10 2:46:10 PM

364

11.51

Chapter 11

At a pressure of 1 atm, water boils at 100°C. Thus, the temperature on the interior of the copper kettle is 100°C and the energy transfer rate through the bottom is W ⎞ 2 ⎛ 102°C − 100°C ⎞ ⎛ ΔT ⎞ ⎛ = 397 P = k A⎜ ⎟⎠ ⎡⎣p ( 0.10 m ) ⎤⎦ ⎜ ⎝ L ⎟⎠ ⎜⎝ m ⋅ °C ⎝ 2.0 × 10 −3 m ⎟⎠ = 1.2 × 10 4 W = 12 kW

11.52

The mass of the water in the heater is ⎛ 3.786 L ⎞ ⎛ 1 m 3 ⎞ kg ⎞ ⎛ = 189 kg m = rV = ⎜ 10 3 3 ⎟ ( 50.0 gal ) ⎜ ⎝ m ⎠ ⎝ 1 gal ⎟⎠ ⎜⎝ 10 3 L ⎟⎠ The energy required to raise the temperature of the water from 20.0°C to 60.0°C is Q = mc ( ΔT ) = (189 kg ) ( 4 186 J kg ) ( 60.0°C − 20.0°C ) = 3.16 × 10 7 J The time required for the water heater to transfer this energy is t=

11.53

Q 3.16 × 10 7 J ⎛ 1 h ⎞ = = 1.83 h P 4 800 J s ⎜⎝ 3 600 s ⎟⎠

The energy conservation equation is Qcold = −Qhot , or mice L f + ⎡( mice + mw ) c w + mcup cCu ⎤ (12°C − 0°C ) = −mPb c Pb(12°C − 98°C ) ⎣ ⎦ This gives ⎛ J ⎞ m Pb ⎜ 128 (86°C ) = ( 0.040 kg ) 3.33 × 105 J kg kg ⋅ °C ⎟⎠ ⎝

(

)

+ ⎡⎣( 0.24 kg ) ( 4 186 J kg ⋅ °C ) + ( 0.100 kg ) ( 387 J kg ⋅ °C ) ⎤⎦ (12°C ) or 11.54

(a)

m Pb = 2.3 kg The net rate of energy transfer by radiation between a body at absolute temperature T and its surroundings at absolute temperature T0 is Pnet = s Ae(T 4 − T04 ) . Hence, with T = 33.0°C = 306 K, T0 = 20.0°C = 293 K, emissivity = e = 0.95, and surface area A = 1.50 m 2 , the net power radiated is W ⎞ 4 4 ⎛ 2 2 Pnet = ⎜ 5.669 6 × 10 −8 ⎟ 1.50 m ( 0.95) ⎡⎣( 306 K ) − ( 293 K ) ⎤⎦ = +1.1 × 10 W 2 ⎝ m ⋅ K4 ⎠

(

(b)

11.55

The positive sign on the net power radiated means that the body is radiating energy away faster than it is absorbing energy from the environment.

The conservation of energy equation is Qcold = −Qhot , or

(m c

w w

This gives

)

+ mcup cglass ( T − 27°C ) = −mCu cCu ( T − 90°C ) T=

T=

56157_11_ch11_p344-370.indd 364

)

(

)

mCu cCu ( 90°C ) + mw cw + mcup cglass ( 27°C ) mw cw + mcup cglass + mCu cCu

, or

( 0.200 )( 387 )( 90°C ) + ⎡⎣( 0.400 ) ( 4 186 ) + ( 0.300 )(837 )⎤⎦ ( 27°C ) = 29 °C ( 0.400 ) ( 4 186 ) + ( 0.300 )(837 ) + ( 0.200 )( 387 )

10/12/10 2:46:14 PM

Energy in Thermal Processes

11.56

(a)

The energy delivered to the heating element (a resistor) is transferred to the liquid nitrogen, causing part of it to vaporize in a liquid-to-gas phase transition. The total energy delivered to the element equals the product of the power and the time interval of 4.0 h.

(b)

The mass of nitrogen vaporized in a 4.0 h period is m=

11.57

365

Q P ⋅ ( Δt ) ( 25 J s ) ( 4.0 h ) ( 3 600 s h ) = 1.8 kg = = Lf 2.01 × 10 5 J kg Lf

Assuming the aluminum-water-calorimeter system is thermally isolated from the environment, Qcold = −Qhot , or

(

)

(

)

(

mAl cAl T f − Ti,Al = −mw cw T f − Ti,w − mcal ccal T f − Ti,cal Since T f = 66.3°C and Ti,cal = Ti,w = 70.0°C, this gives cAl =

) ( mw cw + mcal ccal ) (Ti,w − Tf )

(

mAl T f − Ti,Al

)

, or

⎡ ⎛ ⎛ J ⎞ J ⎞⎤ + ( 0.040 kg ) ⎜ 630 ⎥ ( 70.0 − 66.3) °C ⎢( 0.400 kg ) ⎜ 4 186 ⎟ kg ⋅ °C ⎠ kg ⋅ °C ⎟⎠ ⎦ ⎝ ⎝ J ⎣ = 8.00 × 10 2 cAl = kg ⋅ °C ( 0.200 kg) (66.3 − 27.0 ) °C The variation between this result and the value from Table 11.1 is ⎛ variation ⎞ ⎛ 800 − 900 J kg ⋅ °C ⎞ %=⎜ × 100% = ⎜ ⎟⎠ × 100% = 11.1% 900 J kg ⋅ °C ⎝ accepted value ⎟⎠ ⎝ which is within the 15% tolerance. 11.58

(a)

With a body temperature of T = 37°C + 273 = 310 K and surroundings at temperature T0 = 24°C + 273 = 297 K, the rate of energy transfer by radiation is

(

Pradiation = s Ae T 4 − T04

)

W ⎞ 4 4 ⎛ 2 2 = ⎜ 5.669 6 × 10 −8 ⎟ 2.0 m ( 0.97 ) ⎡⎣( 310 K ) − ( 297 K ) ⎤⎦ = 1.6 × 10 W 2 ⎝ m ⋅ K4 ⎠

(

(b)

The rate of energy transfer by evaporation of perspiration is Pperspiration =

(c)

)

(

)(

)

3 3 Q mLv,perspiration ( 0.40 kg ) 2.43 × 10 kJ kg 10 J kJ = = = 2.7 × 10 2 W Δt Δt 3 600 s

The rate of energy transfer by evaporation from the lungs is kJ ⎞ ⎛ 1 h ⎞ ⎛ 10 3 J ⎞ ⎛ Plungs = ⎜ 38 = 11 W ⎟ ⎝ h ⎠ ⎜⎝ 3 600 s ⎟⎠ ⎜⎝ 1 kJ ⎟⎠

(d)

The excess thermal energy that must be dissipated is kJ ⎞ ⎛ 1 h ⎞ ⎛ 10 3 J ⎞ ⎛ Pexcess = 0.80Pmetabolic = 0.80 ⎜ 2.50 × 10 3 = 5.6 × 10 2 W ⎟ ⎝ h ⎠ ⎜⎝ 3 600 s ⎟⎠ ⎜⎝ 1 kJ ⎟⎠ so the rate energy must be transferred by conduction and convection is

(

)

Pc&c = Pexcess − Pradiation + Pperspiration + Plungs = ( 5.6 − 1.6 − 2.7 − .11) × 10 2 W = 1.2 × 10 2 W

56157_11_ch11_p344-370.indd 365

10/12/10 2:46:17 PM

366

11.59

Chapter 11

The total energy needed is

(

)

Q = mLv = ( 2.00 kg ) 2.00 × 10 4 J kg = 4.00 × 10 4 J and the time required to supply this energy is t= 11.60

Q 4.00 × 10 4 J = = 4.00 × 10 3 P 10.0 J s

⎛ 1 min ⎞ s⎜ = 66.7 min ⎝ 60.0 s ⎟⎠

The energy added to the air in one hour is Q = ( Ptotal ) t = [10 ( 200 W )] ( 3 600 s ) = 7.20 × 10 6 J and the mass of air in the room is

(

m = rV = 1.3 kg m 3

) [(6.0 m )(15.0 m )(3.0 m )] = 3.5 × 10

The change in temperature is ΔT =

kg

Q 7.2 × 10 6 J = = 25°C mc 3.5 × 10 2 kg (837 J kg ⋅ °C )

(

)

T = T0 + ΔT = 20°C + 25°C = 45°C

giving 11.61

2

⎛ T − 30.0°C ⎞ ⎛ 80.0°C − T ⎞ In the steady state, PAu = PAg, or k Au A ⎜ ⎟⎠ ⎟⎠ = k Ag A ⎜⎝ ⎝ L L This gives T=

11.62

(a)

k Au (80.0°C ) + k Ag ( 30.0°C ) k Au + k Ag

=

314 (80.0°C ) + 427 ( 30.0°C ) = 51.2 °C 314 + 427

The rate work is done against friction is P = f ⋅ v = ( 50 N ) ( 40 m s ) = 2.0 × 10 3 J s = 2.0 kW

(b)

In a time interval of 10 s, the energy added to the 10 kg of iron is Q = P ⋅ t = ( 2.0 × 10 3 J s ) (10 s ) = 2.0 × 10 4 J and the change in temperature is ΔT =

11.63

(a)

Q 2.0 × 10 4 J = = 4.5°C mc (10 kg ) ( 448 J kg ⋅ °C )

The energy required to raise the temperature of the brakes to the melting point at 660°C is Q = mc ( ΔT ) = ( 6.00 kg ) ( 900 J kg ⋅ °C ) ( 660°C − 20.0°C ) = 3.46 × 10 6 J The internal energy added to the brakes on each stop is Q1 = ΔKE =

1 1 2 mcar vi2 = (1 500 kg ) ( 25.0 m s ) = 4.69 × 10 5 J 2 2

The number of stops before reaching the melting point is N=

Q 3.46 × 10 6 J = = 7 stops Q1 4.69 × 10 5 J

continued on next page

56157_11_ch11_p344-370.indd 366

10/12/10 2:46:20 PM

Energy in Thermal Processes

(b)

11.64

367

As the car stops, it transforms part of its kinetic energy into internal energy due to air resistance. As soon as the brakes rise above air temperature, they transfer energy by heat to the air. If they reach a high temperature, they transfer energy to the air very quickly.

When liquids 1 and 2 are mixed, the conservation of energy equation is mc1 (17°C − 10°C ) = mc2 ( 20°C − 17°C )

or

⎛ 7⎞ c2 = ⎜ ⎟ c1 ⎝ 3⎠

When liquids 2 and 3 are mixed, energy conservation yields mc3 ( 30°C − 28°C ) = mc2 ( 28°C − 20°C ) Thus, we now know that

c3 = 4 ( 7c1 3)

or

c3 = 4 c2

or

c3 c1 = 28 3

Mixing liquids 1 and 3 will give mc1 ( T − 10°C ) = mc3 ( 30°C − T )

11.65

c1 (10°C ) + c3 ( 30°C ) 10°C + ( 28 3) ( 30°C ) = = 28 °C c1 + c3 1 + ( 28 3)

or

T=

(a)

The internal energy ΔQ added to the volume ΔV of liquid that ﬂows through the calorimeter in time Δt is ΔQ = ( Δm )c ( ΔT ) = r ( ΔV )c ( ΔT ). Thus, the rate of adding energy is ΔQ ⎛ ΔV ⎞ = r c ( ΔT ) ⎜ ⎝ Δt ⎟⎠ Δt where ⎛⎜ ΔV ⎞⎟ is the ﬂow rate through the calorimeter. ⎝ Δt ⎠

(b)

From the result of part (a), the speciﬁc heat is c=

ΔQ Δt 40 J s = 3 r ( ΔT ) ( ΔV Δt ) ( 0.72 g cm ) ( 5.8°C ) ( 3.5 cm 3 s )

⎛ J ⎞ ⎛ 10 3 g ⎞ = 2.7 × 10 3 J kg ⋅ °C = ⎜ 2.7 g ⋅ °C ⎟⎠ ⎜⎝ 1 kg ⎟⎠ ⎝ 11.66

(a)

The surface area of the stove is Astove = Aends + Acylindrical = 2 (p r 2 ) + ( 2p r h ), or side

Astove = 2p ( 0.200 m ) + 2p ( 0.200 m ) ( 0.500 m ) = 0.880 m 2 2

The temperature of the stove is Tstove = 95 ( 400°F − 32.0°F ) = 204°C = 477 K while that of the air in the room is Troom = 95 ( 70.0°F − 32.0°F ) = 21.1°C = 294 K. If the emissivity of the stove is e = 0.920, the net power radiated to the room is 4 4 P = s Astove e ( Tstove − Troom ) 4 4 = ( 5.669 6 × 10 −8 W m 2 ⋅ K 4 ) ( 0.880 m 2 ) ( 0.920 ) ⎡⎣( 477 K ) − ( 294 K ) ⎤⎦

or

P = 2.03 × 10 3 W = 2.03 × 10 3 J s continued on next page

56157_11_ch11_p344-370.indd 367

10/12/10 2:46:23 PM

368

Chapter 11

(b)

The total surface area of the walls and ceiling of the room is A = 4 Awall + Aceiling = 4 [(8.00 ft ) ( 25.0 ft )] + ( 25.0 ft ) = 1.43 × 10 3 ft 2 2

If the temperature of the room is constant, the power lost by conduction through the walls and ceiling must equal the power radiated by the stove. Thus, from thermal conduction equation, P = A ( Th − Tc ) ΣRi , the net R value needed in the walls and ceiling is ΣRi = or 11.67

A ( Th − Tc ) P

=

(1.43 × 10

ft 2 ) ( 70.0°F − 32.0°F ) ⎛ 1 054 J ⎞ ⎛ 1 h ⎞ ⎜⎝ ⎟ 2.03 × 10 3 J s 1 Btu ⎠ ⎜⎝ 3 600 s ⎟⎠ 3

ΣRi = 7.84 ft 2 ⋅ °F ⋅ h Btu

A volume of 1.0 L of water has a mass of m = rV = (10 3 kg m 3 ) (1.0 × 10 −3 m 3 ) = 1.0 kg The energy required to raise the temperature of the water to 100°C and then completely evaporate it is Q = mc ( ΔT ) + mLv , or Q = (1.0 kg ) ( 4 186 J kg ⋅ °C ) (100°C − 20°C ) + (1.0 kg ) ( 2.26 × 10 6 J kg ) = 2.59 × 10 6 J The power input to the water from the solar cooker is ⎡ p ( 0.50 m )2 ⎤ P = ( efficiency ) IA = ( 0.50 ) ( 600 W m 2 ) ⎢ ⎥ = 59 W 4 ⎣ ⎦ so the time required to evaporate the water is t=

11.68

(a)

⎛ 1h ⎞ Q 2.59 × 10 6 J = 12 h = = ( 4.4 × 10 4 s ) ⎜ 59 J s P ⎝ 3 600 s ⎟⎠

From the thermal conductivity equation, P = k A ⎡⎣( Th − Tc ) L ⎤⎦ , the total energy lost by conduction through the insulation during the 24-h period will be Q = P1 (12.0 h ) + P2 (12.0 h ) =

or Q =

( 0.012 0

kA [(37.0°C − 23.0°C) + (37.0°C − 16.0°C)](12.0 h ) L

J s ⋅ m°C ) ( 0.490 m 2 )

9.50 × 10

−2

m

[14.0°C + 21.0°C ](12.0 h ) ⎛⎜⎝

3 600 1h

s⎞ 4 ⎟⎠ = 9.36 × 10 J

The mass of molten wax which will give off this much energy as it solidiﬁes (all at 37°C) is m= (b)

11.69

Q 9.36 × 10 4 J = = 0.457 kg Lf 205 × 10 3 J kg

If the test samples and the inner surface of the insulation are preheated to 37.0°C during the assembly of the box, nothing undergoes a temperature change during the test period. Thus, the masses of the samples and insulation do not enter into the calculation. Only the duration of the test, inside and outside temperatures, along with the surface area, thickness, and thermal conductivity of the insulation need to be known.

The total power radiated by the Sun is P = s AeT 4, where s = 5.669 6 × 10 −8 W m 2 ⋅ K 4 , the emissivity is e = 0.986, the surface area (a sphere) is A = 4p r 2, and the absolute temperature is T = 5 800 K. Thus, P = ( 5.669 6 × 10 −8 W m 2 ⋅ K 4 ) 4p ( 6.96 × 108 m ) ( 0.986 )( 5 800 K ) 2

4

continued on next page

56157_11_ch11_p344-370.indd 368

10/12/10 2:46:26 PM

Energy in Thermal Processes

369

or P = 3.85 × 10 26 W. Thus, the energy radiated each second is E = P ⋅ Δt = ( 3.85 × 10 26 J s )(1.00 s ) = 3.85 × 10 26 J 11.70

We approximate the latent heat of vaporization of water on the skin (at 37°C) by asking how much energy would be needed to raise the temperature of 1.0 kg of water to the boiling point and evaporate it. The answer is C C L37° ≈ cwater ( ΔT ) + L100° = ( 4 186 J kg ⋅ °C ) (100°C − 37°C ) + 2.26 × 10 6 J kg v v

or

C L37° ≈ 2.5 × 10 6 J kg v

Assuming that you are approximately 2.0 m tall and 0.30 m wide, you will cover an area of A = ( 2.0 m ) ( 0.30 m ) = 0.60 m 2 of the beach, and the energy you receive from the sunlight in one hour is Q = IA ( Δt ) = (1 000 W m 2 ) ( 0.60 m 2 ) ( 3 600 s ) = 2.2 × 10 6 J The quantity of water this much energy could evaporate from your body is m=

Q 2.2 × 10 6 J = 0.9 kg ≈ 2.5 × 10 6 J kg L 37° C v

The volume of this quantity of water is V =

m 0.9 kg = 3 ≈ 10 −3 m 3 = 1 L r 10 kg m 3

Thus, you will need to drink almost a liter of water each hour to stay hydrated. Note, of course, that any perspiration that drips off your body does not contribute to the cooling process, so drink up! 11.71

During the ﬁrst 50 minutes, the energy input is used converting m kilograms of ice at 0°C into liquid water at 0°C. The energy required is Q1 = mL f = m ( 3.33 × 10 5 J kg ), so the constant power input must be P=

m ( 3.33 × 10 5 J kg ) Q1 = 50 min ( Δt )1

During the last 10 minutes, the same constant power input raises the temperature of water having a total mass of ( m + 10 kg ) by 2.0°C. The power input needed to do this is P=

Q2 ( m + 10 kg) c ( ΔT ) = ( m + 10 kg) ( 4 186 J kg ⋅ °C) ( 2.0°C) = 10 min ( Δt )2 ( Δt )2

Since the power input is the same in the two periods, we have m ( 3.33 × 10 5 J kg ) 50 min

=

( m + 10 kg) ( 4 186

which simpliﬁes to (8.0 ) m = m + 10 kg

56157_11_ch11_p344-370.indd 369

J kg ⋅ °C ) ( 2.0°C ) 10 min

or

m=

10 kg = 1.4 kg 7.0

10/12/10 2:46:28 PM

370

11.72

Chapter 11

(a)

First, energy must be removed from the liquid water to cool it to 0°C. Next, energy must be removed from the water at 0°C to freeze it, which corresponds to a liquid-to-solid phase transition. Finally, once all the water has frozen, additional energy must be removed from the ice to cool it from 0°C to –8.00°C.

(b)

The total energy that must be removed is Q = Qcool water + Qfreeze + Qcool ice to 0°C

at 0°C

= mw cw 0°C − Ti + mw L f + mw cice T f − 0°C

to –8.00°C

or ⎡⎛ ⎤ J ⎞ J ⎛ J ⎞ Q = ( 75.0 × 10 −3 kg ) ⎢⎜ 4 186 + ⎜ 2 090 −8.00 °C ⎥ −20.0 °C + 3.33 × 10 5 ⎟ ⎟ kg ⋅ °C ⎠ kg ⎝ kg ⋅ °C ⎠ ⎣⎝ ⎦ = 3.25 × 10 4 J = 32.5 kJ 11.73

(a)

In steady state, the energy transfer rate is the same for each of the rods, or PAl = PFe . Thus, ⎛ T − 0°C ⎞ ⎛ 100°C − T ⎞ k Al A ⎜ ⎟ ⎟⎠ = k Fe A ⎜⎝ ⎝ L L ⎠ giving ⎛ k Al ⎞ 238 ⎞ T =⎜ (100°C ) = ⎛⎜⎝ ⎟ (100°C ) = 75.0°C ⎟ 238 + 79.5 ⎠ ⎝ k Al + k Fe ⎠

(b)

If L = 15 cm and A = 5.0 cm 2 , the energy conducted in 30 min is W ⎞ ⎡⎛ −4 2 ⎛ 100°C − 75.0°C ⎞ ⎤ Q = PAl ⋅ t = ⎢⎜ 238 ⎟⎠ 5.0 × 10 m ⎜⎝ ⎟⎠ ⎥ (1 800 s ) ⎝ m ⋅ °C 0.15 m ⎦ ⎣

(

)

= 3.6 × 10 4 J = 36 kJ

56157_11_ch11_p344-370.indd 370

10/12/10 2:46:30 PM

12 The Laws of Thermodynamics QUICK QUIZZES 1.

Choice (b). The magnitude of the work done on a gas during a thermodynamic process is equal to the area under the curve on a PV diagram. Processes in which the volume decreases do positive work on the gas, while processes in which the volume increases do negative work on the gas. The work done on the gas in each of the four processes shown is Wa = − 4.00 × 10 5 J, Wb = + 3.00 × 10 5 J, Wc = − 3.00 × 10 5 J, and Wd = + 4.00 × 10 5 J Thus, the correct ranking (from most negative to most positive) is a, c, b, d.

2.

A is isovolumetric, B is adiabatic, C is isothermal, D is isobaric.

3.

Choice (c). The highest theoretical efﬁciency of an engine is the Carnot efﬁciency given by eC = 1− Tc Th . Hence, the theoretically possible efﬁciencies of the given engines are eA = 1−

700 K 500 K = 0.300, eB = 1− = 0.375, 1 000 K 800 K

and

eC = 1−

300 K = 0.500 600 K

and the correct ranking (from highest to lowest) is C, B, A. 4.

Qr = 0 and Q = 0 in an adiabatic process. If the process was reversible, but not T adiabatic, the entropy of the system could undergo a non-zero change. However, in that case, the entropy of the system’s surroundings would undergo a change of equal magnitude but opposite sign, and the total change of entropy in the universe would be zero. If the process was irreversible, the total entropy of the universe would increase.

5.

The number 7 is the most probable outcome because there are six ways this could occur: 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1. The numbers 2 and 12 are the least probable because they could only occur one way each: either 1-1, or 6-6. Thus, you are six times more likely to throw a 7 than a 2 or 12.

Choice (b). ΔS =

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

For a monatomic ideal gas, the internal energy is U = 32 nRT and the change in internal energy is ΔU = 32 nR ( ΔT ). From the ideal gas law, PV = nRT , observe that nR ( ΔT ) = nRT f − nRTi = Pf V f − Pi Vi . When the pressure is constant, Pf = Pi = P, this reduces to ΔU = 32 ⎡⎣ P V f − Vi ⎤⎦ = 32 P ( ΔV ) , so the change in internal energy for this gas is ΔU = 32 ( 2.00 × 10 5 Pa ) ( +1.50 m 3 ) = 4.50 × 10 5 J, making (c) the correct answer.

(

2.

)

When volume is constant, the work done on the gas is zero, so the ﬁrst law of thermodynamics gives the change in internal energy as ΔU = Q + W = 100 J + 0 = 100 J, and (d) is the correct answer for this question.

371

56157_12_ch12_p371-399.indd 371

10/12/10 2:47:01 PM

372

3.

Chapter 12

The work done by the system on the environment is Wenv = +P ( ΔV ) = ( 70.0 × 10 3 Pa ) ( −0.20 m 3 ) = −14 × 10 3 J = −14 kJ and (c) is the correct choice.

4.

In a cyclic process, the net work done equals the area enclosed by the process curve in a PV diagram. Thus, Wnet = ⎡⎣( 4.00 − 1.00 ) × 10 5 Pa ⎤⎦ ⎡⎣( 2.00 − 1.00 ) m 3 ⎤⎦ + ⎡⎣( 2.00 − 1.00 ) × 10 5 Pa ⎤⎦ ⎡⎣( 3.00 − 2.00 ) m 3 ⎤⎦ = 4.00 × 10 5 J and (d) is the correct answer.

5.

For an adiabatic process, PV g = constant, where g = c p cv = 1.4 for diatomic gases. (See Table 12.1 in the textbook.) Thus, Pf V f1.4 = Pi Vi1.4 , or the ﬁnal pressure will be ⎛V ⎞ Pf = Pi ⎜ i ⎟ ⎝ Vf ⎠

1.4

⎛ 1.00 m 3 ⎞ = (1.00 × 10 Pa ) ⎜ ⎝ 3.50 m 3 ⎟⎠ 5

1.4

= 1.73 × 10 4 Pa

and the correct response is (e). 6.

The work an ideal gas does on the environment during an isothermal expansion is Wenv = nRT ln V f Vi , so for the given process,

(

)

Wenv = ( 3.9 × 10 2 mol ) (8.31 J mol ⋅K )(850 K ) ln ( 2 ) = 1.9 × 10 6 J and (a) is the correct choice. 7.

From conservation of energy, the energy input to the engine must be Qh = Weng + Qc = 15 kJ + 37 kJ = 52 kJ so the efﬁciency is e=

Weng Qh

=

15 kJ = 0.29 or 29% 52 kJ

and the correct choice is (b). 8.

The coefﬁcient of performance of this refrigerator is COP =

9.

10.

Qc W

=

115 kJ = 6.4 18 kJ

which is choice (d).

The internal energy of an ideal gas is a function of the temperature alone. Thus, for the isothermal compression, the internal energy remains constant or choice (d) is true. Work must be done on the gas to compress it into a smaller volume (see Equation 12.10 in the text), so choice (b) is false. If work is done on the gas in compression, but the internal energy is constant, the ﬁrst law of thermodynamics says that energy must be transferred from of the gas by heat, making choice (a) false. Also, since the process is isothermal, choice (c) is false. Thus, the only true choice listed is (d). At a pressure of 1.0 atm, ice melts at absolute temperature TK = 0° + 273.15 = 273.15 K. The thermal energy this block of ice must absorb to fully melt is Qr = mL f = (1.00 kg ) ( 3.33 × 10 5 J kg ) = 3.33 × 10 5 J

56157_12_ch12_p371-399.indd 372

10/12/10 2:47:04 PM

The Laws of Thermodynamics

373

so the change in entropy of the ice is ΔS =

Qr +3.33 × 10 5 J = = +1.22 × 10 3 J K = +1.22 kJ K TK 273.15 K

and (d) is the correct choice. 11.

The maximum theoretical efﬁciency (the Carnot efﬁciency) of a device operating between absolute temperatures Tc < Th is eC = 1− Tc Th . For the given steam turbine, this is eC = 1−

3.0 × 10 2 K = 0.33 or 33% 450 K

and (c) is the correct answer.

12.

By deﬁnition, in an adiabatic process, no energy is transferred to or from the gas by heat. Thus, (c) is a true statement. All other choices are false. In an expansion process, the gas does work on the environment. Since there is no energy input by heat, the ﬁrst law of thermodynamics says that the internal energy of the ideal gas must decrease, meaning the temperature will decrease. Also, in an adiabatic process, PV g = constant, meaning that the pressure must decrease as the volume increases in this adiabatic expansion.

13.

In an isobaric process on an ideal gas, pressure is constant while the gas either expands or is compressed. Since the volume of the gas is changing, work is done either on or by the gas, so choice (b) is a true statement. Also, from the ideal gas law with pressure constant, P ( ΔV ) = nR ( ΔT ) . Thus, the gas must undergo a change in temperature having the same sign as the change in volume. If ΔV > 0, then both ΔT and the change in the internal energy of the gas are positive ( ΔU > 0 ). However, when ΔV > 0, the work done on the gas is negative (W < 0 ) , and the ﬁrst law of thermodynamics says that there must be a positive transfer of energy by heat to the gas (Q = ΔU − W > 0 ) . When ΔV < 0, a similar argument shows that ΔU < 0, W > 0, and Q = ΔU − W < 0. Thus, all of the other listed choices are false statements.

14.

We know that TA ,i = TB ,i , PA ,i = PB ,i , and that VA = VB = constant. From the ideal gas law, n = PV RT , we see that n A = nB . Thus, both choices (c) and (d) are false. Also from the ideal gas law, ΔT = (ΔP)V nR with (ΔP) A > (ΔP) B , we see that (ΔT ) A > (ΔT ) B . However, the ﬁrst law of thermodynamics (ΔU = Q + W = Q + 0 = Q) tells us that (ΔU) A = (ΔU) B since the two gases received equal amounts of thermal energy. Since, for an ideal gas, ΔU = nCv ΔT , the ratio of the molar heat capacities of the two gases must be Cv , A ΔU A n A ΔTA ΔTB = = 2Pi since g > 1 for all ideal gases (see Table 12.1 in the textbook). The work done in an adiabatic process is the area under curve 3, which exceeds that done in either of the other processes. Thus, the correct choice is (b), the adiabatic process involves the most work.

(

56157_12_ch12_p371-399.indd 373

)

10/12/10 2:47:07 PM

374

Chapter 12

16.

The Clausius statement of the second law of thermodynamics says that in any thermodynamics process, reversible or irreversible, the total entropy of the universe must either remain constant (reversible process) or increase (irreversible process). Thus, if in a thermodynamics process, the entropy of a system changes by −6 J K, the entropy of the environment (i.e., the rest of the universe) must increase by +6 J K or more. The correct choice here is (e).

17.

Melting is a constant temperature process, so the change in entropy as the substance melts is given by ΔS = Qr T, where T > 0 is the absolute temperature at which the process occurs and Qr is the energy transferred as heat during the process. Since the substance must absorb energy (Qr > 0) as it melts, we see that ΔS > 0 and the correct answer is choice (b).

18.

With this method of using an air conditioner, the average temperature in the room will increase, and choice (a) is the correct answer. The air conditioner operates on a cyclic process so the change in the internal energy of the refrigerant is zero. Then the conservation of energy gives the thermal energy exhausted to the room as Qh = Qc + Weng , where Qc is the thermal energy the air conditioner removes from the room and Weng is the work done to operate the device. Since Weng > 0, the air conditioner is returning more thermal energy to the room than it is removing, so the average temperature in the room will increase.

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS 2.

(a)

Shaking opens up spaces between the jelly beans. The smaller ones have a chance of falling down into spaces below them. The accumulation of larger ones on top and smaller ones on the bottom implies an increase in order and a decrease in one contribution to the total entropy.

(b)

The second law is not violated and the total entropy of the system increases. The increase in the internal energy of the system comes from the work required to shake the jar of beans (that is, work your muscles must do, with an increase in entropy accompanying the biological process) and also from the small loss of gravitational potential energy as the beans settle together more compactly.

4.

Temperature = a measure of molecular motion. Heat = the process through which energy is transferred between objects by means of random collisions of molecules. Internal energy = energy associated with random molecular motions plus chemical energy, strain potential energy, and an object’s other energy not associated with center of mass motion or location.

6.

A higher steam temperature means that more energy can be extracted from the steam. For a constant temperature heat sink at Tc and steam at Th , the maximum efﬁciency of the power plant varies as eC =

Th − Tc T = 1− c Th Th

and is maximized for high Th . 8.

Assuming an air temperature of 20°C above the surface of the pond, the difference in temperature between the lower layer of the pond and the atmosphere is ΔT = Th − Tc = (100° + 273) − ( 20° + 273) = 80 K and the Carnot efﬁciency is

56157_12_ch12_p371-399.indd 374

em ax = eC =

Th − Tc 80 K = ≈ 21% 373 K Th

10/12/10 2:47:12 PM

The Laws of Thermodynamics

375

10.

A slice of hot pizza cools off. Road friction brings a skidding car to a stop. A cup falls to the ﬂoor and shatters. Any process is irreversible if it looks funny or frightening when shown in a videotape running backward. At fairly low speeds, air resistance is small and the ﬂight of a projectile is nearly reversible.

12.

While the entropy of the sugar solution decreases by some amount, the second law of thermodynamics is not violated if the entropy of the environment increases by an equal or greater amount. Even at essentially constant temperature, energy must be transferred by heat out of the solidifying sugar into the surroundings. This action will increase the entropy of the environment. The water molecules become less ordered as they leave the liquid in the container to mix with the entire atmosphere.

14.

Heat is not a form of energy. Rather, it is the process of transferring energy through microscopic collisions between atoms or molecules. An object having a large mass and/or a high speciﬁc heat may be at a low temperature and still contain more internal energy than does a higher temperature, low-mass object.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

6.1× 10 2 J

(b)

0

(d)

0

(e)

+ 2.0 × 10 2 J

4.

(a)

31 m s

(b)

0.17

6.

(a)

See Solution.

(b)

See Solution.

(c)

The higher pressure during the expansion phase of the process results in more work being done in the process of part (a).

8.

(a)

1.25 MJ

(b)

−1.25 MJ

10.

(a)

–12 MJ

(b)

+12 MJ

12.

(a)

8.12 × 10 4 Pa

(b)

3.65 × 10 4 J

(c)

−5.68 × 10 4 J

(d)

977 K

(e)

1.22 × 10 5 J

(f)

8.55 × 10 4 J

(g)

1.42 × 10 5 J

(h)

additional energy must be transferred to the gas

(i)

1.42 × 10 5 J

(j)

Q = ΔU − W

14.

(a)

−2.7 × 10 5 J

(b)

energy is transferred from the sprinter to the environment by heat

16.

6P0 V0

18.

W = + 32 P0 V0 > 0, ΔU = 0, Q = − 32 P0 V0 < 0

20.

56157_12_ch12_p371-399.indd 375

(c)

− 4.1× 10 2 J

(a)

(a)

–395 J

(b)

285 K; at point A

(c)

37 J

10/12/10 2:47:15 PM

376

22.

Chapter 12

(a)

See Solution.

(b)

(d)

No. From part (c), Q = 52 Wenv, and Wenv is positive for an isobaric expansion process. Thus, the energy transfer by heat must be directed into the gas.

See Solution.

(c)

See Solution.

24.

WBC = 0, QBC < 0, ΔU BC < 0; WCA > 0, QCA < 0, ΔUCA < 0; WAB < 0, QAB > 0, ΔU AB > 0

26.

(a)

WIAF = −76.0 J, WIBF = −101 J, WIF = −88.7 J

(b)

QIAF = 167 J, QIBF = 192 J, QIF = 180 J

28.

(a)

1.66 × 10 −24 mol m 3

30.

600 kJ

32.

(a)

564°C

(b)

No. Any real heat engine operates in an irreversible manner and has an efﬁciency less than the Carnot efﬁciency.

34.

(a)

4 200 J

(b)

0.43 (or 43%)

36.

13.7°C

38.

(a)

7.69 × 108 J

(b)

5.67 × 108 J

40.

(a)

COP = Th

− Tc )

(b)

smaller

42.

(a)

0.051 2 (or 5.12%)

(b)

5.27 × 1012 J

(c)

As fossil-fuel prices rise, this could be an attractive way to use solar energy. However, the potential environmental impact of such an engine would require serious study. The energy output, Qc = Qh − Weng = Qh (1− e ), to the low temperature reservoir (cool water deep in the ocean) could raise the temperature of over a million cubic meters of water by 1°C every hour.

44.

(a)

9.09 kW

(b)

11.9 kJ

46.

(a)

+79 J K

(b)

−79 J K

48.

6.06 kJ K

50.

(a)

one way; 6+6

(b)

6 ways; 1+6, 2+5, 3+4, 4+3, 5+2, and 6+1

52.

(a)

−3.45 J K

(b)

+8.06 J K

(d)

Thermal energy always ﬂows from (Qr < 0) the hot reservoir at Th and into (Qr > 0) the cooler reservoir at Tc . Thus, in the expression for the total change in entropy, ΔS = Qr Tc + Qr Th = Qr Tc − Qr Th , the positive term has the smaller denominator and dominates the sum.

(a)

ΔSh = − Qh Th

(c)

ΔSUniverse = − Qh Th + Qh Tc

54.

56157_12_ch12_p371-399.indd 376

(T

h

(b)

(b)

3.7 × 10 −23 Pa

(c)

(c)

(c)

6.2 × 10116 Pa

17.2

+ 4.61 J K

ΔSc = + Qh Tc

10/12/10 2:47:17 PM

377

The Laws of Thermodynamics

56.

2.35 × 10 3 J

(a)

6.54 h

(b)

0.717 h

(c)

(d)

16.6 lifts

(e)

No, the human body is only about 25% efficient in converting chemical energy to mechanical energy.

58.

3.01

60.

0.55 kg

62.

(a)

251 J

(b)

314 J

(d)

104 J done on the gas

(e)

ΔU = 0 (cyclic process)

64.

(a)

4P0 V0

(b)

4P0 V0

(c)

9.07 kJ

66.

8 × 10 6 J K ⋅s

68.

(a)

1.00 × 10 5 Pa

(b)

1.31× 10 5 J

(c)

3.48 × 10 6 J

(d)

9.28 × 10 3 K; 2.12 × 10 6 Pa

(e)

0.259 kg; 0.494 kg m 3 (f)

2.93 × 10 3 m s

70.

(a)

2.4 × 10 6 J

(b)

1.6 × 10 6 J

(c)

2.8 × 10 2 J

72.

(a)

− 6.00 × 10 5 J

(b)

5.50 × 10 2 K

(c)

1.00 × 10 6 J

(d)

1.60 × 10 6 J

(c)

104 J done by the gas

PROBLEM SOLUTIONS 12.1

(a)

The work done on the gas in this constant pressure process is ⎛ nRT f nRTi ⎞ = −nR T f − Ti W = −P ( ΔV ) = −P ⎜ − ⎝ P P ⎟⎠

(

W = − ( 0.200 mol )(8.31 J mol ⋅K )( 573 K − 293 K ) = − 465 J

or (b) 12.2

(a)

)

The negative sign for work done on the gas indicates that the gas does positive work on its surroundings .

(W ) env

ab

P (atm)

= Pa (Vb − Va ) ⎛ 10 m ⎞ = ⎡⎣3 (1.013 × 10 5 Pa ) ⎤⎦ ( 3.0 L − 1.0 L ) ⎜ ⎝ 1 L ⎟⎠ −3

1

bc

= P (Vc − Vb ) = 0

0

cd

= Pc (Vd − Vc )

(W )

(c)

(W )

env

env

2

= 6.1 × 10 2 J (b)

a

3

b

3

c

d

V (L) 0

1

2

3

⎛ 10 −3 m 3 ⎞ = − 4.1× 10 2 J = ⎡⎣ 2 (1.013 × 10 5 Pa ) ⎤⎦ (1.0 L − 3.0 L ) ⎜ ⎝ 1 L ⎟⎠ continued on next page

56157_12_ch12_p371-399.indd 377

10/13/10 10:40:11 AM

378

Chapter 12

(d)

(W )

da

= P (Va − Vd ) = 0

(e)

(W )

net

= (Wenv )ab + (Wenv )bc + (Wenv )cd + (Wenv )da

env

env

= +6.1× 10 2 J + 0 − 4.1× 10 2 J + 0 = + 2.0 × 10 2 J Note that, to 2 signiﬁcant ﬁgures, this equals the area enclosed within the process diagram given above. 12.3

The constant pressure is P = (1.5 atm )(1.013 × 10 5 Pa atm ) and the work done on the gas is W = −P ( ΔV ) . (a)

ΔV = + 4.0 m 3 and W = −P ( ΔV ) = − (1.5 atm )(1.013 × 10 5 Pa atm ) ( + 4.0 m 3 ) = − 6.1× 10 5 J

(b)

ΔV = − 3.0 m 3, so W = − P ( ΔV ) = − (1.5 atm )(1.013 × 10 5 Pa atm ) ( − 3.0 m 3 ) = + 4.6 × 10 5 J

12.4

(a)

The work done by the gas on the projectile is given by the area under the curve in the PV diagram. This is Wenv = ( triangular area ) + ( rectangular area )

(

) (V

)

(

)

(

)

1 P0 − Pf 2

=

⎛ 1 m3 ⎞ 1 ⎡⎣(11+ 1.0 ) × 10 5 Pa ⎤⎦ ⎡⎣( 40.0 − 8.0 ) cm 3 ⎤⎦ ⎜ 6 = 19 J 2 ⎝ 10 cm 3 ⎟⎠

f

− V0 + Pf V f − V0 =

1 P0 + Pf 2

) (V

=

f

− V0

From the work-energy theorem, Wenv = ΔKE = 12 mv 2 − 0, where Wenv is the work done on the projectile by the gas. Thus, the speed of the emerging projectile is v= (b)

2Wenv = m

2 (19 J ) = 31 m s 40.0 × 10 −3 kg

The air in front of the projectile would exert a retarding force of Fr = Pair A = (1.0 × 10 5 Pa ) ⎡⎣(1.0 cm 2 ) (1 m 2 10 4 cm 2 ) ⎤⎦ = 10 N on the projectile as it moves down the launch tube. The energy spent overcoming this retarding force would be Wspent = Fr ⋅ s = (10 N )( 0.32 m ) = 3.2 J Wspent 3.2 J and the needed fraction is W = 19 J = 0.17 env

12.5

In each case, the work done on the gas is given by the negative of the area under the path on the PV diagram. Along those parts of the path where volume is constant, no work is done. Note that 1 atm = 1.013 × 10 5 Pa and 1 Liter = 10 −3 m 3.

continued on next page

56157_12_ch12_p371-399.indd 378

10/12/10 2:47:24 PM

379

The Laws of Thermodynamics

WIAF = WIA + WAF = −PI (VA − VI ) + 0

(a)

= − ⎡⎣ 4.00 (1.013 × 10 5 Pa ) ⎤⎦ ⎡⎣( 4.00 − 2.00 ) × 10 −3 m 3 ⎤⎦ = − 810 J WIF = − ( triangular area ) − ( rectangular area )

(b)

=−

1 ( PI − PB ) (VF − VB ) − PB (VF − VB ) = − 12 ( PI + PB ) (VF − VB ) 2

1 = − ⎡⎣( 4.000 + 1.00 )(1.013 × 10 5 Pa ) ⎤⎦ ( 4.00 − 2.00 ) × 10 −3 m 3 2 = − 507 J WIBF = WIB + WBF = 0 − PB (VF − VB )

(c)

= − ⎡⎣1.00 (1.013 × 10 5 Pa ) ⎤⎦ ⎡⎣( 4.00 − 2.00 ) × 10 −3 m 3 ⎤⎦ = − 203 J 12.6

The sketches for (a) and (b) are shown below: P1

P

P1

P

P2

P2

V

V (a)

(c) 12.7

V1

V2

(b)

V1

V2

As seen from the areas under the paths in the PV diagrams above, the higher pressure during the expansion phase of the process results in more work done by the gas in (a) than in (b).

With Pf = Pi = P, the ideal gas law gives Pf V f − Pi Vi = P ( ΔV ) = nR ( ΔT ) , so the work done by the gas is

⎛ m ⎞ R T f − Ti Wenv = +P ( ΔV ) = nR ( ΔT ) = ⎜ ⎝ M He ⎟⎠

(

)

If Wenv = 20.0 J, the mass of helium in the gas sample is m= 12.8

(a)

Wenv ( M He )

(

R T f − Ti

)

=

( 20.0 J )( 4.00 g mol ) = 0.0963 g = (8.31 J mol ⋅K )(373 K − 273 K )

The work done by the gas as it expands from point A to point B is given by the area under the PV diagram between these points. Consider the sketch given at the right and observe that this area can be broken into two rectangular areas and two triangular areas. The total area is given by

96.3 mg

P (kPa) 400 B

C

300 200 100

B

D

A A

1

2

3

V (m3) 4

5

6

continued on next page

56157_12_ch12_p371-399.indd 379

10/12/10 2:47:27 PM

380

Chapter 12

Wenv = (100 kPa )( 5.00 m 3 ) + +

1 ( 300 kPa )( 2.00 m 3 ) 2

1 (100 kPa )(1.00 m 3 ) + ( 200 kPa )( 2.00 m 3 ) 2

or Wenv = 1250 ( kPa )( m 3 ) = 1.25 × 10 6 ( N m 2 ) ( m 3 ) = 1.25 MJ (b)

When the volume is decreasing, the work done by the gas is the negative of the area under the PV diagram. Thus, if the gas is compressed from point B to point A along the same path, Wenv = −1.25 MJ

12.9

(a)

From the ideal gas law, nR = PV f T f = PVi Ti . With pressure constant this gives ⎛ Vf ⎞ T f = Ti ⎜ ⎟ = ( 273 K )( 4 ) = 1.09 × 10 3 K ⎝ Vi ⎠

(b)

The work done on the gas is

(

)

(

)

W = −P ( ΔV ) = − PV f − PVi = −nR T f − Ti = −nR ( 4Ti − Ti ) = − (1.00 mol )(8.31 J mol ⋅K )[ 3 ( 273 K )] = − 6.81× 10 3 J = − 6.81 kJ 12.10

(a)

(

)

The work done on the ﬂuid is the negative since V f > Vi of the area under the curve on the PV diagram. Thus, 1 Wi f = − ( 6 × 10 6 Pa − 0 ) ( 2 m 3 − 1 m 3 ) + ⎡⎣( 6 − 2 ) × 10 6 Pa ⎤⎦ ( 2 m 3 − 1 m 3 ) 2

{

}

+ ( 2.00 × 10 6 Pa − 0 ) ( 4 m 3 − 2 m 3 ) Wi f = −12 × 10 6 J = − 12 MJ (b)

(

)

When the system follows the process curve in the reverse direction with V f < Vi , the work done on the ﬂuid equals the area under the process curve, which is the negative of that computed in (a). Thus, W f i = −Wi f = + 12 MJ

12.11

From kinetic theory, the average kinetic energy per molecule is KE m olecule =

3 3⎛ R kB T = ⎜ 2 2 ⎝ NA

⎞ ⎟⎠ T

For a monatomic ideal gas containing N molecules, the total energy associated with random molecular motions is U = N ⋅ KE m olecule =

3⎛ N 2 ⎜⎝ N A

⎞ 3 ⎟⎠ RT = 2 nRT

Since PV = nRT for an ideal gas, the internal energy of a monatomic ideal gas is found to be given by U = 32 PV .

56157_12_ch12_p371-399.indd 380

10/12/10 2:47:28 PM

The Laws of Thermodynamics

12.12

(a)

381

The initial absolute temperature is Ti = 20.0° + 273.15 = 293 K, so the initial pressure is Pi =

nRTi (10.0 mol )(8.31 J mol ⋅K )( 293 K ) = = 8.12 × 10 4 Pa Vi 0.300 m 3

(b)

For a monatomic ideal gas, the internal energy is U = 3nRT 2. Thus, 3 3 U i = nRTi = (10.0 mol )(8.31 J mol ⋅K )( 293 K ) = 3.65 × 10 4 J 2 2

(c)

The work done on the gas in this isobaric expansion is W = −P ( ΔV ) = − (8.12 × 10 4 Pa ) (1.000 m 3 − 0.300 m 3 ) = −5.68 × 10 4 J Pf V f

4

Pa ) (1.00 m 3 )

Tf =

(e)

Uf =

(f)

ΔU = Uf − Ui = 1.22 × 10 5 J − 3.65 × 10 4 J = 8.55 × 10 4 J

(g)

ΔU − W = 8.55 × 10 4 J − ( −5.68 × 10 4 J ) = +1.42 × 10 5 J

nR

=

(8.12 × 10

(d)

(10.0 mol )(8.31 J mol ⋅K )

3 3 nRT f = (10.0 mol )(8.31 J mol ⋅K )( 977 K ) = 1.22 × 10 5 J 2 2

(h)

Since ΔU − W > 0, the increase in the internal energy of the gas exceeds the energy transferred to the gas by work. Thus, additional energy must be transferred to the gas by heat.

(i)

The additional energy that must be transferred to the gas by heat is ΔU − W = +1.42 × 10 5 J

12.13

= 977 K

[See part (g) above.]

(j)

The suggested relationship is Q = ΔU − W , which is a statement of the ﬁrst law of thermodynamics.

(a)

Along the direct path IF, the work done on the gas is W = − ( area under curve ) 1 = − ⎡⎢(1.00 atm )( 4.00 L − 2.00 L ) + ( 4.00 atm − 1.00 atm )( 4.00 L − 2.00 L ) ⎤⎥ 2 ⎣ ⎦ W = − 5.00 atm ⋅ L Thus,

ΔU = Q + W = 418 J − 5.00 atm ⋅ L ⎛ 1.013 × 10 5 Pa ⎞ ⎛ 10 −3 m 3 ⎞ = 418 J − ( 5.00 atm ⋅ L ) ⎜ ⎟⎠ ⎜⎝ 1 L ⎟⎠ = − 88.5 J 1 atm ⎝

(b)

Along path IAF, the work done on the gas is ⎛ 1.013 × 10 5 Pa ⎞ ⎛ 10 −3 m 3 ⎞ W = − ( 4.00 atm )( 4.00 L − 2.00 L ) ⎜ ⎟⎠ ⎜⎝ 1 L ⎟⎠ = −810 J 1 atm ⎝ From the ﬁrst law, Q = ΔU − W = − 88.5 J − ( − 810 J ) = 722 J

56157_12_ch12_p371-399.indd 381

10/12/10 2:47:31 PM

382

12.14

Chapter 12

(a)

We treat the sprinter’s body as a thermodynamic system and apply the ﬁrst law of thermodynamics, ΔU = Q + W. Then, with ΔU = −7.5 × 10 5 J and W = −4.8 × 10 5 J (negative because the sprinter does work on the environment), the energy absorbed as heat is Q = ΔU − W = −7.5 × 10 5 J − ( −4.8 × 10 5 J ) = −2.7 × 10 5 J

(b)

The negative sign in the answer to part (a) means that energy is transferred from the sprinter to the environment by heat .

12.15

12.16

(a)

⎛ 1.013 × 10 5 Pa ⎞ ⎛ 10 −3 m 3 ⎞ W = −P ( ΔV ) = − ( 0.800 atm )( −7.00 L ) ⎜ ⎟⎠ ⎜⎝ 1 L ⎟⎠ = 567 J 1 atm ⎝

(b)

ΔU = Q + W = −400 J + 567 J = 167 J

(

The work done on the gas is the negative since V f > Vi diagram, or

) of the area under the curve on the PV

1 3 W = − ⎡⎢ P0 ( 2V0 − V0 ) + ( 2P0 − P0 ) ( 2V0 − V0 ) ⎤⎥ = − P0 V0 2 2 ⎣ ⎦ From the result of Problem 11 (given as a hint in this problem), ΔU =

3 3 3 3 9 Pf V f − Pi Vi = ( 2P0 ) ( 2V0 ) − P0 V0 = P0 V0 2 2 2 2 2

Thus, from the ﬁrst law, Q = ΔU − W = 12.17

(a)

9 ⎛ 3 ⎞ P0 V0 − ⎜ − P0 V0 ⎟ = 6P0 V0 ⎝ 2 ⎠ 2

The change in the volume occupied by the gas is

(

)

ΔV = V f − Vi = A L f − Li = ( 0.150 m 2 )( −0.200 m ) = −3.00 × 10 −2 m 3 and the work done by the gas is Wenv = +P ( ΔV ) = ( 6 000 Pa ) ( −3.00 × 10 −2 m 3 ) = −180 J (b)

The ﬁrst law of thermodynamics is ΔU = Q + W = −Qoutput − Wenv . Thus, if ΔU = −8.00 J, the energy transferred out of the system by heat is Qoutput = −ΔU − Wenv = − ( −8.00 J ) − ( −180 J ) = + 188 J

12.18

(

)

The work done on the gas is the negative since V f < Vi of the area under the curve on the PV diagram, so

1 3 W = − ⎡⎢ P0 (V0 − 2V0 ) + ( 2P0 − P0 ) (V0 − 2V0 ) ⎤⎥ = + P0 V0 , or W > 0 2 2 ⎣ ⎦

The change in the internal energy of this monatomic ideal gas is 3 3 3 3 Pf V f − Pi Vi = ( 2P0 ) (V0 ) − ( P0 ) ( 2V0 ) = 0 2 2 2 2 3 3 Then, from the ﬁrst law, Q = ΔU − W = 0 − P0 V0 = − P0 V0 , or Q < 0 2 2 ΔU =

56157_12_ch12_p371-399.indd 382

10/12/10 2:47:33 PM

383

The Laws of Thermodynamics

12.19

(a)

W = F ⋅ d = ( 25.0 × 10 3 N ) ⋅ ( 0.130 m ) = 3.25 × 10 3 J = 3.25 kJ

(b)

Since the internal energy of an ideal gas is a function of temperature alone, the change in the internal energy in this isothermal process is ΔU = 0 .

(c)

From the ﬁrst law of thermodynamics, Q = ΔU − W = 0 − ( 3.25 kJ ) = −3.25 kJ

12.20

(d)

If the energy exchanged as heat is Q = 0 while the work done on the gas is positive (W > 0), the ﬁrst law of thermodynamics, ΔU = Q + W = 0 + W > 0, tells us that the internal energy of the system must increase. Since the internal energy of an ideal gas is directly proportional to the absolute temperature, the temperature must increase .

(a)

The work done on the gas as it goes from point A to point C is the negative of the area under the PV diagram between these points. Consider the sketch given at the right and observe that this area can be broken into a rectangular area and a triangular area. The work done on the gas is W = − [( Area 1) + ( Area 2 )] = − ( 0.500 atm )( 6.00 L ) −

P (atm) 0.800

B

0.600 0.400 0.200

2 A

C 1

2

4

6

8

V (L)

1 ( 0.300 atm )( 6.00 L ) 2

⎛ 1.013 × 10 5 N m 2 ⎞ ⎛ 10 −3 m 3 ⎞ = −3.90 atm ⋅ L ⎜ ⎟⎠ ⎜⎝ 1 L ⎟⎠ = −395 J 1 atm ⎝ (b)

The absolute temperature of an ideal gas is given by T = PV nR. Thus, the lowest temperature occurs where the product PV is the smallest. This is seen to be at point A , and the temperature at this point is TA =

(c)

PA VA ( 0.300 atm )( 2.00 L ) = = 285 K nR 0.025 6 mol ) ( 0.082 1 L·atm mol·K ) (

From the ﬁrst law of thermodynamics, ΔU = Q + W = +432 J − 395 J = 37 J

12.21

(a)

In an isothermal process involving an ideal gas, the work done on the gas is W = −Wenv = −nRT ln V f Vi . But, when temperature is constant, the ideal gas law gives Pi Vi = Pf V f = nRT and we may write the work done on the gas as

(

)

⎛ Vf ⎞ ⎛ 1.25 m 3 ⎞ = − 4.58 × 10 4 J W = −Pi Vi ln ⎜ ⎟ = − (1.00 × 10 5 Pa ) ( 0.500 m 3 ) ln ⎜ ⎝ 0.500 m 3 ⎟⎠ ⎝ Vi ⎠ (b)

The change in the internal energy of an ideal gas is ΔU = nCv ( ΔT ) , and for an isothermal process, we have ΔU = 0. Thus, from the ﬁrst law of thermodynamics, the energy transfer by heat in this isothermal expansion is Q = ΔU − W = 0 − ( − 4.58 × 10 4 J ) = + 4.58 × 10 4 J

(c)

56157_12_ch12_p371-399.indd 383

ΔU = 0

[See part (b) above.]

10/12/10 2:47:37 PM

384

12.22

Chapter 12

(a)

From the ideal gas law, Pi Vi = nRTi and Pf V f = nRT f . Thus, if Pi = Pf = P, subtracting these two expressions gives PV f − PVi = nRT f − nRTi , or P ( ΔV ) = nR ( ΔT ) .

(b)

For a monatomic, ideal gas containing N gas atoms, the internal energy is U=N

(

1 2

)

mv 2 = ( nN A ) ( 32 kB T ) = 32 nRT . Thus, the change in internal energy of this gas in a

thermodynamic process is ΔU = 32 nR ( ΔT ). But, using the result of part (a) above, we have, for an isobaric process involving an monatomic ideal gas, ΔU = (c)

3 3 3 nR ( ΔT ) = P ( ΔV ) = Wenv 2 2 2

We recall that the work done on the gas is W = −Wenv, and use the ﬁrst law of thermodynamics to ﬁnd that the energy transferred to the gas by heat to be Q = ΔU − W = ΔU + Wenv =

(d)

3 Wenv + Wenv 2

or

Q=

5 Wenv 2

In an isobaric expansion ( ⇒ ΔV > 0 ), the work done on the environment is Wenv = P ( ΔV ) > 0. Thus, from the result of part (c) above, the energy transfer as heat is Q > 0, meaning that the energy ﬂow is into the gas. Therefore, it is impossible for the gas to exhaust thermal energy in an isobaric expansion.

12.23

(a)

From PV = nRT , and with V f = Vi = V = 0.200 m 3 , we have ⎛ 1.013 × 10 5 N m 2 ⎞ 3 5.00 atm ⎜ ⎟⎠ ( 0.200 m ) 1 atm ⎝ Pi V n= = = 40.6 mol RTi (8.31 J mol·K )(300 K )

(b)

The total heat capacity of the gas is Ctotal = nCv = n ( 3R 2 ) = ( 40.6 mol )(1.50 )(8.31 J mol·K ) = 506 J K

(c)

The work done by the gas during this constant volume process is W = 0 .

(d)

From the ﬁrst law of thermodynamics, ΔU = Q + W = 16.0 kJ + 0 = 16.0 kJ .

(e)

The change in internal energy of a monatomic ideal gas is ΔU = nCv ( ΔT ) = Ctotal ΔT so the change in temperature is ΔT =

ΔU 16.0 kJ 16.0 × 10 3 J = = = 31.6 K Ctotal 506 J K 506 J K

(f)

The ﬁnal temperature is T f = Ti + ΔT = 300 K + 31.6 K = 332 K .

(g)

From the ideal gas law, the ﬁnal pressure of the gas is Pf =

nRT f

=

( 40.6 mol )(8.31 J mol·K )( 332 K )

0.200 m 3 1 atm ⎛ ⎞ = 5.60 × 10 5 Pa ⎜ = 5.53 atm ⎝ 1.013 × 10 5 Pa ⎟⎠

56157_12_ch12_p371-399.indd 384

V

10/12/10 2:47:40 PM

The Laws of Thermodynamics

12.24

385

Volume is constant in process BC, so WBC = 0 . Given that QBC < 0 , the ﬁrst law shows that ΔU BC = QBC + WBC = QBC + 0. Thus, ΔU BC < 0 . For process CA, ΔVCA = VA − VC < 0, so W = −P ( ΔV ) shows that WCA > 0 . Then, given that ΔUCA < 0 , the ﬁrst law gives QCA = ΔUCA − WCA and QCA < 0 . In process AB, the work done on the system is W = − ( area under curve AB ) where 1 2

( area under curve AB ) = PA (VB − VA ) + ( PB − PA ) (VB − VA ) > 0 Hence, WAB < 0 . For the cyclic process, ΔU = ΔU AB + ΔU BC + ΔUCA = 0, so ΔU AB = − ( ΔU BC + ΔUCA ). This gives ΔU AB > 0 , since both ΔU BC and ΔUCA are negative. Finally, from the ﬁrst law, Q = ΔU − W shows that QAB > 0 , since both ΔU AB and − WAB are positive. 12.25

(a)

The original volume of the aluminum is V0 = m rAl , and the change in volume is ΔV = b V0 ( ΔT ) = ( 3a Al ) ( m rAl )( ΔT ) . The work done by the aluminum is then Wenv = +P ( ΔV ) = P ( 3a Al ) ( m rAl )( ΔT ) ⎞ 5.0 kg −1 ⎛ = (1.013 × 10 5 Pa ) 3 ⎡⎣ 24 × 10 −6 ( °C ) ⎤⎦ ⎜ ( 70°C ) = 0.95 J 3 3 ⎟ ⎝ 2.70 × 10 kg m ⎠

(b)

The energy transferred by heat to the aluminum is Q = mcAl ( ΔT ) = ( 5.0 kg ) ( 900 J kg ⋅°C )( 70°C ) = 3.2 × 10 5 J

(c)

The work done on the aluminum is W = −Wenv = − 0.95 J, so the ﬁrst law gives ΔU = Q + W = 3.2 × 10 5 J − 0.95 J = 3.2 × 10 5 J

12.26

(a)

The work done on the gas in each process is the negative of the area under the process curve on the PV diagram. For path IAF, WIAF = WIA + WAF = 0 + WAF , or ⎡ ⎛ 1.013 × 10 5 Pa ⎞ ⎤ ⎡ ⎛ 10 −3 m 3 ⎞ ⎤ 0.500 L WIAF = − ⎢(1.50 atm ) ⎜ ( ) ⎥ ⎢ ⎟⎠ ⎜⎝ 1 L ⎟⎠ ⎥ = − 76.0 J 1 atm ⎝ ⎣ ⎦⎣ ⎦ For path IBF, WIBF = WIB + WBF = WIB + 0, or ⎡ ⎛ 1.013 × 10 5 Pa ⎞ ⎤ ⎡ ⎛ 10 −3 m 3 ⎞ ⎤ WIBF = − ⎢( 2.00 atm ) ⎜ ⎥ ⎢( 0.500 L ) ⎜ ⎥ = −101 J ⎟ 1 atm ⎠⎦⎣ ⎝ ⎝ 1 L ⎟⎠ ⎦ ⎣ For path IF, WIF = WAF − ( triangular area ), or WIF = −76.0 J −

5 ⎤⎡ ⎛ 10 −3 m 3 ⎞ ⎤ 1⎡ 3 ⎛ 1.013 × 10 Pa ⎞ 0.500 L ( ) ⎢( 0.500 m ) ⎜ ⎥ ⎢ ⎟⎠ ⎜⎝ 1 L ⎟⎠ ⎥ = −88.7 J 2⎣ 1 atm ⎝ ⎦⎣ ⎦

continued on next page

56157_12_ch12_p371-399.indd 385

10/12/10 2:47:43 PM

386

Chapter 12

(b)

Using the ﬁrst law, with ΔU = U F − U A = (182 − 91.0 ) J = 91.0 J for each process gives QIAF = ΔU − WIAF = 91.0 J − ( −76.0 J ) = 167 J QIBF = ΔU − WIBF = 91.0 J − ( −101 J ) = 192 J QIF = ΔU − WIF = 91.0 J − ( −88.7 J ) = 180 J

12.27

(a)

For adiabatic processes in ideal gases, Pf V fg = Pi Vig = constant. From the ideal gas law, P = nRT V , so the above expression becomes ⎛ nRT f ⎞ g ⎛ nRTi ⎞ g ⎜ V ⎟ V f = ⎜ V ⎟ Vi ⎝ i ⎠ ⎠ ⎝ f

T f V fg −1 = Ti Vig −1

or

which can be summarized as TV g −1 = C , where C is a constant. (b)

The current radius of the Universe is assumed to be rf = 1.4 × 10 26 m and the temperature is T f = 2.7 K. Since g = 1.67 = 5 3 for monatomic ideal gases (see Table 12.1 in the text), the temperature Ti of the Universe when its radius was ri = 2.0 cm = 2.0 × 10 −2 m must have been ⎛ V fg −1 ⎞ ⎛ Vf ⎞ Ti = T f ⎜ g −1 ⎟ = T f ⎜ ⎟ ⎝ Vi ⎠ ⎝ Vi ⎠

⎛ 1.4 × 10 26 m ⎞ Ti = ( 2.7 K ) ⎜ ⎝ 2 × 10 −2 m ⎟⎠

or 12.28

(a)

g −1

⎛ = Tf ⎜ ⎝

3( 5 3 ) − 3

π rf3 ⎞ 3 ⎟ 4 3 π ri ⎠ 4 3

g −1

⎛ rf ⎞ = Tf ⎜ ⎟ ⎝ ri ⎠

3g − 3

2

⎛ 1.4 × 10 26 m ⎞ = ( 2.7 K ) ⎜ = 1× 10 56 K ⎝ 2 × 10 −2 m ⎟⎠

The number of atoms per mole in any monatomic gas is Avogadro’s number N A = 6.02 × 10 23 atoms mol. Thus, if the density of gas in the Universe is 1 hydrogen atom per cubic meter, the number of moles per unit volume is 1 atom m 3 n 1 atom m 3 = == = 1.66 × 10 −24 mol m 3 V NA 6.02 × 10 23 atoms mol

(b)

With the density of gas found in part (a) and an absolute temperature of T = 2.7 K, the ideal gas law gives the pressure of the Universe as ⎛ n⎞ P = ⎜ ⎟ RT = (1.66 × 10 −24 mol m 3 ) (8.31 J mol ⋅K )( 2.7 K ) = 3.7 × 10 −23 Pa ⎝V⎠

(c)

For an adiabatic expansion, Pi Vig = Pf V fg with g = 1.67 = 5 3 for monatomic ideal gases (see Table 12.1 in the text), so the initial pressure of the Universe is estimated to be g

⎛ ⎛ V fg ⎞ ⎛ Vf ⎞ Pi = Pf ⎜ g ⎟ = Pf ⎜ ⎟ = Pf ⎜ ⎝ Vi ⎠ ⎝ Vi ⎠ ⎝

12.29

g

p rf3 ⎞ ⎛ rf ⎞ = Pf ⎜ ⎟ 3 ⎟ 4 ⎝ ri ⎠ 3 p rf ⎠

4 3

3( 5 3 )

or

⎛ 1.4 × 10 26 m ⎞ Pi = ( 3.7 × 10 −23 Pa ) ⎜ ⎝ 2.0 × 10 −2 m ⎟⎠

giving

5.0 Pi = ⎡⎣( 3.7 × 10 −23 Pa )( 7.0 ) ⎤⎦ (10 27 )

= ( 3.7 × 10 −23 Pa ) ( 7.0 × 10 27 )

5.0

5.0

= ⎡⎣6.2 × 10 −19 Pa ⎤⎦ (10135 ) = 6.2 × 10116 Pa

The net work done by a heat engine operating on the cyclic process shown in Figure P12.29 equals the triangular area enclosed by this process curve. Thus, Weng =

1 ( 6.00 atm − 2.00 atm )( 3.00 m 3 − 1.00 m 3 ) 2

⎛ 1.013 × 10 5 Pa ⎞ 5 = 4.00 atm ⋅ m 3 ⎜ ⎟⎠ = 4.05 × 10 J 1 atm ⎝ = 405 × 10 3 J = 405 kJ

56157_12_ch12_p371-399.indd 386

3g

P (atm) 6.00 4.00 2.00 1.00 2.00 3.00

V (m3)

FIGURE P12.29

10/12/10 2:47:48 PM

387

The Laws of Thermodynamics

12.30

The net work done by a heat engine operating on the cyclic process shown in Figure P12.30 equals the triangular area enclosed by this process curve. This is Weng

P (105 Pa) 6.00

1 = ( base )( altitude ) 2 1 = ⎡⎣( 4.00 − 1.00 ) m 3 ⎤⎦ ⎡⎣( 6.00 − 2.00 ) × 10 5 Pa ⎤⎦ 2

4.00 2.00 1.00 2.00 3.00 4.00

= 6.00 × 10 Pa = 600 kJ 5

12.31

(a)

Tc 298 K = 1− = 0.540 or 54.0% 648 K Th

The absolute temperature of the cold reservoir is Tc = 20° + 273 = 293 K. If the Carnot efﬁciency is to be eC = 0.65, it is necessary that 1−

Tc = 0.65 Th Th =

Thus,

12.33

FIGURE P12.30

The maximum possible efﬁciency for a heat engine operating between reservoirs with absolute temperatures of Tc = 25° + 273 = 298 K and Th = 375° + 273 = 648 K is the Carnot efﬁciency eC = 1−

12.32

V (m3)

or

Tc = 0.35 Th

293 K = 837 K 0.35

or

and

Th =

Tc 0.35

Th = 837 − 273 = 564°C

(b)

No. Any real heat engine will have an efﬁciency less that the Carnot efﬁciency because it operates in an irreversible manner.

(a)

The efﬁciency of a heat engine is e = Weng Qh , where Weng is the work done by the engine and Qh is the energy absorbed from the higher temperature reservoir. Thus, if Weng = Qh 4, the efﬁciency is e = 1 4 = 0.25 or 25% .

(b)

From conservation of energy, the energy exhausted to the lower temperature reservoir is Qc = Qh − Weng. Therefore, if Weng = Qh 4, we have Qc = 3 Qh 4 or Qc Qh = 3 4 .

12.34

(a)

The work done by a heat engine equals the net energy absorbed by the engine, or Weng = Qh − Qc . Thus, the energy absorbed from the high temperature reservoir is Qh = Weng + Qc = 1 800 J + 2 400 J = 4 200 J

(b)

The efﬁciency of the heat engine is e≡

12.35

(a)

Weng Qh

=

1 800 J = 0.43 4 200 J

or

43%

The maximum efﬁciency possible is that of a Carnot engine operating between reservoirs having absolute temperatures of Th = 1 870 + 273 = 2 143 K and Tc = 430 + 273 = 703 K. Th − Tc T 703 K = 1− c = 1− = 0.672 ( or 67.2%) 2 143 K Th Th Weng From e = , we ﬁnd Weng = e Qh = 0.420 (1.40 × 10 5 J ) = 5.88 × 10 4 J Qh eC =

(b)

so

56157_12_ch12_p371-399.indd 387

P=

Weng t

=

5.88 × 10 4 J = 5.88 × 10 4 W = 58.8 kW 1.00 s

10/12/10 2:47:52 PM

388

12.36

Chapter 12

The work done by the engine equals the change in the kinetic energy of the bullet, or 1 1 2 mb v 2f − 0 = ( 2.40 × 10 −3 kg ) ( 320 m s ) = 123 J 2 2

Weng =

Since the efﬁciency of an engine may be written as e=

Weng

Weng

=

Qh

Weng + Qc

where Qc is the exhaust energy from the engine, we ﬁnd that Qc = Weng (1 e − 1). This exhaust energy is absorbed by the 1.80-kg iron body of the gun, so the rise in temperature is ΔT =

12.37

12.38

Qc mgun ciron

Weng

(123 J )(1 0.011 0 − 1)

(1.80 kg) ( 448

Qc

= 13.7°C

1.20 × 10 3 J = 0.294 1.70 × 10 3 J

( or 29.4%)

e≡

(b)

Weng = Qh − Qc = 1.70 × 10 3 J − 1.20 × 10 3 J = 5.00 × 10 2 J

(c)

P=

(a)

The coefﬁcient of performance of a heat pump is COP = Qh W , where Qh is the thermal energy delivered to the warm space and W is the work input required to operate the heat pump. Therefore,

Weng t

=

Qh

= 1−

J kg ⋅°C )

(a)

Qh

= 1−

=

5.00 × 10 2 J = 1.67 × 10 3 W = 1.67 kW 0.300 s

⎡⎛ ⎛ 3 600 s ⎞ ⎤ J⎞ 8 Qh = W ⋅ COP = ( P ⋅ Δt ) ⋅ COP = ⎢⎜ 7.03 × 10 3 8.00 h ⎜ ⎟ ⎟⎠ ⎥ 3.80 = 7.69 × 10 J ⎝ ⎠ s 1 h ⎝ ⎣ ⎦ The energy extracted from the cold space (outside air) is

(

(b)

1 ⎞ ⎛ = Qh ⎜ 1− ⎟ ⎝ COP ⎠ COP 1 ⎞ ⎛ Qc = ( 7.69 × 108 J ) ⎜ 1− = 5.67 × 108 J ⎝ 3.80 ⎟⎠

Qc = Qh − W = Qh − or 12.39

(a) (b)

)

Qh

⎛ 1 y ⎞ kWh ⎞ ⎛ 3.60 × 10 6 J ⎞ ⎛ ⋅ (1 d ) = 4.50 × 10 6 J W = P ⋅ Δt = ⎜ 457 ⎟⎜ ⎜ ⎟ y ⎠ ⎝ 1 kWh ⎠ ⎝ 365.242 d ⎟⎠ ⎝ From the deﬁnition of the coefﬁcient of performance for a refrigerator, ( COP )R = Qc W, the thermal energy removed from the cold space each day is Qc = ( COP )R ⋅W = 6.30 ( 4.50 × 10 6 J ) = 2.84 × 10 7 J

(c)

The water must be cooled 20.0°C before it will start to freeze, so the thermal energy that must be removed from mass m of water to freeze it is Qc = mcw ΔT + mL f . The mass of water that can be frozen each day is then m=

12.40

(a)

Qc cw ΔT + L f

=

2.84 × 10 7 J = 68.2 kg ( 4 186 J kg ⋅°C)( 20.0°C) + 3.33 × 105 J kg

The coefﬁcient of performance of a heat pump is deﬁned as

( COP )hp =

Qh W

=

Qh Qh − Qc

=

1 1− Qc Qh

But when a device operates on the Carnot cycle, Qc Qh = Tc Th . Thus, the coefﬁcient of performance for a Carnot heat pump would be

( COP )hp,C =

Th 1 = 1− Tc Th Th − Tc

continued on next page

56157_12_ch12_p371-399.indd 388

10/12/10 2:47:57 PM

The Laws of Thermodynamics

(b)

From the result of part (a) above, we observe that the COP of a Carnot heat pump would increase if the temperature difference Th − Tc became smaller.

(c)

If Tc = 50° + 273 = 323 K and Th = 70° + 273 = 343 K, the COP of a Carnot heat pump would be

( COP )hp,C = 12.41

389

Th 343 K = = 17.2 Th − Tc 343 K − 323 K

The actual efﬁciency of the engine is e = 1−

Qc Qh

= 1−

300 J = 0.400 500 J

If this is 60.0% of the Carnot efﬁciency, then eC =

e 0.400 2 = = 0.600 0.600 3

Thus, from eC = 1− Tc Th , we ﬁnd Tc 2 1 = 1− eC = 1− = 3 3 Th 12.42

(a)

The Carnot efﬁciency represents the maximum possible efﬁciency. With Th = 20.0°C = 293 K and Tc = 5.00°C = 278 K, this efﬁciency is given by eC = 1−

(b)

( or 5.12%)

The efﬁciency of an engine is e = Weng Qh , so the minimum energy input by heat each hour is Qh

12.43

Tc 278 K = 1− = 0.051 2 293 K Th

m in

=

Weng em ax

6 P ⋅ Δt ( 75.0 × 10 J s ) ( 3 600 s ) = = = 5.27 × 1012 J em ax 0.0512

(c)

As fossil-fuel prices rise, this could be an attractive way to use solar energy. However, the potential environmental impact of such an engine would require serious study. The energy output, Qc = Qh − Weng = Qh (1− e ), to the low temperature reservoir (cool water deep in the ocean) could raise the temperature of over a million cubic meters of water by 1°C every hour.

(a)

We treat the power plant as a heat engine and compute its efﬁciency as e≡

(b)

Weng Qh

=

435 MW = 0.306 1 420 MW

or

30.6%

The work done by a heat engine equals the net energy absorbed by the engine, or Weng = Qh − Qc . Thus, the energy expelled to the low temperature reservoir is Qc = Qh − Weng = 1 420 MW − 435 MW = 985 MW

12.44

(a)

With reservoirs at absolute temperatures of Tc = 80.0 + 273 = 353 K and Th = 350 + 273 = 623 K, the Carnot efﬁciency is eC = 1−

Tc 353 K = 1− = 0.433 623 K Th

(or 43.3%)

continued on next page

56157_12_ch12_p371-399.indd 389

10/12/10 2:48:00 PM

390

Chapter 12

so the maximum power output is Pm ax = (b)

(W ) eng

m ax

t

=

eC Qh t

=

0.433 ( 21.0 kJ ) = 9.09 kW 1.00 s

From e = 1− Qc Qh , the energy expelled by heat each cycle is Qc = Qh (1− e ) = ( 21.0 kJ )(1− 0.433) = 11.9 kJ

12.45

The thermal energy transferred to the room by the water as the water cools from 1.00 × 10 2 °C to 20.0°C is Q = mcw ΔT = ( 0.125 kg ) ( 4 186 J kg ⋅°C )(80.0°C ) = 4.19 × 10 4 J If the room has a constant absolute temperature of T = 20.0° + 273 = 293 K, the increase in the entropy of the room is ΔS =

12.46

(a)

Q 4.19 × 10 4 J = = 143 J K T 293 K

The energy transferred to the ice cube as it melts at a constant temperature of T = 0.0 °C = 273 K is Qice = +mice L f and the change in entropy of the ice cube is ΔSice =

(b)

The energy transferred to the environment during this melting process is Qenv = −Qice = −mice L f and the change in entropy of the environment is ΔSenv =

12.47

5 Qice mice L f ( 0.065 kg ) ( 3.33 × 10 J kg ) = = = + 79 J K T T 273 K

Qenv −mice L f = = − 79 J K T T

The energy transferred from the water by heat, and absorbed by the freezer, is ⎛ J ⎞ = 3.3 × 10 5 J Q = mL f = ( rV ) L f = ⎡⎣(10 3 kg m 3 ) (1.0 × 10 −3 m 3 ) ⎤⎦ ⎜ 3.33 × 10 5 kg ⎟⎠ ⎝ Thus, the change in entropy of the water is (a)

ΔSwater =

( ΔQ ) r

water

T

=

− 3.3 × 10 5 J J = −1.2 kJ K = −1.2 × 10 3 K 273 K

and that of the freezer is (b) 12.48

ΔSfreezer =

( ΔQ ) r

freezer

T

=

+ 3.3 × 10 5 J = +1.2 kJ K 273 K

The energy added to the water by heat is ΔQr = mLv = (1.00 kg ) ( 2.26 × 10 6 J kg ) = 2.26 × 10 6 J so the change in entropy is ΔS =

12.49

ΔQr 2.26 × 10 6 J J = = 6.06 kJ K = 6.06 × 10 3 K T 373 K

The potential energy lost by the log is transferred away by heat, so the energy transferred from the log to the reservoir is ΔQr = mgh. The change in entropy of the reservoir (universe) is then 2 ΔQr mgh ( 70.0 kg ) ( 9.80 m s )( 25.0 m ) ΔS = = = = 57.2 J K T T 300 K

56157_12_ch12_p371-399.indd 390

10/12/10 2:48:03 PM

The Laws of Thermodynamics

12.50

391

(a)

In a game of dice, there is only one way you can roll a 12. You must have a 6 on each die.

(b)

There are six ways to obtain a 7 with a pair of dice. The combinations that yield a 7 are: 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, and 6 + 1. Note that 1 + 6 and 6 + 1 are different combinations in that the 6 occurs on different members of the pair of dice in the two combinations.

12.51

A quantity of energy, of magnitude Q, is transferred from the Sun and added to Earth. Thus, −Q +Q ΔSSun = and ΔSEarth = , so the total change in entropy is TSun TEarth ΔStotal = ΔSEarth + ΔSSun =

Q Q − TEarth TSun

⎞ ⎛ 1 1 = + 3.27 J K = (1 000 J ) ⎜ − ⎝ 290 K 5 700 K ⎟⎠ 12.52

12.53

12.54

56157_12_ch12_p371-399.indd 391

The change in entropy of a reservoir is ΔS = Qr T , where Qr is the energy absorbed (Qr > 0 ) or expelled (Qr < 0 ) by the reservoir, and T is the absolute temperature of the reservoir. (a)

For the hot reservoir:

ΔSh =

−2.50 × 10 3 J = −3.45 J K 725 K

(b)

For the cold reservoir:

ΔSc =

+2.50 × 10 3 J = +8.06 J K 310 K

(c)

For the Universe:

ΔSU = ΔSh + ΔSc = −3.45 J K + 8.06 J K = +4.61 J K

(d)

The magnitudes of the thermal energy transfers, appearing in the numerators, are the same for the two reservoirs, but the cold reservoir necessarily has a smaller denominator. Hence, its positive change dominates.

(a)

The table is shown below. On the basis of the table, the most probable result of a toss is 2 H and 2 T .

End Result

Possible Tosses

Total Number of Same Result

All H 1T, 3H 2T, 2H 3T, 1H All T

HHHH HHHT, HHTH, HTHH, THHH HHTT, HTHT, THHT, HTTH, THTH, TTHH TTTH, TTHT, THTT, HTTT TTTT

1 4 6 4 1

(b)

The most ordered state is the least likely. This is seen to be all H or all T .

(c)

The least ordered state is the most likely. This is seen to be 2H and 2T .

The change in entropy of a reservoir is ΔS = Qr T , where Qr is the energy absorbed (Qr > 0 ) or expelled (Qr < 0 ) by the reservoir, and T is the absolute temperature of the reservoir. − Qh

(a)

For the hot reservoir, Qr = − Qh , and

ΔSh =

(b)

For the cold reservoir, Qr = + Qh , and

ΔSc =

(c)

For the Universe:

ΔSU = ΔSh + ΔSc = −

Th + Qh Tc Qh Th

+

Qh Tc

10/12/10 2:48:06 PM

392

12.55

Chapter 12

Using the metabolic rates from Table 12.4, we ﬁnd the change in the body’s internal energy (i.e., the energy consumed) for each activity, and the total change for the day, as: Metabolic Rate (W)

Activity

Internal Energy Change

Sleeping – 8.0 h

80

ΔU1 = − (80 J s )(8.0 h )( 3 600 s 1 h ) = −2.3 × 10 6 J

Light Chores – 3.0 h

230

ΔU 2 = − ( 230 J s )( 3.0 h ) ( 3 600 s 1 h ) = −2.5 × 10 6 J

Slow Walk – 1.0 h

230

Running – 0.5 h

465

ΔU 3 = − ( 230 J s )(1.0 h )( 3 600 s 1 h ) = −8.3 × 10 5 J

ΔU 4 = − ( 465 J s ) ( 0.50 h )( 3 600 s 1 h ) = −8.4 × 10 5 J ΔU1 + ΔU 2 + ΔU 3 + ΔU 4 = − 6.5 × 10 6 J = − 6.5 MJ

Total 12.56

(a)

At the sleeping rate of 80.0 W, the time required for the body to use the 450 Cal of energy supplied by the bagel is Δt =

(b)

The metabolic rate while working out is P = 80.0 W + 650 W = 7.30 × 10 2 W, and the time to use the energy from the bagel at this rate is Δt =

(c)

ΔU 450 Cal ⎛ 4 186 J ⎞ ⎛ 1 h ⎞ = 6.54 h = ⎜ ⎟ P 80.0 J s ⎝ 1 Cal ⎠ ⎜⎝ 3 600 s ⎟⎠

ΔU 450 Cal = P 7.30 × 10 2 J

⎛ 4 186 J ⎞ ⎛ 1 h ⎞ = 0.717 h ⎜ ⎟ s ⎝ 1 Cal ⎠ ⎜⎝ 3 600 s ⎟⎠

Wper = F ⋅ Δx = mgh = (120 kg ) ( 9.80 m s 2 )( 2.00 m ) = 2.35 × 10 3 J lift

(d)

The additional energy consumed in 1 minute while working out (instead of sleeping) is ΔU = Pincrease (1 min ) = ( 650 J s )( 60.0 s ) = 3.90 × 10 4 J. The number of barbell lifts this should allow in 1 minute is N=

ΔU 3.90 × 10 4 J = = 16.6 Wper 2.35 × 10 3 J lift

(e) 12.57

No. The body is only about 25% efﬁcient in converting chemical energy to mechanical energy.

The maximum rate at which the body can dissipate waste heat by sweating is ⎛ kg ⎞ ⎛ J ⎞⎛ 1 h ⎞ ΔQ ⎛ Δm ⎞ 3 2 430 × 10 3 =⎜ Lv = ⎜ 1.5 ⎟ ⎜ ⎟⎜ ⎟ ⎟⎠ = 1.0 × 10 W h kg Δt ⎝ Δt ⎠ 3 600 s ⎝ ⎝ ⎠⎝ ⎠ If this represents 80% of the maximum sustainable metabolic rate ⎡⎣ i.e., ΔQ Δt = 0.80 ( ΔU Δt )m ax ⎤⎦ , then that maximum rate is

12.58

ΔQ Δt 1.0 × 10 3 W ⎛ ΔU ⎞ = = 1.3 × 10 3 W ⎜⎝ ⎟⎠ = 0.80 Δt m ax 0.80 Operating between reservoirs having temperatures of Th = 100°C = 373 K and Tc = 20°C = 293 K, the theoretical efﬁciency of a Carnot engine is eC = 1−

Tc 293 K = 1− = 0.214 373 K Th

If the temperature of the hotter reservoir is changed to Th′ = 550°C = 823 K, the theoretical efﬁciency of the Carnot engine increases to eC′ = 1−

Tc 293 K = 1− = 0.644 823 K Th′

continued on next page

56157_12_ch12_p371-399.indd 392

10/12/10 2:48:09 PM

The Laws of Thermodynamics

393

The factor by which the efﬁciency has increased is eC′ 0.644 = = 3.01 eC 0.214 12.59

The work output from the engine in an interval of one second is Weng = 1 500 kJ. Since the efﬁciency of an engine may be expressed as e=

Weng Qh

=

Weng Weng + Qc

⎛1 ⎞ ⎛ 1 ⎞ the exhaust energy each second is Qc = Weng ⎜ − 1⎟ = (1 500 kJ ) ⎜ − 1 = 4.5 × 10 3 kJ ⎝e ⎠ ⎝ 0.25 ⎟⎠ The mass of water ﬂowing through the cooling coils each second is m = rV = (10 3 kg m 3 )( 60 L ) (10 −3 m 3 1 L ) = 60 kg so the rise in the temperature of the water is ΔT = 12.60

Qc mcwater

=

4.5 × 10 6 J = 18°C (60 kg) ( 4 186 J kg ⋅°C)

The energy exhausted from a heat engine is Qc = Qh − Weng =

Weng e

⎛1 ⎞ − Weng = Weng ⎜ − 1⎟ ⎝e ⎠

where Qh is the energy input from the high temperature reservoir, Weng is the useful work done, and e = Weng Qh is the efﬁciency of the engine. For a Carnot engine, the efﬁciency is eC = 1− Tc Th = ( Th − Tc ) Th ⎛ Tc ⎞ ⎛ Th ⎞ Qc = Weng ⎜ − 1⎟ = Weng ⎜ ⎝ Th − Tc ⎟⎠ ⎝ Th − Tc ⎠

so we now have

Thus, if Th = 100°C = 373 K and Tc = 20°C=293 K, the energy exhausted when the engine has done 5.0 × 10 4 J of work is 293 K ⎛ ⎞ Qc = ( 5.0 × 10 4 J ) ⎜ = 1.83 × 10 5 J ⎝ 373 K − 293 K ⎟⎠ The mass of ice (at 0°C) this exhaust energy could melt is m= 12.61

(a)

Qc L f , water

=

1.83 × 10 5 J = 0.55 kg 3.33 × 10 5 J kg

The work done by the system in process AB equals the area under this curve on the PV diagram. Thus, Wenv = ( triangular area ) + ( rectangular area ) , or Wenv = [ 12 ( 4.00 atm )( 40.0 L ) Pa ⎞ ⎛ 10 −3 m 3 ⎞ ⎛ + (1.00 atm )( 40.0 L )] ⎜ 1.013 × 10 5 ⎟ ⎝ atm ⎠ ⎜⎝ L ⎟⎠ = 1.22 × 10 4 J = 12.2 kJ Note that the work done on the system is WAB = −Wenv = −12.2 kJ for this process. continued on next page

56157_12_ch12_p371-399.indd 393

10/12/10 2:48:12 PM

394

Chapter 12

(b)

The work done on the system (that is, the work input) for process BC is the negative of the area under the curve on the PV diagram, or Pa ⎞ ⎛ 10 −3 m 3 ⎞ ⎛ WBC = − [(1.00 atm )(10.0 L − 50.0 L )] ⎜ 1.013 × 10 5 ⎟ ⎝ atm ⎠ ⎜⎝ 1 L ⎟⎠ = 4.05 kJ

(c)

The change in internal energy is zero for any full cycle, so the ﬁrst law gives Qcycle = ΔU cycle − Wcycle = 0 − (WAB + WBC + WCA ) = 0 − ( −12.2 kJ + 4.05 kJ + 0 ) = 8.15 kJ

12.62

(a)

From the ﬁrst law, ΔU1→ 3 = Q123 + W123 = + 418 J + ( −167 J ) = 251 J .

(b)

The difference in internal energy between states 1 and 3 is independent of the path used to get from state 1 to state 3.

(c)

Thus,

ΔU1→ 3 = Q143 + W143 = 251 J ,

and

Q143 = 251 J − W143 = 251 J − ( − 63.0 J ) = 314 J

W12341 = W123 + W341 = W123 + ( −W143 ) = −167 J − ( −63.0 J ) = −104 J or 104 J of work is done by the gas in the cyclic process 12341.

(d)

W14321 = W143 + W321 = W143 + ( −W123 ) = −63.0 J − ( −167 J ) = +104 J or 104 J of work is done on the gas in the cyclic process 14321.

12.63

(e)

The change in internal energy is zero for both parts (c) and (d) since both are cyclic processes.

(a)

The change in length, due to linear expansion, of the rod is −1 ΔL = a L 0 ( ΔT ) = ⎡⎣11× 10 −6 ( °C ) ⎤⎦ ( 2.0 m )( 40°C − 20°C ) = 4.4 × 10 −4 m

The load exerts a force F = mg = ( 6 000 kg ) ( 9.80 m s 2 ) = 5.88 × 10 4 N on the end of the rod in the direction of movement of that end. Thus, the work done on the rod is W = F ⋅ ΔL = ( 5.88 × 10 4 N ) ( 4.4 × 10 −4 m ) = 26 J (b)

The energy added by heat is ⎛ J ⎞ Q = mc ( ΔT ) = (100 kg ) ⎜ 448 ( 20 °C) = 9.0 × 105 J kg ⋅°C ⎟⎠ ⎝

(c)

56157_12_ch12_p371-399.indd 394

From the ﬁrst law, ΔU = Q + W = 9.0 × 10 5 J + 26 J = 9.0 × 10 5 J .

10/12/10 2:48:15 PM

The Laws of Thermodynamics

12.64

(a)

395

The work done by the gas during each full cycle equals the area enclosed by the cycle on the PV diagram. Thus Wenv = ( 3P0 − P0 ) ( 3V0 − V0 ) = 4P0 V0

(b)

Since the work done on the gas is W = −Wenv = −4P0 V0 and ΔU = 0 for any cyclic process, the ﬁrst law gives Q = ΔU − W = 0 − ( − 4P0 V0 ) = 4P0 V0

(c)

From the ideal gas law, P0 V0 = nRT0 , so the work done by the gas each cycle is J ⎞ ⎛ Wenv = 4nRT0 = 4 (1.00 mol ) ⎜ 8.31 ⎟ ( 273 K ) ⎝ mol ⋅K ⎠ = 9.07 × 10 3 J = 9.07 kJ

12.65

(a)

The energy transferred to the gas by heat is J ⎞ ⎛ 3 Q = mc ( ΔT ) = (1.00 mol ) ⎜ 20.79 ⎟ (120 K ) = 2.49 × 10 J = 2.49 kJ ⎝ mol ⋅K ⎠

(b)

Treating the neon as a monatomic ideal gas, Equation 12.3b gives the change in internal energy as ΔU = 32 nR(ΔT ) , or ΔU =

(c)

3 J ⎞ 3 (1.00 mol ) ⎛⎜⎝ 8.31 ⎟ (120 K ) = 1.50 × 10 J = 1.50 kJ 2 mol ⋅K ⎠

From the ﬁrst law, the work done on the gas is W = ΔU − Q = 1.50 × 10 3 J − 2.49 × 10 3 J = − 990 J

12.66

Assuming the gravitational potential energy given up by the falling water is transformed into thermal energy when the water hits the bottom of the falls, the rate of thermal energy production is ΔQ ⎛ Δm ⎞ ⎛ ΔV ⎞ =⎜ ⎟ gh = rw ⎜⎝ ⎟ gh Δt ⎝ Δt ⎠ Δt ⎠ Then, if the absolute temperature of the environment is TK = 20.0° + 273 = 293 K, the rate of entropy production is ΔS ΔQ Δt rw ( ΔV Δt ) gh = = TK TK Δt

56157_12_ch12_p371-399.indd 395

or

1 ⎛ 3 kg ⎞ ⎛ m3 ⎞ ⎛ ΔS m⎞ 3 6 5 × 10 = 10 ⎜⎝ 9.8 2 ⎟⎠ ( 50.0 m ) = 8 × 10 ⎜ ⎟ ⎜ 3 ⎟ Δt 293 K ⎝ s ⎠ s m ⎠⎝

and

ΔS = 8 × 10 6 Δt

(N ⋅ m) K ⋅s

( kg ⋅ m s ) ⋅ m 2

K ⋅s

= 8 × 10 6 J K ⋅s

10/12/10 2:48:17 PM

396

12.67

Chapter 12

(a)

TA =

(10.0 × 103 Pa ) (1.00 m3 ) PA VA = nR (10.0 mol )(8.31 J mol ⋅K )

P (kPa) 40.0 30.0

= 1.20 × 10 2 K

20.0

(10.0 × 103 Pa ) (6.00 m3 ) PV TB = B B = nR (10.0 mol )(8.31 J mol ⋅K )

B

1.00 2.00 3.00 4.00 5.00 6.00

= 722 K (b)

A

10.0 V (m3)

As it goes from A to B, the gas is expanding and hence, doing work on the environment. The magnitude of the work done equals the area under the process curve from A to B. We subdivide this area into 2 rectangular and 2 triangular parts: Wenv = ⎡⎣(10.0 × 10 3 Pa )( 6.00 − 1.00 ) m 3 ⎤⎦ + ⎡⎣( 40.0 − 10.0 ) × 10 3 Pa ⎤⎦ (1.00 m 3 ) 1 + 2 ⎡⎢ (1.00 m 3 )( 40.0 − 10.0 ) × 10 3 Pa ⎤⎥ = 1.10 × 10 5 J ⎣2 ⎦

(c)

The change in the internal energy of a monatomic, ideal gas is ΔU = 32 nR ( ΔT ) , so 3 3 nR ( TB − TA ) = (10.0 mol )(8.31 J mol ⋅K )( 722 K − 120 K ) = 7.50 × 10 4 J 2 2

ΔU A→ B = (d)

From the ﬁrst law of thermodynamics, Q = ΔU − W , where W is the work done on the gas. In this case, W = −Wenv = −1.10 × 10 5 J , and Q = ΔU − W = 7.50 × 10 4 J − ( −1.10 × 10 5 J ) = 1.85 × 10 5 J

12.68

(a)

The constant volume occupied by the gases is V = 43 p r 3 = 4p ( 0.500 m ) 3 = 0.524 m 3 and the initial absolute temperature is Ti = 20.0° + 273 = 293 K . 3

To determine the initial pressure, we treat each component of the mixture as an ideal gas and compute the pressure it would exert if it occupied the entire volume of the container. The total pressure exerted by the mixture is then the sum of the partial pressures exerted by the components of the mixture. This gives PH =

nH RT

(b)

PO =

2

V

2

nO RT 2

V

2

and

Pi = PH + PO =

or

Pi =

2

2

nH RT 2

V

+

nO RT 2

V

=

(n

H2

)

+ nO RT 2

V

(14.4 mol + 7.2 mol )(8.31 J mol ⋅K )( 293 K ) 0.524 m 3

= 1.00 × 10 5 Pa

Treating both the hydrogen and oxygen as ideal gases, each with internal energy U = nCv T , where Cv is the molar speciﬁc heat at constant volume, we use Table 12.1 and ﬁnd the initial internal energy of the mixture as U i = U H ,i + U O 2

or

2 ,i

(

)

= nH Cv ,H + nO Cv ,O Ti 2

2

2

2

U i = ⎡⎣(14.4 mol )( 20.4 J mol ⋅K ) + ( 7.2 mol )( 21.1 J mol ⋅K ) ⎤⎦ ( 293 K ) = 1.31× 10 5 J

continued on next page

56157_12_ch12_p371-399.indd 396

10/12/10 2:48:20 PM

The Laws of Thermodynamics

(c)

397

During combustion, this mixture produces 14.4 moles of water (1 mole of water for each mole of hydrogen used) with a conversion of 241.8 kJ of chemical potential energy per mole. Since the volume is constant, no work is done and the additional internal energy generated in the combustion is ΔU = (14.4 mol )( 241.8 kJ mol ) = 3.48 × 10 3 kJ = 3.48 × 10 6 J

(d)

After combustion, the internal energy of the system is U f = U i + ΔU = 1.31× 10 5 J + 3.48 × 10 6 J = 3.61× 10 6 J Treating the steam as an ideal gas, so U = nCv TK , and obtaining the molar heat capacity for water vapor (a polyatomic gas) from Table 12.1, we ﬁnd Tf =

Uf nwater Cv ,water

=

3.61× 10 6 J = 9.28 × 10 3 K (14.4 mol )( 27.0 J mol ⋅K )

and the ﬁnal pressure is Pf = (e)

nwater RT f V

=

(14.4 mol )(8.31 J mol ⋅K ) ( 9.28 × 103 K ) 0.524 m 3

= 2.12 × 10 6 Pa

The total mass of steam present after combustion is msteam = nsteam M water = (14.4 mol )(18.0 × 10 −3 kg mol ) = 0.259 kg rs =

and its density is (f)

Assuming the steam is essentially at rest within the container ( v1 ≈ 0 ) , P2 = 0 (since the steam spews into a vacuum), and y2 = y1 , we use the pressure from part (d) above and Bernoulli’s equation ( P2 + 12 rs v22 + rs gy2 = P1 + 12 rs v12 + rs gy1 ) to ﬁnd the exhaust speed as v2 =

12.69

msteam 0.259 kg = = 0.494 kg m 3 0.524 m 3 V

2P1 = rs

2 ( 2.12 × 10 6 Pa ) 0.494 kg m 3

= 2.93 × 10 3 m s

The work that you have done is ⎛ in ⎞ ⎛ 2.54 × 10 −2 m ⎞ ⎤ step ⎞ ⎡ ⎛ 4.448N ⎞ ⎤ ⎡⎛ Weng = mg ( Δh ) = ⎢(150 lb ) ⎜ 8.00 90.0 30.0 min ( ) ⎟ ⎟⎠ ⎥ ⎜⎝ ⎝ 1 lb ⎟⎠ ⎥⎦ ⎢⎣⎜⎝ step ⎟⎠ ⎜⎝ 1 in min ⎠ ⎣ ⎦ Weng = 3.66 × 10 5 J

or

⎛ 4 186 J ⎞ If the energy input by heat was Qh = ( 600 kcal ) ⎜ = 2.51× 10 6 J , your efﬁciency has ⎝ 1 kcal ⎟⎠ been e=

Weng Qh

=

3.66 × 10 5 J = 0.146 or 14.6% 2.51× 10 6 J

If the actual efﬁciency was e = 0.180 or 18.0%, the actual energy input was Qh

56157_12_ch12_p371-399.indd 397

actual

=

Weng eactual

=

⎛ 1 kcal ⎞ 3.66 × 10 5 J = 485 kcal = ( 2.03 × 10 6 J ) ⎜ 0.180 ⎝ 4 186 J ⎟⎠

10/12/10 2:48:23 PM

398

12.70

Chapter 12

(a)

The energy transferred from the water by heat as it cools is Qh = mc ΔT = ( rV ) c ΔT ⎡⎛ ⎛ 10 3 cm 3 ⎞ ⎤ ⎛ cal ⎞ ⎛ 4.186 J ⎞ g ⎞ = ⎢⎜ 1.0 1.0 L ( ) 3 ⎟ ⎜⎝ 1 L ⎟⎠ ⎥ ⎜⎝ 1.0 g ⋅°C ⎟⎠ ⎜⎝ 1 cal ⎟⎠ ( 570°C − 4.0°C ) ⎝ ⎠ cm ⎣ ⎦ or

(b)

Qh = 2.4 × 10 6 J

The maximum efﬁciency of a heat engine is the Carnot efﬁciency. Thus, eC = 1−

Tc 277 K ( 4.0 + 273) K = 1− = 0.67 = 1− Th 570 + 273 K 843 K ( )

The maximum useful work output is then

(W ) eng

(c)

m ax

= eC Qh = ( 0.67 )( 2.4 × 10 6 J ) = 1.6 × 10 6 J

The energy available from oxidation of the hydrogen sulﬁde in 1.0 L of this water is mol ⎞ J ⎞ ⎡⎛ ⎤⎛ 3 2 U = n ( 310 kJ mol ) = ⎢⎜ 0.90 × 10 −3 ⎟⎠ (1.0 L )⎥ ⎜⎝ 310 × 10 ⎟ = 2.8 × 10 J ⎝ L mol ⎠ ⎣ ⎦

12.71

(a)

With an overall efﬁciency of e = 0.15 and a power output of Pout = 150 MW, the required power input (from burning coal) is Pin =

Pout 150 × 10 6 W = = 1.0 × 10 9 J s e 0.15

The coal used each day is

(1.0 × 109 J s) (8.64 × 104 s d ) = 2.6 × 106 kg Δm Pin = = 3 Δt heat of combustion ⎛ d 3 cal ⎞ ⎛ 10 g ⎞ ⎛ 4.186 J ⎞ ⎜⎝ 7.8 × 10 g ⎟⎠ ⎜⎝ 1 kg ⎟⎠ ⎜⎝ 1 cal ⎟⎠ or (b)

kg ⎞ ⎛ 1 metric ton ⎞ Δm ⎛ = 2.6 × 10 3 metric ton d = ⎜ 2.6 × 10 6 ⎟ d ⎠ ⎜⎝ 10 3 kg ⎟⎠ Δt ⎝

The annual fuel cost is cost = ( coal used yearly ) ⋅ ( rate ) , or cost = ( 2.6 × 10 3 ton d ) ( 365.242 d y ) ( $8.0 ton ) = $7.6 × 10 6 y

(c)

The rate of energy transfer to the river by heat is Pexhaust = Pin − Pout = 1.0 × 10 9 W − 150 × 10 6 W = 8.5 × 108 W Thus, the ﬂow required if the maximum rise in temperature is 5.0°C is flow rate =

56157_12_ch12_p371-399.indd 398

Δmwater Pexhaust 8.5 × 108 J s = = = 4.1× 10 4 kg s Δt cwater ( ΔT ) ( 4 186 J kg ⋅°C )( 5.0°C )

10/12/10 2:48:26 PM

The Laws of Thermodynamics

12.72

(a)

The work done on the gas is the negative of the area under the process curve in a PV diagram. From the sketch at the right, observe that this area consists of a triangular area sitting atop a rectangular area, or W = − ⎡⎣ 12 ( base )( altitude ) + ( width ) ( height ) ⎤⎦

399

P (105 Pa) 4.00 3.00 A

B

2.00 1.00 1.00

2.00

3.00

V (m3)

Thus, 1 W = − ⎡⎢ ( 3.00 − 1.00 ) m 3 ( 4.00 − 2.00 ) × 10 5 Pa + ( 3.00 − 1.00 ) m 3 ( 4.00 − 2.00 ) × 10 5 Pa ⎤⎥ ⎣2 ⎦ = − 6.00 × 10 5 J 5 3 PB VB PA VA PA ( ΔV ) ( 2.00 × 10 Pa )( 3.00 − 1.00 ) m − = 5.50 × 10 2 K = = nR nR nR (87.5 mol )(8.31 J mol ⋅K )

(b)

ΔT = TB − TA =

(c)

The change in internal energy of an ideal gas is ΔU = nCv ΔT . From Table 12.1, which gives the molar speciﬁc heats of diatomic gases, we take the average of the ﬁrst four entries to obtain Cv ≈ 20.8 J mol ⋅K. This then yields ΔU = U B − U A = nCv ΔT = (87.5 mol )( 20.8 J mol ⋅K )( 550 K ) = 1.00 × 10 6 J

(d)

From the ﬁrst law of thermodynamics, ΔU = Q + W , we ﬁnd Q = ΔU − W = 1.00 × 10 6 J − ( −6.00 × 10 5 J ) = 1.60 × 10 6 J

56157_12_ch12_p371-399.indd 399

10/12/10 2:48:28 PM

13 Vibrations and Waves QUICK QUIZZES 1.

Choice (d). To complete a full cycle of oscillation, the object must travel distance 2A to position x = − A and then travel an additional distance 2A returning to the original position at x = + A.

2.

Choice (c). The force producing harmonic oscillation is always directed toward the equilibrium position, and hence, directed opposite to the displacement from equilibrium. The acceleration is in the direction of the force. Thus, it is also always directed opposite to the displacement from equilibrium.

3.

Choice (b). In simple harmonic motion, the force (and hence, the acceleration) is directly proportional to the displacement from equilibrium. Therefore, force and acceleration are both at a maximum when the displacement is a maximum.

4.

Choice (a). The period of an object-spring system is T = 2p m k . Thus, increasing the mass by a factor of 4 will double the period of oscillation.

5.

Choice (c). The total energy of the oscillating system is equal to 12 kA2 , where A is the amplitude of oscillation. Since the object starts from rest at displacement A in both cases, it has the same amplitude of oscillation in both cases.

6.

Choice (d). The expressions for the total energy, maximum speed, and maximum acceleration are E = 12 kA2 , vmax = A k m , and amax = A ( k m ), where A is the amplitude. Thus, all are changed by a change in amplitude. The period of oscillation is T = 2p m k and is unchanged by altering the amplitude.

7.

Choices (c) and (b). An accelerating elevator is equivalent to a gravitational field. Thus, if the elevator is accelerating upward, this is equivalent to an increased effective gravitational field magnitude g, and the period will decrease. Similarly, if the elevator is accelerating downward, the effective value of g is reduced and the period increases. If the elevator moves with constant velocity, the period of the pendulum is the same as that in the stationary elevator.

8.

Choice (a). The clock will run slow. With a longer length, the period of the pendulum will increase. Thus, it will take longer to execute each swing, so that each second according to the clock will take longer than an actual second.

9.

Choice (b). Greater. The value of g on the Moon is about one-sixth the value of g on Earth, so the period of the pendulum on the Moon will be greater than the period on Earth.

400

56157_13_ch13_p400-425.indd 400

10/12/10 2:49:14 PM

Vibrations and Waves

401

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

The wavelength of a wave is the distance from crest to the following crest. Thus, the distance between a crest and the following trough is a half wavelength, giving l = 2 ( 2 m ) = 4 m . The speed of the wave is then v = l f = ( 4 m ) ( 2 Hz ) = 8 m s, and (c) is the correct choice.

2.

When an object undergoes simple harmonic motion, the position as a function of time may be written as x = A cosw t = A cos ( 2p ft ). Comparing this to the given relation, we see that the frequency of vibration is f = 3 Hz, and the period is T = 1 f = 1 3 s, so the correct answer is (c).

3.

In this spring-mass system, the total energy equals the elastic potential energy at the moment the mass is temporarily at rest at x = A = 6 cm (i.e., at the extreme ends of the simple harmonic motion). Thus, E = 12 kA2 and we see that as long as the spring constant k and the amplitude A remain unchanged, the total energy is unchanged. Hence, the energy is still 12 J and (a) is the correct choice.

4.

The energy given the vibratory system equals the elastic potential energy at the extremes of the motion, x = ± A. Thus, E = 12 kA2 and this energy will all be in the form of kinetic energy as the 2 body passes through the equilibrium position, giving 12 mvmax = 12 kA2 or 80.0 N m k = 1.4 m s = ( 0.10 m ) 0.40 kg m and (b) is the correct choice. vmax = A

5.

The frequency of vibration is f=

w 1 = 2p 2p

k m

Thus, increasing the mass by a factor of 9 will decrease the frequency to and the correct answer is (b). 6.

1 3

of its original value,

When the object is at its maximum displacement, the net force (directed back toward the equilibrium position) acting on it has magnitude Fnet = k x max = (8.0 N m ) ( 0.10 m ) = 0.80 N This force will give the mass an acceleration of a = Fnet m = 0.80 N 0.40 kg = 2.0 m s2 , making (d) the correct choice.

7.

The car will continue to compress the spring until all of the car’s original kinetic energy has been converted into elastic potential energy within the spring, i.e., until 12 kx 2 = 12 mvi2 or m 3.0 × 10 5 kg = 0.77 m = ( 2.0 m s ) k 2.0 × 10 6 N m The correct choice is seen to be (a). x = vi

8.

The period of a simple pendulum is T = 2p g , and its frequency is f = 1 T = (1 2p ) g Thus, if the length is doubled so ′ = 2 , the new frequency is

.

g 1 g 1 ⎛ 1 g⎞ f 1 = = = ⎜ ⎟ 2p 2 ⎝ 2p 2 ′ 2p 2 ⎠ and we see that (d) is the correct response. f′ =

56157_13_ch13_p400-425.indd 401

10/12/10 2:49:15 PM

402

9.

Chapter 13

The period of a simple pendulum is T = 2p period will be

g . If the length is changed to ′ = 4 , the new

⎛ ⎞ 4 ′ = 2p = 2 ⎜ 2p = 2T g g g ⎟⎠ ⎝ or the period will be doubled. The correct choice is (e). T ′ = 2p

10.

For a particle executing simple harmonic motion about an equilibrium point x 0 , its position as a function of time is given by x − x 0 = A cos (wt ), and the turning points (i.e., the extremes of the position) are at x = x 0 ± A. That is, the equilibrium position is midway between the turning points, so the correct response is choice (c).

11.

The only false statement among the listed choices is choice (d). At the equilibrium position, x = 0, the elastic potential energy PEs = kx 2 2 is a minimum, and the kinetic energy is a maximum.

12.

In a vertical mass-spring system, the equilibrium position is the point at which the mass will hang at rest on the lower end of the spring. If the mass is raised distance A above this position and released from rest, it will undergo simple harmonic motion, with amplitude A, about the equilibrium position. The upper turning point of the motion is at the point of release, and the lower turning point is distance A below the equilibrium position or distance 2A below the release point. Thus, if the release point is 15 cm above the equilibrium position, the mass drops 30 cm before stopping momentarily and reversing direction. The correct answer is choice (c).

13.

The total energy of the system is constant and equal to the elastic potential energy at x = ± A (where the velocity is zero). That is, KE + PEs = 12 kA2 . Thus, at a location where KE = 2PEs, we have 3PEs = 12 kA2 or

(

⎛1 ⎞ 1 3 ⎜ kx 2 ⎟ = kA2 ⎝2 ⎠ 2

)

and

x2 =

A2 3

or

x=±

A 3

and the correct choice is (c). 14.

When it supports your weight, the center of the diving board flexes down less than the end does when it supports your weight—this is similar to a spring that stretches a smaller distance for the same force: its spring constant is greater because the displacement is smaller. Therefore, the stiffness constant describing the center of the board is greater than the stiffness constant describing the end. Thus, the frequency, given by f =

1 2p

k m

is greater when you bounce at the center of the board. Choice (a) is correct.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

Friction. This includes both air resistance and damping within the spring.

4.

Each half-spring will have twice the spring constant of the full spring, as shown by the following argument. The force exerted by a spring is proportional to the separation of the coils as the spring is extended. Imagine that we extend a spring by a given distance and measure the distance between coils. We then cut the spring in half. If one of the half-springs is now extended by the same distance, the coils will be twice as far apart as they were for the complete spring. Thus, it takes twice as much force to stretch the half-spring, from which we conclude that the half-spring has a spring constant which is twice that of the complete spring.

56157_13_ch13_p400-425.indd 402

10/12/10 2:49:18 PM

Vibrations and Waves

403

6.

No. The period of vibration is T = 2p g and g is smaller at high altitude. Therefore, the period is longer on the mountain top and the clock will run slower.

8.

Shorten the pendulum to decrease the period between ticks.

10.

The speed of the pulse is v = F m , so increasing the tension F in the hose increases the speed of the pulse. Filling the hose with water increases the mass per unit length m, and will decrease the speed of the pulse.

12.

As the temperature increases, the length of the pendulum will increase due to thermal expansion, and with a greater length, the period of the pendulum increases. Thus, it takes longer to execute each swing, so that each second according to the clock will take longer than an actual second. Consequently, the clock will run slow.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

1.54 cm

4.

(a)

1.1 × 10 2 N

(b)

The graph should be a straight line passing through the origin with a positive slope of 1.0 × 10 3 N m.

6.

(a)

327 N

(b)

1.25 × 10 3 N

8.

(a)

0.625 J

(b)

0.791 m s

10.

(a)

575 N m

(b)

46.0 J

12.

2.23 m s

14.

(a)

E = 12 kA2

(b)

16.

(a)

4.58 N

(b)

0.125 J

(d)

1.00 m s

(e)

the speed would be lower

(f)

The numeric value of the coefficient of kinetic friction would be required.

mv 2 = kx 2

(c)

x=±A

(c)

18.3 m s2

3

(b)

0.78 m s

(c)

18 m s2

0.628 m s

(b)

0.500 Hz

(c)

3.14 rad s

(a)

1.89 Hz

(b)

33.7 N m

(c)

0.118 m

(a)

3.9 × 10 5 N m

(b)

2.2 Hz

0.15 J

18.

(a)

20.

3.06 m s

22.

(a)

24. 26.

56157_13_ch13_p400-425.indd 403

1 2

10/12/10 2:49:21 PM

404

Chapter 13

(a)

0.25 s

(b)

4.0 Hz

(d)

21 ms

30.

(a)

±A 3 2

(b)

±A

32.

(a)

250 N m

(b)

22.4 rad s , 3.56 Hz , 0.281 s

(c)

0.313 J

(d)

5.00 cm

(f)

0.919 cm

(g)

+1.10 m s , − 4.59 m s2

34.

(a)

59.6 m

(b)

37.5 s

36.

1.001 5

38.

1.66 × 10 −2 kg ⋅ m 2

40.

(a)

3.65 s

(b)

42.

(a)

2.00 cm

(d) 44.

(a)

46.

5.72 mm

48.

(a)

50.

219 N

52.

(a)

The units of the first T are seconds, the units of the second are newtons.

(b)

The first T is period of time; the second is force of tension.

28.

5.2 cm

2

(e)

1.12 m s , 25.0 m s2

6.41 s

(c)

4.24 s

(b)

4.00 s

(c)

p 2 rad s

p cm s

(e)

4.93 cm s2

(f)

x = ( 2.00 cm ) sin (p t 2 )

0.357 Hz

(b)

0.985 m s

(b)

0.25 Hz

0.20 Hz

54.

1.64 m s2

56.

7.07 m s

58.

586 m s

60.

(a)

v = 2nL t

(b)

F = 4n 2 ML t 2

62.

(a)

0.25 m

(b)

0.47 N m

(d)

0.12 m s

64.

(a)

5.10 ms

(b)

1.75 m

66.

0.990 m

56157_13_ch13_p400-425.indd 404

(c)

(c)

0.23 m

10/12/10 2:49:23 PM

Vibrations and Waves

68.

(a)

100 m s

(b)

374 J

70.

(a)

19.8 m s

(b)

8.95 m

72.

(a) and (b) See Solution for proofs.

74.

1.3 cm s

76.

(a)

ΣFy = −ky f − mg = mv 2 (L − y f )

(b)

mv 2 = 2mg L − y f − ky 2f

(c)

y f = −0.110 m

(

)

(d)

405

greater than

PROBLEM SOLUTIONS 13.1

(a)

Taking to the right as positive, the spring force acting on the block at the instant of release is Fs = −kA = − (130 N m ) ( +0.13 m ) = −17 N or 17 N to the left

(b)

At this instant, the acceleration is a=

13.2

Fs −17 N = = −28 m s2 m 0.60 kg

13.4

a = 28 m s2 to the left

The force compressing the spring is the weight of the object. Thus, the spring will be compressed a distance of

(

)

2 F mg ( 2.30 kg ) 9.80 m s = = = 1.54 × 10 −2 m = 1.54 cm k k 1.46 × 10 3 N m Assuming the spring obeys Hooke’s law, the magnitude of the force required to displace the end a distance x from the equilibrium position (by either compressing or stretching the spring) is F = k x , where k is the force constant of the spring.

x =

13.3

or

( F = k x = (137 N m ) ( 7.36 × 10

) m ) = 10.1 N

(a)

If x = − 4.80 cm, the required force is F = k x = (137 N m ) 4.80 × 10 −2 m = 6.58 N

(b)

If x = + 7.36 cm, the required force is

(a)

The spring constant is k =

(

Fs x

=

−2

mg 50 N = = 1.0 × 10 3 N m x 5.0 × 10 −2 m

)

F = Fs = kx = 1.0 × 10 3 N m ( 0.11 m ) = 1.1 × 10 2 N (b)

13.5

The graph will be a straight line passing through the origin with a slope equal to k = 1.0 × 10 3 N m.

When the system is in equilibrium, the tension in the spring F = k x must equal the weight of the object. Thus,

(

)

−2 k x ( 47.5 N m ) 5.00 × 10 m k x = mg giving m = = = 0.242 kg g 9.80 m s2

56157_13_ch13_p400-425.indd 405

10/12/10 2:49:26 PM

406

13.6

Chapter 13

(a)

The free-body diagram of the point in the center of the string is given at the right. From this, we see that

→

35.0°

T

→

ΣFx = 0 ⇒ F − 2T sin 35.0° = 0 or

(b)

T=

(a)

F 375 N = = 327 N 2 sin 35.0° 2 sin 35.0°

→

T

35.0°

Since the bow requires an applied horizontal force of 375 N to hold the string at 35.0° from the vertical, the tension in the spring must be 375 N when the spring is stretched 30.0 cm. Thus, the spring constant is k=

13.7

F

F 375 N = = 1.25 × 10 3 N m x 0.300 m

When the block comes to equilibrium, ΣFy = −ky0 − mg = 0, giving y0 = −

(

)

(10.0 kg) 9.80 m s2 mg =− = − 0.206 m k 475 N m

or the equilibrium position is 0.206 m below the unstretched position of the lower end of the spring. (b)

When the elevator (and everything in it) has an upward acceleration of a = 2.00 m s2, applying Newton’s second law to the block gives ΣFy = −k ( y0 + y ) − mg = ma y

or

ΣFy = ( −ky0 − mg ) − ky = ma y

where y = 0 at the equilibrium position of the block. Since −ky0 − mg = 0 [see part (a)], this becomes −ky = ma and the new position of the block is y= or

13.8

ma y

(10.0 kg) ( +2.00

− 475 N m

m s2

) = − 4.21 × 10

−2

m = − 4.21 cm

4.21 cm below the equilibrium position .

(c)

When the cable breaks, the elevator and its contents will be in free-fall with a y = −g. The new “equilibrium” position of the block is found from ΣFy = −ky0′ − mg = m ( −g ), which yields y0′ = 0. When the cable snapped, the block was at rest relative to the elevator at distance y0 + y = 0.206 m + 0.042 1 m = 0.248 m below the new “equilibrium” position. Thus, while the elevator is in free-fall, the block will oscillate with amplitude = 0.248 m about the new “equilibrium” position, which is the unstretched position of the spring’s lower end.

(a)

The work required to stretch the spring equals the elastic potential energy of the spring in the stretched condition, or W=

(b)

1 2 1 kx = 5.00 × 10 2 N m 5.00 × 10 −2 m 2 2

(

)(

)

2

= 0.625 J

In the initial condition, the spring-block system is at rest (KEi = 0) with elastic potential energy of PEs,i = 0.625 J. Since the spring force is conservative, conservation of energy gives KE f + PEs, f = KEi + PEs,i = 0.625 J. Thus, when the block is at the equilibrium position (PEs, f = 0), we have KE f = 12 mv 2f = 0.625 J, or vf =

56157_13_ch13_p400-425.indd 406

−k

=

2 ( 0.625 J ) = m

2 ( 0.625 J ) = 0.791 m s 2.00 kg

10/12/10 2:49:29 PM

Vibrations and Waves

13.9

(a)

407

Assume the rubber bands obey Hooke’s law. Then, the force constant of each band is k=

Fs 15 N = 1.5 × 10 3 N m = x 1.0 × 10 −2 m

Thus, when both bands are stretched 0.20 m, the total elastic potential energy is 2 ⎛1 ⎞ PEs = 2 ⎜ kx 2 ⎟ = 1.5 × 10 3 N m ( 0.20 m ) = 60 J ⎝2 ⎠

(

(b)

)

Conservation of mechanical energy gives ( KE + PEs ) f = ( KE + PEs )i , or 1 2 mv + 0 = 0 + 60 J 2

13.10

13.11

v=

so

2 ( 60 J ) = 49 m s 50 × 10 −3 kg

Fm ax 230 N = = 575 N m x m ax 0.400 m

(a)

k=

(b)

work done = PEs =

1 2 1 2 kx = ( 575 N m )( 0.400 ) = 46.0 J 2 2

From conservation of mechanical energy,

( KE + PE

g

+ PEs

) = ( KE + PE f

g

+ PEs

)

i

or 0 + mgh f + 0 = 0 + 0 +

1 2 kxi , 2

giving k= 13.12

2 mgh f x

2 i

=

(

) m)

2 ( 0.100 kg ) 9.80 m s2 ( 0.600 m )

( 2.00 × 10

−2

2

(

Conservation of mechanical energy, KE + PEg + PEs

= 2.94 × 10 3 N m

) = ( KE + PE i

g

)

+ PEs , gives f

1 2 1 mvi + 0 + 0 = 0 + 0 + kx 2f , 2 2 k 5.00 × 10 6 N m xi = 3.16 × 10 −2 m = 2.23 m s m 1000 kg An unknown quantity of mechanical energy is converted into internal energy during the collision. Thus, we apply conservation of momentum from just before to just after the collision and obtain mvi + M ( 0 ) = ( M + m ) V, or the speed of the block and embedded bullet just after collision is V = (m M + m)vi . We now use conservation of mechanical energy, (KE + PEs ) f = (KE + PEs )i , from just after collision until the block comes to rest. This gives 0 + 12 kx 2f = 12 ( M + m ) V 2 + 0, or or

13.13

xf = V yielding

13.14

(a)

(

vi =

M+m ⎛ m ⎞ = vi ⎜ ⎝ M + m ⎟⎠ k xf =

(10.0 × 10

−3

)

M+m = k

kg ( 300 m s )

( 2.01 kg) (19.6

)

N m)

mvi

( M + m) k

= 0.478 m

At either of the turning points, x = ± A, the constant total energy of the system is momentarily stored as elastic potential energy in the spring. Thus, E = 12 kA2 .

(b)

56157_13_ch13_p400-425.indd 407

When the object is distance x from the equilibrium position, the elastic potential energy is PEs = kx 2 2 and the kinetic energy is KE = mv 2 2. At the position where KE = 2PEs , it is necessary that 1 2 1 2 ⎛1 ⎞ mv = 2 ⎜ kx 2 ⎟ or mv = kx 2 ⎝2 ⎠ 2 2 continued on next page

10/12/10 2:49:33 PM

408

Chapter 13

(c)

When KE = 2PEs , conservation of energy gives E = KE + PEs = 2 ( PEs ) + PEs = 3PEs, or 1 2 ⎛1 ⎞ kA = 3 ⎜ kx 2 ⎟ ⎝2 ⎠ 2

13.15

(a)

A 3

1 2 mvmax = E = 47.0 J 2

2E 2 ( 47.0 J ) = = 7.90 kg 2 vmax ( 3.45 m s )2

At any position, the constant total energy is E = KE + PEs = 12 mv 2 + 12 kx 2 , so at x = 0.160 m

(

)

2 ( 47.0 J ) − 1.63 × 10 3 N m ( 0.160 m )

2E − kx 2 = m

7.90 kg

2

= 2.57 m s

At x = 0.160 m, where v = 2.57 m s, the kinetic energy is KE =

(f)

x=±

If, at the equilibrium position, v = vmax = 3.45 m s, the mass of the block is

v= (e)

or

At the equilibrium position ( x = 0 ) , the spring is momentarily in its relaxed state and PEs = 0, so all of the energy is in the form of kinetic energy. This gives

m= (d)

k A2 2 3k 2

2E 2 ( 47.0 J ) = = 1.63 × 10 3 N m 2 x max ( 0.240 m )2

KE x = 0 = (c)

x=±

At maximum displacement from equilibrium, all of the energy is in the form of elastic 2 potential energy, giving E = 12 kx max , and k=

(b)

⇒

1 2 1 2 mv = ( 7.90 kg ) ( 2.57 m s ) = 26.1 J 2 2

At x = 0.160 m, where KE = 26.1 J, the elastic potential energy is PEs = E − KE = 47.0 J − 26.1 J = 20.9 J or alternately:

(g)

PEs =

1 2 1 2 kx = 1.63 × 10 3 N m ( 0.160 m ) = 20.9 J 2 2

(

)

At the first turning point (for which x < 0 since the block started from rest at x = +0.240 m and has passed through the equilibrium at x = 0), all of the remaining energy is in the form of elastic potential energy, so 1 2 kx = E − Eloss = 47.0 J − 14.0 J = 33.0 J 2 x=−

and 13.16

2 ( 33.0 J ) 2 ( 33.0 J ) =− = − 0.201 m k 1.63 × 10 3 N m

(

)

(a)

F = k x = (83.8 N m ) 5.46 × 10 −2 m = 4.58 N

(b)

E = PEs =

(c)

While the block was held stationary at x = 5.46 cm, ΣFx = −Fs + F = 0, or the spring force was equal in magnitude and oppositely directed to the applied force. When the applied force is suddenly removed, there is a net force Fs = 4.58 N directed toward the equilibrium position acting on the block. This gives the block an acceleration having magnitude a =

1 2 1 kx = (83.8 N m ) 5.46 × 10 −2 m 2 2

(

)

2

= 0.125 J

Fs 4.58 N = = 18.3 m s2 m 0.250 kg

continued on next page

56157_13_ch13_p400-425.indd 408

10/12/10 2:49:37 PM

Vibrations and Waves

(d)

409

At the equilibrium position, PEs = 0, so the block has kinetic energy KE = E = 0.125 J and speed 2E 2 ( 0.125 J ) = = 1.00 m s m 0.250 kg If the surface was rough, the block would spend energy overcoming a retarding friction force as it moved toward the equilibrium position, causing it to arrive at that position with a lower speed than that computed above. v=

(e)

(f)

13.17

Computing a numeric value for this lower speed requires knowledge of the coefficient of kinetic friction between the block and surface.

(

From conservation of mechanical energy, KE + PEg + PEs 2 2 2 1 1 1 2 mv + 0 + 2 kx = 0 + 0 + 2 kA , or

(

k 2 A − x2 m

v= (a)

(19.6 N m ) ( 0.040 m )2 ( 0.40 kg)

k 2 A = m

x=A

)

1 4

k 2 A m

A2, or

3 3 = 3.5 cm = ( 4.0 cm ) 2 2

1 2 1 2 kA = ( 250 N m ) ( 0.035 m ) = 0.15 J 2 2

2 The maximum speed occurs at the equilibrium position where PEs = 0. Thus, E = 12 mvmax , or

vmax =

2E k 250 N m =A = ( 0.035 m ) = 0.78 m s m m 0.50 kg

The acceleration is a = ΣF m = −kx m. Thus, a = amax at x = −x max = − A. amax =

56157_13_ch13_p400-425.indd 409

0.26 m s

KE = 0 at x = A, so E = KE + PEs = 0 + 12 kA2 , or the total energy is E=

(c)

0.26 m s

If v = 12 vmax, then

This gives A2 − x 2 =

(b)

= 0.28 m s

(19.6 N m ) ⎡( 0.040 m )2 − + 0.015 m 2 ⎤ = ( )⎦ ( 0.40 kg) ⎣

(

(a)

i

When x = + 0.015 m,

k 2 1 A − x2 = m 2

13.18

)

+ PEs , we have

)

(19.6 N m ) ⎡( 0.040 m )2 − − 0.015 m 2 ⎤ = ( )⎦ ( 0.40 kg) ⎣

v= (d)

g

When x = − 0.015 m, v=

(c)

f

The speed is a maximum at the equilibrium position, x = 0. vmax =

(b)

) = ( KE + PE

⎛ 250 N m ⎞ −k ( − A ) ⎛ k ⎞ =⎜ ⎟ A=⎜ ( 0.035 m ) = 18 m s2 ⎝ m⎠ m ⎝ 0.50 kg ⎟⎠

10/12/10 2:49:42 PM

410

13.19

Chapter 13

The maximum speed occurs at the equilibrium position and is vmax = A k m. Thus, kA2 (16.0 N m ) ( 0.200 m ) = = 4.00 kg 2 vmax ( 0.400 m s )2 2

m= and

(

)

Fg = mg = ( 4.00 kg ) 9.80 m s2 = 39.2 N 13.20

v=

⎛ 10.0 N m ⎞ k 2 ⎡( 0.250 m )2 − ( 0.125 m )2 ⎤ = 3.06 m s A − x2 = ⎜ −3 ⎟ ⎦ m ⎝ 50.0 × 10 kg ⎠ ⎣

13.21

(a)

PEs,i =

(b)

Since the surface is frictionless, the total energy of the block-spring system is constant. Thus, KE + PEs = KEi + PEs,i = 0 + 1.53 J. At the equilibrium position, PEs = 0, so the 2 kinetic energy must be KEmax = 12 mvmax = 1.53 J, which yields

(

)

1 2 1 kxi = (850 N m ) 6.00 × 10 −2 m 2 2

(

vmax = (c)

2KEmax = m

)

2

= 1.53 J

2 (1.53 J ) = 1.75 m s 1.00 kg

At x = xi 2 = 3.00 cm, the elastic potential energy is PEs = 12 kx 2, and the conservation of energy gives KE + PEs = E, or 12 mv 2 + 12 kx 2 = E and

(

13.22

(a) (b) (c)

13.23

13.24

2 (1.53 J ) − (850 N m ) 3.00 × 10 −2 m 2E − kx 2 v= = m 1.00 kg 2p r 2p ( 0.200 m ) vt = = = 0.628 m s T 2.00 s 1 1 f = = = 0.500 Hz T 2.00 s 2p 2p w= = = 3.14 rad s T 2.00 s

The angle of the crank pin is q = w t. Its x-coordinate is x = A cosq = A cosw t, where A is the distance from the center of the wheel to the crank pin. The displacement of the piston from its zero position (i.e., its location when q = wt = p 2) is the same as that of the crankpin, x(t) = A coswt. This is of the correct form to describe simple harmonic motion. Hence, one must conclude that the motion is indeed simple harmonic. (a) (b)

= 1.51 m s

A

x⫽0 x ⫽0

x(t)

x(t)

The period of oscillation of an object-spring system is T = 2p m k , so the force constant is 2 4p 2 m 4p ( 0.238 kg ) = = 33.7 N m 2 T ( 0.528 s )2

At the turning points (x = ± A) in the oscillation, all of the energy is temporarily stored as elastic potential energy, or E = kA2 2. Thus, A=

56157_13_ch13_p400-425.indd 410

2

␻t

1 1 f = = = 1.89 Hz T 0.528 s

k= (c)

)

2E = k

2 ( 0.234 J ) = 0.118 m 33.7 N m

10/12/10 2:49:46 PM

Vibrations and Waves

13.25

411

The spring constant is found from

(

)

2 Fs mg ( 0.010 kg ) 9.80 m s = 2.5 N m k= = = x 3.9 × 10 −2 m x

When the object attached to the spring has mass m = 25 g, the period of oscillation is m = 2p k

T = 2p 13.26

(a)

The springs compress 0.80 cm when supporting an additional load of m = 320 kg. Thus, the spring constant is k=

(b)

(a)

(

)

2 ΔFs mg ( 320 kg ) 9.80 m s = 3.9 × 10 5 N m = = Δx 0.80 × 10 −2 m Δx

When the empty car, M = 2.0 × 10 3 kg, oscillates on the springs, the frequency will be f =

13.27

0.025 kg = 0.63 s 2.5 N m

1 1 = T 2p

3.9 × 10 5 N m = 2.2 Hz 2.0 × 10 3 kg

k 1 = M 2p

The period of oscillation is T = 2p m k , where k is the spring constant and m is the mass of the object attached to the end of the spring. Hence, 0.250 kg = 1.0 s 9.5 N m If the cart is released from rest when it is 4.5 cm from the equilibrium position, the amplitude of oscillation will be A = 4.5 cm = 4.5 × 10 −2 m. The maximum speed is then given by T = 2p

(b)

vmax = Aw = A (c)

)

9.5 N m = 0.28 m s 0.250 kg

When the cart has a displacement of x = 2.0 cm from the equilibrium position, its speed will be v=

13.28

(

k = 4.5 × 10 −2 m m

(

)

k 2 A − x2 = m

9.5 N m ⎡( 0.045 m )2 − ( 0.020 m )2 ⎤ = 0.25 m s ⎦ 0.250 kg ⎣

The general expression for the position as a function of time for an object undergoing simple harmonic motion with x = 0 at t = 0 is x = A sin (wt ). Thus, if x = ( 5.2 cm ) sin (8.0p ⋅ t ) , we have that the amplitude is A = 5.2 cm and the angular frequency is w = 8.0p rad s. (a)

The period is T=

(b)

2p 2p = = 0.25 s w 8.0p s−1

The frequency of motion is f=

1 1 = = 4.0 s−1 = 4.0 Hz T 0.25 s

(c)

As discussed above, the amplitude of the motion is A = 5.2 cm .

(d)

Note: For this part, your calculator should be set to operate in radians mode. If x = 2.6 cm, then ⎛ x⎞ ⎛ 2.6 cm ⎞ wt = sin −1 ⎜ ⎟ = sin −1 ⎜ = sin −1 ( 0.50 ) = 0.52 radians ⎝ A⎠ ⎝ 5.2 cm ⎟⎠ continued on next page

56157_13_ch13_p400-425.indd 411

10/12/10 2:49:50 PM

412

Chapter 13

and t= 13.29

(a)

0.52 rad 0.52 rad = 2.1 × 10 −2 s = 21 × 10 −3 s = 21 ms = w 8.0p rad s

At the equilibrium position, the total energy of the system is in the form of kinetic energy 2 and 12 mvmax = E , so the maximum speed is vmax =

(b)

2 4p 2 m 4p ( 0.326 kg ) = = 206 N m T2 ( 0.250 s )2

At the turning points, x = ± A, the total energy of the system is in the form of elastic potential energy, or E = 12 kA2 , giving the amplitude as A=

13.30

2 ( 5.83 J ) = 5.98 m s 0.326 kg

The period of an object-spring system is T = 2p m k , so the force constant of the spring is k=

(c)

2E = m

2E = k

2 ( 5.83 J ) = 0.238 m 206 N m

For a system executing simple harmonic motion, the total energy may be written as 2 E = KE + PEs = 12 mvmax = 12 kA2 , where A is the amplitude and vmax is the speed at the 2 equilibrium position. Observe from this expression, that we may write vmax = kA2 m. (a)

2 If v = 12 vmax , then E = 12 mv 2 + 12 kx 2 = 12 mvmax becomes 2 ⎞ 1 2 1 2 1 ⎛ vmax m⎜ + kx = mvmax 2 ⎝ 4 ⎟⎠ 2 2

and gives x2 = (b)

3 ⎛ m⎞ 2 3 ⎛ m ⎞ ⎡ k 2 ⎤ 3A2 ⎜⎝ ⎟⎠ vmax = ⎜⎝ ⎟⎠ ⎢ A ⎥ = 4 k 4 k ⎣m ⎦ 4

x=±

A 3 2

and

x=±

A 2

If the elastic potential energy is PEs = 12 E, we have 1 2 1 ⎛ 1 2⎞ kx = ⎜ kA ⎟ ⎠ 2 2⎝2

13.31

or

or

x2 =

A2 2

Note: Your calculator must be in radians mode for part (a) of this problem. (a)

The angular frequency of this oscillation is w = k m and the displacement at time t is x = A coswt . At t = 3.50 s, the spring force will be F = −kx = −kA cos(wt), or ⎡⎛ 5.00 N m ⎞ ⎤ N⎞ ⎛ F = − ⎜ 5.00 3.50 s )⎥ = −11.0 N, 3.00 m ) cos ⎢⎜ ( ( ⎟ ⎟ ⎝ m⎠ ⎢⎣⎝ 2.00 kg ⎠ ⎥⎦ or

(b)

F = 11.0 N directed to the left

The time required for one complete oscillation is T = 2p w = 2p m k . Hence, the number of oscillations made in 3.50 s is N=

56157_13_ch13_p400-425.indd 412

Δt Δt 3.50 s = = T 2p 2p m k

5.00 N m = 0.881 2.00 kg

10/12/10 2:49:54 PM

Vibrations and Waves

13.32

F 7.50 N = 250 N m = x 3.00 × 10 −2 m

(a)

k=

(b)

w=

250 N m = 22.4 rad s , 0.500 kg

k = m f =

w 1 = 2p 2p T=

and (c)

250 N m = 3.56 Hz , 0.500 kg

At t = 0, v = 0 and x = 5.00 × 10 −2 m, so the total energy of the oscillator is 1 2 1 2 1 mv + kx = 0 + ( 250 N m ) 5.00 × 10 −2 m 2 2 2

(

When x = A, v = 0 so E = KE + PEs = 0 + Thus,

(e)

k 1 = m 2p

1 1 = = 0.281 s f 3.56 Hz

E = KE + PEs = (d)

413

A=

2E = k

At x = 0, KE = amax =

)

2

= 0.313 J

1 2 kA . 2

2 ( 0.313 J ) = 5.00 × 10 −2 m = 5.00 cm 250 N/m

1 2 mvmax = E, 2

2 ( 0.313 J ) = 1.12 m s 0.500 kg

2E = m

vmax =

or

(

)

−2 Fmax k A ( 250 N m ) 5.00 × 10 m = = = 25.0 m s2 m m 0.500 kg

Note: To solve parts (f) and (g), your calculator should be set in radians mode. (f)

At t = 0.500 s, Equation 13.14a gives the displacement as ⎡ 250 N m ⎤ x = A cos (w t ) = A cos t k m = ( 5.00 cm ) cos ⎢( 0.500 s ) ⎥ = 0.919 cm 0.500 kg ⎦ ⎣

(

(g)

)

From Equation 13.14b, the velocity at t = 0.500 s is

(

v = − Aw sin (w t ) = − A k m sin t k m = − ( 5.00 × 10 −2 m )

)

⎡ 250 N m 250 N m ⎤ sin ⎢( 0.500 s ) ⎥ = + 1.10 m s 0.500 kg 0.500 kg ⎦ ⎣

and from Equation 13.14c, the acceleration at this time is

(

a = − Aw 2 cos (w t ) = − A ( k m ) cos t k m

)

⎡ ⎛ 250 N m ⎞ 250 N m ⎤ 2 = − 5.00 × 10 −2 m ⎜ cos ⎢( 0.500 s ) ⎥ = − 4.59 m s ⎟ 0.500 kg ⎝ 0.500 kg ⎠ ⎣ ⎦

(

13.33

)

From Equation 13.6, v = ± Hence,

v = ±w

)

(

)

A2 − A2 cos2 (w t ) = ±w A 1 − cos2 (w t ) = ± w A sin (wt )

From Equation 13.2, a = −

56157_13_ch13_p400-425.indd 413

(

k 2 A − x 2 = ± w 2 A2 − x 2 m

k x = −w 2 ⎡⎣ A cos (w t ) ⎤⎦ = −w 2 A cos (w t ) m

10/12/10 2:49:58 PM

414

13.34

Chapter 13

(a)

The height of the tower is almost the same as the length of the pendulum. From T = 2p L g , we obtain

(

)

9.80 m s2 (15.5 s ) gT2 L= = = 59.6 m 4p 2 4p 2 2

(b)

On the Moon, where g = 1.67 m s2, the period will be

(a)

L 59.6 m = 2p = 37.5 s g 1.67 m s2 The period is the time for one complete oscillation. Hence, T = 2p

13.35

T= (b)

or

T = 1.46 s

The period of oscillation of a simple pendulum is T = 2p gravity must be g=

13.36

2.00 min ⎛ 60 s ⎞ 120 s ⎜⎝ ⎟= 1 min ⎠ 82.0 82

g , so the local acceleration of

4p 2 4p 2 ( 0.520 m ) 2 = 2 = 9.59 m s 2 T (120 s 82.0 )

The period in Tokyo is TT = 2p

LT gT

and the period in Cambridge is TC = 2p

LC . gC

We know that TT = TC = 2.000 s, from which we see that LT LC = gT gC 13.37

or

gC LC 0.994 2 = = = 1.001 5 gT LT 0.992 7

(a)

The period of the pendulum is T = 2p L g . Thus, on the Moon where the free-fall acceleration is smaller, the period will be longer and the clock will run slow .

(b)

The ratio of the pendulum’s period on the Moon to that on Earth is TMoon 2p L gMoon = = TEarth 2p L gEarth

gEarth gMoon

Hence, the pendulum of the clock on Earth makes gEarth gMoon “ticks” while the clock on the Moon is making 1.00 “tick.” After the Earth clock has ticked off 24.0 h and again reads 12:00 midnight, the Moon clock will have ticked off

( 24.0 h )

gMoon 1.63 m s2 ⎛ 60 min ⎞ = ( 24.0 h ) = 9.79 h = 9 h + ( 0.79 h ) ⎜ = 9 h + 47 min 2 ⎝ 1 h ⎟⎠ gEarth 9.80 m s

and will read 9 : 47 AM . 13.38

The coat hanger acts as a physical pendulum and its period of oscillation is T = 2p I mgd , where d is the distance from the pivot to the center of mass. Thus, the moment of inertia about the axis perpendicular to the plane of oscillation and passing through the pivot must be 2

2

⎛ T ⎞ ⎛ 1.25 s ⎞ I = mgd ⎜ = ( 0.238 kg ) 9.80 m s2 ( 0.180 m ) ⎜ = 1.66 × 10 −2 kg ⋅ m 2 ⎟ ⎝ 2p ⎠ ⎝ 2p ⎟⎠

56157_13_ch13_p400-425.indd 414

(

)

10/12/10 2:50:02 PM

Vibrations and Waves

13.39

From T = 2π L g , the length of a pendulum with period T is L =

(9.8 m s ) (1.0 s)

On Earth, with T = 1.0 s , L =

4p 2

(3.7 m s ) (1.0 s) on Mars, L =

If T = 1.0 s

= 0.25 m = 25 cm

2

2

(b)

gT2 . 4p 2

2

2

(a)

415

4p 2

= 0.094 m = 9.4 cm

(c) and (d) The period of an object on a spring is T = 2p m k , which is independent of the local free-fall acceleration. Thus, the same mass will work on Earth and on Mars. This mass is k T 2 (10 N m ) (1.0 s ) = = 0.25 kg 4p 2 4p 2 2

m= 13.40

The apparent free-fall acceleration is the vector sum of the actual free-fall acceleration and the negative of the elevator’s acceleration. To see this, consider an object that is hanging from a vertical string in the elevator and appears to be at rest to the elevator passengers. These passengers believe the tension in the string is the negative of the object’s weight, or T = −m g apparent , where g apparent is the apparent free-fall acceleration in the elevator. An observer located outside the elevator applies Newton’s second law to this object by writing ΣF = T + m g = ma e , where a e is the acceleration of the elevator and all its contents. Thus, T = m ( a e − g ) = −m g apparent , which gives g apparent = g − a e . (a)

If we choose downward as the positive direction, then a e = − 5.00 m s 2 in this case and g apparent = ( 9.80 + 5.00 ) m s 2 = +14.8 m s 2 (downward). The period of the pendulum is T = 2p

(b)

5.00 m = 3.65 s 14.8 m s 2

L = 2p gapparent

Again choosing downward as positive, a e = 5.00 m s 2 and g apparent = ( 9.80 − 5.00 ) m s 2 = + 4.80 m s 2 (downward) in this case. The period is now given by T = 2p

(c)

L gapparent

5.00 m = 6.41 s 4.80 m s 2

= 2p

→

If a e = 5.00 m s 2 horizontally, the vector sum g apparent = g − a e is as shown in the sketch at the right. The magnitude is gapparent =

(5.00 m s ) + (9.80 m s ) 2

2

2

2

= 11.0 m s 2 ,

–ae

→

g

→ g apparent

and the period of the pendulum is T = 2p

56157_13_ch13_p400-425.indd 415

L = 2p gapparent

5.00 m = 4.24 s 11.0 m s 2

10/12/10 2:50:05 PM

416

13.41

Chapter 13

(a)

(b)

The distance from the bottom of a trough to the top of a crest is twice the amplitude of the wave. Thus, 2A = 8.26 cm and A = 4.13 cm .

8.26 cm

The horizontal distance from a crest to a trough is a half wavelength. Hence, 5.20 cm

l 2 = 5.20 cm and l = 10.4 cm (c)

The period is T=

(d)

FIGURE P13.41

1 1 = = 5.56 × 10 −2 s f 18.0 s −1

The wave speed is v = l f = (10.4 cm )(18.0 s −1 ) = 187 cm s = 1.87 m s

13.42

(a)

The amplitude is the magnitude of the maximum displacement from equilibrium ( at x = 0 ) . Thus, A = 2.00 cm .

(b)

The period is the time for one full cycle of the motion. Therefore, T = 4.00 s .

(c)

x (cm) 2.00 1.00 0.00

2

3

4

5

6

t (s)

–1.00

The period may be written as T = 2p w , so the angular frequency is

–2.00 FIGURE P13.42

2p 2p p w= = = rad s T 4.00 s 2 (d)

1

The total energy may be expressed as E = 12 mvm2 ax = 12 kA2 . Thus, vm ax = A k m , and since w = k m , this becomes vm ax = w A and yields ⎛π ⎞ vm ax = ωA = ⎜ rad s⎟ ( 2.00 cm ) = p cm s ⎝2 ⎠

(e)

The spring exerts maximum force, F = k x , when the object is at maximum distance from equilibrium, i.e., at x = A = 2.00 cm. Thus, the maximum acceleration of the object is am ax =

(f)

Fm ax m

2

=

kA ⎛p ⎞ =w2A= ⎜ rad s⎟ ( 2.00 cm ) = 4.93 cm s 2 ⎝2 ⎠ m

The general equation for position as a function of time for an object undergoing simple harmonic motion with t = 0 when x = 0 is x = Asin (wt ) . For this oscillator, this becomes ⎛p ⎞ x = ( 2.00 cm ) sin ⎜ t ⎟ ⎝2 ⎠

56157_13_ch13_p400-425.indd 416

10/12/10 2:50:09 PM

Vibrations and Waves

13.43

(a)

The speed of propagation for a wave is the product of its frequency and its wavelength, v = l f . Thus, the frequency must be f=

13.44

417

v 3.00 × 108 m s = = 5.45 × 1014 Hz l 5.50 × 10 −7 m T=

1 1 = 1.83 × 10 −15 s = f 5.45 × 1014 Hz

(b)

The period is

(a)

The frequency of a transverse wave is the number of crests that pass a given point each second. Thus, if 5.00 crests pass in 14.0 seconds, the frequency is f=

(b)

5.00 = 0.357 s −1 = 0.357 Hz 14.0 s

The wavelength of the wave is the distance between successive maxima or successive minima. Thus, l = 2.76 m and the wave speed is v = l f = ( 2.76 m )( 0.357 s −1 ) = 0.985 m s

13.45

The speed of the wave is v=

Δx 425 cm = = 42.5 cm s Δt 10.0 s

and the frequency is number of vibrations occurring each second, or f = 40.0 vib 30.0 s. Thus, 13.46

l=

From v = l f , the wavelength (and size of smallest detectable insect) is l=

13.47

v 42.5 cm s ( 42.5 cm s )(30.0 s ) = 31.9 cm = = f 40.0 vib 30.0 s 40.0 vib

v 343 m s = = 5.72 × 10 −3 m = 5.72 mm f 60.0 × 10 3 Hz

The frequency of the wave (that is, the number of crests passing the cork each second) is f = 2.00 s −1 and the wavelength (distance between successive crests) is l = 8.50 cm. Thus, the wave speed is v = l f = (8.50 cm )( 2.00 s −1 ) = 17.0 cm s = 0.170 m s and the time required for the ripples to travel 10.0 m over the surface of the water is Δt =

13.48

(a)

Δx 10.0 m = = 58.8 s v 0.170 m s

When the boat is at rest in the water, the speed of the wave relative to the boat is the same as the speed of the wave relative to the water, v = 4.0 m s. The frequency detected in this case is f=

v 4.0 m s = = 0.20 Hz l 20 m

continued on next page

56157_13_ch13_p400-425.indd 417

10/12/10 2:50:13 PM

418

Chapter 13

(b)

Taking eastward as positive, v wave,boat = v wave,water − v boat,water (see the discussion of relative velocity in Chapter 3 of the textbook) gives v wave,boat = + 4.0 m s − ( −1.0 m s ) = + 5.0 m s

and

vboat,wave = v wave,boat = 5.0 m s

Thus, f= 13.49

vboat,wave 5.0 m s = = 0.25 Hz λ 20 m

The down and back distance is 4.00 m + 4.00 m = 8.00 m. The speed is then v=

d total 4 (8.00 m ) = 40.0 m s = F m = t 0.800 s

m=

m 0.200 kg = = 5.00 × 10 −2 kg m L 4.00 m

Now,

so F = m v 2 = ( 5.00 × 10 −2 kg m ) ( 40.0 m s ) = 80.0 N 2

13.50

The speed of the wave is v=

Δx 20.0 m = = 25.0 m s Δt 0.800 s

and the mass per unit length of the rope is m = m L = 0.350 kg m. Thus, from v = F m , we obtain F = v 2 m = ( 25.0 m s ) ( 0.350 kg m ) = 219 N 2

13.51

v=

F = m

13.52

(a)

f=

1 350 N = 5.20 × 10 2 m s 5.00 × 10 −3 kg m 1 T

→ T=

v= (b) 13.53

(a)

T m

1 f

→ [T ] =

→ T = m v2

1 1 = −1 = T [f] T

→ [T ] = [m ][v 2 ] =

units are seconds M L2 ML ⋅ = 2 L T2 T

units are newtons

The first T is period of time; the second is force of tension. The mass per unit length is m=

m 0.0600 kg = = 1.20 × 10 −2 kg m L 5.00 m

From v = F m , the required tension in the string is F = v 2 m = ( 50.0 m s ) (1.20 × 10 −2 kg m ) = 30.0 N 2

(b)

56157_13_ch13_p400-425.indd 418

v=

F 8.00 N = = 25.8 m s m 1.20 × 10 −2 kg m

10/12/10 2:50:15 PM

Vibrations and Waves

13.54

419

The mass per unit length of the wire is m=

m 4.00 × 10 −3 kg = = 2.50 × 10 −3 kg m, L 1.60 m

and the speed of the pulse is v=

L 1.60 m = = 44.3 m s Δt 0.0361 s

The tension in the wire is F = mg = m v 2 , so the lunar acceleration of gravity must be −3 v 2 m ( 44.3 m s ) ( 2.50 × 10 kg m ) = 1.64 m s 2 = m 3.00 kg 2

g= 13.55

The period of the pendulum is T = 2p L g , so the length of the string is 2 gT 2 ( 9.80 m s ) ( 2.00 s ) = = 0.993 m 4p 2 4p 2 2

L=

The mass per unit length of the string is then m=

m 0.060 0 kg kg = = 6.04 × 10 −2 L 0.993 m m

When the pendulum is vertical and stationary, the tension in the string is

(

)

F = M ball g = ( 5.00 kg ) 9.80 m s2 = 49.0 N and the speed of transverse waves in it is v= 13.56

If m1 = m1 L is the mass per unit length for the first string, then m 2 = m2 L = m1 2L = m1 2 is that of the second string. Thus, with F2 = F1 = F, the speed of waves in the second string is v2 =

13.57

49.0 N = 28.5 m s 6.04 × 10 −2 kg m

F = m

(a)

⎛ F ⎞ 2F = 2⎜ ⎟ = 2 v1 = 2 ( 5.00 m s ) = 7.07 m s m1 ⎝ m1 ⎠

The tension in the string is F = mg = ( 3.00 kg ) ( 9.80 m s 2 ) = 29.4 N. Then, from v = F m , the mass per unit length is m=

(b)

F = m2

F 29.4 N = = 5.10 × 10 −2 kg m 2 v ( 24.0 m s )2

When m = 2.00 kg, the tension is F = mg = ( 2.00 kg ) ( 9.80 m s 2 ) = 19.6 N and the speed of transverse waves in the string is v=

56157_13_ch13_p400-425.indd 419

F = m

19.6 N = 19.6 m s 5.10 × 10 −2 kg m

10/12/10 2:50:19 PM

420

13.58

Chapter 13

If the tension in the wire is F, the tensile stress is Stress = F A, so the speed of transverse waves in the wire may be written as v=

F = m

A ⋅ Stress = m L

Stress m (A ⋅ L)

But A ⋅ L = V = volume, so m ( A ⋅ L ) = r = density. Thus, v = Stress r . Taking the density of steel to be equal to that of iron, the maximum speed of waves in the wire is (Stress)max = rsteel

vmax = 13.59

(a)

The speed of transverse waves in the line is v = F m , with m = m L being the mass per unit length. Therefore, v=

13.60

2.70 × 10 9 Pa = 586 m s 7.86 × 10 3 kg m 3

F = m

F = m L

FL = m

(12.5 N )( 38.0 m ) 2.65 kg

= 13.4 m s

(b)

The worker could throw an object, such as a snowball, at one end of the line to set up a pulse, and use a stopwatch to measure the time it takes a pulse to travel the length of the line. From this measurement, the worker would have an estimate of the wave speed, which in turn can be used to estimate the tension.

(a)

In making n round trips along the length of the line, the total distance traveled by the pulse is Δx = n ( 2L ) = 2nL. The wave speed is then v=

(b)

Δx 2nL = t t

From v = F m as the speed of transverse waves in the line, the tension is 2

2 2 4n 2 ML ⎛ M ⎞ ⎛ 2nL ⎞ ⎛ M ⎞ ⎛ 4n L ⎞ F = mv 2 = ⎜ ⎟ ⎜ = = ⎜⎝ ⎟⎠ ⎜ 2 ⎟ ⎝ L ⎠ ⎝ t ⎟⎠ L ⎝ t ⎠ t2

13.61

(a)

Constructive interference produces the maximum amplitude Am′ ax = A1 + A2 = 0.30 m + 0.20 m = 0.50 m

(b)

Destructive interference produces the minimum amplitude Amin ′ = A1 − A2 = 0.30 m − 0.20 m = 0.10 m

13.62

We are given that x = A cos(w t ) = ( 0.25 m ) cos( 0.4p t ) . (a)

By inspection, the amplitude is seen to be A = 0.25 m .

(b)

The angular frequency is w = 0.4p rad s. But w = k m , so the spring constant is k = mw 2 = ( 0.30 kg ) ( 0.4p rad s ) = 0.47 N m 2

continued on next page

56157_13_ch13_p400-425.indd 420

10/12/10 2:50:22 PM

Vibrations and Waves

(c)

421

Note: Your calculator must be in radians mode for part (c). At t = 0.30 s, x = ( 0.25 m ) cos ⎡⎣( 0.4p rad s ) ( 0.30 s )⎤⎦ = 0.23 m

(d)

From conservation of mechanical energy, the speed at displacement x is given by v =w

A2 − x 2 . Thus, at t = 0.30 s, when x = 0.23 m, the speed is

v = ( 0.4 p rad s ) 13.63

( 0.25 m )2 − ( 0.23 m )2 = 0.12 m s

The period of a vibrating object-spring system is T = 2p w = 2p m k , so the spring constant is

(a)

2 4p 2 m 4p ( 2.00 kg ) = = 219 N m T2 ( 0.600 s )2

k=

If T = 1.05 s for mass m2, this mass is

(b)

kT 2 ( 219 N m ) (1.05 s ) = = 6.12 kg 4p 2 4p 2 2

m2 = 13.64

(a)

The period is the reciprocal of the frequency, or T=

13.65

1 1 = = 5.10 × 10 −3 s = 5.10 ms f 196 s −1

vsound 343 m s = = 1.75 m f 196 s −1

(b)

l=

(a)

The period of a simple pendulum is T = 2p T1 = 2p

g

= 2p

0.700 m = 1.68 s 9.80 m s 2

The period of an object-spring system is T = 2p m k , so if the period of the second

(b)

system is T2 = T1 , then 2p m k = 2p mg

k= 13.66

=

(1.20 kg) (9.80

0.700 m

m s2 )

g and the spring constant is = 16.8 N m

Since the spring is “light,” we neglect any small amount of energy lost in the collision with the spring, and apply conservation of mechanical energy from when the block first starts until it comes to rest again. This gives

( KE + PE Thus,

56157_13_ch13_p400-425.indd 421

g , so the period of the first system is

x m ax =

g

+ PEs

) = ( KE + PE f

2mghi = k

g

)

+ PEs , or 0 + 0 + i

1 2 kx m ax = 0 + 0 + mghi 2

2 ( 0.500 kg ) ( 9.80 m s 2 ) ( 2.00 m ) 20.0 N m

= 0.990 m

10/12/10 2:50:25 PM

422

13.67

Chapter 13

Choosing PEg = 0 at the initial height of the 3.00-kg object, conservation of mechanical energy

(

gives KE + PEg + PEs

) = ( KE + PE f

g

speed of the object after falling distance x. (a)

(a)

1 2

mv 2 + mg ( −x ) + 12 kx 2 = 0, where v is the

2 2 mg 2 ( 3.00 kg ) ( 9.80 m s ) = 588 N m = 0.100 m x m ax

When x = 5.00 cm = 0.050 0 m, the energy equation gives v = 2gx − kx 2 m , or v = 2 ( 9.80 m s 2 ) ( 0.050 0 m ) −

13.68

i

When v = 0, the non-zero solution to the energy equation from above gives 2 1 2 kx m ax = mgx m ax , or k=

(b)

)

+ PEs , or

(588

N m ) ( 0.050 0 m ) = 0.700 m s 3.00 kg 2

We apply conservation of mechanical energy from just after the collision until the block comes to rest. Conservation of energy gives ( KE + PEs ) f = ( KE + PEs )i or 0 + 12 k x 2f = 12 MV 2 + 0. The speed of the block just after the collision is then V=

k x 2f M

=

(900

N m ) ( 0.050 0 m ) = 1.50 m s 1.00 kg 2

Now, we apply conservation of momentum from just before impact to immediately after the collision. This gives m ( vbullet )i + 0 = m ( vbullet ) f + MV , or

(v (b)

bullet

)

⎛ 1.00 kg ⎞ ⎛ M⎞ = ( vbullet )i − ⎜ ⎟ V = 400 m s − ⎜ (1.5 m s ) = 100 m s ⎝ m⎠ 5.00 × 10 −3 kg ⎟⎠ ⎝

f

The mechanical energy converted into internal energy during the collision is 2 2 ΔE = KEi − ΣKE f = 12 m ( vbullet )i − 12 m ( vbullet ) f − 12 MV 2 , or ΔE =

1 (5.00 × 10−3 kg) ⎡⎣( 400 m s)2 − (100 m s)2 ⎤⎦ − 12 (1.00 kg) (1.50 m s)2 2

ΔE = 374 J 13.69

The maximum acceleration of the oscillating system is

→

n

am ax = w A = ( 2p f ) A 2

2

The friction force, fs , acting between the two blocks must be capable of accelerating block B at this rate. When block B is on the verge of slipping, fs = ( fs )m ax = m s n = m s mg = mam ax and we must have am ax = ( 2p f ) A = m s g 2

Thus, A =

56157_13_ch13_p400-425.indd 422

B → fs → → Fg ⫽ m g

( 0.600 ) ( 9.80 m s2 ) ms g = = 6.62 × 10 −2 m = 6.62 cm 2 2 ( 2p f ) [ 2p (1.50 Hz )]

10/12/10 2:50:29 PM

Vibrations and Waves

13.70

(a)

When the gun is fired, the energy initially stored as elastic potential energy in the spring is transformed into kinetic energy of the bullet. Assuming no loss of energy, we have 2 2 1 1 2 mv = 2 kx i , or v = xi

(b)

423

k 9.80 N m = ( 0.200 m ) = 19.8 m s m 1.00 × 10 −3 kg

From Δy = v0 y t + 12 a y t 2 , the time required for the pellet to drop 1.00 m to the floor, starting with v0 y = 0, is t=

2 ( Δy ) = ay

2 ( −1.00 m ) = 0.452 s −9.80 m s 2

The range (horizontal distance traveled during the flight) is then Δx = v0 x t = (19.8 m s ) ( 0.452 s ) = 8.95 m 13.71

(a)

The force diagram at the right shows the forces acting on the balloon when it is displaced distance s = Lq along the circular arc it follows. The net force tangential to this path is

Equilibrium position

→

⫹y

B ␪

s ⫽ L␪

Fnet = ΣFx = −B sinq + mg sinq = − ( B − mg ) sinq

L

For small angles, sinq ≈ q = s L. Also, mg = ( rHe V ) g and the buoyant force is B = ( rair V ) g. Thus, the net restoring force acting on the balloon is ⫹x

␪ →

mg

⎡ ( r − rHe ) Vg ⎤ Fnet ≈ − ⎢ air ⎥s L ⎢⎣ ⎥⎦

→

T

␪

Observe that this is in the form of Hooke’s law, F = −k s, with k = ( rair − rHe ) Vg L. Thus, the motion will be simple harmonic . (b)

The period of this simple harmonic motion is given by T=

1 2p = = 2p f w

m = 2p k

⎛ rHe ⎞ L rHe V = 2p ⎜ ⎝ rair − rHe ⎟⎠ g ( rair − rHe ) Vg L

This yields ⎛ ⎞ ( 3.00 m ) 0.179 kg m 3 T = 2p ⎜ = 1.40 s 3 3 ⎟ ⎝ 1.29 kg m − 0.179 kg m ⎠ ( 9.80 m s 2 ) 13.72

(a)

When the object is given some small upward displacement, the net restoring force exerted on it by the rubber bands is Fnet = ΣFy = −2 F sinq , where tanq =

y L

For small displacements, the angle q will be very small. Then sinq ≈ tanq = y L, and the net restoring force is ⎛ y⎞ ⎛ 2F⎞ Fnet = −2 F ⎜ ⎟ = − ⎜ y ⎝ L⎠ ⎝ L ⎟⎠ continued on next page

56157_13_ch13_p400-425.indd 423

10/12/10 2:50:32 PM

424

Chapter 13

The net restoring force found in part (a) is in the form of Hooke’s law F = −ky, with k = 2 F L . Thus, the motion will be simple harmonic, and the angular frequency is

(b)

k = m

w= 13.73

2F mL

Newton’s law of gravitation is GMm ⎛4 ⎞ F = − 2 , where M = r ⎜ p r 3 ⎟ ⎝3 ⎠ r

m M

⎛4 ⎞ F= − ⎜ p r Gm ⎟ r ⎝3 ⎠

Thus,

F

r RE

which is of Hooke’s law form, F = −k r , with k= 13.74

4 p r Gm 3

The inner tip of the wing is attached to the end of the spring and always moves with the same speed as the end of the vibrating spring. Thus, its maximum speed is vinner, m ax = vspring, m ax = A

k 4.7 × 10 −4 N m = ( 0.20 cm ) = 0.25 cm s m 0.30 × 10 −3 kg

Treating the wing as a rigid bar, all points in the wing have the same angular velocity at any instant in time. As the wing rocks on the fulcrum, the inner tip and outer tips follow circular paths of different radii. Since the angular velocities of the tips are always equal, we may write v v w = outer = inner . The maximum speed of the outer tip is then router rinner ⎛r ⎞ ⎛ 15.0 mm ⎞ vouter, m ax = ⎜ outer ⎟ vinner, m ax = ⎜ ( 0.25 cm s ) = 1.3 cm s ⎝ 3.00 mm ⎟⎠ r ⎝ inner ⎠ 13.75

500 N m = 15.8 rad s 2.00 kg

k = m

(a)

w=

(b)

Apply Newton’s second law to the block while the elevator is accelerating:

ΣFy = Fs − mg = may With Fs = kx and a y = g 3 , this gives kx = m ( g + g 3), or x= 13.76

(a)

2 4 mg 4 ( 2.00 kg ) ( 9.80 m s ) = = 5.23 × 10 −2 m = 5.23 cm 3k 3 ( 500 N m )

Note that as the spring passes through the vertical position, the object is moving in a circular arc of radius L − y f . Also, observe that the y-coordinate of the object at this point must be negative y f < 0 , so the spring is stretched and exerting an upward tension force of magnitude greater than the object’s weight. This is necessary so the object experiences a net

(

)

continued on next page

56157_13_ch13_p400-425.indd 424

10/12/10 2:50:36 PM

Vibrations and Waves

425

force toward the pivot to supply the needed centripetal acceleration in this position. This is summarized by Newton’s second law applied to the object at this point, stating ΣFy = −ky f − mg = (b)

Conservation of energy requires that E = KEi + PEg ,i + PEs ,i = KE f + PEg , f + PEs , f , or E = 0 + mgL + 0 =

1 2 1 mv + mgy f + ky 2f 2 2

(

)

mv 2 = 2mg L − y f − ky 2f

reducing to (c)

mv 2 L − yf

From the result of part (a), observe that

(

mv 2 = − L − y f

) ( ky

f

+ mg

)

Substituting this into the result from part (b) gives

(

) (

2mg L − y f = − L − y f

) ( ky

f

)

+ mg + ky 2f

After expanding and regrouping terms, this becomes

( 2k ) y 2f + ( 3mg − kL ) y f + ( −3mgL ) = 0 which is a quadratic equation ay 2f + by f + c = 0, with a = 2k = 2 (1 250 N m ) = 2.50 × 10 3 N m

(

)

b = 3mg − kL = 3 ( 5.00 kg ) 9.80 m s2 − (1 250 N m ) (1.50 m ) = −1.73 × 10 3 N

(

)

and c = −3mgL = −3 ( 5.00 kg ) 9.80 m s2 (1.50 m ) = −221 N ⋅ m Applying the quadratic formula, keeping only the negative solution [see the discussion in part (a)] gives

(

) ( −1.73 × 10 ) − 4 ( 2.50 × 10 ) ( −221) 2 ( 2.50 × 10 )

3 −b − b 2 − 4ac − −1.73 × 10 − yf = = 2a

or (d)

56157_13_ch13_p400-425.indd 425

3 2

3

3

y f = −0.110 m

Because the length of this pendulum varies and is longer throughout its motion than a simple pendulum of length L, its period will be greater than that of a simple pendulum.

10/12/10 2:50:38 PM

14 Sound QUICK QUIZZES 1.

Choice (c). The speed of sound in air is given by v = ( 331 m s ) T 273 K . Thus, increasing the absolute temperature, T, will increase the speed of sound. Changes in frequency, amplitude, or air pressure have no effect on the speed of sound.

2.

Choice (c). The distance between you and the buzzer is increasing. Therefore, the intensity at your location is decreasing. As the buzzer falls, it moves away from you with increasing speed. This causes the detected frequency to decrease.

3.

Choice (b). The speed of sound increases in the warmer air, while the speed of the sound source (the plane) remains constant. Therefore, the ratio of the speed of the source to that of sound (that is, the Mach number) decreases.

4.

Choices (b) and (e). A string fastened at both ends can resonate at any integer multiple of the fundamental frequency. Of the choices listed, only 300 Hz and 600 Hz are integer multiples of the 150 Hz fundamental frequency.

5.

Choice (d). In the fundamental mode, an open pipe has a node at the center and antinodes at each end. The fundamental wavelength of the open pipe is then twice the length of the pipe and the fundamental frequency is fopen = v 2L. When one end of the pipe is closed, the fundamental mode has a node at the closed end and an antinode at the open end. In this case, the fundamental wavelength is four times the length of the pipe and the fundamental frequency is fclose = v 4L.

6.

Choice (a). The change in the length of the pipe, and hence the fundamental wavelength, is negligible. As the temperature increases, the speed of sound in air increases and this causes an increase in the fundamental frequency, f0 = v l0.

7.

Choice (b). Since the beat frequency is steadily increasing, you are increasing the difference between the frequency of the string and the frequency of the tuning fork. Thus, your action is counterproductive and you should reverse course by loosening the string.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

All sound waves travel at the same speed in air of a given temperature. Thus, v = l f = l2 f2, giving ⎛ l⎞ ⎛ l ⎞ f = 2f f2 = ⎜ ⎟ f = ⎜ ⎝ l 2 ⎟⎠ ⎝ l2 ⎠ and the correct choice is (b).

2.

The Celsius temperature on this day was TC = sound in the air was

5 9

(T

F

− 32 ) =

5 134 − 32 ) = 56.7°C. The 9(

speed of

426

56157_14_ch14_p426-456.indd 426

10/12/10 2:51:07 PM

Sound

v = ( 331 m s ) 1+

427

TC 56.7 = 364 m s = ( 331 m s ) 1+ 273 273

and the correct answer is (c). 3.

The speed of sound in a ﬂuid is given in Equation 14.1 as v = B r , where B is the bulk modulus of the ﬂuid and r is its density. The speed of sound in ethyl alcohol is found to be v=

B = r

1.0 × 10 9 Pa = 1.1× 10 3 m s 0.806 × 10 3 kg m 3

and (a) is the correct choice. 4.

Note that the value given for the speed of sound in aluminum in Table 14.1 is for “bulk media” only and does not apply to the thin, solid rod at issue here. Thus, we turn to Equation 14.3, and use data from Tables 9.1 and 9.2 to obtain vAl =

YAl = rAl

7.0 × 1010 Pa = 5.1× 10 3 m s 2.7 × 10 3 kg m 3

This shows choice (e) to be the correct answer. 5.

The relation between the decibel level and the sound intensity is b = 10 ⋅ log ( I I 0 ), where the reference intensity is I 0 = 1.0 × 10 −12 W m 2 and log refers to the base 10 logarithm. Thus, if b = 105 dB, the sound intensity is I = I 0 ⋅10 b

10

= (1.0 × 10 −12 W m 2 ) ⋅1010.5 = 3.16 × 10 −2 W m 2

and (d) is the correct answer. 6.

In a uniform medium, the intensity of sound varies inversely with the square of the distance from the source. (See Equation 14.8 in the textbook.) Thus, if the distance from the source is tripled, the new sound intensity will be one-ninth of its original value, making (a) the correct choice.

7.

The apparent frequency fO detected by an observer from a source emitting sound of frequency fS is given by ⎛ v + vO ⎞ fO = fS ⎜ ⎝ v − vS ⎟⎠ where v is the speed of sound in air, vO is the velocity of the observer relative to the air and is positive if the observer moves toward the source, while vS is the velocity of the source relative to the air and is positive if the source moves toward the observer. In this case, we have ⎡ 343 m s + ( −30.0 m s ) ⎤ 3 fO = (1.00 × 10 3 Hz ) ⎢ ⎥ = 1.07 × 10 m s 343 m s − +50.0 m s ( ) ⎦ ⎣ so the correct choice is (d).

8.

56157_14_ch14_p426-456.indd 427

When a sound wave travels from air into water, several properties will change. The wave speed will increase as the wave crosses the boundary into the water, the spacing between crests (the wavelength) will increase since crests move away from the boundary faster than they move up to the boundary, and the sound intensity in the water will be less than it was in air because

10/12/10 2:51:09 PM

428

Chapter 14

some sound is reﬂected by the water surface. However, the frequency (number of crests passing each second) will be unchanged, since a crest moves away from the boundary every time a crest arrives at the boundary. Among the listed choices, the only correct statement is choice (d). 9.

The number of beats per second (the beat frequency) equals the difference in the frequencies of the two tuning forks. Thus, if the beat frequency is 5 Hz and one fork is known to have a frequency of 245 Hz, the frequency of the second fork could be either f2 = 245 Hz − 5 Hz = 240 Hz or f2 = 245 Hz + 5 Hz = 250 Hz. This means that the best answer for the question is choice (e), since choices (a) and (d) are both possibly correct.

10.

At resonance, a tube closed at one end and open at the other forms a standing wave pattern with a node at the closed end and antinode at the open end. In the fundamental mode (or ﬁrst harmonic), the length of the tube closed at one end is a quarter wavelength ( L = l1 4 or l1 = 4L ). Therefore, for the given tube, l1 = 4 ( 0.580 m ) = 2.32 m and the fundamental frequency is f1 =

v 343 m s = = 148 Hz l1 2.32 m

and the correct answer is choice (a). 11.

When the two ends of a pipe are alike (either both open or both closed), all harmonics (integer multiples of the fundamental frequency) are present among the resonant frequencies of the pipe. However, in a pipe closed at one end and open at the other, only the odd harmonics (i.e., only odd integer multiples of the fundamental frequency) are resonant frequencies for the pipe. In the case of the given pipe, the fundamental frequency is f1 = 150 Hz, and both f2 = 2 f1 = 300 Hz (an even multiple) and f3 = 3 f1 = 450 Hz (an odd multiple) are resonant frequencies. Thus, the pipe must have either two open ends or two closed ends, and the correct choice is (b).

12.

The ambulance driver, sitting at a ﬁxed distance from the siren, hears the actual frequency emitted by the siren. However, the distance between you and the siren is decreasing, so you will detect a frequency higher than the actual 500 Hz. Choice (c) is the correct answer.

13.

The speed of sound in air, at atmospheric pressure, is determined by the temperature of the air and does not depend on the frequency of the sound. Sound from siren A will have a wavelength that is half the wavelength of the sound from B, but the speed of the sound (the product of frequency times wavelength) will be the same for the two sirens. The correct choice is (e).

14.

In the fundamental mode (ﬁrst harmonic), a pipe open at both ends has antinodes at each end and a node at the center. The wavelength of this harmonic is l1 = 2L and the resonant frequency is fopen = v 2L. If one end of the pipe is now closed, the fundamental mode will have a node at the closed end with an antinode at the open end. The wavelength of the ﬁrst harmonic is l1′ = 4L and the resonant frequency is fclosed = v 4L. Thus, we see that fopen = 2 fclosed and the correct choice is (b).

15.

Doubling the power output of the source will double the intensity of the sound at the observer’s location. The original decibel level of the sound is b = 10 ⋅ log ( I I 0 ). After doubling the power output and intensity, the new decibel level will be b ′ = 10 ⋅ log ( 2I I 0 ) = 10 ⋅ log ⎡⎣ 2 ( I I 0 ) ⎤⎦ = 10 ⋅ ⎡⎣ log 2 + log ( I I 0 ) ⎤⎦ = 10 ⋅ log 2 + b so the increase in decibel level is b ′ − b = 10 ⋅ log 2 = 3.0 dB, making (c) the correct answer.

56157_14_ch14_p426-456.indd 428

10/12/10 2:51:11 PM

Sound

429

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The resonant frequency depends on the length of the pipe. Thus, changing the length of the pipe will cause different frequencies to be emphasized in the resulting sound. The shorter the pipe, the higher the fundamental resonance frequency.

4.

The speed of light is so high that the arrival of the ﬂash is practically simultaneous with the lightning discharge. Thus, the delay between the ﬂash and the arrival of the sound of thunder is the time sound takes to travel the distance separating the lightning from you. By counting the seconds between the ﬂash and thunder and knowing the approximate speed of sound in air, you have a rough measure of the distance to the lightning bolt.

6.

A vibrating string is not able to set very much air into motion when vibrated alone. Thus it will not be very loud. If it is placed on the instrument, however, the string’s vibration sets the sounding board of the guitar into vibration. A vibrating piece of wood is able to move a lot of air, and the note is louder.

8.

A beam of electromagnetic waves of known frequency is sent toward a speeding car, which reﬂects the beam back to a detector in the police car. The amount the returning frequency has been shifted depends on the velocity of the oncoming car.

10.

Consider the level of ﬂuid in the bottle to be adjusted so that the air column above it resonates at the ﬁrst harmonic. This is given by f = v 4L. This equation indicates that as the length L of the column increases (ﬂuid level decreases), the resonant frequency decreases.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

1× 1011 Pa

4.

0.196 s

6.

l = 17 m for f = 20 Hz; l = 1.7 cm for f = 20 000 Hz

8.

18.6 m

10.

(a)

1.0 × 10 3 W m 2

(b)

The intensity level is 1 000 times that at the threshold of pain.

12.

150 dB

14.

(a)

5.0 × 10 −17 W

(b) 5.0 × 10 −5 W

16.

(a)

0.316 W m 2

(b) 1.58 W m 2

(d)

104 dB

(e) 1.26 × 10 6 m

(f)

The sound intensity falls as the sound wave travels farther from the source until it is much lower than the ambient noise level and is drowned out.

(a)

1.32 × 10 −4 W m 2

18.

56157_14_ch14_p426-456.indd 429

(b)

(c) 2.47 × 10 −2 W m 2

81.2 dB

10/12/10 2:51:13 PM

430

Chapter 14

20.

(a)

7.96 × 10 −2 W/m 2

(b)

109 dB

22.

(a)

I A IB = 2

(b)

I A IC = 5

24.

(a)

10.0 kHz

(b)

3.33 kHz

26.

32.0 m s

28.

(a)

3.29 m s

(b) Yes, the bat gains on the insect at a rate of 1.71 m s.

30.

(a)

2.16 × 10 −2 m s

(b)

2 000 029 Hz

32.

(a)

fO = fS ⎡⎣( v + vO ) ( v − vS ) ⎤⎦

(b)

the yellow submarine

(c) 2 000 058 Hz (c)

the red submarine

increases the time (period), decreases the frequency

(e)

negative

(f)

decreases the time (period), increases the frequency

(g)

positive

5.30 × 10 3 Hz

34.

(a)

0.227 m

36.

1.43 m

38.

823.8 N

40.

1.00 cm toward the nut

42.

120 Hz

46.

2.82 m

(d)

(h)

44.

(c)

(a)

n fA = A 2L A

(b)

TA mA

(d)

TB TA = 9 16

(a)

m = 4.9 × 10 −3 kg m

(c)

no standing wave will form

0.454 m

2

(b)

fB = f A 2

⎛ n ⎞ (c) TB = ⎜ A ⎟ TA ⎝ n + 1⎠ A

(b)

2

(b)

4.29 cm

48.

9.00 kHz

50.

(a)

536 Hz

52.

(a)

fn = n ( 0.085 8 Hz ) n = 1, 2, 3, …

(b) Yes. The tunnel can resonate at many closely spaced frequencies, and the sound would be greatly ampliﬁed.

54.

(a)

56.

29.7 cm

58.

3.98 Hz

56157_14_ch14_p426-456.indd 430

f1 = 50.0 Hz

(b)

open at only one end

(c)

1.72 m

10/12/10 2:51:15 PM

Sound

60.

2.95 cm

62.

∼ 1 000 mosquitoes

64.

(a)

66.

21.4 m

68.

1 204 Hz

70.

7.8 m

72.

(a)

74.

1.95 m s

76.

(a)

65.0 dB

0.655 m

(b)

67.8 dB

(b)

13.4°C

(c)

431

69.6 dB

(b) 3.6 × 10 2 Hz

32.2°C

PROBLEM SOLUTIONS 14.1

(a)

We ignore the time required for the lightning ﬂash to arrive. Then, the distance to the lightning stroke is d ≈ vsound ⋅ Δt = ( 343 m s )(16.2 s ) = 5.56 × 10 3 m = 5.56 km

(b)

14.2

No. Since vlight >> vsound, the time required for the ﬂash of light to reach the observer is negligible in comparison to the time required for the sound to arrive, and knowledge of the actual value of the speed of light is not needed.

The speed of longitudinal waves in a ﬂuid is v = B r . Considering the Earth’s crust to consist of a very viscous ﬂuid, our estimate of the average bulk modulus of the material in Earth’s crust is B = rv 2 = ( 2 500 kg m 3 ) ( 7 × 10 3 m s ) = 1× 1011 Pa 2

14.3

The Celsius temperature is TC = air is v = ( 331 m s ) 1+

14.4

(T

F

− 32 ) =

5 114 − 32 ) = 9(

45.6°C and the speed of sound in the

TC 45.6 = 358 m s = ( 331 m s ) 1+ 273 273

The speed of sound in seawater at 25°C is 1 533 m s. Therefore, the time for the sound to reach the sea ﬂoor and return is t=

14.5

5 9

2d 2 (150 m ) = = 0.196 s v 1 533 m s

Since the sound had to travel the distance between the hikers and the mountain twice, the time required for a one-way trip was 1.50 s. The distance the sound traveled to the mountain was d = ( 343 m s )(1.50 s ) = 515 m

56157_14_ch14_p426-456.indd 431

10/12/10 2:51:17 PM

432

14.6

Chapter 14

At T = 27°C, the speed of sound in air is v = ( 331 m s ) 1+

TC 27 = 347 m s = ( 331 m s ) 1+ 273 273

The wavelength of the 20 Hz sound is l=

v 347 m s = = 17 m f 20 Hz

and that of the 20 000 Hz is l=

347 m s = 1.7 × 10 −2 m = 1.7 cm 20 000 Hz

Thus, range of wavelengths of audible sounds at 27°C is 1.7 cm to 17 m . 14.7

At T = 27.0°C, the speed of sound in air is v27 = ( 331 m s ) 1+

TC 27.0 = 347 m s = ( 331 m s ) 1+ 273 273

and at T = 10.0°C, the speed is v10 = ( 331 m s ) 1+

TC 10.0 = 337 m s = ( 331 m s ) 1+ 273 273

Since v = λ f , the change in wavelength will be Δl = 14.8

v10 v27 v10 − v27 (337 − 347) m s − = = = −2.5 × 10 −3 m = −2.5 mm f f f 4.00 × 10 3 Hz

At a temperature of T = 10.0°C, the speed of sound in air is v = ( 331 m s ) 1+

TC 10.0 = 337 m s = ( 331 m s ) 1+ 273 273

The elapsed time between when the stone was released and when the sound is heard is the sum of the time t1 required for the stone to fall distance h and the time t2 required for sound to travel distance h in air on the return up the well. That is, t1 + t2 = 2.00 s. The distance the stone falls, gt 2 starting from rest, in time t1 is h= 1 2 h t2 = = 2.00 s − t1 Also, the time for the sound to travel back up the well is v Combining these two equations yields

(g

2v ) t12 = 2.00 s − t1

With v = 337 m s and g = 9.80 m s 2 , this becomes

(1.45 × 10

−2

s −1 ) t12 + t1 − 2.00 s = 0

Applying the quadratic formula yields one positive solution of t1 = 1.95 s, so the depth of the well is 2 gt12 ( 9.80 m s )(1.95 s ) = = 18.6 m 2 2 2

h=

56157_14_ch14_p426-456.indd 432

10/12/10 2:51:19 PM

Sound

14.9

433

Because the speed of sound in air is vair = 343 m s while its speed in the steel rail is

(a)

vsteel = 5 950 m s, the pulse traveling in the steel rail arrives first . (b)

The difference in times when the two pulses reach the microphone at the opposite end of the rail is Δt =

14.10

⎞ ⎛ L L 1 1 = 2.34 × 10 −2 s = 23.4 ms − = (8.50 m ) ⎜ − vair vsteel ⎝ 343 m s 5 950 m s ⎟⎠

The decibel level, b , of a sound is given b = 10 log ( I I 0 ), where I is the intensity of the sound, and I 0 = 1.0 × 10 −12 W m 2 is the reference intensity. Therefore, if b = 150 dB, the intensity is

(a)

I = I 0 × 10 b

= (1.0 × 10 −12 W m 2 ) × 1015 = 1.0 × 10 3 W m 2

The threshold of pain is I = 1 W m 2 and the answer to part (a) is 1 000 times greater than this, explaining why some airport employees must wear hearing protection equipment.

(b) 14.11

10

If the intensity of this sound varied inversely with the square of the distance from the source 2 1 2 = 4 800 km from the source is given by

( I = constant r ), the ratio of the intensities at distances r =161 km and r 2

⎛ 161 km ⎞ ⎞ ⎛ r1 ⎞ I 2 ⎛ constant ⎞ ⎛ r12 =⎜ ⎟ =⎜ =⎜ 2 ⎟ ⎜ ⎟ r2 I1 ⎝ ⎝ 4 800 km ⎟⎠ ⎠ ⎝ constant ⎠ ⎝ r2 ⎠

2

The difference in the decibel levels at distances r1 and r2 from this source was then 2 ⎛ I I0 ⎞ ⎛I ⎞ ⎛I ⎞ ⎛ I2 ⎞ ⎛ 161 km ⎞ = 10 ⋅ log b 2 − b1 = 10 ⋅ log ⎜ 2 ⎟ − 10 ⋅ log ⎜ 1 ⎟ = 10 ⋅ log ⎜ 2 ⋅ = 10 ⋅ log ⎟ ⎜⎝ 4 800 km ⎟⎠ ⎜⎝ I ⎟⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎝ I 0 I1 ⎠ 1

or

b 2 − b1 = −29.5 dB. This gives the decibel level on Rodriguez Island as b 2 = b1 − 29.5 dB = 180 dB − 29.5 dB ≈ 151 dB

14.12

The decibel level due to the ﬁrst siren is ⎛ 100.0 W m 2 ⎞ = 140 dB b1 = 10 ⋅ log ⎜ ⎝ 1.0 × 10 −12 W m 2 ⎟⎠ Thus, the decibel level of the sound from the ambulance is b 2 = b1 + 10 dB = 140 dB + 10 dB = 150 dB

14.13

In terms of their intensities, the difference in the decibel level of two sounds is ⎛ I I0 ⎞ ⎛I ⎞ ⎛I ⎞ ⎛ I2 ⎞ b 2 − b1 = 10 ⋅ log ⎜ 2 ⎟ − 10 ⋅ log ⎜ 1 ⎟ = 10 ⋅ log ⎜ 2 ⋅ ⎟ = 10 ⋅ log ⎜ ⎟ ⎝ I1 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎝ I 0 I1 ⎠ Thus,

I2 b = 10( I1

2 − b1

)

10

or

I 2 = I1 × 10(

b 2 − b 1 ) 10

If b 2 − b1 = 30.0 dB and I1 = 3.0 × 10 −11 W m 2, then I 2 = ( 3.0 × 10 −11 W m 2 ) × 10 3.00 = 3.0 × 10 −8 W m 2

56157_14_ch14_p426-456.indd 433

10/12/10 2:51:23 PM

434

14.14

Chapter 14

The sound power incident on the eardrum is P = IA, where I is the intensity of the sound and A = 5.0 × 10 −5 m 2 is the area of the eardrum. (a)

At the threshold of hearing, I = 1.0 × 10 −12 W m 2 , and P = (1.0 × 10 −12 W m 2 ) ( 5.0 × 10 −5 m 2 ) = 5.0 × 10 −17 W

(b)

At the threshold of pain, I = 1.0 W m 2, and P = (1.0 W m 2 ) ( 5.0 × 10 −5 m 2 ) = 5.0 × 10 −5 W

14.15

The decibel level b = 10 log ( I I 0 ), where I 0 = 1.00 × 10 −12 W m 2. (a) (b)

If b = 100 dB, then log ( I I 0 ) = 10, giving I = 1010 I 0 = 1.00 × 10 −2 W m 2 . If all three toadﬁsh sound at the same time, the total intensity of the sound produced is I ′ = 3I = 3.00 × 10 −2 W m 2, and the decibel level is ⎛ 3.00 × 10 −2 W m 2 ⎞ b ′ = 10 log ⎜ ⎝ 1.00 × 10 −12 W m 2 ⎟⎠ = 10 log ⎡⎣( 3.00 )(1010 ) ⎤⎦ = 10 [ log ( 3.00 ) + 10 ] = 105 dB

14.16

(a)

From the deﬁning equation of the decibel level, b = 10 ⋅ log ( I I 0 ), we solve for the intensity as I = I 0 ⋅10 b 10 and ﬁnd that I = (1.0 × 10 −12 W m 2 ) ⋅10115 10 = 1.0 × 10 −12 +11.5 W m 2 = 10 −0.5 W m 2 = 0.316 W m 2

(b)

If 5 trumpets are sounded together, the total intensity of the sound is I 5 = 5I1 = 5 ( 0.316 W m 2 ) = 1.58 W m 2

(c)

If the sound propagates uniformly in all directions, the intensity varies inversely as the square of the distance from the source, I = constant r 2 , and we ﬁnd that I 2 ⎛ r1 ⎞ = I1 ⎜⎝ r2 ⎟⎠

(d) (e)

56157_14_ch14_p426-456.indd 434

2

or

2 ⎛r ⎞ ⎛ 1.0 m ⎞ I 2 = I1 ⎜ 1 ⎟ = (1.58 W m 2 ) ⎜ = 2.47 × 10 −2 W m 2 ⎟ ⎝ ⎠ r 8.0 m ⎝ 2⎠

⎛I ⎞ ⎛ 2.47 × 10 −2 W m 2 ⎞ = 104 dB b row 1 = 10 ⋅ log ⎜ row 1 ⎟ = 10 ⋅ log ⎜ ⎝ 1.0 × 10 −12 W m 2 ⎟⎠ ⎝ I0 ⎠ The intensity of sound at the threshold of hearing is I 0 = 1.0 × 10 −12 W m 2 , and from the 2 discussion and result of part (c), we have I 0 I 2 = ( r2 r0 ) , and with the intensity being I 2 = 2.47 × 10 −2 W m 2 at distance r2 = 8.0 m, the distance at which the intensity would be I 0 = 1.0 × 10 −12 W m 2 is r0 = r2

(f)

2

I2 2.47 × 10 −2 W m 2 = (8.0 m ) = 1.26 × 10 6 m I0 1.0 × 10 −12 W m 2

The sound intensity level falls as the sound wave travels farther from the source until it is much lower than the ambient noise level and is drowned out.

10/12/10 2:51:27 PM

Sound

14.17

435

The intensity of a spherical sound wave at distance r from a point source is I = Pav 4p r 2, where Pav is the average power radiated by the source. Thus, at distances r1 = 5.0 m and r2 = 10 km = 10 4 m, the intensities of the sound wave radiating out from the elephant are I1 = Pav 4p r12 and I 2 = Pav 4p r22 2 giving I 2 = ( r1 r2 ) I1. From the deﬁning equation, b = 10 log ( I I o ), the intensity level of the sound at distance r2 from the elephant is seen to be 2 ⎡⎛ r ⎞ 2 I ⎤ ⎛ I2 ⎞ ⎛I ⎞ ⎛I ⎞ ⎛ r1 ⎞ ⎛r ⎞ 1 1 ⎥ = 10 log ⎜ ⎟ + 10 log ⎜ 1 ⎟ = 20 log ⎜ 1 ⎟ + 10 log ⎜ 1 ⎟ b 2 = 10 log ⎜ ⎟ = 10 log ⎢⎜ ⎟ ⎝ r2 ⎠ ⎝ r2 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎢⎣⎝ r2 ⎠ I 0 ⎥⎦

or 14.18

(a)

⎛ 5.0 m ⎞ b 2 = 20 log ⎜ 4 ⎟ + b 1 = −66 dB + 103 dB = 37 dB ⎝ 10 m ⎠ The intensity of the musical sound (b = 80 dB) is I m usic = I 0 10 β 10 = I 0 (108.0 ), and that produced by the crying baby (b = 75 dB) is I baby = I 0 (10 7.5 ). Thus, the total intensity of the sound engulﬁng you is I = I m usic + I baby = I 0 (108.0 + 10 7.5 ) = (1.0 × 10 −12 W m 2 ) (1.32 × 108 ) = 1.32 × 10 −4 W m 2

(b)

The combined sound level is b = 10 log ( I I 0 ) = 10 log (1.32 × 108 ) = 81.2 dB

14.19

(a)

The intensity of sound at 10 km from the horn (where b = 50 dB) is I = I 0 10 β 10 = (1.0 × 10 −12 W m 2 )10 5.0 = 1.0 × 10 −7 W m 2 Thus, from I = P 4p r 2, the power emitted by the source is P = 4p r 2 I = 4p (10 × 10 3 m ) (1.0 × 10 −7 W m 2 ) = 4p × 101 W = 1.3 × 10 2 W 2

(b)

At r = 50 m, the intensity of the sound will be I=

P 1.3 × 10 2 W = = 4.1× 10 −3 W m 2 2 2 4p r 4p ( 50 m )

and the sound level is ⎛ I ⎞ ⎛ 4.1× 10 −3 W m 2 ⎞ = 10 log ( 4.1 × 10 9 ) = 96 dB b = 10 log ⎜ ⎟ = 10 log ⎜ ⎝ 1.0 × 10 −12 W m 2 ⎟⎠ ⎝ I0 ⎠ 14.20

P 100 W = = 7.96 × 10 −2 W m 2 2 4p r 2 4p (10.0 m )

(a)

I=

(b)

⎛ I ⎞ ⎛ 7.96 × 10 −2 W m 2 ⎞ b = 10 log ⎜ ⎟ = 10 log ⎜ ⎝ 1.00 × 10 −12 W m 2 ⎟⎠ ⎝ I0 ⎠ = 10 log ( 7.96 × 1010 ) = 109 dB

continued on next page

56157_14_ch14_p426-456.indd 435

10/12/10 2:51:31 PM

436

Chapter 14

At the threshold of pain (b = 120 dB), the intensity is I = 1.00 W m 2. Thus, from I = P 4p r 2, the distance from the speaker is

(c)

14.21

100 W = 2.82 m 4p (1.00 W m 2 )

P = 4p I

r=

The sound level for intensity I is b = 10 log ( I I 0 ). Therefore, ⎛ I I0 ⎞ ⎛I ⎞ ⎛I ⎞ ⎛ I2 ⎞ b 2 − b1 = 10 log ⎜ 2 ⎟ − 10 log ⎜ 1 ⎟ = 10 log ⎜ 2 ⋅ ⎟ = 10 log ⎜ ⎟ ⎝ I1 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎝ I 0 I1 ⎠ Since I = P 4p r 2 = ( P 4p ) r 2 , the ratio of intensities is I 2 ⎛ P 4p ⎞ ⎛ r12 ⎞ r12 = = I1 ⎜⎝ r22 ⎟⎠ ⎜⎝ P 4p ⎟⎠ r22 2

2

14.22

2

The intensity at distance r from the source is I = (a)

I A rB2 (100 m ) + (100 m ) = = = 2 I B rA2 (100 m )2

(b)

I A rC2 (100 m ) + ( 200 m ) = = = 5 I C rA2 (100 m )2

2

2

( P 4p ) P = 2 r2 4p r

2

2

2

Source rA=100 m

Thus,

⎛ r2 ⎞ ⎛r ⎞ ⎛r ⎞ b 2 − b1 = 10 log ⎜ 12 ⎟ = 10 log ⎜ 1 ⎟ = 20 log ⎜ 1 ⎟ ⎝r ⎠ ⎝r ⎠ ⎝r ⎠

rC B

A

14.23

rB

100 m

100 m

C

When a stationary observer ( vO = 0 ) hears a moving source, the observed frequency is ⎛ v + vO ⎞ ⎛ v ⎞ = fS ⎜ fO = fS ⎜ ⎟ ⎝ v − vS ⎠ ⎝ v − vS ⎟⎠ (a)

When the train is approaching, vS = + 40.0 m s, and

(f ) O

approach

⎞ ⎛ 343 m s = 362 Hz = ( 320 Hz ) ⎜ ⎝ 343 m s − 40.0 m s ⎟⎠

After the train passes and is receding, vS = − 40.0 m s, and

(f ) O

⎤ ⎡ 343 m s = ( 320 Hz ) ⎢ ⎥ = 287 Hz ⎣ 343 m s − ( − 40.0 m s ) ⎦

recede

Thus, the frequency shift that occurs as the train passes is ΔfO = ( fO )recede − ( fO )approach = −75 Hz, or it is a 75 Hz drop (b)

As the train approaches, the observed wavelength is l=

(f ) O

56157_14_ch14_p426-456.indd 436

v

approach

=

343 m s = 0.948 m 362 Hz

10/12/10 2:51:35 PM

Sound

14.24

437

The general expression for the observed frequency of a sound when the source and/or the observer are in motion is ⎛ v + vO ⎞ fO = fS ⎜ ⎝ v − vS ⎟⎠ Here, v is the velocity of sound in air, vO is the velocity of the observer, vS is the velocity of the source, and fS is the frequency that would be detected if both the source and observer were stationary. (a)

If fS = 5.00 kHz and the observer is stationary ( vO = 0 ), the frequency detected when the source moves toward the observer at half the speed of sound ( vS = + v 2 ) is ⎛ v+0 ⎞ = ( 5.00 kHz )( 2 ) = 10.0 kHz fO = ( 5.00 kHz ) ⎜ ⎝ v − v 2 ⎟⎠

(b)

When fS = 5.00 kHz and the source moves away from a stationary observer at half the speed of sound ( vS = − v 2 ), the observed frequency is ⎛ v+0 ⎞ ⎛ 2⎞ = ( 5.00 kHz ) ⎜ ⎟ = 3.33 kHz fO = ( 5.00 kHz ) ⎜ ⎝ 3⎠ ⎝ v + v 2 ⎟⎠

14.25

⎛ v + vO ⎞ . Since each train moves toward the Both source and observer are in motion, so fO = fS ⎜ ⎝ v − vS ⎟⎠ other, vO > 0 and vS > 0. The speed of the source (train 2) is vS = 90.0

km ⎛ 1 000 m ⎞ ⎛ 1 h ⎞ ⎜ ⎟⎜ ⎟ = 25.0 m s h ⎝ 1 km ⎠ ⎝ 3 600 s ⎠

and that of the observer (train 1) is vO = 130 km h = 36.1 m s. Thus, the observed frequency is ⎛ 343 m s + 36.1 m s ⎞ = 596 Hz fO = ( 500 Hz ) ⎜ ⎝ 343 m s − 25.0 m s ⎟⎠ 14.26

(a)

Since the observer hears a reduced frequency, the source and observer are getting farther apart. Hence, the cyclist is behind the car .

(b)

With the cyclist (observer) behind the car (source) and both moving in the same direction, the observer moves toward the source ( vO > 0 ) while the source moves away from the observer ( vS < 0 ). Thus, vO = + vcyclist = + vcar 3 and vS = − vcar , where vcar is the speed of the car. The observed frequency is ⎡ v + vcar 3 ⎛ v + vO ⎞ ⎢ = f fO = fS ⎜ S ⎝ v − vS ⎟⎠ ⎢⎣ v − − vcar

(

)

⎤ ⎛ v + vcar 3 ⎞ ⎥ = fS ⎜ ⎟, ⎥⎦ ⎝ v + vcar ⎠

giving ⎛ 343 m s + vcar 3 ⎞ 415 Hz = ( 440 Hz ) ⎜ ⎟ ⎝ 343 m s + vcar ⎠

56157_14_ch14_p426-456.indd 437

and

vcar = 32.0 m s

10/12/10 2:51:38 PM

438

14.27

Chapter 14

With the train approaching the stationary observer (vO = 0) at speed vt , the source velocity is vS = + vt and the observed frequency is ⎛ 343 m s ⎞ 465 Hz = fS ⎜ ⎟ ⎝ 343 m s − vt ⎠

[1]

As the train recedes, the source velocity is vS = − vt and the observed frequency is ⎛ 343 m s ⎞ 441 Hz = fS ⎜ ⎟ ⎝ 343 m s + vt ⎠

[2]

Dividing Equation [1] by [2] gives 465 343 m s + vt = 441 343 m s − vt and solving for the speed of the train yields vt = 9.09 m s . 14.28

(a)

We let the speed of the insect be vbug and the speed of the bat be vbat = 5.00 m s, and break the action into two steps. In the ﬁrst step, the bat is the sound source ﬂying toward the observer (the insect), so vS = + vbat , while the insect (observer) is ﬂying away from the source, making vO = − vbug . If f0 is the actual frequency sound emitted by the bat, the frequency detected (and reﬂected) by the moving insect is

(

⎡v+ − v ⎛ v + vO ⎞ bug ⎢ = f freflect = f0 ⎜ 0 ⎟ ⎢ v − + vbat ⎝ v − vS ⎠ ⎣

(

) ⎤⎥

or

) ⎥⎦

⎛ v − vbug freflect = f0 ⎜ ⎜⎝ v − vbat

⎞ ⎟ ⎟⎠

In the second step of the action, the insect acts as a sound source, reﬂecting a wave of frequency freflect back to the bat which acts as a moving observer. Since the source (insect) is moving away from the observer, vS = − vbug , and the observer (bat) is moving toward the source (insect) giving vO = + vbat . The frequency of the return sound received by the bat is then

(

⎡ v+ + v ⎛ v + vO ⎞ bat ⎢ = f freturn = freflect ⎜ reflect ⎟ ⎢ ⎝ v − vS ⎠ v − − vbug ⎣

(

) ⎤⎥

) ⎥⎦

or

⎛ v+ v bat freturn = freflect ⎜ ⎜⎝ v + vbug

⎞ ⎟ ⎟⎠

Combing the results of the two steps gives ⎛ v − vbug freturn = f0 ⎜ ⎜⎝ v − vbat

⎞⎛ v+ v bat ⎟⎜ ⎟⎠ ⎜⎝ v + vbug

⎞ ⎟ ⎟⎠

or ⎛ 343 m s − vbug 40.4 kHz = ( 40.0 kHz ) ⎜ ⎜⎝ 343 − 5.00

⎞ ⎛ 343 + 5.00 ⎟⎜ ⎟⎠ ⎜⎝ 343 m s + vbug

⎞ ⎟ ⎟⎠

This reduces to

(

⎛ 40.0 ⎞ ⎛ 348 ⎞ 343 m s + vbug = ⎜ 343 m s − vbug ⎝ 40.4 ⎟⎠ ⎜⎝ 338 ⎟⎠

)

continued on next page

56157_14_ch14_p426-456.indd 438

10/12/10 2:51:42 PM

Sound

439

or ⎡⎛ 40.0 ⎞ ⎛ 348 ⎞ ⎤ ⎡⎛ 40.0 ⎞ ⎛ 348 ⎞ ⎤ ⎢⎜⎝ 40.4 ⎟⎠ ⎜⎝ 338 ⎟⎠ + 1⎥ vbug = ( 343 m s ) ⎢⎜⎝ 40.4 ⎟⎠ ⎜⎝ 338 ⎟⎠ − 1⎥ ⎣ ⎦ ⎣ ⎦ and yields vbug = 3.29 m s . (b) 14.29

Yes, the bat is gaining on the insect at a rate of 5.00 m s − 3.29 m s = 1.71 m s .

For a source receding from a stationary observer, ⎛ v fO = fS ⎜ ⎜⎝ v − − vS

(

)

⎞ ⎛ v ⎞ ⎟ = fS ⎜ ⎟ ⎟⎠ ⎝ v + vS ⎠

Thus, the speed the falling tuning fork must reach is ⎛ f ⎞ ⎛ 512 Hz ⎞ vS = v ⎜ S − 1⎟ = ( 343 m s ) ⎜ − 1 = 19.1 m s ⎝ 485 Hz ⎟⎠ ⎝ fO ⎠ The distance it has fallen from rest before reaching this speed is vS2 − 0 (19.1 m s ) − 0 = = 18.6 m 2 ay 2 ( 9.80 m s 2 ) 2

Δy1 =

The time required for the 485 Hz sound to reach the observer is t=

Δy1 18.6 m = = 0.054 2 s 343 m s v

During this time the fork falls an additional distance Δy2 = vS t +

1 2 1 2 a y t = (19.1 m s ) ( 0.054 2 s ) + ( 9.80 m s 2 ) ( 0.054 2 s ) = 1.05 m 2 2

The total distance fallen before the 485 Hz sound reaches the observer is Δy = Δy1 + Δy2 = 18.6 m + 1.05 m = 19.7 m 14.30

(a)

⎛ 115 min ⎞ = 12.0 rad s w = 2p f = 2p ⎜ ⎝ 60.0 s min ⎟⎠ and for harmonic motion, vm ax = w A = (12.0 rad s ) (1.80 × 10 −3 m ) = 2.16 × 10 −2 m s

(b)

The heart wall is a moving observer (vO = + vm ax ) and the detector a stationary source, so the maximum frequency reﬂected by the heart wall is

(f ) wall

m ax

⎛ v + vm ax ⎞ ⎛ 1 500 + 0.021 6 ⎞ = ( 2 000 000 Hz ) ⎜ = fS ⎜ ⎟ ⎟⎠ = 2 000 029 Hz v 1 500 ⎝ ⎝ ⎠

continued on next page

56157_14_ch14_p426-456.indd 439

10/12/10 2:51:45 PM

440

Chapter 14

(c)

Now, the heart wall is a moving source (vS = + vm ax ) and the detector a stationary observer. The observed frequency of the returning echo is ⎛ ⎞ ⎞ ⎛ v 1 500 = 2 000 058 Hz = ( 2 000 029 Hz ) ⎜ fecho = ( fwall )m ax ⎜ ⎟ ⎝ 1 500 − 0.021 6 ⎟⎠ ⎝ v − vm ax ⎠

14.31

(a)

For a plane traveling at Mach 3.00, the half-angle of the conical wave front is

x

⎛v ⎞ ⎛ 1 ⎞ q = sin −1 ⎜ sound ⎟ = sin −1 ⎜ ⎝ 3.00 ⎟⎠ v ⎝ plane ⎠ The distance the plane has moved when the wave front reaches the observer is x = h tanq , or

t ⫽0

u h

h Observer hears Observer the “boom”

Observer Obser ver

a

20.0 km x= = 56.6 km tan ⎡⎣sin −1 (1 3.00 ) ⎤⎦

u

t ⫽?

b FIGURE P14.31

The time required for the plane to travel this distance, and hence the time when the shock wave reaches the observer, is t=

14.32

x vplane

=

x 56.6 × 10 3 m = = 56.3 s 3.00vsound 3.00 ( 335 m s )

(b)

The plane is 56.6 km farther along as computed above.

(a)

From Equation 14.12 in the textbook, ⎛ v + vO ⎞ fO = fS ⎜ ⎝ v − vS ⎟⎠ where fS is the frequency emitted by the source, fO is the frequency detected by the observer, v is the speed of the wave in the propagating medium, vO is the velocity of the observer relative to the medium, and vS is the velocity of the source relative to the propagating medium.

(b)

The yellow submarine is the source or emitter of the sound waves.

(c)

The red submarine is the observer or receiver of the sound waves.

(d)

The motion of the observer away from the source tends to increase the time observed between arrivals of successive pressure maxima. This effect tends to cause an increase in the observed period and a decrease in the observed frequency.

(e)

In this case, the sign of vO should be negative to decrease the numerator in Equation 14.12, and thereby decrease the calculated observed frequency.

(f)

The motion of the source toward the observer tends to decrease the time between the arrival of successive pressure maxima, decreasing the observed period , and increasing the observed frequency.

continued on next page

56157_14_ch14_p426-456.indd 440

10/12/10 2:51:48 PM

Sound

(g)

(h) 14.33

441

In this case, the sign of vS should be positive to decrease the denominator in Equation 14.12, and thereby increase the calculated observed frequency. ⎡ 1 533 m s + ( −3.00 m s ) ⎤ ⎛ v + vO ⎞ 3 = ( 5.27 × 10 3 Hz ) ⎢ fO = fS ⎜ ⎥ = 5.30 × 10 Hz ⎟ v − v 1 533 m s − +11.0 m s ( )⎦ ⎝ S ⎠ ⎣ x

The wavelength of the waves generated by the speakers is l=

v 343 m s = = 0.429 m f 800 Hz

1.25 m x

Speaker 1

Speaker 2

For the waves from the two speakers to interfere destructively at some point, the difference in the path lengths from the speakers to that point must be an odd multiple of a half-wavelength. Thus, along the line connecting the two speakers, destructive interference (and minima in amplitude) occur where

(1.25 m − x ) − x = ( 2n + 1) or where This gives:

l 2

x = 0.625 m − ( 2n + 1) n=0

where n is any integer

l ⎛ 2n + 1 ⎞ = 0.625 m − ⎜ ( 0.429 m ) ⎝ 4 ⎟⎠ 4

⇒ x = 0.518 m

n = −1 ⇒ x = 0.732 m

n = 1 ⇒ x = 0.303 m

n = −2

⇒ x = 0.947 m

n=2

n = −3

⇒ x = 1.16 m

⇒ x = 0.089 m

Thus, minima occur at distances of 0.089 m, 0.303 m, 0.518 m, 0.732 m, 0.947 m, and 1.16 m from either speaker. 14.34

14.35

v 343 m s The wavelength of the sound emitted by the speaker is l = = = 0.454 m, and a half f 756 Hz wavelength is l 2 = 0.227 m. (a)

If a condition of constructive interference currently exists, this can be changed to a case of destructive interference by adding a distance of l 2 = 0.227 m to the path length through the upper arm.

(b)

To move from a case of constructive interference to the next occurrence of constructive interference, one should increase the path length through the upper arm by a full wavelength, or by l = 0.454 m .

At point D, the distance of the ship from point A is = d 2 + (800 m ) = 2

800 m

B

( 600 m )2 + (800 m )2 = 1 000 m

Since destructive interference occurs for the ﬁrst time when the ship reaches D, it is necessary that − d = l 2, or l = 2 ( − d ) = 2 (1 000 m − 600 m ) = 800 m

56157_14_ch14_p426-456.indd 441

A

d

D

10/12/10 2:51:53 PM

The speakers emit sound of wavelength

Δy

v 343 m s l= = = 0.762 m f 450 Hz l 2 = 0.381 m

so

r1

1.50 m

14.36

Chapter 14

3.00 m

442

r2

Initially, Δy = 0, and r1 = r2 =

O

8.00 m

(1.50 m )2 + (8.00 m )2 = 8.14 m

To create destructive interference at point O, we move the top speaker upward distance Δy from its original location until we have r1 − r2 = l 2. Since this did not change r2, we must now have r1 = r2 + l 2 = 8.14 m + 0.381 m = 8.52 m But, after moving the speaker, this gives r1 =

14.37

(1.50 m + Δy ) + (8.00 m ) 2

2

or

(1.50 m + Δy )

Thus,

Δy = 8.59 m 2 − 1.50 m = 1.43 m

2

= 8.52 m

= (8.52 m ) − (8.00 m ) = 8.59 m 2 2

2

d

The wavelength of the sound is l= (a)

v 343 m s = = 0.500 m f 686 Hz

At the ﬁrst relative maximum (constructive interference),

r

x

r = x + l = x + 0.500 m Using the Pythagorean theorem, r 2 = x 2 + d 2, or

FIGURE P14.37 (modified)

( x + 0.500 m )2 = x 2 + ( 0.700 m )2 giving x = 0.240 m . (b)

At the ﬁrst relative minimum (destructive interference), r = x + l 2 = x + 0.250 m Therefore, the Pythagorean theorem yields

( x + 0.250 m )2 = x 2 + ( 0.700 m )2 or

56157_14_ch14_p426-456.indd 442

x = 0.855 m

10/12/10 2:51:57 PM

Sound

14.38

443

In the fundamental mode of vibration, the wavelength of waves in the wire is l = 2 L = 2 ( 0.700 0 m ) = 1.400 m If the wire is to vibrate at f = 261.6 Hz, the speed of the waves must be v = l f = (1.400 m )( 261.6 Hz ) = 366.2 m s The mass per unit length of the wire is m=

m 4.300 × 10 −3 kg = = 6.143 × 10 −3 kg m L 0.700 0 m

and the required tension is given by v = F m as F = v 2 m = ( 366.2 m s ) ( 6.143 × 10 −3 kg m ) = 823.8 N 2

14.39

In the third harmonic, the string of length L forms a standing wave of three loops, each of length l 2 = L 3. The wavelength of the wave is then

l/2

l/4

2L 16.0 m l= = ≈ 5.33 m 3 3 (a)

0

3l/2

l

3l /4

5l/4

The nodes in this string, ﬁxed at each end, will occur at distances of 0, l 2 = 2.67 m, l = 5.33 m, and 3l 2 = 8.00 m from the end. Antinodes occur halfway between each pair of adjacent nodes, or at distances of l 4 = 1.33 m, 3l 4 = 4.00 m, and 5l 4 = 6.67 m from the end.

(b)

The linear density is m=

m 40.0 × 10 −3 kg = = 5.00 × 10 −3 kg m L 8.00 m

and the wave speed is v=

F = m

49.0 N = 99.0 m s 5.00 × 10 −3 kg m

Thus, the frequency is 14.40

f=

v 99.0 m s = = 18.6 Hz l 5.33 m

In the fundamental mode, the distance from the ﬁnger of the cellist to the far end of the string is one-half of the wavelength for the transverse waves in the string. Thus, when the string resonates at 449 Hz, l = 2 ( 68.0 cm − 20.0 cm ) = 96.0 cm The speed of transverse waves in the string is therefore v = l f = ( 0.960 m )( 449 Hz ) = 431 m s continued on next page

56157_14_ch14_p426-456.indd 443

10/12/10 2:52:00 PM

444

Chapter 14

For a resonance frequency of 440 Hz, the wavelength would be v 431 m s = = 0.980 m = 98.0 cm f ′ 440 Hz

l′ =

To produce this tone, the cellist should position her ﬁnger at a distance of x= L−

l 98.0 cm = 68.0 cm − = 19.0 cm 2 2

from the nut. Thus, she should move her finger 1.00 cm toward the nut. 14.41

When the string vibrates in the ﬁfth harmonic (i.e., in ﬁve equal segments) at a frequency of f5 = 630 Hz, we have L = 5(l5 2), or the wavelength is l5 = 2L 5. The speed of transverse waves in the string is then v = l5 f5 = (2L 5) f5 For the string to vibrate in three segments (i.e., third harmonic), the wavelength must be such that L = 3(l3 2) or l3 = 2L 3. The new frequency would then be f3 =

14.42

v (2L 5) f5 3 3 = = f5 = ( 630 Hz ) = 378 Hz l3 5 2L 3 5

If a wire of length is ﬁxed at both ends, the wavelength of the fundamental mode of vibration is l1 = 2 . The speed of transverse waves in the wire is v = F m , where F is the tension in the wire and m is the mass per unit length of the wire. The fundamental frequency for the wire is then f1 =

v 1 = l1 2

F 1 ⎛ 1⎞ = ⎜ ⎟ m 2⎝ ⎠

F m

If we have two wires with the same mass per unit length, one of length L and under tension F while the second has length 2L and tension 4F, the ratio of the fundamental frequencies of the two wires is f1, long f1, short

=

1 2

(1 2L ) (1 L ) 1 2

4F m 1 = 4 =1 2 F m

or the two wires have the same fundamental frequency of vibration. If this frequency is f1 = 60 Hz, then the frequency of the second harmonic for both wires is f2 = 2 f1 = 2 ( 60 Hz ) = 120 Hz 14.43

m=

m 25.0 × 10 −3 kg = = 1.85 × 10 −2 kg m L 1.35 m

(a)

The linear density is

(b)

In a string ﬁxed at both ends, the fundamental mode has a node at each end and a single antinode in the center, so that L = l 2, or l = 2L = 2 (1.10 m ) = 2.20 m. Then, the desired wave speed in the wire is v = l f = ( 2.20 m )( 41.2 Hz ) = 90.6 m s

(c)

The speed of transverse waves in a string is v = F m , so the required tension is F = m v 2 = (1.85 × 10 −2 kg m ) ( 90.6 m s ) = 152 N 2

continued on next page

56157_14_ch14_p426-456.indd 444

10/12/10 2:52:02 PM

Sound

(d)

l = 2L = 2 (1.10 m ) = 2.20 m

(e)

The wavelength of the longitudinal sound waves produced in air by the vibrating string is lair =

14.44

445

(a)

[See part (b) above.]

vair 343 m s = = 8.33 m 41.2 Hz f

A string ﬁxed at each end forms standing wave patterns with a node at each end and an integer number of loops, each loop of length l 2, with an antinode at its center. Thus, L = n ( l 2 ) or l = 2L n. If the string has tension T and mass per unit length m, the speed of transverse waves is v = l f = T m . Thus, when the string forms a standing wave of n loops (and hence n antinodes), the frequency of vibration is f=

(b)

T m v n = = l 2L n 2L

⇒

fA =

nA 2L A

TA mA

Assume the length is doubled, LB = 2L A, and a new standing wave is formed having nB = n A and TB = TA . Then nB 2LB

fB = (c)

T m

nA TB = m A 2 ( 2L A )

1⎛ n TA = ⎜ A m A 2 ⎝ 2L A

TA ⎞ f = A ⎟ mA ⎠ 2

Solving the general result obtained in part (a) for the tension in the string gives T = 4m f 2 L2 n 2. Thus, if fB = f A, LB = L A, and nB = n A + 1, we ﬁnd 2

⎛ 4m A f A2 L2A ⎞ ⎛ n ⎞ 4m A fB2 L2B 4m A f A2 L2A n A2 n A2 TB = = = TA = ⎜ A ⎟ TA = 2 2 ⎜ 2 2 2 ⎟ nB n ⎝ nA + 1⎠ ⎠ ( n A + 1) ( n A + 1) ( n A + 1) ⎝ A (d)

If now we have fB = 3 f A, LB = L A 2, and nB = 2n A, then TB =

14.45

(a)

2 2 4m A fB2 L2B 4m A ( 9 f A ) ( L A 4 ) 9 ⎛ 4m A f A2 L2A ⎞ 9 = = ⎜ ⎟⎠ = 16 TA 16 ⎝ nB2 n A2 ( 4nA2 )

or

TB 9 = TA 16

From the sketch at the right, notice that when d = 2.00 m, L=

5.00 m − d = 1.50 m, 2

d L

and ⎛ d / 2⎞ q = sin −1 ⎜ = 41.8° ⎝ L ⎟⎠ Then evaluating the net vertical force on the lowest bit of string, gives the tension in the string as F= (b)

→

F

q

→

F →

mg

ΣFy = 2F cosq − mg = 0

(12.0 kg) (9.80 m s2 ) mg = = 78.9 N 2 cosq 2 cos ( 41.8° )

The speed of transverse waves in the string is v=

F = m

78.9 N = 2.81× 10 2 m s 0.001 00 kg m continued on next page

56157_14_ch14_p426-456.indd 445

10/12/10 2:52:06 PM

446

Chapter 14

For the pattern shown, 3 ( l 2 ) = d, so l =

2d 4.00 m = 3 3

Thus, the frequency is f= 14.46

(a)

2 v 3 ( 2.81× 10 m s ) = = 2.11× 10 2 Hz l 4.00 m

For a standing wave of 6 loops, 6 ( l 2 ) = L, or l = L 3 = (2.0 m) 3. The speed of the waves in the string is then ⎛ 2.0 m ⎞ v=l f =⎜ (150 Hz ) = 1.0 × 10 2 m s ⎝ 3 ⎟⎠ Since the tension in the string is F = mg = ( 5.0 kg ) ( 9.80 m s 2 ) = 49 N, v = F m gives m=

(b)

F 49 N = = 4.9 × 10 −3 kg m 2 v 2 (1.0 × 10 2 m )

If m = 45 kg, then F = ( 45 kg ) ( 9.80 m s 2 ) = 4.4 × 10 2 N, and 4.4 × 10 2 N = 3.0 × 10 2 m s 4.9 × 10 −3 kg m

v=

Thus, the wavelength will be l=

v 3.0 × 10 2 m s = = 2.0 m f 150 Hz

and the number of loops is (c)

n=

2.0 m L = = 2 l 2 1.0 m

If m = 10 kg, the tension is F = (10 kg ) ( 9.80 m s 2 ) = 98 N, and v= Then, l =

and n =

98 N = 1.4 × 10 2 m s 4.9 × 10 −3 kg m v 1.4 × 10 2 m s = = 0.93 m f 150 Hz

2.0 m L = is not an integer, l 2 0.47 m

so no standing wave will form . 14.47

The speed of transverse waves in the string is v=

F 50.000 N = = 70.711 m s m 1.000 0 × 10 −2 kg m

continued on next page

56157_14_ch14_p426-456.indd 446

10/12/10 2:52:11 PM

Sound

447

The fundamental wavelength is l 1 = 2 L = 1.200 0 m and its frequency is f1 =

v 70.711 m s = = 58.926 Hz l1 1.200 0 m

The harmonic frequencies are then fn = nf1 = n ( 58.926 Hz ), with n being an integer The largest one under 20 000 Hz is f339 = 19 976 Hz = 19.976 kHz . 14.48

The distance between adjacent nodes (and between adjacent antinodes) is one-quarter of the circumference. d NN = d AA = so and

l 20.0 cm = = 5.00 cm 2 4

l = 10.0 cm = 0.100 m, f=

v 900 m s = = 9.00 × 10 3 Hz = 9.00 kHz l 0.100 m

The singer must match this frequency quite precisely for some interval of time to feed enough energy into the glass to crack it. 14.49

Assuming an air temperature of T = 37°C = 310 K, the speed of sound inside the pipe is v = ( 331 m s )

TK 310 = ( 331 m s ) = 353 m s 273 K 273

In the fundamental resonant mode, the wavelength of sound waves in a pipe closed at one end is l = 4L. Thus, for the whooping crane l = 4 ( 5.0 ft ) = 2.0 × 101 ft and 14.50

(a)

f=

v ( 353 m s ) ⎛ 3.281 ft ⎞ = ⎜ ⎟ = 58 Hz l 2.0 × 101 ft ⎝ 1 m ⎠

In the fundamental resonant mode of a pipe open at both ends, the distance between antinodes is d AA = l 2 = L. Thus, l = 2 L = 2 ( 0.320 m ) = 0.640 m

(b) 14.51

f=

d AA =

l 1 ⎛ v ⎞ 1 ⎛ 343 m s ⎞ = 4.29 × 10 −2 m = 4.29 cm = = 2 2 ⎜⎝ f ⎟⎠ 2 ⎜⎝ 4 000 Hz ⎟⎠

Hearing would be best at the fundamental resonance, so l = 4L = 4 ( 2.8 cm ) and

56157_14_ch14_p426-456.indd 447

v 343 m s = = 536 Hz l 0.640 m

and

f=

v 343 m s ⎛ 100 cm ⎞ 3 = ⎜ ⎟ = 3.1× 10 Hz = 3.1 kHz l 4 ( 2.8 cm ) ⎝ 1 m ⎠

10/12/10 2:52:14 PM

448

14.52

14.53

Chapter 14

(a)

To form a standing wave in the tunnel, open at both ends, one must have an antinode at each end, a node at the middle of the tunnel, and the length of the tunnel must be equal to an integral number of half-wavelengths [L = n ( l 2 ) or l = 2L n]. The resonance frequencies of the tunnel are then vsound ⎛ ⎞ 343 m s 343 m s in air fn = = = n⎜ ⎟ = n ( 0.085 8 Hz ) n = 1, 2, 3, … 3 2L n ln ⎝ 2 ( 2.00 × 10 m ) ⎠

(b)

It would be good to make such a rule. Any car horn would produce several closely spaced resonance frequencies of the air in the tunnel, so the sound would be greatly ampliﬁed. Other drivers might be frightened directly into dangerous behavior or might blow their horns also.

(a)

The fundamental wavelength of the pipe open at both ends is l1 = 2 L = v f1. Since the speed of sound is 331 m s at 0°C, the length of the pipe is L=

(b)

v 331 m s = = 0.552 m 2 f1 2 ( 300 Hz )

At T = 30°C = 303 K, v = ( 331 m s )

TK 303 = ( 331 m s ) = 349 m s 273 273

and f1 = 14.54

(a)

Observe from Equations 14.18 and 14.19 in the textbook that the difference between successive resonance frequencies is constant, regardless of whether the pipe is open at both ends or is closed at one end. Thus, the resonance frequencies of 650 Hz or less for this pipe must be 650 Hz, 550 Hz, 450 Hz, 350 Hz, 250 Hz, 150 Hz, and 50.0 Hz, with the lowest or fundamental frequency being f1 = 50.0 Hz .

(b)

Note, from the list given above, the resonance frequencies are only the odd multiples of the fundamental frequency. This is a characteristic of a pipe that is open at only one end and closed at the other.

(c)

The length of a pipe with an antinode at the open end and a node at the closed end is onequarter of the wavelength of the fundamental frequency, so the length of this pipe must be L=

14.55

v v 349 m s = = = 316 Hz l1 2 L 2 ( 0.552 m )

l1 vsound 343 m s = = = 1.72 m 4 ( 50.0 Hz ) 4 4 f1

In a string ﬁxed at both ends, the length of the string is equal to a half-wavelength of the fundamental resonance frequency, so l1 = 2L. The fundamental frequency may then be written as f1 =

v 1 = l1 2L

F = m

F 4L2 m

If a second identical string with tension F ′ < F is struck, the fundamental frequency of vibration would be f1′ =

⎛ F ⎞ F′ F′ F′ = ⎜ 2 ⎟ = f1 4L m F 4L2 m F ⎝ ⎠

continued on next page

56157_14_ch14_p426-456.indd 448

10/12/10 2:52:17 PM

Sound

449

When the two strings are sounded together, the beat frequency heard will be ⎛ ⎛ 5.40 × 10 2 N ⎞ F′ ⎞ 2 = 1.10 × 10 fbeat = f1 − f1′ = f1 ⎜ 1− Hz 1− ( ) ⎜ ⎟ = 5.64 beats s 6.00 × 10 2 N ⎠ F ⎟⎠ ⎝ ⎝ 14.56

By shortening her string, the second violinist increases its fundamental frequency. Thus, f1′ = f1 + fbeat = (196 + 2.00 ) Hz = 198 Hz. Since the tension and the linear density are both identical for the two strings, the speed of transverse waves, v = F m , has the same value for both strings. Therefore, l′1 f1′ = l1 f1, or l′1 = l1 ( f1 f1′) . The fundamental wavelength of a string ﬁxed at both ends is l = 2L, and this yields ⎛ f ⎞ ⎛ 196 ⎞ = 29.7 cm L ′ = L ⎜ 1 ⎟ = ( 30.0 cm ) ⎜ ⎝ 198 ⎟⎠ ⎝ f1′⎠

14.57

The commuter, stationary relative to the station and the ﬁrst train, hears the actual source frequency fO,1 = fS = 180 Hz from the ﬁrst train. The frequency the commuter hears from the second train, moving relative to the station and commuter, is given by

(

)

fO,2 = fS ± fbeat = 180 Hz ± 2 Hz = 178 Hz or 182 Hz

(

)

This stationary observer ( vO = 0 ) hears the lower frequency fO,2 = 178 Hz if the second train is moving away from the station vS = − vS , so fO = fS [(v + v O ) (v − vS )] gives the speed of the receding second train as

(

)

⎛ 343 m s + 0 178 Hz = (180 Hz ) ⎜ ⎜⎝ 343 m s − − vS

(

)

⎞ ⎛ 343 m s + 0 ⎞ ⎟ = (180 Hz ) ⎜ ⎟ ⎟⎠ ⎝ 343 m s + vS ⎠

or ⎛ 180 Hz ⎞ 343 m s + vS = ( 343 m s ) ⎜ ⎝ 178 Hz ⎟⎠

and

⎡⎛ 180 Hz ⎞ ⎤ vS = ( 343 m s ) ⎢⎜ ⎟ − 1⎥ = 3.85 m s ⎣⎝ 178 Hz ⎠ ⎦

so one possibility for the second train is vS = 3.85 m s away from the station .

( )

)

The other possibility is that the second train is moving toward the station vS = + vS and the commuter is detecting the higher of the possible frequencies fO,2 = 182 Hz . In this case, fO = fS [(v + v O ) (v − vS )] yields

(

⎛ 343 m s + 0 ⎞ 182 Hz = (180 Hz ) ⎜ ⎟ ⎝ 343 m s − vS ⎠ or

and

⎛ 180 Hz ⎞ 343 m s − vS = ( 343 m s ) ⎜ ⎝ 182 Hz ⎟⎠

⎡ ⎛ 180 Hz ⎞ ⎤ vS = ( 343 m s ) ⎢1− ⎜ ⎟ ⎥ = 3.77 m s ⎣ ⎝ 182 Hz ⎠ ⎦

In this case, the velocity of the second train is vS = 3.77 m s toward the station . 14.58

The temperatures of the air in the two pipes are T1 = 27°C = 300 K and T2 = 32°C = 305 K. The speed of sound in the two pipes is v1 = ( 331 m s )

T1 273 K

and v2 = ( 331 m s )

T2 273 K

Since the pipes have the same length, the fundamental wavelength, l = 4L, is the same for them. Thus, from f = v l , the ratio of their fundamental frequencies is seen to be f2 f1 = v2 v1, which gives f2 = f1 ( v2 v1 ). continued on next page

56157_14_ch14_p426-456.indd 449

10/12/10 2:52:20 PM

450

Chapter 14

The beat frequency produced is then ⎛ T ⎞ ⎛v ⎞ fbeat = f2 − f1 = f1 ⎜ 2 − 1⎟ = f1 ⎜ 2 − 1⎟ ⎝ v1 ⎠ ⎝ T1 ⎠ or 14.59

(a)

⎛ 305 K ⎞ fbeat = ( 480 Hz ) ⎜ − 1⎟ = 3.98 Hz ⎝ 300 K ⎠ First consider the wall a stationary observer receiving sound from an approaching source having velocity va. The frequency received and reﬂected by the wall is freflect = fS [v (v − va )]. Now consider the wall as a stationary source emitting sound of frequency freflect to an observer approaching at velocity va . The frequency of the echo heard by the observer is ⎛ v ⎞ ⎛ v + va ⎞ ⎛ v + va ⎞ ⎛ v + va ⎞ = fS ⎜ fecho = freflect ⎜ ⎟⎠ = fS ⎜ ⎜⎝ ⎟ ⎝ v ⎟⎠ v ⎝ v − va ⎠ ⎝ v − va ⎟⎠ Thus, the beat frequency between the tuning fork and its echo is ⎛ 2va ⎞ ⎛ v + va ⎞ ⎛ 2 (1.33) ⎞ = ( 256 Hz ) ⎜ fbeat = fecho − fS = fS ⎜ − 1⎟ = fS ⎜ = 1.99 Hz ⎟ ⎝ 343 − 1.33 ⎟⎠ ⎝ v − va ⎠ ⎝ v − va ⎠

(b)

When the student moves away from the wall, va changes sign so the frequency of the echo heard is fecho = fS [(v − va ) (v + va )]. The beat frequency is then ⎛ ⎛ 2 va ⎞ v − va ⎞ = fS ⎜ fbeat = fS − fecho = fS ⎜ 1− ⎟ ⎟ v + va ⎠ ⎝ ⎝ v + va ⎠ giving

va =

v fbeat 2 fS − fbeat

The receding speed needed to observe a beat frequency of 5.00 Hz is va = 14.60

(343 m s )(5.00 Hz ) = 3.38 m s 2 ( 256 Hz ) − 5.00 Hz

The extra sensitivity of the ear at 3 000 Hz appears as downward dimples on the curves in Figure 14.29 of the textbook. At T = 37°C = 310 K, the speed of sound in air is v = ( 331 m s )

TK 310 = 353 m s = ( 331 m s ) 273 273

Thus, the wavelength of 3 000 Hz sound is l=

v 353 m s = = 0.118 m f 3 000 Hz

For the fundamental resonant mode in a pipe closed at one end, the length required is L=

56157_14_ch14_p426-456.indd 450

l 0.118 m = = 0.0295 m = 2.95 cm 4 4

10/12/10 2:52:25 PM

Sound

14.61

451

At normal body temperature of T = 37.0°C, the speed of sound in air is v = ( 331 m s ) 1+

TC 37.0 = ( 331 m s ) 1+ 273 273

and the wavelength of sound having a frequency of f = 20 000 Hz is v (331 m s ) 1+ 37.0 = 1.76 × 10 −2 m = 1.76 cm = f ( 20 000 Hz ) 273

l=

Thus, the diameter of the eardrum is 1.76 cm . 14.62

From the deﬁning equation for the decibel level, b = 10 log(I I 0 ), the intensity of sound having a decibel level b is

(

I = 10 b

10

)I

0

Thus, the intensity of a 40 dB sound is I 40 = (10 4.0 )I 0, while that of a 70 dB sound is I 70 = (10 7.0 )I 0. Since the combined intensity of sound from a swarm of n mosquitoes is I swarm = nI 40, we must require that I swarm = nI 40 = I 70 or

n=

7.0 I 70 (10 ) I 0 = = 10 3.0 = 1 000 I 40 (10 4.0 ) I 0

We conclude that the swarm should contain ∼ 1 000 mosquitoes to yield a 70 dB sound. 14.63

(a)

With a decibel level of 103 dB, the intensity of the sound at 1.60 m from the speaker is found from b = 10 ⋅ log ( I I 0 ) as I = I 0 ⋅10 b

10

= (1.00 × 10 −12 W m 2 ) ⋅1010.3 = 1.00 × 10 −1.7 W m 2

If the speaker broadcasts equally well in all directions, the intensity (power per unit area) at 1.60 m from the speaker is uniformly distributed over a spherical wave front of radius r = 1.60 m centered on the speaker. Thus, the power radiated is P = IA = I ( 4p r 2 ) = (1.00 × 10 −1.7 W m 2 ) 4p (1.60 m ) = 0.642 W 2

14.64

Poutput

efficiency =

(a)

At point C, the distance from speaker A is rA =

Pinput

=

0.642 W = 0.004 3 or 0.43% 150 W

(b)

C

( 3.00 m )2 + ( 4.00 m )2 = 5.00 m

and the intensity of the sound from this speaker is IA =

PA 1.00 × 10 −3 W = 2 4p rA2 4p ( 5.00 m )

rA

4.00 m

rB

A

= 3.18 × 10 −6 W m 2

B 3.00 m

2.00 m

The sound level at C due to speaker A alone is then ⎛I ⎞ ⎛ 3.18 × 10 −6 W m 2 ⎞ = 65.0 dB b A = 10 ⋅ log ⎜ A ⎟ = 10 ⋅ log ⎜ ⎝ 1.00 × 10 −12 W m 2 ⎟⎠ ⎝ I0 ⎠ continued on next page

56157_14_ch14_p426-456.indd 451

10/12/10 2:52:28 PM

452

Chapter 14

(b)

The distance from point C to speaker B is rB = ( 2.00 m ) + ( 4.00 m ) = 4.47 m and the intensity of the sound from this speaker alone is 2

IB =

2

PB 1.50 × 10 −3 W = = 5.97 × 10 −6 W m 2 2 4p rB2 4p ( 4.47 m )

The sound level at C due to speaker B alone is therefore ⎛I ⎞ ⎛ 5.97 × 10 −6 W m 2 ⎞ = 67.8 dB b B = 10 ⋅ log ⎜ B ⎟ = 10 ⋅ log ⎜ ⎝ 1.00 × 10 −12 W m 2 ⎟⎠ ⎝ I0 ⎠ (c)

If both speakers are sounded together, the total sound intensity at point C is I AB = I A + I B = 3.18 × 10 −6 W m 2 + 5.97 × 10 −6 W m 2 = 9.15 × 10 −6 W m 2 and the total sound level in decibels is ⎛I ⎞ ⎛ 9.15 × 10 −6 W m 2 ⎞ = 69.6 dB b AB = 10 ⋅ log ⎜ AB ⎟ = 10 ⋅ log ⎜ ⎝ 1.00 × 10 −12 W m 2 ⎟⎠ ⎝ I0 ⎠

14.65

We assume that the average intensity of the sound is directly proportional to the number of cars passing each minute. If the sound level in decibels is b = 10 ⋅ log ( I I 0 ), the intensity of the sound is I = I 0 ⋅10 b 10, so the average intensity in the afternoon, when 100 cars per minute are passing, is I100 = I 0 ⋅1080.0 10 = (1.00 × 10 −12 W m 2 ) ⋅108.00 = 1.00 × 10 −4 W m 2 The expected average intensity at night, when only 5 cars pass per minute, is given by the ratio I 5 I100 = 5 100 =1 20, or I5 =

I100 1.00 × 10 −4 W m 2 = = 5.00 × 10 −6 W m 2 20 20

and the expected sound level in decibels is ⎛I ⎞ ⎛ 5.00 × 10 −6 W m 2 ⎞ = 67.0 dB b 5 = 10 ⋅ log ⎜ 5 ⎟ = 10 ⋅ log ⎜ ⎝ 1.00 × 10 −12 W m 2 ⎟⎠ ⎝ I0 ⎠ 14.66

The well will act as a pipe closed at one end (the bottom) and open at the other (the top). The resonant frequencies are the odd integer multiples of the fundamental frequency, or fn = ( 2n − 1) f1 , where n = 1, 2, 3, …. Thus, if fn and fn +1 are two successive resonant frequencies, their difference is

(

)

fn +1 − fn = [ 2 ( n + 1) − 1] f1 − ( 2n − 1) f1 = 2n + 2 − 1 − 2n + 1 f1 = 2 f1 In this case, we have 60.0 Hz − 52.0 Hz = 2 f1, giving the fundamental frequency for the well as f1 = 4.00 Hz. In the fundamental mode, the well (pipe closed at one end) forms a standing wave pattern with a node at the bottom and the ﬁrst antinode at the top, making the depth of the well d=

56157_14_ch14_p426-456.indd 452

l1 1 ⎛ vsound ⎞ 1 ⎛ 343 m s ⎞ = = ⎜ ⎟ = 21.4 m 4 4 ⎜⎝ f1 ⎟⎠ 4 ⎝ 4.00 Hz ⎠

10/12/10 2:52:31 PM

Sound

14.67

453

If r1 and r2 are the distances of the two observers from the speaker, the intensities of the sound at their locations are I1 =

P 4p r12

I2 =

and

P 4p r22

where P is the power output of the speaker. The difference in the decibel levels for the two observers is 2

⎛I ⎞ ⎛I ⎞ ⎛I ⎞ ⎛ r2 ⎞ ⎛r ⎞ ⎛r ⎞ b1 − b 2 = 10 log ⎜ 1 ⎟ − 10 log ⎜ 2 ⎟ = 10 log ⎜ 2 ⎟ = 10 log ⎜ 22 ⎟ = 10 log ⎜ 2 ⎟ = 20 log ⎜ 2 ⎟ ⎝ I1 ⎠ ⎝ r1 ⎠ ⎝ r1 ⎠ ⎝ r1 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ Since b1 = 80 dB and b 2 = 60 dB, we ﬁnd that 80 − 60 = 20 log(r2 r1 ). This yields log(r2 r1 ) = 1.0 We also know that

and

r2 r1 = 10

r2 = 10r1

[1]

r1 + r2 = 36.0 m

Substituting Equation [1] into [2] gives: Then, Equation [2] yields 14.68

or

[2] 11r1 = 36.0 m

or

r1 = 36.0 m 11 ≈ 3.3 m

r2 = 36.0 m − r1 = 36.0 m − 3.3 m = 32.7 m

We take toward the east as the positive direction, so the velocity of the sea water relative to Earth is v W E = −10.0 km h. The velocity of the trailing ship, which is the sound source (S), relative the propagation medium (sea water) is then ⎛ 0.278 m s ⎞ = 20.6 m s v SW = v SE − v W E = + 64.0 km h − ( −10.0 km h ) = +74.0 km h ⎜ ⎝ 1 km h ⎟⎠ The velocity of the leading ship, the observer (O) in this case, relative to the water is ⎛ 0.278 m s ⎞ = 15.3 m s v OW = v OE − v W E = + 45.0 km h − ( −10.0 km h ) = +55.0 km h ⎜ ⎝ 1 km h ⎟⎠ With the source moving toward the observer, but the observer moving away from the source, the frequency detected by the observer is given by Equation 14.12 as ⎛ v + (− v OW ) ⎞ ⎛ v + vO ⎞ = fS ⎜ f0 = fS ⎜ ⎟ ⎟ ⎝ v − vS ⎠ ⎝ v − (+ v SW ) ⎠ The speed of sound in sea water is v = 1 533 m s (Table 14.1) and the frequency emitted by the source is fS = 1 200.0 Hz, so the observed frequency should be ⎛ 1 533 m s − 15.3 m s ⎞ = 1 204 Hz f0 = (1 200.0 Hz ) ⎜ ⎝ 1 533 m s − 20.6 m s ⎟⎠

14.69

This situation is very similar to the fundamental resonance of an organ pipe that is open at both ends. The wavelength of the standing waves in the crystal is l = 2 t, where t is the thickness of the crystal, and the frequency is f=

56157_14_ch14_p426-456.indd 453

v 3.70 × 10 3 m s = = 2.62 × 10 5 Hz = 262 kHz l 2 ( 7.05 × 10 −3 m )

10/12/10 2:52:34 PM

454

14.70

Chapter 14

The distance from the window ledge to the man’s head is Δy = d − h = 20.0 m − 1.75 m = 18.3 m The time for a warning to travel this distance is t1 = (18.3 m) (343 m s) = 0.0534 s, so the total time needed to receive the warning and react is t1′ = t1 + 0.300 s = 0.353 s. The elapsed time when the pot, starting from rest, reaches the level of the man’s head is t2 =

2 ( Δy ) = ay

2 ( −18.3 m ) = 1.93 s −9.80 m s 2

Thus, the latest the warning should be sent is at t = t2 − t1′ = 1.93 s − 0.353 s = 1.58 s into the fall. At this time, the pot has fallen Δy =

1 2 1 2 g t = ( 9.80 m s 2 )(1.58 s ) = 12.2 m 2 2

and is 20.0 m − 12.2 m = 7.8 m above the sidewalk. 14.71

On the weekend, there are one-fourth as many cars passing per minute as on a weekday. Thus, the intensity, I 2 , of the sound on the weekend is one-fourth of that, I1, on a weekday. The difference in the decibel levels is therefore ⎛I ⎞ ⎛I ⎞ ⎛I ⎞ b1 − b 2 = 10 log ⎜ 1 ⎟ − 10 log ⎜ 2 ⎟ = 10 log ⎜ 1 ⎟ = 10 log(4) = 6 dB ⎝ I2 ⎠ ⎝ Io ⎠ ⎝ Io ⎠ so, b 2 = b1 − 6 dB = 70 dB − 6 dB = 64 dB

14.72

(a)

At T = 20°C = 293 K, the speed of sound in air is v = ( 331 m s ) 1+

TC 20.0 = 343 m s = ( 331 m s ) 1+ 273 273

The ﬁrst harmonic or fundamental of the ﬂute (a pipe open at both ends) is given by l1 = 2L =

v 343 m s = = 1.31 m f1 261.6 Hz

Therefore, the length of the ﬂute is L= (b)

λ1 1.31 m = = 0.655 m 2 2

In the colder room, the length of the ﬂute, and hence its fundamental wavelength, is essentially unchanged (that is, l1′ = l1 = 1.31 m). However, the speed of sound, and thus the frequency of the fundamental, will be lowered. At this lower temperature, the frequency must be f1′ = f1 − fbeat = 261.6 Hz − 3.00 Hz = 258.6 Hz The speed of sound in this room is v ′ = l1′ f1′ = (1.31 m )( 258.6 Hz ) = 339 m s From v = ( 331 m s ) 1+ TC 273, the temperature in the colder room is given by 2 ⎡⎛ ⎤ ⎡⎛ 339 m s ⎞ 2 ⎤ ⎞ v T = ( 273°C ) ⎢⎜ − 1 = 273°C ( ) ⎥ ⎢⎜ ⎟ − 1⎥ = 13.4°C ⎟ ⎢⎣⎝ 331 m s ⎠ ⎢⎣⎝ 331 m s ⎠ ⎦⎥ ⎦⎥

56157_14_ch14_p426-456.indd 454

10/12/10 2:52:38 PM

Sound

14.73

The maximum speed of the oscillating block and speaker is k 20.0 N m = ( 0.500 m ) = 1.00 m s m 5.00 kg

vm ax = Aw = A (a)

When the speaker moves away from the stationary observer, the source velocity is vS = −vm ax and the minimum frequency heard is

(f ) O

(b)

m in

⎛ ⎞ ⎛ v ⎞ 343 m s = 439 Hz = ( 440 Hz ) ⎜ = fS ⎜ ⎝ 343 m s + 1.00 m s ⎟⎠ ⎝ v + vm ax ⎟⎠

When the speaker (sound source) moves toward the stationary observer, then vS = +vm ax and the maximum frequency heard is

(f ) O

14.74

455

m ax

⎛ ⎞ ⎛ v ⎞ 343 m s = 441 Hz = ( 440 Hz ) ⎜ = fS ⎜ ⎟ ⎝ 343 m s − 1.00 m s ⎟⎠ ⎝ v − vm ax ⎠

The speed of transverse waves in the wire is F = m

vT =

F⋅L = m

( 400 N )( 0.750 m ) 2.25 × 10 −3 kg

= 365 m s

When the wire vibrates in its third harmonic, l = 2L 3 = 0.500 m, so the frequency of the vibrating wire and the sound produced by the wire is fS =

vT 365 m s = = 730 Hz 0.500 m l

Since both the wire and the wall are stationary, the frequency of the wave reﬂected from the wall matches that of the waves emitted by the wire. Thus, as the student approaches the wall at speed vO , he approaches one stationary source and recedes from another stationary source, both emitting frequency fS = 730 Hz. The two frequencies that will be observed are

(f ) O

1

⎛ v + vO ⎞ ⎛ v − vO ⎞ and ( fO )2 = fS ⎜ = fS ⎜ ⎟ ⎟ ⎝ v ⎠ ⎝ v ⎠

The beat frequency is

so 14.75

(

⎛ v + vO − v − vO fbeat = ( fO )1 − ( fO )2 = fS ⎜ ⎜⎝ v

) ⎞⎟ = 2 f ⎟⎠

S

vO

v

⎛ f ⎞ ⎡ 8.30 Hz ⎤ vO = ⎜ beat ⎟ v = ⎢ ⎥ ( 343 m s ) = 1.95 m s ⎝ 2 fS ⎠ ⎣ 2 ( 730 Hz ) ⎦

The speeds of the two types of waves in the rod are vlong =

Y = r

Y Y ( A⋅ L) = and vtrans = mV m

Thus, if vlong = 8 vtrans , we have

F = m

F = m L

F⋅L m

Y ( A⋅ L) ⎛ F⋅L⎞ = 64 ⎜ , or the required tension is ⎝ m ⎟⎠ m

2 10 2 ⎡ −2 ⎤ Y ⋅ A ( 6.80 × 10 N m ) ⎣p ( 0.200 × 10 m ) ⎦ = 1.34 × 10 4 N F= = 64 64

56157_14_ch14_p426-456.indd 455

10/12/10 2:52:41 PM

456

14.76

Chapter 14

(a)

For the fundamental mode of a pipe open at both ends, L = l1 2 or the wavelength of the waves traveling through the air in the pipe is l1 = 2L = 2(0.500 m) = 1.00 m. If the frequency of this fundamental mode is f1 = 350 Hz, the speed of sound waves within the pipe must be v = λ1 f1 = (1.00 m)(350 Hz) = 350 m s From v = (331 m s) 1+ TC 273, the Celsius temperature of the air in the pipe is 2 ⎤ ⎡⎛ ⎡⎛ 350 m s ⎞ 2 ⎤ ⎞ v TC = ( 273°C ) ⎢⎜ − 1 = 273°C ( ) ⎥ ⎢⎜ ⎟ − 1⎥ = 32.2°C ⎟ ⎢⎣⎝ 331 m s ⎠ ⎥⎦ ⎢⎣⎝ 331 m s ⎠ ⎥⎦

(b)

If the temperature rises to T ′ = T + 20.0°C = 52.2°C, the speed of sound in the air will be v ′ = (331 m s) 1+ TC′ 273 = (331 m s) 1+ 52.2 273, and the new length of the pipe will be L ′ = L0 [1+ a (ΔT )] = (0.500 m)[1 + (19 × 10 −6 °C−1 )(20.0°C)] The new fundamental wavelength is l1′ = 2 L ′, and the new fundamental resonance frequency will be f1′ =

56157_14_ch14_p426-456.indd 456

(331 m s) 1+ 52.2 273 v′ = = 3.6 × 10 2 Hz l1′ 2(0.500 m)[1 + (19 × 10 −6 °C−1 )(20.0°C)]

10/12/10 2:52:44 PM

15 Electric Forces and Electric Fields QUICK QUIZZES 1.

Choice (b). Object A must have a net charge because two neutral objects do not attract each other. Since object A is attracted to positively-charged object B, the net charge on A must be negative.

2.

Choice (b). By Newton’s third law, the two objects will exert forces having equal magnitudes but opposite directions on each other.

3.

Choice (c). The electric ﬁeld at point P is due to charges other than the test charge. Thus, it is unchanged when the test charge is altered. However, the direction of the force this ﬁeld exerts on the test change is reversed when the sign of the test charge is changed.

4.

Choice (a). If a test charge is at the center of the ring, the force exerted on the test charge by charge on any small segment of the ring will be balanced by the force exerted by charge on the diametrically opposite segment of the ring. The net force on the test charge, and hence the electric ﬁeld at this location, must then be zero.

5.

Choices (c) and (d). The electron and the proton have equal magnitude charges of opposite signs. The forces exerted on these particles by the electric ﬁeld have equal magnitude and opposite directions. The electron experiences an acceleration of greater magnitude than does the proton because the electron’s mass is much smaller than that of the proton.

6.

Choice (a). The ﬁeld is greatest at point A because this is where the ﬁeld lines are closest together. The absence of lines at point C indicates that the electric ﬁeld there is zero.

7.

Choice (c). When a plane area A is in a uniform electric ﬁeld E, the ﬂux through that area is Φ E = EA cosq , where q is the angle the electric ﬁeld makes with the line normal to the plane of A. If A lies in the xy-plane and E is in the z-direction, then q = 0° and Φ E = EA = ( 5.00 N C ) ( 4.00 m 2 ) = 20.0 N ⋅ m 2 C.

8.

Choice (b). If q = 60° in Quick Quiz 15.7 above, then Φ E = EA cosq which yields Φ E = ( 5.00 N C ) ( 4.00 m 2 ) cos ( 60° ) = 10.0 N ⋅ m 2 C.

9.

Choice (d). Gauss’s law states that the electric ﬂux through any closed surface is equal to the net enclosed charge divided by the permittivity of free space. For the surface shown in Figure 15.28, the net enclosed charge is Q = −6 C, which gives Φ E = Q ∈0 = − ( 6 C ) ∈0.

10.

Choices (b) and (d). Since the net ﬂux through the surface is zero, Gauss’s law says that the net change enclosed by that surface must be zero as stated in (b). Statement (d) must be true because there would be a net ﬂux through the surface if more lines entered the surface than left it (or vise-versa).

1

68719_15_ch15_p001-029.indd 1

1/7/11 2:28:37 PM

2

Chapter 15

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

To balance the weight of the ball, the magnitude of the upward electric force must equal the magnitude of the downward gravitation force, or qE = mg, which gives E=

−3 2 mg ( 5.0 × 10 kg ) ( 9.80 m s ) = = 1.2 × 10 4 N C q 4.0 × 10 −6 C

and the correct choice is (b). 2.

The magnitude of the electric ﬁeld at distance r from a point charge q is E = ke q r 2 , so E=

(8.99 × 10

9

N ⋅ m 2 C2 ) (1.60 × 10 −19 C )

(5.29 × 10

−11

m)

2

= 5.14 × 1011 N C ∼ 1012 N C

making (e) the best choice for this question. 3.

The magnitude of the electric force between two protons separated by distance r is F = ke e 2 r 2 , so the distance of separation must be r=

ke e 2 = F

(8.99 × 10

9

N ⋅ m 2 C2 ) (1.60 × 10 −19 C ) 2.3 × 10 −26 N

2

= 0.10 m

and (a) is the correct choice. 4.

The ball is made of a metal, so free charges within the ball will very quickly rearrange themselves to produce electrostatic equilibrium at all points within the ball. As soon as electrostatic equilibrium exists inside the ball, the electric ﬁeld is zero at all points within the ball. Thus, the correct choice is (c).

5.

Choosing the surface of the box as the closed surface of interest and applying Gauss’s law, the net electric ﬂux through the surface of the box is found to be ΦE =

Qinside ( 3.0 − 2.0 − 7.0 + 1.0 ) × 10 −9 C = = −5.6 × 10 2 N ⋅ m 2 C ∈0 8.85 × 10 −12 C2 N ⋅ m 2

meaning that (b) is the correct choice. 6.

From Newton’s second law, the acceleration of the electron will be ax =

−19 3 Fx qEx ( −1.60 × 10 C ) (1.00 × 10 N C ) = = = −1.76 × 1014 m s 2 m m 9.11 × 10 −31 kg

The kinematics equation vx2 = v02 x + 2ax ( Δx ), with vx = 0, gives the stopping distance as − ( 3.00 × 10 6 m s ) −v 2 Δx = 0 x = = 2.56 × 10 −2 m = 2.56 cm 2ax 2 ( −1.76 × 1014 m s 2 ) 2

so (a) is the correct response for this question.

68719_15_ch15_p001-029.indd 2

1/7/11 2:28:38 PM

Electric Forces and Electric Fields

7.

3

The displacement from the −4.00 nC charge at point ( 0, 1.00 ) m to the point ( 4.00, − 2.00 ) m

(

(

)

)

has components rx = x f − xi = +4.00 m and ry = y f − yi = −3.00 m, so the magnitude of this

(

)

displacement is r = r + r = 5.00 m and its direction is q = tan −1 ry rx = −36.9°. The 2 x

2 y

x-component of the electric ﬁeld at point ( 4.00, − 2.00 ) m is then Ex = E cosq =

(8.99 × 109 N ⋅ m 2 C2 ) ( −4.00 × 10−9 C) cos ( −36.9°) = −1.15 N C ke q cosq = r2 ( 5.00 m )2

and the correct response is (d). 8.

The magnitude of the electric force between charges Q1i and Q2 i, separated by distance ri , is Fi = ke Q1i Q2 i ri 2 . If changes are made so Q1 f = Q1i , Q2 f = Q2 i 3, and rf = 2ri , the magnitude of the new force will be Ff =

ke Q1 f Q2 f rf2

=

ke Q1i (Q2 i 3)

( 2r )

2

i

=

1 ⎛ ke Q1i Q2 i ⎞ 1 = Fi 2 3 ( 2 ) ⎜⎝ ri 2 ⎟⎠ 12

so choice (a) is the correct answer for this question. 9.

Each of the situations described in choices (a) through (d) displays a high degree of symmetry, and as such, readily lends itself to the use of Gauss’s law to determine the electric ﬁelds generated. Thus, the best answer for this question is choice (e), stating that Gauss’s law can be readily applied to ﬁnd the electric ﬁeld in all of these contexts.

10.

When a charged insulator is brought near a metallic object, free charges within the metal move around, causing the metallic object to become polarized. Within the metallic object, the center of charge for the type of charge opposite to that on the insulator will be located closer to the charged insulator than will the center of charge for the same type as that on the insulator. This causes the attractive force between the charged insulator and the opposite type of charge in the metal to exceed the magnitude of the repulsive force between the insulator and the same type of charge in the metal. Thus, the net electric force between the insulator and the metallic object is one of attraction, and choice (b) is the correct answer.

11.

The outer regions of the atoms in your body and the atoms making up the ground both contain negatively charged electrons. When your body is in close proximity to the ground, these negatively charged regions exert repulsive forces on each other. Since the atoms in the solid ground are rigidly locked in position and cannot move away from your body, this repulsive force prevents your body from penetrating the ground. The best response for this question is choice (e).

12.

The positive charge +2Q makes a contribution to the electric ﬁeld at the upper right corner that is directed away from this charge in the direction of the arrow labeled (a). The magnitude of this contribution is E+ = ke (2Q) / 2s 2, where s is the length of a side of the square. Each of the negative charges makes a contribution of magnitude E− Q = ke Q s 2 directed back toward that charge. The vector sum of these two contributions due to negative charges has magnitude E− = 2E− Q cos 45° = 2ke Q s 2

FIGURE MCQ15.12

and is directed along the diagonal of the square in the direction of the arrow labeled (d). Since E− > E+ , the resultant electric ﬁeld at the upper right corner of the square is in the direction of arrow (d) and has magnitude E = E− − E+ = ( 2 − 1)ke Q s 2 . The correct answer to the question is choice (d).

68719_15_ch15_p001-029.indd 3

1/7/11 2:28:40 PM

4

Chapter 15

13.

If the positive charge +2Q at the lower left corner of the square in the above ﬁgure were removed, the ﬁeld contribution E+ discussed above would be eliminated. This would leave only E− = 2ke Q s 2 as the resultant ﬁeld at the upper right corner. This has a larger magnitude than the resultant ﬁeld E found above, making choice (a) the correct answer.

14.

Metal objects normally contain equal amounts of positive and negative charge and are electrically neutral. The positive charges in both metals and nonmetals are bound up in the nuclei of the atoms and cannot move about or be easily removed. However, in metals, some of the negative charges (the outer or valence electrons in the atoms) are quite loosely bound, can move about rather freely, and are easily removed from the metal. When a metal object is given a positive charge, this is accomplished by removing loosely bound electrons from the metal rather than by adding positive charge to it. Taking away the electrons to leave a net positive charge behind very slightly decreases the mass of the coin. Thus, choice (d) is the best choice for this question.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

Electrons are more mobile than protons and are more easily freed from atoms than the protons which are tightly bound within the nuclei of the atoms.

4.

Conducting shoes are worn to avoid the build up of a static charge on them as the wearer walks. Rubber-soled shoes acquire a charge by friction with the ﬂoor and could discharge with a spark, possibly causing an explosive burning situation, where the burning is enhanced by the oxygen.

6.

No. Object A might have a charge opposite in sign to that of B, but it also might be neutral. In this latter case, object B causes object A to be polarized, pulling charge of the sign opposite the change on B toward the near face of A and pushing an equal amount of charge of the same sign as that on B toward the far face. Then, due to difference in distances, the force of attraction exerted by B on the induced charge of opposite sign is slightly larger than the repulsive force exerted by B on the induced charge of like sign. Therefore, the net force on A is toward B.

8.

(a)

Yes. The positive charges create electric ﬁelds that extend in all directions from those charges. The total ﬁeld at point A is the vector sum of the individual ﬁelds produced by the charges at that point.

(b)

No, because there are no ﬁeld lines emanating from or converging on point A.

(c)

No. There must be a charged object present to experience a force.

10.

Electric ﬁeld lines start on positive charges and end on negative charges. Thus, if the fair-weather ﬁeld is directed into the ground, the ground must have a negative charge.

12.

To some extent, a television antenna will act as a lightning rod on the house. If the antenna is connected to the Earth by a heavy wire, a lightning discharge striking the house may pass through the metal support rod and be safely carried to the Earth by the ground wire.

68719_15_ch15_p001-029.indd 4

1/7/11 2:28:42 PM

Electric Forces and Electric Fields

14.

16.

(a)

If the charge is tripled, the ﬂux through the surface is also tripled because the net ﬂux is proportional to the charge inside the surface.

(b)

The ﬂux remains constant when the volume changes because the surface surrounds the same amount of charge, regardless of its volume.

(c)

The ﬂux does not change when the shape of the closed surface changes.

(d)

The ﬂux through the closed surface remains unchanged as the charge inside the surface is moved to another location inside that surface.

(e)

The ﬂux is zero because the charge inside the surface is zero. All of these conclusions are arrived at through an understanding of Gauss’s law.

5

All of the constituents of air are nonpolar except for water. The polar water molecules in the air quite readily “steal” charge from a charged object, as any physics teacher trying to perform electrostatics demonstrations in the summer well knows. As a result—it is difﬁcult to accumulate large amounts of excess charge on an object in a humid climate. During a North American winter, the cold, dry air allows accumulation of signiﬁcant excess charge, giving the possibility for shocks caused by static electricity sparks.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

1.57 N directed to the left

4.

(a)

6.

2.25 × 10 −9 N m

8.

4.33ke q 2 a 2 to the right and 45° above the horizontal

0.115 N

(b)

1.25 cm

10.

F6 m C = 46.7 N to the left; F1.5 m C = 157 N to the right; F−2 m C = 111 N to the left

12.

5.15 × 10 3 N m

14.

16.7 mC

16.

(a)

0

(b)

30.0 N

(c)

21.6 N

(d)

17.3 N

(e)

–13.0 N

(f)

17.3 N

(g)

17.0 N

(h)

FR = 24.3 N at 44.5° above the + x-axis

(b)

40.0 N to the left 5.27 × 10 5 m s

18.

(a) 2.00 × 10 7 N C to the right

20.

(a)

5.27 × 1013 m s 2

(b)

22.

(a)

See Solution.

(b) 1.4 × 10 3 N C

24.

(a) 2.19 × 10 5 N C at 85.2° below the + x-axis

(c)

7.5 × 10 −2 N

(b) The electric ﬁeld would be unchanged, but the force would double in magnitude.

68719_15_ch15_p001-029.indd 5

1/7/11 2:28:43 PM

6

Chapter 15

26.

4 (a) 1.80 × 10 N C to the right

28.

1.88 × 10 3 N C at 4.40° below the + x-axis

30.

(a)

32.

(a), (b), and (c)

34.

(a)

q1 q2 = −1 3

(b)

(b)

9.00 × 10 −5 N to the left

q1 < 0, q2 > 0

See Solution for Sketches. (b)

E = 0 at the center of the triangle.

(c) 1.73ke q a 2

(b)

positive y-direction

36.

(a)

(b) 1.8 × 10 6 N m

38.

r ∼ 10 −6 m

40.

1.38 × 10 3 N ⋅ m 2 C

42.

(a)

44.

(a) 6.55 × 10 5 N ⋅ m 2 C

(b) 1.64 × 10 5 N ⋅ m 2 C

46.

(a) 1.92 × 10 7 N ⋅ m 2 C

(b) 3.20 × 10 6 N ⋅ m 2 C

See Solution.

zero

3.38 nC

(c) 1.1 × 10 5 N m

(b) The sphere contains a net positive charge in a spherically symmetric distribution.

(c) The answer to (b) would change because of the loss of symmetry; however, the answer to (a) would be unchanged since Qinside is unaltered. 48.

The electric ﬁeld is everywhere parallel to the z-axis with the following magnitudes: (a)

50.

Ez = +s 2 ∈0

(b)

Ez = +3s 2 ∈0

(c)

Ez = −s 2 ∈0

(a) 2.94 × 10 5 N C (b) The magnitude of the ﬁeld is independent of distance from the sheet, provided that distance is small compared to the dimensions of the sheet.

52.

+5.25 mC

54.

E = K ed in the direction of the electron’s motion.

56.

(a)

0.307 s

58.

(a)

For 0 < x < 1.00 m, the ﬁeld contributions made by the two charges are in the same direction and cannot add to zero.

(b) Yes, neglecting gravity causes a 2.28% error.

(b) Any point for x < 0 is nearer the larger charge, so the ﬁeld contributions made by the two charges cannot have equal magnitudes and add to zero.

68719_15_ch15_p001-029.indd 6

(c)

Points for x > 1.00 m are nearer the smaller charge and the ﬁeld contributions by the two charges are in opposite directions. Thus, it is possible for them to add to zero.

(d)

E = 0 at x = +9.47 m

1/7/11 2:28:44 PM

7

Electric Forces and Electric Fields

60.

0.951 m

62.

1.98 mC

64.

(a)

(b) 1.67 × 10 −7 s for q = 37.0°; 2.21 × 10 −7 s for q = 53.0°

37.0° or 53.0°

PROBLEM SOLUTIONS 15.1

(a)

From Coulomb’s law, F = ke

Q1Q2 , we have r2

F = (8.99 × 10 9 N ⋅ m 2 C2 ) (b) 15.2

( 7.50 × 10

−9

C ) ( 4.20 × 10 −9 C )

(1.80 m )2

= 8.74 × 10 −8 N

Since these are like charges (both positive), the force is repulsive .

Particle A exerts a force toward the right on particle B. By Newton’s third law, particle B will then exert a of equal magnitude force toward the left on particle A. The ratio of the ﬁnal magnitude of the force to the original magnitude of the force is Ff Fi

=

⎛r ⎞ =⎜ i ⎟ 2 ke q1 q2 ri ⎝ rf ⎠ ke q1 q2 rf2

2

2

so

2 ⎛r ⎞ ⎛ 13.7 mm ⎞ Ff = Fi ⎜ i ⎟ = ( 2.62 N ) ⎜ = 1.57 N ⎟ ⎝ 17.7 mm ⎠ ⎝ rf ⎠

The ﬁnal vector force that B exerts on A is 1.57 N directed to the left . 15.3

(a)

When the balls are an equilibrium distance apart, the tension in the string equals the magnitude of the repulsive electric force between the balls. Thus, F=

or

15.4

ke q ( 2q ) = 2.50 N ⇒ r2

q=

( 2.50 N )( 2.00 m )2

q2 =

2 (8.99 × 10 9 N ⋅ m 2 C2 )

( 2.50 N ) r 2 2ke

= 2.36 × 10 −5 C = 23.6 mC

(b)

The charges induce opposite charges in the bulkheads, but the induced charge in the bulkhead near ball B is greater, due to B’s greater charge. Therefore, the system moves slowly towards the bulkhead closer to ball B .

(a)

The gravitational force exerted on the upper sphere by the lower one is negligible in comparison to the gravitational force exerted by the Earth and the downward electrical force exerted by the lower sphere. Therefore,

ΣFy = 0 ⇒ T − mg − Fe = 0 or

T = mg +

ke q1 q2 d2

T = ( 7.50 × 10

−3

T m, q1 d

mg

Fe m, q2

9 2 2 −9 −9 m ⎞ (8.99 × 10 N ⋅ m C ) ( 32.0 × 10 C ) ( 58.0 × 10 C ) ⎛ kg ) ⎜ 9.80 2 ⎟ + 2 ⎝ s ⎠ ( 2.00 × 10 −2 m )

giving T = 0.115 N

continued on next page

68719_15_ch15_p001-029.indd 7

1/7/11 2:28:47 PM

8

Chapter 15

ΣFy = 0 → Fe =

(b)

ke q1 q2 = T − mg, d2

ke q1 q2 T − mg

d=

and

Thus, if T = 0.180 N, d=

15.5

(8.99 × 10

N ⋅ m 2 C2 ) ( 32.0 × 10 −9 C ) ( 58.0 × 10 −9 C ) = 1.25 × 10 −2 m = 1.25 cm m ⎛ ⎞ 0.180 N − ( 7.50 × 10 −3 kg ) ⎜ 9.80 2 ⎟ ⎝ s ⎠

2 2 (8.99 × 109 N ⋅ m 2 C2 ) ⎡⎣ 4 (1.60 × 10 −19 C) ⎤⎦ ke ( 2e ) F= = = 36.8 N 2 r2 (5.00 × 10 −15 m )

(a)

The mass of an alpha particle is m = 4.002 6 u, where 1 u = 1.66 × 10 −27 kg is the uniﬁed mass unit. The acceleration of either alpha particle is then

(b)

a= 15.6

9

F 36.8 N = = 5.54 × 10 27 m s 2 m 4.002 6 (1.66 × 10 −27 kg )

The attractive force between the charged ends tends to compress the molecule. Its magnitude is −19 2 ⎛ k (1e ) N ⋅ m 2 ⎞ (1.60 × 10 C ) F= e 2 = ⎜ 8.99 × 10 9 = 4.89 × 10 −17 N 2 r C2 ⎟⎠ ( 2.17 × 10 −6 m ) ⎝ 2

The compression of the “spring” is x = ( 0.010 0 ) r = ( 0.010 0 ) ( 2.17 × 10 −6 m ) = 2.17 × 10 −8 m , so the spring constant is k = 15.7

F 4.89 × 10 −17 N = = 2.25 × 10 −9 N m . x 2.17 × 10 −8 m

In the new equilibrium position,

ΣFy = Fs − Fe = kd − Fe = 0.

Fs = kd

Fs

Thus,

k q q r 2 ke q1 q2 k= = e 1 2 = d d d ⋅ r2

or

k=

(8.99 × 10

9

q1

N ⋅ m 2 C2 ) ( 0.800 × 10 −6 C ) ( −0.600 × 10 −6 C )

(3.50 × 10

−2

m ) ( 5.00 × 10

−2

m)

r

Fe

2

q2

k = 49.3 N m 15.8

See the sketch at the right. The magnitudes of the forces are F1 = F2 = ke

q ( 2q ) q ( 3q ) q2 q2 = 2k and F = k = 1.50k e 3 e e 2 a2 a2 a2 a 2

(

)

The components of the resultant force on charge q are Fx = F1 + F3 cos 45° = ( 2 + 1.50 cos 45° ) ke

q2 q2 = 3.06ke 2 2 a a

continued on next page

68719_15_ch15_p001-029.indd 8

1/7/11 2:28:49 PM

Electric Forces and Electric Fields

and

Fy = F1 + F3 sin 45° = ( 2 + 1.50 sin 45° ) ke

9

q2 q2 = 3.06ke 2 2 a a

⎛ q2 ⎞ q2 The magnitude of the resultant force is Fe = Fx2 + Fy2 = 2 ⎜ 3.06ke 2 ⎟ = 4.33ke 2 a ⎠ a ⎝ ⎛ Fy ⎞ and it is directed at q = tan −1 ⎜ ⎟ = tan −1 (1.00 ) = 45° above the horizontal . ⎝ Fx ⎠ 15.9

(a)

The spherically symmetric charge distributions behave as if all charge was located at the centers of the spheres. Therefore, the magnitude of the attractive force is F=

(b)

When the spheres are connected by a conducting wire, the net charge qnet = q1 + q2 = −6.0 × 10 −9 C will divide equally between the two identical spheres. Thus, the force is now F= or

15.10

12 × 10 −9 C ) (18 × 10 −9 C ) ke q1 q2 9 2 2 ( = 8.99 × 10 N ⋅ m C = 2.2 × 10 −5 N ( ) r2 ( 0.30 m )2

ke ( qnet 2 ) r2

2

−9 ⎛ N ⋅ m 2 ⎞ ( −6.0 × 10 C ) = ⎜ 8.99 × 10 9 2 C2 ⎟⎠ 4 ( 0.30 m ) ⎝

2

F = 9.0 × 10 −7 N (repulsion)

The forces are as shown in the sketch below.

F1 =

F2 =

F3 =

2 6.00 × 10 −6 C ) (1.50 × 10 −6 C ) ke q1 q2 ⎛ 9 N⋅m ⎞ ( = 8.99 × 10 = 89.9 N 2 ⎜⎝ r122 C2 ⎟⎠ (3.00 × 10 −2 m )

ke q1 q3 r132 ke q2 q3 r232

−6 −6 ⎛ N ⋅ m 2 ⎞ ( 6.00 × 10 C ) ( 2.00 × 10 C ) = ⎜ 8.99 × 10 9 = 43.2 N 2 C2 ⎟⎠ ⎝ (5.00 × 10 −2 m ) −6 −6 ⎛ N ⋅ m 2 ⎞ (1.50 × 10 C ) ( 2.00 × 10 C ) = ⎜ 8.99 × 10 9 = 67.4 N 2 C2 ⎟⎠ ⎝ ( 2.00 × 10−2 m )

The net force on the 6 mC charge is F6 m C = F1 − F2 = 46.7 N to the left . The net force on the 1.5 mC charge is F1.5 m C = F1 + F3 = 157 N to the right . The net force on the −2 mC charge is F−2 m C = F2 + F3 = 111 N to the left .

68719_15_ch15_p001-029.indd 9

1/7/11 2:28:51 PM

10

15.11

Chapter 15

In the sketch at the right, FR is the resultant of the forces F6 and F3 that are exerted on the charge at the origin by the 6.00 nC and the –3.00 nC charges, respectively. −9 −9 ⎛ N ⋅ m 2 ⎞ ( 6.00 × 10 C ) ( 5.00 × 10 C ) F6 = ⎜ 8.99 × 10 9 = 3.00 × 10 −6 N 2 2 ⎟ C ⎠ ⎝ ( 0.300 m ) −9 −9 ⎛ N ⋅ m 2 ⎞ ( 3.00 × 10 C ) ( 5.00 × 10 C ) F3 = ⎜ 8.99 × 10 9 = 1.35 × 10 −5 N C2 ⎟⎠ ⎝ ( 0.100 m )2

The resultant is FR = or 15.12

(F ) + (F ) 2

6

3

2

⎛F ⎞ = 1.38 × 10 −5 N at q = tan −1 ⎜ 3 ⎟ = 77.5° ⎝ F6 ⎠

FR = 1.38 × 10 −5 N at 77.5° below the − x-axis

At equilibrium, ΣFx = Fe − Fs = 0 kx =

Thus,

or

q1

Fs = Fe Fs = kx

ke q1 q2 d2

Fe

q2

d

and the force constant of the spring is k=

9 2 2 −6 −6 ke q1 q2 (8.99 × 10 N ⋅ m C ) ( 2.70 × 10 C ) (8.60 × 10 C ) = 2 xd 2 (5.00 × 10 −3 m ) (9.00 × 10 −2 m )

k = 5.15 × 10 3 N m 15.13

Please see the sketch at the right. F1 =

(8.99 × 10

or

F1 = 0.288 N

F2 =

(8.99 × 10

or

F2 = 0.503 N

9

N ⋅ m 2 C2 ) ( 2.00 × 10 −6 C ) ( 4.00 × 10 −6 C )

( 0.500 m )2

9

N ⋅ m 2 C2 ) ( 2.00 × 10 −6 C ) ( 7.00 × 10 −6 C )

( 0.500 m )2

The components of the resultant force acting on the 2.00 mC charge are: Fx = F1 − F2 cos 60.0° = 0.288 N − ( 0.503 N ) cos 60.0° = 3.65 × 10 −2 N and Fy = −F2 sin 60.0° = − ( 0.503 N ) sin 60.0° = − 0.436 N The magnitude and direction of this resultant force are

continued on next page

68719_15_ch15_p001-029.indd 10

1/7/11 2:28:53 PM

Electric Forces and Electric Fields

F = Fx2 + Fy2 = at 15.14

( 0.036 5 N ) + ( 0.436 N ) 2

2

11

= 0.438 N

⎛ Fy ⎞ ⎛ −0.436 N ⎞ = −85.2° or 85.2° below the + x-axis q = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎝ 0.036 5 N ⎟⎠ ⎝ Fx ⎠

At equilibrium, Fe = Fs or

Q

Fe

ke Q 2 = kx = k ( L − Li ) L2

Fs

Q

L

Thus, Q=

k ( L − Li ) L2

=

ke

(100

N m ) ( 0.500 m − 0.400 m ) ( 0.500 m ) 8.99 × 10 9 N ⋅ m 2 C2

2

Q = 1.67 × 10 −5 C = 16.7 mC 15.15

Consider the free-body diagram of one of the spheres given at the right. Here, T is the tension in the string and Fe is the repulsive electrical force exerted by the other sphere. ΣFy = 0 ⇒ T cos 5.0° = mg,

or T =

mg cos 5.0°

ΣFx = 0 ⇒ Fe = T sin 5.0° = mg tan 5.0° At equilibrium, the distance separating the two spheres is r = 2 L sin 5.0°. Thus, Fe = mg tan 5.0° becomes q = ( 2 L sin 5.0° )

mg tan 5.0° ke

= [ 2 ( 0.300 m ) sin 5.0° ] 15.16

ke q 2 = mg tan 5.0° and yields ( 2 L sin 5.0°)2

( 0.20 × 10

−3

kg ) ( 9.80 m s 2 ) tan 5.0°

8.99 × 10 9 N ⋅ m 2 C2

In the sketch at the right, rB C = and

FAC

( 4.00 m ) + ( 3.00 m ) = 5.00 m 2

2

(0, 3.00 m) C

⎛ 3.00 m ⎞ q = tan −1 ⎜ = 36.9° ⎝ 4.00 m ⎟⎠

(a)

(F )

(b)

(F )

AC

AC

x

= 0

y

= ke

= 7.2 nC

1.00 × 10−4 C q FBC

(0, 0) A 3.00 × 10−4 C

qA qC 2 rAC

= (8.99 × 10 9 N ⋅ m 2 C2 )

(3.00 × 10

−4

rBC q

−6.00 × 10−4 C B (4.00 m, 0)

C ) (1.00 × 10 −4 C )

( 3.00 m )2

= 30.0 N

continued on next page

68719_15_ch15_p001-029.indd 11

1/7/11 2:28:55 PM

12

15.17

Chapter 15

qB qC

= (8.99 × 10 9 N ⋅ m 2 C2 )

(6.00 × 10

−4

(c)

FB C = ke

(d)

(F )

x

= FB C cosq = ( 21.6 N ) cos ( 36.9° ) = 17.3 N

(e)

(F )

y

= − FB C sinq = − ( 21.6 N ) sin ( 36.9° ) = −13.0 N

(f)

(F ) = (F ) + (F )

x

= 0 + 17.3 N = 17.3 N

(g)

(F ) = (F ) + (F )

y

= 30.0 − 13.0 N = 17.0 N

(h)

FR =

BC

BC

R

R

rB2C

AC

x

AC

y

BC

x

BC

y

(F ) + (F ) 2

R

x

R

2 y

=

C ) (1.00 × 10 −4 C )

( 5.00 m )2

= 21.6 N

(17.3 N )2 + (17.0 N )2 = 24.3 N

and

⎡ ( FR ) y j = tan −1 ⎢ ⎢⎣ ( FR )x

or

FR = 24.3 N at 44.5° above the + x-axis

⎤ ⎛ 17.0 N ⎞ ⎥ = tan −1 ⎜ = 44.5° ⎝ 17.3 N ⎟⎠ ⎥⎦

In order to suspend the object in the electric ﬁeld, the electric force exerted on the object by the ﬁeld must be directed upward and have a magnitude equal to the weight of the object. Thus, Fe = qE = mg, and the magnitude of the electric ﬁeld must be

(

)

3.80 g ( 9.80 m s 2 ) ⎛ 1 kg ⎞ mg 3 E= = ⎜ 3 ⎟ = 2.07 × 10 N C q 18 × 10 −6 C ⎝ 10 g ⎠ The electric force on a negatively charged object is in the direction opposite to that of the electric ﬁeld. Since the electric force must be directed upward, the electric ﬁeld must be directed downward. 15.18

(a)

Taking to the right as positive, the resultant electric ﬁeld at point P is given by ER = E1 + E3 − E2 =

ke q1 ke q3 ke q2 + 2 − 2 r12 r3 r2

⎛ N ⋅ m 2 ⎞ ⎡ 6.00 × 10 −6 C 2.00 × 10 −6 C 1.50 × 10 −6 C ⎤ = ⎜ 8.99 × 10 9 + − ⎢ ⎥ 2 C2 ⎟⎠ ⎢⎣ ( 0.020 0 m ) ⎝ ( 0.030 0 m )2 ( 0.010 0 m )2 ⎥⎦ This gives ER = + 2.00 × 10 7 N C or (b)

F = qE R = ( −2.00 × 10 −6 C ) ( 2.00 × 10 7 N C ) = −40.0 N or

68719_15_ch15_p001-029.indd 12

E R = 2.00 × 10 7 N C to the right

F = 40.0 N to the left

1/7/11 2:28:58 PM

13

Electric Forces and Electric Fields

15.19

The force on a negative charge is opposite to the direction of the electric ﬁeld and has magnitude F = q E. Thus, F = −6.00 × 10 −6 C ( 5.25 × 10 5 N C ) = 3.15 N

15.20

and

F = 3.15 N due north

(a)

The magnitude of the force on the electron is F = q E = eE, and the acceleration is a=

(b) 15.21

(a)

−19 F eE (1.60 × 10 C ) ( 300 N C ) = = = 5.27 × 1013 m s 2 me me 9.11 × 10 −31 kg

v = v0 + at = 0 + ( 5.27 × 1013 m s 2 ) (1.00 × 10 −8 s ) = 5.27 × 10 5 m s y

The electrical force must be directed up the incline and have a magnitude equal to the tangential component of the gravitational force.

n

Q,

Fe

x

m

ΣFx = 0 ⇒ Fe − mg sinq = 0

(b)

q

or

Fe = Q E = mg sinq

E=

−3 2 mg sinq ( 5.40 × 10 kg ) ( 9.80 m s ) sin 25.0° = = 3.19 × 10 3 N C Q 7.00 × 10 −6 C

and

E = mg sinq Q

q mg

Since Fe must be directed up the incline and the electrical force on a negative charge is directed opposite to the ﬁeld, it is necessary to have the electric ﬁeld directed down the incline. E = 3.19 × 10 3 N C down the incline

Thus, 15.22

(a) Fe q

T

mg

(b)

ΣF

y

=0

⇒

Fe sinq − mg = 0

or

Fe =

mg sinq

Since Fe = qE, this gives E=

(c)

ΣF

x

and

68719_15_ch15_p001-029.indd 13

(5.8 × 10−3 kg) (9.8 m s2 ) = 1.4 × 103 N C mg = q sinq (68 × 10−6 C) sin 37°

=0

T=

⇒

Fe cosq − T = 0

(5.8 × 10

−3

kg ) ( 9.8 m s 2 )

tan 37°

or

mg ⎛ mg ⎞ T = Fe cosq = ⎜ cosq = ⎝ sinq ⎟⎠ tanq

= 7.5 × 10 −2 N

1/7/11 2:29:00 PM

14

15.23

15.24

Chapter 15

(a)

a=

(b)

t=

−19 F qE (1.60 × 10 C ) ( 640 N C ) = = = 6.12 × 1010 m s 2 1.673 × 10 −27 kg m mp

Δv 1.20 × 10 6 m s = = 1.96 × 10 −5 s = 19.6 ms a 6.12 × 1010 m s 2 v 2f − v02

(1.20 × 10 m s) − 0 = = 2 ( 6.12 × 10 m s ) 2

6

(c)

Δx =

(d)

KE f =

(a)

Please refer to the solution of Problem 15.13 earlier in this chapter. There it is shown that the resultant electric force experienced by the 2.00 mC located at the origin is F = 0.438 N at 85.2° below the + x-axis. Since the electric ﬁeld at a location is deﬁned as the force per unit charge experienced by a test charge placed in that location, the electric ﬁeld at the origin in the charge conﬁguration of Figure P15.13 is

2a

2

11.8 m

2 1 1 m p v 2f = (1.673 × 10 −27 kg )(1.20 × 10 6 m s ) = 1.20 × 10 −15 J 2 2

E= (b)

10

F 0.438 N = at − 85.2° = 2.19 × 10 5 N C at 85.2° below the + x-axis q0 2.00 × 10 −6 C

The electric force experienced by the charge at the origin is directly proportional to the magnitude of that charge. Thus, doubling the magnitude of this charge would double the magnitude of the electric force. However, the electric ﬁeld is the force per unit charge and the field would be unchanged if the charge was doubled. This is easily seen in the calculation of part (a) above. Doubling the magnitude of the charge at the origin would double both the numerator and the denominator of the ratio F q0 , but the value of the ratio (i.e., the electric ﬁeld) would be unchanged.

15.25

r=

(

E y = E1 sin 45.0° + 0 = (a)

+Q

d d d = = 2 cos 45.0° 2 2 2 2

E2 r

45.0°

+Q

2d

ke Q ke Q ke Q ⎛ 2 ⎞ ke Q kQ ⎛ 2⎞ = 2 − 2e − 2 ⎜ 2 ⎟ ⎜⎝ 2 ⎟⎠ = 1 − 2 d 2 d r ⎝ 2 ⎠ d d 2 ( )

kQ ⎛ 2⎞ ke Q ⎛ 2 ⎞ = 2e = 2 ⎜ ⎟ r ⎝ 2 ⎠ ( d 2 ) ⎜⎝ 2 ⎟⎠

Observe the ﬁgure at the right: kQ E1 = E2 = e 2 r

45.0° d P

45.0°

)

Ex = E2 − E1 cos 45.0° =

15.26

E1

From the ﬁgure at the right, observe that 2r cos 45.0° + d = 2d. Thus,

k q and E3 = e 2 r

E y = E1 sin 30.0° − E2 sin 30.0° = 0 Ex = E1 cos 30.0° + E2 cos 30.0° − E3

(

2

)

ke Q d2 Q = +3.00 nC r= 4.0 0c

y

m

E3

q = −2.00 nC r

P

E1 30.0° 30.0° x E2

Q = +3.00 nC

continued on next page

68719_15_ch15_p001-029.indd 14

1/7/11 2:29:02 PM

Electric Forces and Electric Fields

Ex =

k q k 2ke Q cos 30.0° − e 2 = 2e ( 2Q cos 30.0° − q ) 2 r r r

(8.99 × 10 N ⋅ m C ) ⎡ 2 3.00 × 10 ( ( 4.00 × 10 m ) ⎣ 9

=

15

2

2

−9

2

−2

C ) cos 30.0° − 2.00 × 10 −9 C ⎤⎦

Ex = +1.80 × 10 4 N C Then,

Fe = q ′E = ( −5.00 × 10 −9 C ) (1.80 × 10 4 N C ) = −9.00 × 10 −5 N

(b)

or 15.27

E = E x2 + E y2 = 1.80 × 10 4 N C and E = 1.80 × 10 4 N C to the right

Fe = 9.00 × 10 −5 N to the left

If the resultant ﬁeld is zero, the contributions from the two charges must be in opposite directions and also have equal magnitudes. Choose the line connecting the charges as the x-axis, with the origin at the –2.5 µC charge. Then, the two contributions will have opposite directions only in the regions x < 0 and x > 1.0 m. For the magnitudes to be equal, the point must be nearer the smaller charge. Thus, the point of zero resultant ﬁeld is on the x-axis at x < 0. Requiring equal magnitudes gives k q ke q1 = e 22 2 r1 r2 Thus,

(1.0 m + d )

6.0 mC 2.5 mC = 2 d (1.0 m + d )2

or 2.5 =d 6.0

Solving for d yields d = 1.8 m, 15.28

or

1.8 m to the left of the − 2.5 mC charge

The altitude of the triangle is h = ( 0.500 m ) sin 60.0° = 0.433 m and the magnitudes of the ﬁelds due to each of the charges are 9 2 2 −9 ke q1 (8.99 × 10 N ⋅ m C ) ( 3.00 × 10 C ) E1 = 2 = h ( 0.433 m )2

= 144 N C E2 =

ke q2 (8.99 × 109 N ⋅ m 2 C2 ) (8.00 × 10 −9 C) = 1.15 × 103 N C = r22 ( 0.250 m )2

continued on next page

68719_15_ch15_p001-029.indd 15

1/7/11 2:29:04 PM

16

Chapter 15

E3 =

and

ke q3 r32

=

(8.99 × 10

9

N ⋅ m 2 C2 ) ( 5.00 × 10 −9 C )

( 0.250 m )2

Thus, ΣEx = E2 + E3 = 1.87 × 10 3 N C giving

( ΣE ) + ( ΣE ) 2

ER = and

(

2

y

ΣE y = −E1 = −144 N C

= 1.88 × 10 3 N C

)

q = tan −1 ΣE y ΣEx = tan −1 ( −0.0769 ) = −4.40°

Hence 15.29

x

and

= 719 N C

E R = 1.88 × 10 3 N C at 4.40° below the +x-axis

From the symmetry of the charge distribution, students should recognize that the resultant electric ﬁeld at the center is ER = 0 If one does not recognize this intuitively, consider: E R = E1 + E 2 + E 3

15.30

Ex = E1 x − E2 x =

and

E y = E1 y + E2 y − E3 =

Thus,

ER = Ex2 + E y2 = 0

(b)

68719_15_ch15_p001-029.indd 16

ke q k q k q sin 30° + e 2 sin 30° − e 2 = 0 r r r2

The magnitude of q2 is three times the magnitude of q1 because 3 times as many lines emerge from q2 as enter q1. q2 = 3 q1 (a)

15.31

ke q k q cos 30° − e 2 cos 30° = 0 2 r r

so

Then,

q1 q2 = −1 3

q2 > 0 because lines emerge from it, and q1 < 0 because lines terminate on it.

Note in the sketches at the right that electric ﬁeld lines originate on positive charges and terminate on negative charges. The density of lines is twice as great for the −2 q charge in (b) as it is for the 1q charge in (a).

1/7/11 2:29:06 PM

Electric Forces and Electric Fields

15.32

Rough sketches for these charge conﬁgurations are shown below.

15.33

(a)

The sketch for (a) is shown at the right. Note that four times as many lines should leave q1 as emerge from q2 although, for clarity, this is not shown in this sketch.

(b)

The ﬁeld pattern looks the same here as that shown for (a) with the exception that the arrows are reversed on the ﬁeld lines.

(a)

The electric ﬁeld in the plane of the charges has the general appearance shown at the right:

(b)

It is zero at the center of the triangle , where (by symmetry) one can see that the three charges individually produce ﬁelds that cancel out. In addition to the center of the triangle, the electric ﬁeld lines in the second ﬁgure to the right indicate three other points near the middle of each leg of the triangle where E = 0, but they are more difﬁcult to ﬁnd mathematically.

(c)

Using the sketch at the right, observe that E1 = E2 = ke q a 2 and the resultant ﬁeld at point P is E = E1 + E 2. Thus,

15.34

17

Ex = E1 cos 60.0° − E2 cos 60.0° = 0 and

E y = E1 sin 60.0° + E2 sin 60.0° =

2ke q sin 60.0° a2

The magnitude of the resultant field is then E = Ex2 + E y2 = 0 + E y2 = E y =

1.73ke q a2 continued on next page

68719_15_ch15_p001-029.indd 17

1/7/11 2:29:08 PM

18

Chapter 15

Since Ex = 0 and E y > 0 , the resultant ﬁeld at point P is directed vertically upward in the positive y-direction.

(d)

15.35

15.36

(a)

Zero net charge on each surface of the sphere.

(b)

The negative charge lowered into the sphere repels − 5mC on the outside surface, and leaves + 5 mC on the inside surface of the sphere.

(c)

The negative charge lowered inside the sphere neutralizes the inner surface, leaving zero charge on the inside . This leaves − 5mC on the outside surface of the sphere.

(d)

When the object is removed, the sphere is left with − 5.00 mC on the outside surface and zero charge on the inside .

(a)

The dome is a closed conducting surface. Therefore, the electric ﬁeld is zero everywhere inside it. At the surface and outside of this spherically symmetric charge distribution, the ﬁeld is as if all the charge were concentrated at the center of the sphere.

(b)

At the surface, E=

Outside the spherical dome, E = ke q r 2. Thus, at r = 4.0 m,

(c)

E= 15.37

9 2 2 −4 ke q (8.99 × 10 N ⋅ m C ) ( 2.0 × 10 C ) = = 1.8 × 10 6 N C R2 (1.0 m )2

(8.99 × 10

9

N ⋅ m 2 C2 ) ( 2.0 × 10 −4 C )

( 4.0 m )2

= 1.1 × 10 5 N C

For a uniformly charged sphere, the ﬁeld is strongest at the surface. Thus,

Em ax =

ke qm ax , R2

6 R 2 Em ax ( 2.0 m ) ( 3.0 × 10 N C ) = = 1.3 × 10 −3 C ke 8.99 × 10 9 N ⋅ m 2 C2 2

or 15.38

qm ax =

If the weight of the drop is balanced by the electric force, then mg = q E = eE, so the mass of the drop must be m=

−19 4 eE (1.6 × 10 C ) ( 3 × 10 N C ) = ≈ 5 × 10 −16 kg g 9.8 m s 2

But m = rV = r ( 4p r 3 3) and the radius of the drop is r = [ 3m 4pr ] , which gives 13

⎡ 3 ( 5 × 10 −16 kg ) ⎤ r=⎢ ⎥ 3 ⎢⎣ 4p (858 kg m ) ⎥⎦

13

68719_15_ch15_p001-029.indd 18

= 5.2 × 10 −7 m

or

r ∼ 1 mm

1/7/11 2:29:09 PM

Electric Forces and Electric Fields

15.39

(a)

(b)

Fe = ma = (1.67 × 10 −27 kg ) (1.52 × 1012 m s 2 ) = 2.54 × 10 −15 N in the direction of the acceleration, or radially outward. The direction of the ﬁeld is the direction of the force on a positive charge (such as the proton). Thus, the ﬁeld is directed radially outward . The magnitude of the ﬁeld is E=

15.40

19

Fe 2.54 × 10 −15 N = = 1.59 × 10 4 N C q 1.60 × 10 −19 C normal line

When an electric ﬁeld of magnitude E is incident on a surface of area A, the ﬂux through the surface is q

Φ E = EA cosq

E 65.0°

where q is the angle between E and the line normal to the surface. Thus, in this case, q = 90.0° − 65.0° = 25.0° and 15.41

Φ E = ( 435 N C ) ( 3.50 m 2 ) cos 25.0° = 1.38 × 10 3 N ⋅ m 2 C

The area of the rectangular plane is A = ( 0.350 m ) ( 0.700 m ) = 0.245 m 2 . (a)

When the plane is parallel to the yz-plane, its normal line is parallel to the x-axis and makes an angle q = 0° with the direction of the ﬁeld. The ﬂux is then Φ E = EA cosq = ( 3.50 × 10 3 N C ) ( 0.245 m 2 ) cos 0° = 858 N ⋅ m 2 C

15.42

(b)

When the plane is parallel to the x-axis, q = 90° and Φ E = 0 .

(c)

Φ E = EA cosq = ( 3.50 × 10 3 N C ) ( 0.245 m 2 ) cos 40.0° = 657 N ⋅ m 2 C

(a)

Gauss’s law states that the electric ﬂux through any closed surface equals the net charge enclosed divided by ∈0 . We choose to consider a closed surface in the form of a sphere, centered on the center of the charged sphere and having a radius inﬁnitesimally larger than that of the charged sphere. The electric ﬁeld then has a uniform magnitude and is perpendicular to our surface at all points on that surface. The ﬂux through the chosen closed surface is therefore Φ E = EA = E ( 4p r 2 ) , and Gauss’s law gives Qinside = ∈0 Φ E = 4p ∈0 Er 2 = 4p (8.85 × 10 −12 C2 N ⋅ m 2 ) ( 575 N C ) ( 0.230 m ) = 3.38 × 10 −9 C = 3.38 nC 2

(b)

68719_15_ch15_p001-029.indd 19

Since the electric ﬁeld displays spherical symmetry, you can conclude that the charge distribution generating that ﬁeld is spherically symmetric . Also, since the electric ﬁeld lines are directed outward away from the sphere, the sphere must contain a net positive charge .

1/7/11 2:29:12 PM

20

15.43

15.44

Chapter 15

From Gauss’s law, the electric ﬂux through any closed surface is equal to the net charge enclosed divided by ∈0 . Thus, the ﬂux through each surface (with a positive ﬂux coming outward from the enclosed interior and a negative ﬂux going inward toward that interior) is For S1:

Φ E = Qnet ∈0 = ( +Q − 2Q ) ∈0 = −Q ∈0

For S2:

Φ E = Qnet ∈0 = ( +Q − Q ) ∈0 = 0

For S3:

Φ E = Qnet ∈0 = ( −2Q + Q − Q ) ∈0 = −2Q ∈0

For S4:

Φ E = Qnet ∈0 = ( 0 ) ∈0 = 0

(a)

From Gauss’s law, the total ﬂux through this closed surface is ΦE =

(b)

Qenclosed +5.80 × 10 −6 C = = 6.55 × 10 5 N ⋅ m 2 C ∈0 8.85 × 10 −12 C2 N ⋅ m 2

When the charge is located in the center of this tetrahedron, we have total symmetry and the ﬂux is the same through each of the four surfaces. Thus, the ﬂux through any one face is Φ1 face =

15.45

Φ E 6.55 × 10 5 N ⋅ m 2 C = = 1.64 × 10 5 N ⋅ m 2 C 4 4

We choose a spherical Gaussian surface, concentric with the charged spherical shell and of radius r. Then, ΣEA cosq = E ( 4p r 2 ) cos 0° = 4p r 2 E . (a)

For r > a (that is, outside the shell), the total charge enclosed by the Gaussian surface is Q = +q − q = 0. Thus, Gauss’s law gives 4p r 2 E = 0, or E = 0 .

(b)

Inside the shell, r < a, and the enclosed charge is Q = +q. Therefore, from Gauss’s law, 4p r 2 E =

kq q q , or E = = e2 2 r ∈0 4p ∈0 r

The ﬁeld for r < a is E = ke q r 2 directed radially outward . 15.46

(a)

The surface of the cube is a closed surface which surrounds a total charge of Q = 1.70 × 10 2 mC. Thus, by Gauss’s law, the electric ﬂux through the full surface of the cube is ΦE =

(b)

Qinside 1.70 × 10 −4 C = = 1.92 × 10 7 N ⋅ m 2 C ∈0 8.85 × 10 −12 C2 N ⋅ m 2

Since the charge is located at the center of the cube, the six faces of the cube are symmetrically positioned around the location of the charge. Thus, one-sixth of the ﬂux passes through each of the faces, or Φ face =

(c)

68719_15_ch15_p001-029.indd 20

Φ E 1.92 × 10 7 N ⋅ m 2 C = = 3.20 × 10 6 N ⋅ m 2 C 6 6

The answer to part (b) would change because the charge could now be at different distances from each face of the cube, but the answer to part (a) would be unchanged because the ﬂux through the entire closed surface depends only on the total charge inside the surface.

1/7/11 2:29:14 PM

Electric Forces and Electric Fields

15.47

21

Note that with the point charge −2.00 nC positioned at the center of the spherical shell, we have complete spherical symmetry in this situation. Thus, we can expect the distribution of charge on the shell, as well as the electric ﬁelds both inside and outside of the shell, to also be spherically symmetric. (a)

We choose a spherical Gaussian surface, centered on the center of the conducting shell, with radius r = 1.50 m < a as shown at the right. Gauss’s law gives Φ E = EA = E ( 4p r 2 ) =

so

E=

(8.99 × 10

9

Qinside Qinside kQ or E = = e center ∈0 4p ∈0 r 2 r2

N ⋅ m 2 C2 ) ( −2.00 × 10 −9 C )

(1.50 m )2

and E = −7.99 N C . The negative sign means that the ﬁeld is radial inward . (b)

All points at r = 2.20 m are in the range a < r < b, and hence are located within the conducting material making up the shell. Under conditions of electrostatic equilibrium, the ﬁeld is E = 0 at all points inside a conducting material.

(c)

If the radius of our Gaussian surface is r = 2.50 m > b, Gauss’s law (with total spheriQinside kQ cal symmetry) leads to E = = e inside just as in part (a). However, now 2 4p ∈0 r r2 Qinside = Qshell + Qcenter = +3.00 nC − 2.00 nC = +1.00 nC. Thus, we have

(8.99 × 10 E=

9

N ⋅ m 2 C2 ) ( +1.00 × 10 −9 C )

( 2.50 m )2

= +1.44 N C

with the positive sign telling us that the ﬁeld is radial outward at this location. (d)

Under conditions of electrostatic equilibrium, all excess charge on a conductor resides entirely on its surface. Thus, the sum of the charge on the inner surface of the shell and that on the outer surface of the shell is Qshell = +3.00 nC. To see how much of this is on the inner surface, consider our Gaussian surface to have a radius r that is inﬁnitesimally larger than a. Then, all points on the Gaussian surface lie within the conducting material, meaning that E = 0 at all points and the total ﬂux through the surface is Φ E = 0. Gauss’s law then states that Qinside = Qinner + Qcenter = 0, or surface

Qinner = − Qcenter = − ( −2.00 nC ) = +2.00 nC surface

The charge on the outer surface must be Qouter = Qshell − Qinner = 3.00 nC − 2.00 nC = +1.00 nC surface

15.48

surface

Please review Example 15.8 in your textbook. There it is shown that the electric ﬁeld due to a nonconducting plane sheet of charge parallel to the xy-plane has a constant magnitude given by Ez = s sheet 2 ∈0, where s sheet is the uniform charge per unit area on the sheet. This ﬁeld is everywhere perpendicular to the xy-plane, is directed away from the sheet if it has a positive charge density, and is directed toward the sheet if it has a negative charge density.

continued on next page

68719_15_ch15_p001-029.indd 21

1/7/11 2:29:16 PM

22

Chapter 15

In this problem, we have two plane sheets of charge, both parallel to the xy-plane and separated by a distance of 2.00 cm. The upper sheet has charge density s sheet = −2s , while the lower sheet has s sheet = +s . Taking upward as the positive z-direction, the ﬁelds due to each of the sheets in the three regions of interest are:

Region z 2.00 cm

Lower sheet (at z = 0)

Upper sheet (at z = 2.00 cm)

Electric Field

Electric Field

Ez = −

+s s =− 2 ∈0 2 ∈0

Ez = +

−2s s =+ ∈0 2 ∈0

Ez = +

+s s =+ 2 ∈0 2 ∈0

Ez = +

−2s s =+ ∈0 2 ∈0

Ez = +

+s s =+ 2 ∈0 2 ∈0

Ez = −

−2s s =− ∈0 2 ∈0

When both plane sheets of charge are present, the resultant electric ﬁeld in each region is the vector sum of the ﬁelds due to the individual sheets for that region.

15.49

(a)

For z < 0:

Ez = Ez ,lower + Ez ,upper = −

s s s + = + 2 ∈0 ∈0 2 ∈0

(b)

For 0 < z < 2.00 cm:

Ez = Ez ,lower + Ez ,upper = +

s s 3s + = + 2 ∈0 ∈0 2 ∈0

(c)

For z > 2.00 cm:

Ez = Ez ,lower + Ez ,upper = +

s s s − = − 2 ∈0 ∈0 2 ∈0

The radius of each sphere is small in comparison to the distance to the nearest neighboring charge (the other sphere). Thus, we shall assume that the charge is uniformly distributed over the surface of each sphere and, in its interaction with the other charge, treat it as though it were a point charge. In this model, we then have two identical point charges, of magnitude 35.0 mC, separated by a total distance of 310 m (the length of the cord plus the radius of each sphere). Each of these charges repels the other with a force of magnitude

(35.0 × 10 −3 C) = 115 N Q2 Fe = ke 2 = (8.99 × 10 9 N ⋅ m 2 C2 ) 2 r (3.10 × 10 2 m ) 2

Thus, to counterbalance this repulsion and hold each sphere in equilibrium, the cord must have a tension of 115 N , so it will exert a 115 N on that sphere, directed toward the other sphere. 15.50

(a)

As shown in Example 15.8 in the textbook, the electric ﬁeld due to a nonconducting plane sheet of charge has a constant magnitude of E = s 2 ∈0 , where s is the uniform charge per unit area on the sheet. The direction of the ﬁeld at all locations is perpendicular to the plane sheet and directed away from the sheet if s is positive, and toward the sheet if s is negative. Thus, if s = +5.20 mC m 2, the magnitude of the electric ﬁeld at all distances greater than zero from the plane (including the distance of 8.70 cm) is E=

s +5.20 × 10 −6 C m 2 = = 2.94 × 10 5 N C 2 ∈0 2 (8.85 × 10 −12 C2 N ⋅ m 2 )

continued on next page

68719_15_ch15_p001-029.indd 22

1/7/11 2:29:19 PM

Electric Forces and Electric Fields

(b) 15.51

23

The ﬁeld does not vary with distance as long as the distance is small compared with the dimensions of the sheet.

The three contributions to the resultant electric ﬁeld at the point of interest are shown in the sketch at the right. The magnitude of the resultant ﬁeld is ER = −E1 + E2 + E3 ER = −

ke q1 ke q2 ke q3 + 2 + 2 = ke r12 r2 r3

⎡ q1 q3 ⎤ q2 ⎢− 2 + 2 + 2 ⎥ r2 r3 ⎥⎦ ⎢⎣ r1

⎛ N ⋅ m 2 ⎞ ⎡ 4.0 × 10 −9 C 5.0 × 10 −9 C 3.0 × 10 −9 C ⎤ ER = ⎜ 8.99 × 10 9 + + ⎢− ⎥ 2 C2 ⎟⎠ ⎢⎣ ( 2.5 m ) ⎝ (1.2 m )2 ⎥⎦ ( 2.0 m )2 ER = + 24 N C or E R = 24 N C in the +x-direction 15.52

Consider the force diagram shown at the right. ΣFy = 0 ⇒ T cosq = mg or T =

mg cosq

⎛ mg ⎞ ΣFx = 0 ⇒ Fe = T sinq = ⎜ sinq = mg tanq ⎝ cosq ⎟⎠ Since Fe = qE, we have qE = mg tanq , or q = mg tanq E

( 2.00 × 10 q= 15.53

(a)

−3

kg ) ( 9.80 m s 2 ) tan15.0°

1.00 × 10 3 N C

= +5.25 × 10 −6 C = + 5.25 mC

At a point on the x-axis, the contributions by the two charges to the resultant ﬁeld have equal magnitudes given by E 1 = E2 = ke q r 2. The components of the resultant ﬁeld are ⎛ k q⎞ ⎛ k q⎞ E y = E1 y − E2 y = ⎜ e2 ⎟ sinq − ⎜ e2 ⎟ sinq = 0 ⎝ r ⎠ ⎝ r ⎠ ⎡ k ( 2q ) ⎤ ⎛ k q⎞ ⎛ k q⎞ Ex = E1 x + E2 x = ⎜ e2 ⎟ cosq + ⎜ e2 ⎟ cosq = ⎢ e 2 ⎥ cosq ⎝ r ⎠ ⎝ r ⎠ ⎣ r ⎦

and

Since

cosq b r b b = 2 = 3 = 2 , the resultant ﬁeld is 3 2 2 r r r ( a + b2 )

ER =

ke ( 2q ) b

(a

2

+ b2 )

3 2

in the +x-direction

continued on next page

68719_15_ch15_p001-029.indd 23

1/7/11 2:29:22 PM

24

Chapter 15

Note that the result of part (a) may be written as ER =

(b)

ke (Q ) b

(a

2

total charge in the charge distribution generating the ﬁeld.

+ b2 )

3 2

, where Q = 2q is the

In the case of a uniformly charged circular ring, consider the ring to consist of a very large number of pairs of charges uniformly spaced around the ring. Each pair consists of two identical charges located diametrically opposite each other on the ring. The total charge of pair number i is Qi . At a point on the axis of the ring, this pair of charges generates an elec3 2 tric ﬁeld contribution that is parallel to the axis and has magnitude Ei = ke b Qi ( a 2 + b 2 ) . The resultant electric ﬁeld of the ring is the summation of the contributions by all pairs of charges, or ⎡ kb ER = ΣEi = ⎢ 2 e 2 3 2 ⎢ (a + b ) ⎣

⎤ ke b Q ⎥ ΣQi = 3 2 2 ⎥ (a + b2 ) ⎦

where Q = ΣQi is the total charge on the ring. ER = 15.54

(a

ke Q b 2

+ b2 )

3 2

in the +x-direction

It is desired that the electric ﬁeld exert a retarding force on the electrons, slowing them down and bringing them to rest. For the retarding force to have maximum effect, it should be anti-parallel to the direction of the electron’s motion. Since the force an electric ﬁeld exerts on negatively charged particles (such as electrons) is in the direction opposite to the ﬁeld, the electric field should be in the direction of the electron’s motion. The work a retarding force of magnitude Fe = q E = eE does on the electrons as they move distance d is W = Fe d cos180° = −Fe d = −eEd. The work-energy theorem (W = ΔKE ) then gives −eEd = KE f − KEi = 0 − K and the magnitude of the electric ﬁeld required to stop the electrons in distance d is E=

15.55

−K −ed

E = K ed

or

F2

Consider the sketch at the right and observe: ⎛d⎞ q = tan ⎜ ⎟ = 45.0° ⎝d⎠

+Q

−1

Fx = F1 x + F2 x = +

68719_15_ch15_p001-029.indd 24

r

d

+2Q

Thus,

or

F1

r = d2 + d2 = d 2

and

and

q

d

−Q

ke −Q Q ke Q 2 k Q2 ⎛ 2 ⎞ 2 ⎛ ke Q 2 ⎞ = + 0.354 = + sin 45.0° + 0 = + e 2 ⎜ 2 2 4 ⎜⎝ d ⎟⎠ r 2d ⎝ 2 ⎟⎠ d2

Fy = F1 y + F2 y = −

ke −Q Q ke (Q ) ( 2Q ) ke Q 2 ⎛ 2 ⎞ 2ke Q 2 + cos 45.0° + = − r2 d2 2d 2 ⎜⎝ 2 ⎟⎠ d2

⎛ ke Q 2 2 ⎞ ke Q 2 Fy = ⎜ 2 − = +1.65 2 4 ⎟⎠ d d2 ⎝

1/7/11 2:29:24 PM

Electric Forces and Electric Fields

15.56

(a)

25

The downward electrical force acting on the ball is Fe = qE = ( 2.00 × 10 −6 C ) (1.00 × 10 5 N C ) = 0.200 N The total downward force acting on the ball is then F = Fe + mg = 0.200 N + (1.00 × 10 −3 kg ) ( 9.80 m s 2 ) = 0.210 N Thus, the ball will behave as if it were in a modiﬁed gravitational ﬁeld where the effective free-fall acceleration is F 0.210 N = = 210 m s 2 m 1.00 × 10 −3 kg

" g" =

The period of the pendulum will be L 0.500 m = 2p = 0.307 s "g" 210 m s 2

T = 2p (b)

Yes . The force of gravity is a signiﬁcant portion of the total downward force acting on the ball. Without gravity, the effective acceleration would be "g" =

Fe 0.200 N = = 200 m s 2 m 1.00 × 10 −3 kg 0.500 m = 0.314 s 200 m s 2

giving T = 2p

a 2.28% difference from the correct value with gravity included. 15.57

The sketch at the right gives a force diagram of the positively 2 charged sphere. Here, F1 = ke q r 2 is the attractive force exerted by the negatively charged sphere, and F2 = qE is exerted by the electric ﬁeld. ΣFy = 0 ⇒ T cos10° = mg or T =

mg cos10° 2

ΣFx = 0 ⇒ F2 = F1 + T sin10° or qE =

ke q + mg tan10° r2

At equilibrium, the distance between the two spheres is r = 2 ( L sin10° ). Thus, E=

=

ke q mg tan10° + 2 q 4 ( L sin10° )

(8.99 × 10

9

N ⋅ m 2 C2 ) ( 5.0 × 10 −8 C )

4 [( 0.100 m ) sin10° ]

or the needed electric ﬁeld strength is

68719_15_ch15_p001-029.indd 25

2

+

( 2.0 × 10

−3

kg ) ( 9.80 m s 2 ) tan10°

(5.0 × 10

−8

C)

E = 4.4 × 10 5 N C

1/7/11 2:29:26 PM

26

15.58

Chapter 15

(a)

At any point on the x-axis in the range 0 < x < 1.00 m, the contributions made to the resultant electric ﬁeld by the two charges are both in the positive x-direction. Thus, it is not possible for these contributions to cancel each other and yield a zero ﬁeld.

(b)

Any point on the x-axis in the range x < 0 is located closer to the larger magnitude charge ( q = 5.00 mC) than the smaller magnitude charge ( q = 4.00 mC). Thus, the contribution to the resultant electric ﬁeld by the larger charge will always have a greater magnitude than the contribution made by the smaller charge. It is not possible for these contributions to cancel to give a zero resultant ﬁeld.

(c)

If a point is on the x-axis in the region x > 1.00 m, the contributions made by the two charges are in opposite directions. Also, a point in this region is closer to the smaller magnitude charge than it is to the larger charge. Thus, there is a location in this region where the contributions of these charges to the total ﬁeld will have equal magnitudes and cancel each other.

(d)

When the contributions by the two charges cancel each other, their magnitudes must be equal. That is, ke

(5.00 mC) = k ( 4.00 mC ) x2

e

( x − 1.00 m )2

Thus, the resultant ﬁeld is zero at 15.59

or x − 1.00 m = +x 4 5

x=

1.00 m = + 9.47 m 1− 4 5

The two spheres have charges q1 and q2 = 2q1, so the repulsive force that one exerts on the other has magnitude Fe = ke q1 q2 r 2 = 2ke q12 r 2 .

.

From Figure P15.59 in the textbook, observe that the distance separating the two spheres is r = 3.00 cm + 2 [( 5.00 cm ) sin10.0° ] = 4.74 cm = 0.047 4 m From the force diagram of one sphere given on the right, observe that ΣFy = 0 ⇒ T cos10.0° = mg or T = mg cos10.0° ⎛ mg ⎞ and ΣFx = 0 ⇒ Fe = T sin10.0° = ⎜ sin10.0° = mg tan10.0° ⎝ cos10.0° ⎟⎠ Thus, 2ke q12 r 2 = mg tan10.0° or

q1 =

giving and

68719_15_ch15_p001-029.indd 26

mgr 2 tan10.0° = 2ke

( 0.0150 kg) (9.80

m s 2 ) ( 0.0474 m ) tan10.0° 2

2 (8.99 × 10 9 N ⋅ m 2 C2 )

q1 = 5.69 × 10 −8 C as the charge on one sphere,

q2 = 2q1 = 1.14 × 10 −7 C as the charge on the other sphere.

1/7/11 2:29:28 PM

Electric Forces and Electric Fields

15.60

d = 1.50 m

Consider the sketch at the right. The electrical forces acting on the third charge, Q, have magnitudes q1 = 3q

k q Q 3k qQ F1 = e 21 = e 2 x x F2 =

and

27

F2 Q

F1

x

d−x

q2 = q

ke q2 Q k qQ = e 2 ( d − x ) ( d − x )2

The bead with charge Q is in equilibrium when F1 = F2, or 3 ke qQ x

2

15.61

ke qQ

( d − x )2

3(d − x ) = x 2 2

giving and

=

x=

3d = 1+ 3

3 (d − x ) = x

or

3 (1.50 m ) = 0.951 m 1+ 3

Because of the spherical symmetry of the charge distribution, any electric ﬁeld present will be radial in direction. If a ﬁeld does exist at distance R from the center, it is the same as if the net charge located within r ≤ R were concentrated as a point charge at the center of the inner sphere. Charge located at r > R does not contribute to the ﬁeld at r = R. (a)

At r = 1.00 cm, E = 0 since static electric ﬁelds cannot exist within conducting materials.

(b)

The net charge located at r ≤ 3.00 cm is Q = +8.00 mC. Thus, at r = 3.00 cm, E=

=

ke Q r2

(8.99 × 10

9

N ⋅ m 2 C2 ) (8.00 × 10 −6 C )

(3.00 × 10

−2

m)

2

= 7.99 × 10 7 N C ( outward )

(c)

At r = 4.50 cm, E = 0 , since this is located within conducting materials.

(d)

The net charge located at r ≤ 7.00 cm is Q = + 4.00 mC. Thus, at r = 7.00 cm,

continued on next page

68719_15_ch15_p001-029.indd 27

1/7/11 2:29:30 PM

28

Chapter 15

ke Q r2

E=

(8.99 × 10

= 15.62

9

N ⋅ m 2 C2 ) ( 4.00 × 10 −6 C )

( 7.00 × 10 −2 m )

2

= 7.34 × 10 6 N C ( outward )

Consider the free-body diagram of the rightmost charge given at the right. ΣFy = 0 ⇒ T cosq = mg

T = mg cosq

or

ΣFx = 0 ⇒ Fe = T sinq = ( mg cosq ) sinq = mg tanq

and

Fe =

But,

ke q 2 ke q 2 ke q 2 ke q 2 5ke q 2 + = + = 2 2 r12 r22 ( L sinq ) ( 2L sinq ) 4L2 sin 2 q

5ke q 2 = mg tanq 4L2 sin 2 q

Thus,

q=

or

4L2 mg sin 2 q tanq 5ke

If q = 45.0°, m = 0.100 kg, and L = 0.300 m, then 4 ( 0.300 m ) ( 0.100 kg ) ( 9.80 m s 2 ) sin 2 ( 45.0° ) tan ( 45.0° ) 2

q= or 15.63

(a)

5 (8.99 × 10 9 N ⋅ m 2 C2 )

q = 1.98 × 10 −6 C = 1.98 mC When an electron (negative charge) moves distance Δx in the direction of an electric ﬁeld, the work done on it is W = Fe ( Δx ) cosq = eE ( Δx ) cos180° = −eE ( Δx )

(

)

From the work-energy theorem Wnet = KE f − KEi with KE f = 0, we have −eE ( Δx ) = −KEi , (b)

or

KEi 1.60 × 10 −17 J = = 1.00 × 10 3 N C e ( Δx ) (1.60 × 10 −19 C ) ( 0.100 m )

v − v0 0 − 2 ( KEi ) m = = − eE m a

t=

2m ( KEi ) eE

2 ( 9.11 × 10 −31 kg ) (1.60 × 10 −17 J )

(1.60 × 10

−19

C ) (1.00 × 10 3 N C )

= 3.37 × 10 −8 s = 33.7 ns

After bringing the electron to rest, the electric force continues to act on it, causing the electron to accelerate in the direction opposite to the field at a rate of a=

68719_15_ch15_p001-029.indd 28

E=

The magnitude of the retarding force acting on the electron is Fe = eE, and Newton’s second law gives the acceleration as a = − Fe m = − eE m. Thus, the time required to bring the electron to rest is t=

(c)

or

−19 3 eE (1.60 × 10 C ) (1.00 × 10 N C ) = = 1.76 × 1014 m s 2 m 9.11× 10 −31 kg

1/7/11 2:29:32 PM

Electric Forces and Electric Fields

15.64

(a)

29

The acceleration of the protons is downward (in the direction of the ﬁeld) and ay =

−19 Fe eE (1.60 × 10 C ) ( 720 N C ) = = = 6.90 × 1010 m s 2 m m 1.67 × 10 −27 kg

The time of ﬂight for the proton is twice the time required to reach the peak of the arc, or ⎛ v ⎞ 2v sinq 0y t = 2tpeak = 2 ⎜ ⎟= 0 ⎜⎝ a y ⎟⎠ ay The horizontal distance traveled in this time is ⎛ 2v sinq ⎞ v 2 sin 2q R = v0 x t = ( v0 cosq ) ⎜ 0 ⎟= 0 ⎜⎝ ⎟⎠ ay ay Thus, if R = 1.27 × 10 −3 m, we must have sin 2q =

ay R v02

(6.90 × 10

10

=

m s 2 ) (1.27 × 10 −3 m )

(9 550

m s)

2

= 0.961

giving 2q = 73.9° or 2q = 180° − 73.9° = 106°. Hence, q = 37.0° or 53.0° (b)

68719_15_ch15_p001-029.indd 29

The time of ﬂight for each possible angle of projection is: For q = 37.0°:

t=

2v0 sinq

For q = 53.0°:

t=

2v0 sinq

ay

ay

=

2 ( 9 550 m s ) sin 37.0° = 1.67 × 10 −7 s 6.90 × 1010 m s 2

=

2 ( 9 550 m s ) sin 53.0° = 2.21× 10 −7 s 6.90 × 1010 m s 2

1/7/11 2:29:35 PM

16 Electrical Energy and Capacitance QUICK QUIZZES 1.

Choice (b). The field exerts a force on the electron, causing it to accelerate in the direction opposite to that of the field. In this process, electrical potential energy is converted into kinetic energy of the electron. Note that the electron moves to a region of higher potential, but because the electron has negative charge this corresponds to a decrease in the potential energy of the electron.

2.

Choice (a). The electron, a negatively charged particle, will move toward the region of higher electric potential. Because of the electron’s negative charge, this corresponds to a decrease in electrical potential energy.

3.

Choice (b). Charged particles always tend to move toward positions of lower potential energy. The electrical potential energy of a charged particle is PE = qV and, for positively-charged particles, this decreases as V decreases. Thus, a positively-charged particle located at x = A would move toward the left.

4.

Choice (d). For a negatively-charged particle, the potential energy (PE = qV ) decreases as V increases. A negatively charged particle would oscillate around x = B which is a position of minimum potential energy for negative charges.

5.

Choice (d). If the potential is zero at a point located a finite distance from charges, negative charges must be present in the region to make negative contributions to the potential and cancel positive contributions made by positive charges in the region.

6.

Choice (c). Both the electric potential and the magnitude of the electric field decrease as the distance from the charged particle increases. However, the electric flux through the balloon does not change because it is proportional to the total charge enclosed by the balloon, which does not change as the balloon increases in size.

7.

Choice (a). From the conservation of energy, the final kinetic energy of either particle will be given by

(

)

(

)

KE f = KEi + PEi − PE f = 0 + qVi − qV f = −q V f − Vi = −q ( ΔV ) For the electron, q = −e and ΔV = +1 V giving KE f = − ( −e ) ( +1 V ) = +1 eV. For the proton, q = +e and ΔV = −1 V, so KE f = − ( e ) ( −1 V ) = +1 eV, the same as that of the electron. 8.

Choice (c). The battery moves negative charge from one plate and puts it on the other. The first plate is left with excess positive charge whose magnitude equals that of the negative charge moved to the other plate.

9.

(a) C decreases. (d) ΔV increases.

(b) Q stays the same. (c) (e) The energy stored increases.

E stays the same.

30

68719_16_ch16_p030-061.indd 30

1/7/11 2:32:36 PM

Electrical Energy and Capacitance

31

Because the capacitor is removed from the battery, charges on the plates have nowhere to go. Thus, the charge on the capacitor plates remains the same as the plates are pulled apart. Because E = s ∈0 = (Q A) ∈0 , the electric field is constant as the plates are separated. Because ΔV = Ed and E does not change, ΔV increases as d increases. Because the same charge is stored at a higher potential difference, the capacitance (C = Q ΔV ) has decreased. Because energy stored = Q 2 2C and Q stays the same while C decreases, the energy stored increases. The extra energy must have been transferred from somewhere, so work was done. This is consistent with the fact that the plates attract one another, and work must be done to pull them apart. 10.

(a) C increases. (d) ΔV remains the same.

(b) Q increases. (c) (e) The energy stored increases.

E stays the same.

The presence of a dielectric between the plates increases the capacitance by a factor equal to the dielectric constant. Since the battery holds the potential difference constant while the capacitance increases, the charge stored (Q = CΔV ) will increase. Because the potential difference and the distance between the plates are both constant, the electric field (E = ΔV d) will stay the same. The battery maintains a constant potential difference. With ΔV constant while capacitance increases, the stored energy [energy stored = 12 C(ΔV )2 ] will increase. 11.

Choice (a). Increased random motions associated with an increase in temperature make it more difficult to maintain a high degree of polarization of the dielectric material. This has the effect of decreasing the dielectric constant of the material, and in turn, decreasing the capacitance of the capacitor.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

The change in the potential energy of the proton is equal to the negative of the work done on it by the electric field. Thus,

(

)

ΔPE = −W = −qE x ( Δx ) = − +1.6 × 10 −19 C (850 N C ) ( 2.5 m − 0 ) = −3.4 × 10 −16 J and (b) is the correct choice for this question. 2.

Because electric forces are conservative, the kinetic energy gained is equal to the decrease in electrical potential energy, or

(

)

KE = −PE = −q ( ΔV ) = − ( −1 e ) +1.00 × 10 4 V = +1.00 × 10 4 eV so the correct choice is (a). 3.

In a uniform electric field, the change in electric potential is ΔV = −E x ( Δx ), giving Ex = −

( (

) )

V f − Vi ΔV (190 V − 120 V ) =− =− = −35 V m = −35 N C Δx ( 5.0 m − 3.0 m ) x f − xi

and it is seen that the correct choice is (d).

68719_16_ch16_p030-061.indd 31

1/7/11 2:32:37 PM

32

4.

Chapter 16

From conservation of energy, KE f + PE f = KEi + PEi , or speed of the nucleus is vB = vA2 +

(

=

1

2

mvB2 = 1 2 mvA2 + qVA − qVB , the final

2q(VA − VB ) m

(

)

2 ⎡⎣ 2 1.60 × 10 −19 C (1.50 − 4.00 ) × 10 3 V ⎤⎦ 6.20 × 10 m s + = 3.78 × 10 5 m s 6.63 × 10 −27 kg 5

)

2

Thus, the correct answer is choice (b). 5.

In a series combination of capacitors, the equivalent capacitance is always less than any individual capacitance in the combination, meaning that choice (a) is false. Also, for a series combination of capacitors, the magnitude of the charge is the same on all plates of capacitors in the combination, making both choices (d) and (e) false. The potential difference across the capacitance Ci is ΔVi = Q Ci , where Q is the common charge on each capacitor in the combination. Thus, the largest potential difference (voltage) appears across the capacitor with the least capacitance, making choice (b) the correct answer.

6.

The total potential at a point due to a set of point charges qi is V = ∑ kqi ri i

where ri is the distance from the point of interest to the location of the charge qi. Note that in this case, the point at the center of the circle is equidistant from the 4 point charges located on the rim of the circle. Note also that q2 + q3 + q4 = ( +1.5 − 1.0 − 0.5) mC = 0, so we have ke q1 ke q2 ke q3 ke q4 ke k kq + + + = ( q1 + q2 + q3 + q4 ) = e ( q1 + 0 ) = e 1 = V1 r r r r r r r 4 = 4.5 × 10 V or the total potential at the center of the circle is just that due to the first charge alone, and the correct answer is choice (b). Vcenter =

7.

With the given specifications, the capacitance of this parallel-plate capacitor will be C=

(

)(

)(

)

2 −12 C2 N ⋅ m 2 1.00 cm 2 ⎛ 1 m 2 ⎞ k ∈0 A 1.00 × 10 8.85 × 10 = ⎜⎝ 10 4 cm 2 ⎟⎠ 1.00 × 10 −3 m d

= 8.85 × 10 −11 F = 88.5 × 10 −12 F = 88.5 pF and the correct choice is (a). 8.

Keeping the capacitor connected to the battery means that the potential difference between the plates is kept at a constant value equal to the voltage of the battery. Since the capacitance of a parallel-plate capacitor is C = k ∈0 A d, doubling the plate separation d, while holding other characteristics of the capacitor constant, means the capacitance will be decreased by a factor of 2. The energy stored in a capacitor may be expressed as U = 12 C(ΔV )2, so when the potential difference ΔV is held constant while the capacitance is decreased by a factor of 2, the stored energy decreases by a factor of 2, making (c) the correct choice for this question.

9.

When the battery is disconnected, there is no longer a path for charges to use in moving onto or off of the plates of the capacitor. This means that the charge Q is constant. The capacitance of a

68719_16_ch16_p030-061.indd 32

1/7/11 2:32:39 PM

Electrical Energy and Capacitance

33

parallel-plate capacitor is C = k ∈0 A d and the dielectric constant is k ≈ 1 when the capacitor is air filled. When a dielectric with dielectric constant k = 2 is inserted between the plates, the capacitance is doubled C f = 2Ci . Thus, with Q constant, the potential difference between the plates, ΔV = Q C, is decreased by a factor of 2, meaning that choice (a) is a true statement. The electric field between the plates of a parallel-plate capacitor is E = ΔV d and decreases when ΔV decreases, making choice (e) false and leaving (a) as the only correct choice for this question.

(

)

10.

Once the capacitor is disconnected from the battery, there is no path for charges to move onto or off of the plates, so the charges on the plates are constant, and choice (e) can be eliminated. The capacitance of a parallel-plate capacitor is C = k ∈0 A d, so the capacitance decreases when the plate separation d is increased. With Q constant and C decreasing, the energy stored in the capacitor, U = Q 2 2C, increases, making choice (a) false and choice (b) true. The potential difference between the plates, ΔV = Q C = Q ⋅ d k ∈0 A, increases and the electric field between the plates, E = ΔV d = Q k ∈0 A, is constant. This means that both choices (c) and (d) are false and leaves choice (b) as the only correct response.

11.

Capacitances connected in parallel all have the same potential difference across them and the equivalent capacitance, Ceq = C1 + C2 + C3 + … , is larger than the capacitance of any one of the capacitors in the combination. Thus, choice (c) is a true statement. The charge on a capacitor is Q = C(ΔV ), so with ΔV constant, but the capacitances different, the capacitors all store different charges that are proportional to the capacitances, making choices (a), (b), (d), and (e) all false. Therefore, (c) is the only correct answer.

12.

For a series combination of capacitors, the magnitude of the charge is the same on all plates of capacitors in the combination. Also, the equivalent capacitance is always less than any individual capacitance in the combination. Therefore, choice (a) is true while choices (b) and (c) are both false. The potential difference across a capacitor is ΔV = Q C, so with Q constant, capacitors having different capacitances will have different potential differences across them, with the largest potential difference being across the capacitor with the smallest capacitance. This means that choices (d) and (e) are false, and choice (f) is true. Thus, both choices (a) and (f) are true statements.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The potential energy between a pair of point charges separated by distance R is PE = ke q1q2 R. Thus, the potential energy for each of the four systems is: (a)

PEa = ke

Q ( 2Q ) Q2 = 2ke r r

(b)

PEb = ke

( −Q ) ( −Q ) = k

(c)

PEc = ke

Q ( −Q ) 1 Q2 = − ke 2 2r r

(d)

PEd = ke

( −Q ) ( −2Q ) = k

r

2r

Q2 r

e

e

Q2 r

Therefore, the correct ranking from largest to smallest is (a) > (b) = (d) > (c). 4.

To move like charges together from an infinite separation, at which the potential energy of the system of two charges is zero, requires work to be done on the system by an outside agent. Hence energy is stored, and potential energy is positive. As charges with opposite signs move together from an infinite separation, energy is released, and the potential energy of the set of charges becomes negative.

6.

A sharp point on a charged conductor would produce a large electric field in the region near the point. An electric discharge could most easily take place at the point.

68719_16_ch16_p030-061.indd 33

1/7/11 2:32:40 PM

34

8.

Chapter 16

There are eight different combinations that use all three capacitors in the circuit. These combinations and their equivalent capacitances are: ⎛ 1 1 1⎞ All three capacitors in series: Ceq = ⎜ + + ⎟ C C C ⎝ 1 2 3⎠

−1

All three capacitors in parallel: Ceq = C1 + C2 + C3 One capacitor in series with a parallel combination of the other two: −1

−1

⎛ 1 ⎛ 1 ⎛ 1⎞ 1 ⎞ 1 1⎞ Ceq = ⎜ + ⎟ , Ceq = ⎜ + ⎟ , Ceq = ⎜ + ⎟ ⎝ C1 + C2 C3 ⎠ ⎝ C3 + C1 C2 ⎠ ⎝ C2 + C3 C1 ⎠

−1

One capacitor in parallel with a series combination of the other two: ⎛ CC ⎞ ⎛ CC ⎞ ⎛ CC ⎞ Ceq = ⎜ 1 2 ⎟ + C3, Ceq = ⎜ 3 1 ⎟ + C2, Ceq = ⎜ 2 3 ⎟ + C1 C + C C + C ⎝ 1 ⎝ 3 ⎝ C 2 + C3 ⎠ 2⎠ 1⎠ 10.

12.

(a)

If the wires are disconnected from the battery and not allowed to touch each other or another object, the charge on the plates is unchanged.

(b)

If, after being disconnected from the battery, the wires are connected to each other, electrons will rapidly flow from the negatively charged plate to the positively charged plate to leave the capacitor uncharged with both plates neutral.

The primary choice would be the dielectric. You would want to choose a dielectric that has a large dielectric constant and dielectric strength, such as strontium titanate, where k ≈ 233 (Table 16.1). A convenient choice could be thick plastic or Mylar. Secondly, geometry would be a factor. To maximize capacitance, one would want the individual plates as close as possible, since the capacitance is proportional to the inverse of the plate separation—hence the need for a dielectric with a high dielectric strength. Also, one would want to build, instead of a single parallel-plate capacitor, several capacitors in parallel. This could be achieved through “stacking” the plates of the capacitor. For example, you can alternately lay down sheets of a conducting material, such as aluminum foil, sandwiched between your sheets of insulating dielectric. Making sure that none of the conducting sheets are in contact with their nearest neighbors, connect every other plate together as illustrated in the figure below.

Dielectric

Conductor This technique is often used when “home-brewing” signal capacitors for radio applications, as they can withstand huge potential differences without flashover (without either discharge between plates around the dielectric or dielectric breakdown). One variation on this technique is to sandwich together flexible materials such as aluminum roof flashing and thick plastic, so the whole product can be rolled up into a “capacitor burrito” and placed in an insulating tube, such as a PVC pipe, and then filled with motor oil (again to prevent flashover).

68719_16_ch16_p030-061.indd 34

1/7/11 2:32:43 PM

Electrical Energy and Capacitance

14.

35

The material of the dielectric may be able to withstand a larger electric field than air can withstand before breaking down to pass a spark between the capacitor plates.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a)

6.16 × 10 −17 N

(b)

3.69 × 1010 m s 2 in the direction of the electric field

(c)

7.38 cm

4.

6.67 × 1011 electrons

6.

(a) 1.10 × 10 −2 N to the right

8.

(d)

− 49.5 V

(a)

− 2.31 kV

(b) 1.98 × 10 −3 J

(c)

−1.98 × 10 −3 J

(b)

Protons would require a greater potential difference.

(b)

+10.8 kV

(c) ΔVp ΔVe = − mp me 10.

40.2 kV

12.

(a)

14.

−9.08 J

16.

(a)

See Solution.

(b) 3ke q a

18.

(a)

See Solution.

(b)

2⎞ 1 ⎛ V = ( 22.5 V ⋅ m ) ⎜ − ⎟ ⎝ 1.20 m − x x ⎠

(c)

−37.5 V

(d)

x = 0.800 m

(a)

Conservation of energy alone yields one equation with two unknowns.

(b)

Conservation of linear momentum

(c)

v p = 1.05 × 10 7 m s , vα = 2.64 × 10 6 m s

20.

+5.39 kV

22.

5.4 × 10 5 V

24.

(a) V = 4 2keQ a

26.

(a)

28.

(a) 1.00 mF

30.

31.0 Å

32.

1.23 kV

68719_16_ch16_p030-061.indd 35

3.00 mF

(c)

See Solution.

(b) W = 4 2ke qQ a (b) 36.0 mC (b) 100 V

1/7/11 3:11:16 PM

36

34.

Chapter 16

(a) 17.0 mF (c)

(b)

9.00 V

45.0 mC on C1 , 108 mC on C2

36.

3.00 pF and 6.00 pF

38.

(a)

6.00 mF

(d)

Q4 = 144 mC, Q2 = 72.0 mC, Qrightmost = 216 mC

(b) 12.0 mF

(c)

432 mC

(e) Q24 = Q8 = 216 mC

branch

(f) 40.

9.00 V

(a) 2C

(g)

27.0 V

(b)

Q1 > Q3 > Q2

(c)

ΔV1 > ΔV2 = ΔV3

(d) Q1 and Q3 increase, Q2 decreases 42.

(a)

6.04 mF

44.

(a)

5.96 mF

(b) 83.6 mC

(b) 89.4 mC on the 20.0 mF capacitor, 63.0 mC on the 6.00 mF capacitor, 26.3 mC on the 15.0 mF capacitor, and 26.3 mC on the 3.00 mF capacitor 46.

(a) Ceq = 12.0 mF, Estored,total = 8.64 × 10 −4 J (b)

Estored,1 = 5.76 × 10 −4 J, Estored,2 = 2.88 × 10 −4 J It will always be true that Estored,1 + Estored,2 = Estored,total.

(c)

5.66 V; C2 , with the largest capacitance, stores the most energy.

48.

9.79 kg

50.

(a) 13.3 nC

52.

1.04 m

54.

0.443 mm

56.

(a)

13.5 mJ

(b)

Estored,2 = 3.60 mJ, Estored,3 = 5.40 mJ, Estored,4 = 1.80 mJ, Estored,6 = 2.70 mJ

(b)

272 nC

(c) The energy stored in the equivalent capacitance equals the sum of the energies stored in the individual capacitors. 1 Cp ± 2

1 2 1 C p − C pCs , C2 = C p ∓ 4 2

58.

C1 =

60.

(a) 1.8 × 10 4 V (d)

62.

68719_16_ch16_p030-061.indd 36

1 2 C p − C pCs 4

(b)

−3.6 × 10 4 V

(b)

See Solution.

(c)

−1.8 × 10 4 V

−5.4 × 10 −2 J

(a) C =

ab ke ( b − a )

1/7/11 2:32:46 PM

Electrical Energy and Capacitance

64.

k = 2.33

66.

(a)

(d) 68.

(a)

2ke q d 5

(b)

4ke q 2 d 5

(b)

4.4 mm

37

4ke q 2 d 5

(c)

8ke q 2 md 5 0.1 mm

PROBLEM SOLUTIONS 16.1

(a)

Because the electron has a negative charge, it experiences a force in the direction opposite to the field and, when released from rest, will move in the negative x-direction. The work done on the electron by the field is

(

)

(

W = Fx ( Δx ) = ( qE x ) Δx = −1.60 × 10 −19 C ( 375 N C ) −3.20 × 10 −2 m

)

= 1.92 × 10 −18 J (b)

The change in the electric potential energy is the negative of the work done on the particle by the field. Thus, ΔPE = −W = −1.92 × 10 −18 J

(c)

Since the Coulomb force is a conservative force, conservation of energy gives ΔKE + ΔPE = 0, or KE f = 12 me v 2f = KEi − ΔPE = 0 − ΔPE, and vf =

16.2

(

9.11× 10 −31 kg

)=

2.05 × 10 6 m s in the − x-direction

)

(a)

F = qE = 1.60 × 10 −19 C ( 385 N C ) = 6.16 × 10 −17 N

(b)

a=

(c)

Δx = v0 t +

F 6.16 × 10 −17 N = = 3.69 × 1010 m s 2 in the direction of the electric field m p 1.67 × 10 −27 kg

(

)(

1 2 1 at = 0 + 3.69 × 1010 m s 2 2.00 × 10 −6 s 2 2

= 7.38 × 10 16.3

(

−2 −1.92 × 10 −18 J

−2 ( ΔPE ) = me

−2

)

2

m = 7.38 cm

The work done by the agent moving the charge out of the cell is Winput = −Wfield = − ( −ΔPEe ) = +q ( ΔV )

(

)(

)

= 1.60 × 10 −19 C + 90 × 10 −3 J C = 1.4 × 10 −20 J 16.4

Assuming the sphere is isolated, the excess charge on it is uniformly distributed over its surface. Under this spherical symmetry, the electric field outside the sphere is the same as if all the excess charge on the sphere were concentrated as a point charge located at the center of the sphere.

continued on next page

68719_16_ch16_p030-061.indd 37

1/7/11 2:32:50 PM

38

Chapter 16

Thus, at r = 8.00 cm > Rsphere = 5.00 cm, the electric field is E = ke Q r 2. The required charge then has magnitude Q = Er 2 ke, and the number of electrons needed is

(

)(

)

2

1.50 × 10 5 N C 8.00 × 10 −2 m Q Er 2 n= = 6.67 × 1011 electrons = = e ke e 8.99 × 10 9 N ⋅ m 2 C2 1.60 × 10 −19 C

(

)(

)

ΔV 25 × 10 3 J C = = 1.7 × 10 6 N C d 1.5 × 10 −2 m

16.5

E=

16.6

(a)

F = qE = +40.0 × 10 −6 C ( +275 N C ) = 1.10 × 10 −2 N directed toward the right

(b)

WAB = F ( Δx ) cosq = 1.10 × 10 −2 N ( 0.180 m ) cos 0° = 1.98 × 10 −3 J

(c)

ΔPE = −WAB = −1.98 × 10 −3 J

(d)

ΔV = VB − VA =

(a)

E=

(b)

F= qE=

(c)

W = F ⋅ s cosq

16.7

(

)

(

)

ΔPE −1.98 × 10 −3 J = = − 49.5 V q +40.0 × 10 −6 C

ΔV 600 J C = = 1.13 × 10 5 N C d 5.33 × 10 −3 m

(

)

1.60 × 10 −19 C ( 600 J C ) q ΔV = = 1.80 × 10 −14 N d 5.33 × 10 −3 m

(

)

= 1.80 × 10 −14 N ⎡⎣( 5.33 − 2.90 ) × 10 −3 m ⎤⎦ cos 0° = 4.37 × 10 −17 J 16.8

(a)

Using conservation of energy, ΔKE + ΔPE = 0, with KE f = 0 since the particle is “stopped,” we have 1 1 ⎛ ⎞ ΔPE = −ΔKE = − ⎜ 0 − me vi2 ⎟ = + 9.11 × 10 −31 kg 2.85 × 10 7 m s ⎝ ⎠ 2 2

(

)(

)

2

= +3.70 × 10 −16 J The required stopping potential is then ΔV = (b)

Being more massive than electrons, protons traveling at the same initial speed will have more initial kinetic energy and require a greater magnitude stopping potential .

(c)

Since ΔVstopping = ΔPE q = ( −ΔKE ) q = − mv 2 2 q, the ratio of the stopping potential for a proton to that for an electron having the same initial speed is

(

ΔVp ΔVe

68719_16_ch16_p030-061.indd 38

ΔPE +3.70 × 10 −16 J = = −2.31 × 10 3 V = −2.31 kV q −1.60 × 10 −19 C

=

− mp vi2 2(+e) − me vi2 2(−e)

)

= − mp me

1/7/11 2:32:51 PM

Electrical Energy and Capacitance

16.9

(a)

We use conservation of energy, Δ ( KE ) + Δ ( PEs ) + Δ ( PEe ) = 0, recognizing that Δ(KE) = 0 since the block is at rest at both the beginning and end of the motion. The change in the elastic potential energy is 2 given by Δ ( PEs ) = 12 kx max − 0, where x max is the maximum stretch of the spring. The change in the electrical potential energy is the negative of the work the electric field does, Δ ( PEe ) = −W = −Fe (Δx) = − (QE ) x max. 2 Thus, 0 + 12 kx max − (QE ) x max = 0, which yields x max =

(b)

39

(

)(

)

−6 4 2QE 2 35.0 × 10 C 4.86 × 10 V m = = 4.36 × 10 −2 m = 4.36 cm k 78.0 N m

At equilibrium, ΣF = Fs + Fe = 0, or − kx eq + QE = 0. Therefore, x eq =

QE 1 = x max = 2.18 cm k 2

The amplitude is the distance from the equilibrium position to each of the turning points ( at x = 0 and x = 4.36 cm ), so A = 2.18 cm = xmax 2 . (c)

2 From conservation of energy, Δ ( KE ) + Δ ( PEs ) + Δ ( PEe ) = 0, we have 0 + 12 kx max + QΔV = 0. Since x max = 2A, this gives 2 kx max k ( 2A ) =− 2Q 2Q

2

ΔV = − 16.10

ΔV = −

or

2kA2 Q

Using Δy = v0 y t + 12 a y t 2 for the full flight gives 0 = v0 y t f + 12 a y t 2f , or a y = −2v0 y t f , where t f is the full time of the flight. Then, using v 2y = v02 y + 2a y (Δy) for the upward part of the flight gives

( Δy )max =

0 − v02 y 2a y

=

(

−v02 y

2 −2 v0 y t f

)

=

v0 y t f 4

=

( 20.1 m s ) ( 4.10 s ) = 20.6 m 4

From Newton’s second law, ay =

ΣFy m

=

−mg − qE qE ⎞ ⎛ = −⎜g + ⎟ ⎝ m m⎠

Equating this to the earlier result gives a y = − ( g + qE m ) = −2v0 y t f , so the electric field strength is ⎤ ⎛ 2.00 kg ⎞ ⎡ 2 ( 20.1 m s ) ⎤ ⎛ m ⎞ ⎡ 2v0 y E=⎜ ⎟⎢ − g⎥ = ⎜ − 9.80 m s 2 ⎥ = 1.95 × 10 3 N C ⎟ ⎢ −6 ⎝ q ⎠ ⎢⎣ t f ⎥⎦ ⎝ 5.00 × 10 C ⎠ ⎣ 4.10 s ⎦

(

)

Thus, ( ΔV )max = ( Δymax ) E = ( 20.6 m ) 1.95 × 10 3 N C = 4.02 × 10 4 V = 40.2 kV 16.11

(

)(

)

(

)(

)

(a)

VA =

8.99 × 10 9 N ⋅ m 2 C2 −1.60 × 10 −19 C ke q = = −5.75 × 10 −7 V 0.250 × 10 −2 m rA

(b)

VB =

8.99 × 10 9 N ⋅ m 2 C2 −1.60 × 10 −19 C ke q = = −1.92 × 10 −7 V 0.750 × 10 −2 m rB

continued on next page

68719_16_ch16_p030-061.indd 39

1/7/11 2:32:54 PM

40

Chapter 16

(

)

ΔV = VB − VA = −1.92 × 10 −7 V − −5.75 × 10 −7 V = +3.83 × 10 −7 V (c)

16.12

(a)

No . The original electron will be repelled by the negatively charged particle which suddenly appears at point A. Unless the electron is fixed in place, it will move in the opposite direction, away from points A and B, thereby lowering the potential difference between these points. VA = ∑

⎛ −15.0 × 10 −9 C 27.0 × 10 −9 C ⎞ ke qi = 8.99 × 10 9 N ⋅ m 2 C2 ⎜ + = +5.39 kV ri ⎝ 2.00 × 10 −2 m 2.00 × 10 −2 m ⎟⎠

VB = ∑

⎛ −15.0 × 10 −9 C 27.0 × 10 −9 C ⎞ ke qi = 8.99 × 10 9 N ⋅ m 2 C2 ⎜ + = +10.8 kV ri ⎝ 1.00 × 10 −2 m 1.00 × 10 −2 m ⎟⎠

i

(b)

i

16.13

(a)

(

)

(

)

Calling the 2.00 mC charge q3, V =∑ i

⎛q q ke qi q = ke ⎜ 1 + 2 + 2 3 2 ⎜⎝ r1 r2 ri r1 + r2

⎞ ⎟ ⎟⎠

⎛ ⎛ N ⋅ m 2 ⎞ 8.00 × 10 −6 C 4.00 × 10 −6 C ⎜ = ⎜ 8.99 × 10 9 + + C2 ⎟⎠ ⎜ 0.060 0 m 0.030 0 m ⎝ ⎝

2.00 × 10 −6 C

( 0.060 0 )2 + ( 0.030 0 )2

⎞ ⎟ m ⎟⎠

V = 2.67 × 10 6 V (b)

Replacing 2.00 × 10 −6 C by − 2.00 × 10 −6 C in part (a) yields V = 2.13 × 10 6 V

16.14

(

)

W = q ( ΔV ) = q V f − Vi , and V f = 0 since the final location of the 8.00 mC is an infinite distance from other charges. The potential, due to the other charges, at the initial location of the 8.00 mC is Vi = ke ( q1 r1 + q2 r2 ) . Thus, the required energy for the move is ⎡ ⎛ q q ⎞⎤ W = q ⎢ 0 − ke ⎜ 1 + 2 ⎟ ⎥ ⎝ r1 r2 ⎠ ⎥⎦ ⎢⎣ ⎛ ⎛ N ⋅ m 2 ⎞ 2.00 × 10 −6 C ⎜ + = − 8.00 × 10 −6 C ⎜ 8.99 × 10 9 C2 ⎟⎠ ⎜ 0.030 0 m ⎝ ⎝

(

)

4.00 × 10 −6 C

( 0.030 0 )2 + ( 0.060 0 )2

⎞ ⎟ m ⎟⎠

W = − 9.08 J 16.15

(a)

V =∑ i

(b)

ke qi ⎛ N ⋅ m 2 ⎞ ⎛ 5.00 × 10 −9 C 3.00 × 10 −9 C ⎞ = ⎜ 8.99 × 10 9 − = 103 V ri C2 ⎟⎠ ⎜⎝ 0.175 m 0.175 m ⎟⎠ ⎝

(

)(

)

2 5.00 × 10 −9 C − 3.00 × 10 −9 C ke qi q2 ⎛ 9 N⋅m ⎞ PE = = ⎜ 8.99 × 10 = − 3.85 × 10 − 7 J 2 ⎟ r12 C ⎠ 0.350 m ⎝

The negative sign means that positive work must be done to separate the charges by an infinite distance (that is, bring them up to a state of zero potential energy).

68719_16_ch16_p030-061.indd 40

1/7/11 2:32:57 PM

41

Electrical Energy and Capacitance

16.16

(a)

At the center of the triangle, each of the identical charges produce a field contribution of magnitude E1 = ke q a 2 . The three contributions are oriented as shown at the right and the components of the resultant field are:

y E1

E1

30°

30° x

E x = ΣE x = +E1 cos 30° − E1 cos 30° = 0 E y = ΣE y = +E1 sin 30° − E1 + E1 sin 30° = 0 E1

Thus, the resultant field has magnitude E = E x2 + E y2 = 0 (b)

The total potential at the center of the triangle is V = ΣVi = Σ

(c)

16.17

ke qi ke q ke q ke q 3ke q = + + = ri a a a a

Imagine a test charge placed at the center of the triangle. Since the field is zero at the center, the test charge will experience no electrical force at that point. The fact that the potential is not zero at the center means that work would have to be done by an external agent to move a test charge from infinity to the center.

The Pythagorean theorem gives the distance from the midpoint of the base to the charge at the apex of the triangle as r3 =

( 4.00 cm )2 − (1.00 cm )2 = 15 cm = 15 × 10 −2 m

Then, the potential at the midpoint of the base is V = ∑ ke qi ri, or ⎛ V = ⎜ 8.99 × 10 9 ⎝

(

) (

i

) (

)

−9 −7.00 × 10 −9 C +7.00 × 10 −9 C ⎞ N ⋅ m ⎞ ⎛ −7.00 × 10 C + + ⎜ ⎟ C2 ⎟⎠ ⎝ 0.010 0 m 0.010 0 m 15 × 10 −2 m ⎠ 2

= −1.10 × 10 4 V = − 11.0 kV 16.18

(a)

See the sketch below: 1000

y

500 x ⫺1.2

⫺0.8

⫺0.4

0

0.4

0.8

1.2

1.6

2

⫺500

⫺1000

continued on next page

68719_16_ch16_p030-061.indd 41

1/7/11 2:32:59 PM

42

Chapter 16

(b)

At the point (x, 0), where 0 < x < 1.20 m, the potential is V =∑ i

ke qi ke ( −2q ) ke q 1 2⎞ ⎛ = − + = ke q ⎜ ⎝ 1.20 m − x x ⎟⎠ ri x 1.20 m − x

or ⎛ N ⋅ m2 ⎞ 1 2⎞ 1 2⎞ ⎛ ⎛ V = ⎜ 8.99 × 10 9 = ( 22.5 V ⋅ m ) ⎜ − − 2.50 × 10 −9 C ⎜ ⎝ 1.20 m − x x ⎟⎠ ⎝ 1.20 m − x x ⎟⎠ C2 ⎟⎠ ⎝

(

(c)

)

At x = + 0.600 m, the potential is 2 22.5 V ⋅ m 1 ⎛ ⎞ V = ( 22.5 V ⋅ m ) ⎜ − =− = −37.5 V ⎝ 1.20 m − 0.600 m 0.600 m ⎟⎠ 0.600 m

(d)

When 0 < x < 1.20 m and V = 0, we have 1 (1.20 m − x) − 2 x = 0, or x = 2.40 m − 2x. This yields x = 2.40 m 3 = 0.800 m .

16.19

(a)

When the charge configuration consists of only the two protons ( q1 and q2 in the sketch ), the potential energy of the configuration is

(

)(

8.99 × 10 9 N ⋅ m 2 C2 1.60 × 10 −19 C kqq PEa = e 1 2 = r12 6.00 × 10 −15 m

)

2

or PEa = 3.84 × 10 −14 J (b)

When the alpha particle ( q3 in the sketch ) is added to the configuration, there are three distinct pairs of particles, each of which possesses potential energy. The total potential energy of the configuration is now

( )

⎛ ke 2e 2 ⎞ ke q1q2 ke q1q3 ke q2 q3 + + = PEa + 2 ⎜ ⎟ r12 r13 r23 ⎝ r13 ⎠ where use has been made of the facts that q1q3 = q2 q3 = e ( 2e ) = 2e 2 and PEb =

r13 = r23 = ( 3.00 fm ) + ( 3.00 fm ) = 18.0 fm = 18.0 × 10 −15 m. Also, note that the first term in this computation is just the potential energy computed in part (a). Thus, 2

PEb = PEa +

4ke e 2 r13

= 3.84 × 10 (c)

2

−14

J+

(

)(

4 8.99 × 10 9 N ⋅ m 2 C2 1.60 × 10 −19 C 18.0 × 10 −15 m

)

2

= 2.55 × 10 −13 J

If we start with the three-particle system of part (b) and allow the alpha particle to escape to infinity [thereby returning us to the two-particle system of part (a)], the change in electric potential energy will be ΔPE = PEa − PEb = 3.84 × 10 −14 J − 2.55 × 10 −13 J = −2.17 × 10 −13 J

(d)

Conservation of energy, ΔKE + ΔPE = 0, gives the speed of the alpha particle at infinity in the situation of part (c) as 12 ma va2 − 0 = −ΔPE, or

continued on next page

68719_16_ch16_p030-061.indd 42

1/7/11 2:33:00 PM

43

Electrical Energy and Capacitance

va = (e)

6.64 × 10

−27

kg

) = 8.08 × 10

6

ms

When, starting with the three-particle system, the two protons are both allowed to escape to infinity, there will be no remaining pairs of particles and hence no remaining potential energy. Thus, ΔPE = 0 − PEb = −PEb, and conservation of energy gives the change in kinetic energy as ΔKE = −ΔPE = +PEb. Since the protons are identical particles, this increase in kinetic energy is split equally between them giving KEproton = 12 m p v 2p = 12 ( PEb ), or

16.20

(

−2 −2.17 × 10 −13 J

−2 ( ΔPE ) = ma

PEb 2.55 × 10 −13 J = = 1.24 × 10 7 m s mp 1.67 × 10 −27 kg

vp =

(a)

If a proton and an alpha particle, initially at rest 4.00 fm apart, are released and allowed to recede to infinity, the final speeds of the two particles will differ because of the difference in the masses of the particles. Thus, attempting to solve for the final speeds by use of conservation of energy alone leads to a situation of having one equation with two unknowns , and does not permit a solution.

(b)

In the situation described in part (a) above, one can obtain a second equation with the two unknown final speeds by using conservation of linear momentum. Then, one would have two equations which could be solved simultaneously for both unknowns.

(c)

From conservation of energy: ⎡⎣ ma v + m p v = 2 a

2 p

2ke qa q p ri

(

=

1 2

)

ma va2 + 12 m p v 2p − 0 ⎤⎦ + ⎡⎣ 0 − ke qa q p ri ⎤⎦ = 0, or

(

)(

)(

2 8.99 × 10 9 N ⋅ m 2 C2 3.20 × 10 −19 C 1.60 × 10 −19 C 4.00 × 10

−15

)

m

yielding ma va2 + m p v 2p = 2.30 × 10 −13 J

[1]

From conservation of linear momentum, ma va + m p v p = 0

or

⎛ mp ⎞ va = ⎜ vp ⎝ ma ⎟⎠

[2]

Substituting Equation [2] into Equation [1] gives 2

⎛ mp ⎞ 2 ma ⎜ v p + m p v 2p = 2.30 × 10 −13 J ⎝ ma ⎟⎠

or

⎛ mp ⎞ 2 −13 ⎜⎝ m + 1⎟⎠ m p v p = 2.30 × 10 J a

and vp =

2.30 × 10 −13 J

(m

p

)

ma + 1 m p

=

(1.67 × 10

−27

2.30 × 10 −13 J = 1.05 × 10 7 m s 6.64 × 10 −27 +1 1.67 × 10 −27 kg

)(

)

Then, Equation [2] gives the final speed of the alpha particle as ⎛ mp ⎞ ⎛ 1.67 × 10 −27 kg ⎞ va = ⎜ 1.05 × 10 7 m s = 2.64 × 10 6 m s vp = ⎜ ⎟ ⎝ 6.64 × 10 −27 kg ⎟⎠ ⎝ ma ⎠

(

68719_16_ch16_p030-061.indd 43

)

1/7/11 2:33:03 PM

44

16.21

Chapter 16

(a)

Conservation of energy gives ⎛1 1⎞ KE f = KEi + PEi − PE f = 0 + ke q1q2 ⎜ − ⎟ ⎝ ri rf ⎠

(

)

With q1 = +8.50 nC, q2 = −2.80 nC, ri = 1.60 mm, and rf = 0.500 mm, this becomes ⎛ N ⋅ m2 ⎞ 1 1 ⎛ ⎞ KE f = ⎜ 8.99 × 10 9 − 8.50 × 10 −9 C −2.80 × 10 −9 C ⎜ ⎟ 2 −6 −6 ⎟ ⎝ C ⎠ 1.60 × 10 m 0.500 × 10 m ⎠ ⎝

(

When r = rf = 0.500 mm and KE = KE f = 0.294 J, the speed of the sphere having mass m = 8.00 mg = 8.00 × 10 −6 kg is vf =

16.22

)

KE f = 0.294 J

yielding (b)

)(

(

2 KE f m

)=

2 ( 0.294 J ) = 271 m s 8.00 × 10 −6 kg

The excess charge on the metal sphere will be uniformly distributed over its surface. In this spherically symmetric situation, the electric field and the electric potential outside the sphere is the same as if all the excess charge were concentrated as a point charge at the center of the sphere. Thus, for points outside the sphere, E = ke

Q r2

V = ke

and

Q = E ⋅r r

If the sphere has a radius of r = 18 cm = 0.18 m and the air breaks down when E = 3.0 × 10 6 V m, the electric potential at the surface of the sphere when breakdown occurs is

(

)

V = 3.0 × 10 6 V m ( 0.18 m ) = 5.4 × 10 5 V 16.23

From conservation of energy, ( KE + PEe ) f = ( KE + PEe )i , which gives 0 + keQq rf = 12 ma vi2 + 0, or rf =

rf = 16.24

(a)

2 keQq 2 ke ( 79e )( 2e ) = ma vi2 ma vi2

(

)

(

2 8.99 × 10 9 N ⋅ m 2 C2 (158 ) 1.60 × 10 −19 C

(6.64 × 10

)(

−27

kg 2.00 × 10 m s 7

)

2

)

2

= 2.74 × 10 −14 m

The distance from any one of the corners of the square to the point at the center is one half the length of the diagonal of the square, or r=

diagonal = 2

a2 + a2 a 2 a = = 2 2 2

Since the charges have equal magnitudes and are all the same distance from the center of the square, they make equal contributions to the total potential. Thus, Vtotal = 4Vsingle = 4 charge

keQ kQ Q = 4 e = 4 2ke a r a 2

continued on next page

68719_16_ch16_p030-061.indd 44

1/7/11 2:33:05 PM

Electrical Energy and Capacitance

45

The work required to carry charge q from infinity to the point at the center of the square is equal to the increase in the electric potential energy of the charge, or

(b)

Q⎞ qQ ⎛ W = PEcenter − PE∞ = qVtotal − 0 = q ⎜ 4 2ke ⎟ = 4 2ke ⎝ a⎠ a 16.25

(

)

(a)

6 2 C2 ⎞ 1.0 × 10 m A ⎛ −12 C = ∈0 = ⎜ 8.85 × 10 = 1.1 × 10 −8 F d ⎝ N ⋅ m 2 ⎟⎠ 800 m ( )

(b)

Qmax = C ( ΔV )max = C ( Emax d ) = ∈0

(

A Emax d = ∈0 AEmax d

(

)(

)

)(

)

= 8.85 × 10 −12 C2 N ⋅ m 2 1.0 × 10 6 m 2 3.0 × 10 6 N C = 27 C 16.26

16.27

C=

(b)

Q = C ( ΔV ) = ( 3.00 mF )(12.0 V ) = 36.0 mC

(a)

The capacitance of this air-filled ( dielectric constant, k = 1.00 ) parallel-plate capacitor is C=

16.28

16.29

16.30

Q 27.0 mC = = 3.00 mF ΔV 9.00 V

(a)

(

)(

)

−12 C2 N ⋅ m 2 2.30 × 10 −4 m 2 k ∈0 A (1.00 ) 8.85 × 10 = = 1.36 × 10 −12 F = 1.36 pF 1.50 × 10 −3 m d

(

)

(b)

Q = C ( ΔV ) = 1.36 × 10 −12 F (12.0 V ) = 1.63 × 10 −11 C = 16.3 × 10 −12 C = 16.3 pC

(c)

E=

ΔV 12.0 V = = 8.00 × 10 3 V m = 8.00 × 10 3 N C d 1.50 × 10 −3 m

(a)

C=

Q 10.0 mC = = 1.00 mF V 10.0 V

(b)

V=

Q 100 mC = = 100 V C 1.00 mF

(a)

E=

ΔV 20.0 V = = 1.11 × 10 4 V m = 11.1 kV m toward the negative plate d 1.80 × 10 −3 m

(b)

C=

8.85 × 10 −12 C2 N ⋅ m 2 7.60 × 10 −4 m 2 ∈0 A = = 3.74 × 10 −12 F = 3.74 pF 1.80 × 10 −3 m d

(c)

Q = C ( ΔV ) = 3.74 × 10 −12 F ( 20.0 V ) = 7.48 × 10 −11 C = 74.8 pC on one plate and −74.8 pC on the other plate.

(

)(

(

)

)

C = ∈0 A d, so d=

(

)(

)

8.85 × 10 −12 C2 N ⋅ m 2 21.0 × 10 −12 m 2 ∈0 A = = 3.10 × 10 −9 m 60.0 × 10 −15 F C

⎛ 1Å ⎞ d = 3.10 × 10 −9 m ⎜ −10 ⎟ = 31.0 Å ⎝ 10 m ⎠

(

68719_16_ch16_p030-061.indd 45

)

1/7/11 2:33:08 PM

46

16.31

Chapter 16

(a)

Assuming the capacitor is air-filled (k = 1), the capacitance is C=

16.32

(

)(

)

8.85 × 10 −12 C2 N ⋅ m 2 0.200 m 2 ∈0 A = = 5.90 × 10 −10 F d 3.00 × 10 −3 m

(

)

(b)

Q = C ( ΔV ) = 5.90 × 10 −10 F ( 6.00 V ) = 3.54 × 10 −9 C

(c)

E=

ΔV 6.00 V = = 2.00 × 10 3 V m = 2.00 × 10 3 N C d 3.00 × 10 −3 m

(d)

s =

Q 3.54 × 10 −9 C = = 1.77 × 10 −8 C m 2 A 0.200 m 2

(e)

Increasing the distance separating the plates decreases the capacitance, the charge stored, and the electric field strength between the plates. This means that all of the previous answers will be decreased .

ΣFy = 0 ⇒ T cos15.0° = mg or T =

mg cos15.0°

ΣFx = 0 ⇒ qE = T sin15.0° = mg tan15.0° or

E=

mg tan15.0° q

ΔV = Ed =

mgd tan15.0° q

(350 × 10 ΔV = 16.33

(a)

−6

)(

)

kg 9.80 m s 2 ( 0.040 0 m ) tan15.0° 30.0 × 10

−9

C

= 1.23 × 10 3 V = 1.23 kV

Capacitors in a series combination store the same charge, Q = Ceq (ΔV ), where Ceq is the equivalent capacitance and ΔV is the potential difference maintained across the series combination. The equivalent capacitance for the given series combination is 1 Ceq = 1 C1 + 1 C2, or Ceq = C1C2 (C1 + C2 ), giving Ceq =

( 2.50 mF ) (6.25 mF ) = 1.79 mF 2.50 mF + 6.25 mF

and the charge stored on each capacitor in the series combination is Q = Ceq ( ΔV ) = (1.79 mF ) ( 6.00 V ) = 10.7 mC (b)

68719_16_ch16_p030-061.indd 46

When connected in parallel, each capacitor has the same potential difference, ΔV = 6.00 V, maintained across it. The charge stored on each capacitor is then For C1 = 2.50 mF:

Q1 = C1 ( ΔV ) = ( 2.50 mF )( 6.00 V ) = 15.0 mC

For C2 = 6.25 mF:

Q2 = C2 ( ΔV ) = ( 6.25 mF )( 6.00 V ) = 37.5 mC

1/7/11 2:33:10 PM

Electrical Energy and Capacitance

16.34

47

(a)

Ceq = C1 + C2 = 5.00 mF + 12.0 mF = 17.0 mF

(b)

In a parallel combination, the full potential difference maintained between the terminals of the battery exists across each capacitor. Thus, ΔV1 = ΔV2 = ΔVbattery = 9.00 V

(c)

Q1 = C1 ( ΔV1 ) = ( 5.00 mF )( 9.00 V ) = 45.0 mC Q2 = C2 ( ΔV2 ) = (12.0 mF )( 9.00 V ) = 108 mC

16.35

(a)

First, we replace the parallel combination between points b and c by its equivalent capacitance, Cbc = 2.00 mF + 6.00 mF = 8.00 mF. Then, we have three capacitors in series between points a and d. The equivalent capacitance for this circuit is therefore 1 1 1 3 1 = + + = Ceq Cab Cbc Ccd 8.00 mF giving

(b)

Ceq =

8.00 mF = 2.67 mF 3

The charge stored on each capacitor in the series combination is Qab = Qbc = Qcd = Ceq ( ΔVad ) = ( 2.67 mF )( 9.00 V ) = 24.0 mC Then, note that ΔVbc = Qbc Cbc = 24.0 mC 8.00 mF = 3.00 V. The charge on each capacitor in the original circuit is:

(c)

On the 8.00 mF between a and b:

Q8 = Qab = 24.0 mC

On the 8.00 mF between c and d:

Q8 = Qcd = 24.0 mC

On the 2.00 mF between b and c:

Q2 = C2 ( ΔVbc ) = ( 2.00 mF )( 3.00 V ) = 6.00 mC

On the 6.00 mF between b and c:

Q6 = C6 ( ΔVbc ) = ( 6.00 mF )( 3.00 V ) = 18.0 mC

Note that ΔVab = Qab Cab = 24.0 mC 8.00 mF = 3.00 V, and that ΔVcd = Qcd Ccd = 24.0 mC 8.00 mF = 3.00 V. We earlier found that ΔVbc = 3.00 V, so we conclude that the potential difference across each capacitor in the circuit is ΔV8 = ΔV2 = ΔV6 = ΔV8 = 3.00 V

16.36

Cparallel = C1 + C2 = 9.00 pF ⇒ 1 Cseries

=

1 1 + C1 C2

⇒

Cseries =

Thus, using Equation [1],

C1 = 9.00 pF − C2

[1]

C1C2 = 2.00 pF C1 + C2

Cseries =

(9.00 pF − C2 ) C2 (9.00 pF − C2 ) + C2

= 2.00 pF which reduces to

continued on next page

68719_16_ch16_p030-061.indd 47

1/7/11 2:33:13 PM

48

Chapter 16

C22 − ( 9.00 pF ) C2 + 18.0 ( pF ) = 0 , or ( C2 − 6.00 pF ) (C2 − 3.00 pF ) = 0 2

Therefore, either C2 = 6.00 pF and, from Equation [1], C1 = 3.00 pF or C2 = 3.00 pF

and

C1 = 6.00 pF.

We conclude that the two capacitances are 3.00 pF and 6.00 pF . 16.37

(a)

3.00 mF 6.00 mF

The equivalent capacitance of the series combination in the upper branch is 1

=

Cupper or

1 1 2 +1 + = 3.00 mF 6.00 mF 6.00 mF 2.00 mF 4.00 mF

Cupper = 2.00 mF

Likewise, the equivalent capacitance of the series combination in the lower branch is 1

=

Clower

1 1 2 +1 + = 2.00 mF 4.00 mF 4.00 mF

or

90.0 V

Clower = 1.33 mF

These two equivalent capacitances are connected in parallel with each other, so the equivalent capacitance for the entire circuit is Ceq = Cupper + Clower = 2.00 mF + 1.33 mF = 3.33 mF (b)

Note that the same potential difference, equal to the potential difference of the battery, exists across both the upper and lower branches. The charge stored on each capacitor in the series combination in the upper branch is Q3 = Q6 = Qupper = Cupper ( ΔV ) = ( 2.00 mF )( 90.0 V ) = 180 mC and the charge stored on each capacitor in the series combination in the lower branch is Q2 = Q4 = Qlower = Clower ( ΔV ) = (1.33 mF )( 90.0 V ) = 120 mC

(c)

16.38

(a)

The potential difference across each of the capacitors in the circuit is: ΔV2 =

Q2 120 mC = = 60.0 V C2 2.00 mF

ΔV4 =

Q4 120 mC = = 30.0 V C4 4.00 mF

ΔV3 =

Q3 180 mC = = 60.0 V C3 3.00 mF

ΔV6 =

Q6 180 mC = = 30.0 V C6 6.00 mF

The equivalent capacitance of the series combination in the rightmost branch of the circuit is 1 Cright or

=

1 1 1+ 3 + = 24.0 mF 8.00 mF 24.0 mF

Figure P16.38

Cright = 6.00 mF

continued on next page

68719_16_ch16_p030-061.indd 48

1/7/11 2:33:16 PM

Electrical Energy and Capacitance

(b)

49

The equivalent capacitance of the three capacitors now connected in parallel with each other and with the battery is Ceq = 4.00 mF + 2.00 mF + 6.00 mF = 12.0 mF

(c)

The total charge stored in this circuit is

Diagram 1

Qtotal = Ceq ( ΔV ) = (12.0 mF )( 36.0 V ) or (d)

Qtotal = 432 mC

The charges on the three capacitors shown in Diagram 1 are:

Diagram 2

Q4 = C4 ( ΔV ) = ( 4.00 mF )( 36.0 V ) = 144 mC Q2 = C2 ( ΔV ) = ( 2.00 mF )( 36.0 V ) = 72 mC Qright = Cright ( ΔV ) = ( 6.00 mF )( 36.0 V ) = 216 mC Yes. Q4 + Q2 + Qright = Qtotal as it should. (e)

The charge on each capacitor in the series combination in the rightmost branch of the original circuit (Figure P16.38) is Q24 = Q8 = Qright = 216 mC

(f)

ΔV24 =

(g)

ΔV8 =

Q24 216 mC = = 9.00 V C24 24.0 mF

Q8 216 mC = = 27.0 V C8 8.00 mF

Note that ΔV8 + ΔV24 = ΔV = 36.0 V as it should.

16.39

The circuit may be reduced in steps as shown above. Using Figure 3,

Qac = ( 4.00 mF )( 24.0 V ) = 96.0 mC

Then, in Figure 2, ( ΔV )ab =

Qac 96.0 mC = = 16.0 V Cab 6.00 mF

continued on next page

68719_16_ch16_p030-061.indd 49

1/7/11 2:33:19 PM

50

Chapter 16

and

( ΔV )bc = ( ΔV )ac − ( ΔV )ab = 24.0 V − 16.0 V = 8.00 V

Finally, using Figure 1, Q5 = ( 5.00 mF )( ΔV )ab = 80.0 mC

Q1 = C1 ( ΔV )ab = (1.00 mF )(16.0 V ) = 16.0 mC , Q8 = (8.00 mF )( ΔV )bc = 64.0 mC , 16.40

(a)

Q4 = ( 4.00 mF )( ΔV )bc = 32.0 mC

and

Consider the simplification of the circuit as shown below: C1

+

ΔV

C1

a

C3

C2

−

ΔV

+

a

ΔV

Cab

−

b

+ −

Ceq

b

Since C2 and C3 are connected in parallel, Cab = C2 + C3 = C + 5C = 6C. Now observe that C1 and Cab are connected in series, giving 1 1 1 = + Ceq C1 Cab (b)

or

Ceq =

C1Cab ( 3C )( 6C ) = = 2C C1 + Cab 3C + 6C

Since capacitors C1 and Cab are connected in series, Q1 = Qab = Qeq = Ceq ( ΔV ) = 2C ( ΔV )

(c)

Qab 2C ( ΔV ) ΔV = = , giving Cab 6C 3

Then,

ΔVab =

Also,

Q3 = C3 ( ΔVab ) =

Q2 = C2 ( ΔVab ) =

5C ( ΔV ) . Therefore, 3

C ( ΔV ) 3

Q1 > Q3 > Q2

Since capacitors C1 and Cab are in series with the battery, ΔV1 = ΔV − ΔVab = ΔV −

ΔV 2 = ΔV 3 3

Also, with capacitors C2 and C3 in parallel between points a and b, ΔV2 = ΔV3 = ΔVab = Thus, (d)

ΔV 3

ΔV1 > ΔV2 = ΔV3

Consider the following steps: (i)

Increasing C3 while C1 and C2 remain constant will increase Cab = C2 + C3. ⎛ Cab ⎞ , will increase. Therefore, the equivalent capacitance, Ceq = C1 ⎜ ⎝ C1 + Cab ⎟⎠

continued on next page

68719_16_ch16_p030-061.indd 50

1/7/11 2:33:22 PM

Electrical Energy and Capacitance

(ii)

Since C1 and Cab are in series, Q1 = Qab = Ceq ( ΔV ). Thus, Q1 will increase as Ceq increases. Also, Qab experiences the same increase.

(iii)

Because ΔV1 = Q1 C1, an increase in Q1 causes ΔV1 to increase and causes ΔVab = ΔV − ΔV1 to decrease. Thus, since Q2 = C2 ( ΔVab ), Q2 will decrease .

51

(iv) With capacitors C2 and C3 in parallel between points a and b, we have Qab = Q2 + Q3 or Q3 = Qab − Q2. Thus, with Qab increasing [see Step (ii)] while Q2 is decreasing [see Step (iii)], we see that Q3 will increase . 16.41

(a)

From Q = C ( ΔV ), Q25 = ( 25.0 mF )( 50.0 V ) = 1.25 × 10 3 mC = 1.25 mC and

(b)

Q40 = ( 40.0 mF )( 50.0 V ) = 2.00 × 10 3 mC = 2.00 mC

Since the negative plate of one capacitor was connected to the positive plate of the other, the net charge stored in the new parallel combination is Q = Q40 − Q25 = 2.00 × 10 3 mC − 1.25 × 10 3 mC = 750 mC The two capacitors, now in parallel, have a common potential difference ΔV across them. The new charges on each of the capacitors are Q25′ = C1 ( ΔV ) and Q40′ = C2 ( ΔV ). Thus, Q25′ =

⎛ 25 mF ⎞ C1 5 Q40′ = Q40′ Q40′ = ⎜ ⎟ 8 C2 ⎝ 40 mF ⎠

and the total change now stored in the combination may be written as 5 13 Q = Q40′ + Q25′ = Q40′ + Q40′ = Q40′ = 750 mC 8 8 giving Q40′ = (c)

The potential difference across each capacitor in the new parallel combination is ΔV =

16.42

(a)

8 ( 750 mC) = 462 mC and Q25′ = Q − Q40′ = ( 750 − 462) mC = 288 mC 13

Q Q 750 mC = = = 11.5 V Ceq C1 + C2 65.0 mF

The original circuit reduces to a single equivalent capacitor in the steps shown below.

eq

continued on next page

68719_16_ch16_p030-061.indd 51

1/7/11 2:33:25 PM

52

Chapter 16

⎛ 1 1 ⎞ Cs = ⎜ + ⎟ ⎝ C1 C2 ⎠

−1

⎛ 1 1 ⎞ =⎜ + ⎝ 5.00 mF 10.0 mF ⎟⎠

−1

= 3.33 mF

C p1 = Cs + C3 + Cs = 2 ( 3.33 mF ) + 2.00 mF = 8.66 mF C p 2 = C2 + C2 = 2 (10.0 mF ) = 20.0 mF ⎛ 1 1 ⎞ Ceq = ⎜ + ⎟ ⎝ C p1 C p 2 ⎠ (b)

−1

⎛ ⎞ 1 1 =⎜ + ⎝ 8.66 mF 20.0 mF ⎟⎠

−1

= 6.04 mF

The total charge stored between points a and b is Qtotal = Ceq ( ΔV )ab = ( 6.04 mF )( 60.0 V ) = 362 mC Then, looking at the third figure, observe that the charges of the series capacitors of that figure are Q p1 = Q p 2 = Qtotal = 362 mC. Thus, the potential difference across the upper parallel combination shown in the second figure is

( ΔV ) p1 =

Q p1 C p1

=

362 mC = 41.8 V 8.66 mF

Finally, the charge on C3 is Q3 = C3 ( ΔV ) p1 = ( 2.00 mF )( 41.8 V ) = 83.6 mC 16.43

From Q = C ( ΔV ), the initial charge of each capacitor is Q1 = (1.00 mF )(10.0 V ) = 10.0 mC

and

Q2 = ( 2.00 mF ) ( 0 ) = 0

After the capacitors are connected in parallel, the potential difference across one is the same as that across the other. This gives ΔV =

Q1′ Q2′ = 1.00 mF 2.00 mF

or

Q2′ = 2 Q1′

[1]

From conservation of charge, Q1′ + Q2′ = Q1 + Q2 = 10.0 mC. Then, substituting from Equation [1], this becomes Q1′ + 2 Q1′ = 10.0 mC , giving

Q2′ = 20 mC 3 = 6.67 mC

Finally, from Equation [1], 16.44

(a)

Q1′ = 10 mC 3 = 3.33 mC

We simplify the circuit in stages as shown below:

15.0 mF 3.00 mF

Cs

a

b

a

b

20.0 mF 6.00 mF

20.0 mF

a

Cp

b

a

Ceq

b

20.0 mF

6.00 mF

continued on next page

68719_16_ch16_p030-061.indd 52

1/7/11 2:33:28 PM

Electrical Energy and Capacitance

1 1 1 = + Cs 15.0 mF 3.00 mF

or

Cs =

53

(15.0 mF ) (3.00 mF ) = 2.50 mF 15.0 mF + 3.00 mF

C p = Cs + 6.00 mF = 2.50 mF + 6.00 mF = 8.50 mF 1 1 1 = + Ceq C p 20.0 mF (b)

or

Ceq =

(8.50 mF ) ( 20.0 mF ) = 8.50 mF + 20.0 mF

5.96 mF

Q20 = QC = Qeq = Ceq ( ΔVab ) = ( 5.96 mF )(15.0 V ) = 89.4 mC p

ΔVC =

QC

p

Cp

p

(

=

89.4 mC = 10.5 V 8.50 mF

) = (6.00 mF )(10.5 V ) = 63.0 mC = Q = Q = C ( ΔV ) = ( 2.50 mF )(10.5 V ) = 26.3 mC

so

Q6 = C6 ΔVC

and

Q15

3

p

s

s

Cp

The charges are 89.4 mC on the 20 mF capacitor, 63.0 mC on the 6 mF capacitor, and 26.3 mC on both the 15 mF and 3 mF capacitors. 16.45

Energy stored =

16.46

(a)

Q2 1 1 2 2 = C ( ΔV ) = ( 4.50 × 10 −6 F )(12.0 V ) = 3.24 × 10 −4 J 2 2C 2

The equivalent capacitance of a series combination of C1 and C2 is 1 1 2 +1 1 = + = Ceq 18.0 mF 36.0 mF 36.0 mF

or

Ceq = 12.0 mF

When this series combination is connected to a 12.0-V battery, the total stored energy is Total energy stored = (b)

1 1 2 2 Ceq ( ΔV ) = (12.0 × 10 −6 F )(12.0 V ) = 8.64 × 10 −4 J 2 2

The charge stored on each of the two capacitors in the series combination is Q1 = Q2 = Qtotal = Ceq ( ΔV ) = (12.0 mF )(12.0 V ) = 144 mC = 1.44 × 10 −4 C and the energy stored in each of the individual capacitors is

(1.44 × 10−4 C) = 5.76 × 10−4 J Q2 Energy stored in C1 = 1 = 2C1 2 (18.0 × 10 −6 F ) 2

(1.44 × 10−4 C) = 2.88 × 10−4 J Q2 Energy stored in C2 = 2 = 2C2 2 ( 36.0 × 10 −6 F ) 2

and

Energy stored in C1 + Energy stored in C2 = 5.76 × 10 −4 J + 2.88 × 10 −4 J = 8.64 × 10 −4 J, which is the same as the total stored energy found in part (a). This must be true if the computed equivalent capacitance is truly equivalent to the original combination.

continued on next page

68719_16_ch16_p030-061.indd 53

1/7/11 2:33:31 PM

54

Chapter 16

(c)

If C1 and C2 had been connected in parallel rather than in series, the equivalent capacitance would have been Ceq = C1 + C2 = 18.0 mF + 36.0 mF = 54.0 mF. If the total energy stored ⎡⎣ 12 Ceq ( ΔV )2 ⎤⎦ in this parallel combination is to be the same as was stored in the original series combination, it is necessary that ΔV =

2 (8.64 × 10 −4 J )

2 ( Total energy stored ) = Ceq

54.0 × 10 −6 F

= 5.66 V

Since the two capacitors in parallel have the same potential difference across them, the 2 energy stored in the individual capacitors ⎡⎣ 12 C ( ΔV ) ⎤⎦ is directly proportional to their capacitances. The larger capacitor, C2 , stores the most energy in this case.

16.47

(a)

The energy initially stored in the capacitor is

( Energy stored )

1

(b)

=

Qi2 1 1 2 2 = Ci ( ΔV )i = ( 3.00 mF )( 6.00 V ) = 54.0 mJ 2 2Ci 2

When the capacitor is disconnected from the battery, the stored charge becomes isolated with no way off the plates. Thus, the charge remains constant at the value Qi as long as the capacitor remains disconnected. Since the capacitance of a parallel-plate capacitor is C = k ∈0 A d, when the distance d separating the plates is doubled, the capacitance is decreased by a factor of 2 C f = Ci 2 = 1.50 mF . The stored energy (with Q unchanged) becomes

(

( Energy stored ) (c)

2

⎛ Q2 ⎞ Qi2 Qi2 = = 2 ⎜ i ⎟ = 2 ( Energy stored )1 = 108 mJ 2C f 2 (Ci 2 ) ⎝ 2Ci ⎠

=

When the capacitor is reconnected to the battery, the potential difference between the plates is reestablished at the original value of ΔV = ( ΔV )i = 6.00 V, while the capacitance remains at C f = Ci 2 = 1.50 mF. The energy stored under these conditions is

( Energy stored )

3

16.48

)

=

1 1 2 2 C f ( ΔV )i = (1.50 mF )( 6.00 V ) = 27.0 mJ 2 2

The energy transferred to the water is W=

8 1 ⎡1 ⎤ = ( 50.0 C )(1.00 × 10 V ) = 2.50 × 10 7 J Q ΔV ( ) ⎥⎦ 100 ⎢⎣ 2 200

Thus, if m is the mass of water boiled away, W = m ⎡⎣c ( ΔT ) + Lv ⎤⎦ becomes ⎡⎛ ⎤ J ⎞ 2.50 × 10 7 J = m ⎢⎜ 4186 (100°C − 30.0°C ) + 2.26 × 106 J kg ⎥ ⎟ kg ⋅°C ⎠ ⎣⎝ ⎦ giving 16.49

(a)

m=

2.50 × 10 7 J = 9.79 kg ⎡⎣ 2.93 × 10 5 J kg + 2.26 × 10 6 J kg ⎤⎦

Note that the charge on the plates remains constant at the original value, Q0 , as the dielectric is inserted. Thus, the change in the potential difference, ΔV = Q C, is due to a change in capacitance alone. The ratio of the final and initial capacitances is Cf Ci

=

k ∈0 A d =k ∈0 A d

and

Cf Ci

=

Q0 ( ΔV ) f Q0 ( ΔV )i

=

( ΔV )i 85.0 V = = 3.40 ( ΔV ) f 25.0 V

continued on next page

68719_16_ch16_p030-061.indd 54

1/7/11 2:33:34 PM

Electrical Energy and Capacitance

55

Thus, the dielectric constant of the inserted material is k = 3.40 , and the material is probably nylon (see Table 16.1).

16.50

(b)

If the dielectric only partially filled the space between the plates, leaving the remaining space air-filled, the equivalent dielectric constant would be somewhere between k = 1.00 (air) and k = 3.40. The resulting potential difference would then lie somewhere between ( ΔV )i = 85.0 V and ( ΔV ) f = 25.0 V.

(a)

If the maximum electric field that can exist between the plates before breakdown (i.e., the dielectric strength) is Em ax, the maximum potential difference across the plates is ΔVm ax = Em ax ⋅ d, where d is the plate separation. The maximum change on either plate then has magnitude Qm ax = C ( ΔVm ax ) = C ( Em ax ⋅ d ) Since the capacitance of a parallel-plate capacitor is C = k ∈0 A d, the maximum charge is ⎛ k ∈0 A ⎞ Qm ax = ⎜ Em ax ⋅ d = k ∈0 AEm ax ⎝ d ⎟⎠

(

)

The area of each plate is A = 5.00 cm 2 = 5.00 × 10 −4 m 2, and when air is the dielectric, k = 1.00 and Em ax = 3.00 × 10 6 V m (see Table 16.1). Thus, Qm ax = (1.00 )(8.85 × 10 −12 C N ⋅ m 2 ) ( 5.00 × 10 −4 m 2 ) ( 3.00 × 10 6 V m ) = 1.33 × 10 −8 C = 13.3 nC (b)

If the dielectric is now polystyrene (k = 2.56 and Em ax = 24.0 × 10 6 V m ), then Qm ax = ( 2.56 )(8.85 × 10 −12 C N ⋅ m 2 ) ( 5.00 × 10 −4 m 2 ) ( 24.0 × 10 6 V m ) = 2.72 × 10 −7 C = 272 nC

16.51

(a)

The dielectric constant for Teflon® is k = 2.1, so the capacitance is C=

−12 2 2 −4 2 k ∈0 A ( 2.1) (8.85 × 10 C N ⋅ m ) (175 × 10 m ) = d 0.040 0 × 10 −3 m

C = 8.1× 10 −9 F = 8.1 nF (b)

For Teflon®, the dielectric strength is Em ax = 60 × 10 6 V m, so the maximum voltage is ΔVm ax = Em ax d = ( 60 × 10 6 V m ) ( 0.040 0 × 10 −3 m

)

ΔVm ax = 2.4 × 10 3 V = 2.4 kV 16.52

Before the capacitor is rolled, the capacitance of this parallel-plate capacitor is C=

k ∈0 A k ∈0 ( w × L ) = d d

where A is the surface area of one side of a foil strip. Thus, the required length is 9.50 × 10 −8 F ) ( 0.025 0 × 10 −3 m ) ( C ⋅d L= = = 1.04 m k ∈0 w ( 3.70 )(8.85 × 10 −12 C2 N ⋅ m 2 ) ( 7.00 × 10 −2 m )

68719_16_ch16_p030-061.indd 55

1/7/11 2:33:36 PM

56

16.53

Chapter 16

(a)

V=

m 1.00 × 10 −12 kg = = 9.09 × 10 −16 m 3 r 1100 kg m 3

Since V = 4p r 3 3, the radius is r = [ 3V 4p ] , and the surface area is 13

3V ⎤ A = 4p r = 4p ⎡⎢ ⎣ 4p ⎥⎦ 2

2 3

⎡ 3 ( 9.09 × 10 −16 m 3 ) ⎤ = 4p ⎢ ⎥ 4p ⎢⎣ ⎥⎦

2 3

= 4.54 × 10 −10 m 2

−12 2 2 −10 2 k ∈0 A ( 5.00 )(8.85 × 10 C N ⋅ m ) ( 4.54 × 10 m ) = = 2.01× 10 −13 F d 100 × 10 −9 m

(b)

C=

(c)

Q = C ( ΔV ) = ( 2.01× 10 −13 F ) (100 × 10 −3 V ) = 2.01 × 10 −14 C and the number of electronic charges is n=

16.54

(

)

For a parallel-plate capacitor, C = k ∈0 A d and Q = s A = C ( ΔV ). Thus, s A = k ∈0 A d ( ΔV ), and d = (k ∈0 s )( ΔV ). With air as the dielectric material (k = 1.00 ), the separation of the plates must be d=

16.55

Q 2.01× 10 −14 C = = 1.26 × 10 5 e 1.60 × 10 −19 C

(1.00 )(8.85 × 10 −12 C2 N ⋅ m 2 )(150 V )

(3.00 × 10

−10

C cm 2 ) (10 4 cm 2 1 m 2 )

= 4.43 × 10 −4 m = 0.443 mm

Since the capacitors are in series, the equivalent capacitance is given by d d + d 2 + d3 1 1 1 d 1 d = + + = 1 + 2 + 3 = 1 Ceq C1 C2 C3 ∈0 A ∈0 A ∈0 A ∈0 A

16.56

∈0 A where d = d1 + d 2 + d3 d

or

Ceq =

(a)

Please refer to the solution of Problem 16.37 where the following results were obtained: Ceq = 3.33 mF

Q3 = Q6 = 180 mC

3.00 mF 6.00 mF

Q2 = Q4 = 120 mC 2.00 mF 4.00 mF

The total energy stored in the full circuit is then

( Energy stored )

total

=

1 1 2 2 Ceq ( ΔV ) = ( 3.33 × 10 −6 F )( 90.0 V ) 2 2 90.0 V

= 1.35 × 10 −2 J = 13.5 × 10 −3 J = 13.5 mJ (b)

The energy stored in each individual capacitor is

(120 × 10 −6 C) = 3.60 × 10 −3 J = 3.60 mJ Q2 = 2 = 2C2 2 ( 2.00 × 10 −6 F ) 2

For 2.00 mF :

( Energy stored )

For 3.00 mF:

(180 × 10 −6 C) = 5.40 × 10 −3 J = 5.40 mJ Q2 ( Energy stored )3 = 3 = 2C3 2 ( 3.00 × 10 −6 F )

2

2

continued on next page

68719_16_ch16_p030-061.indd 56

1/7/11 2:33:39 PM

57

Electrical Energy and Capacitance

120 × 10 −6 C ) ( Q42 = = = 1.80 × 10 −3 J = 1.80 mJ 2C4 2 ( 4.00 × 10 −6 F ) 2

For 4.00 mF:

( Energy stored )

For 6.00 mF:

( Energy stored )6 =

4

(180 × 10 −6 C) = 2.70 × 10−3 J = 2.70 mJ Q62 = 2C6 2 ( 6.00 × 10 −6 F ) 2

(c)

The total energy stored in the individual capacitors is Energy stored = ( 3.60 + 5.40 + 1.80 + 2.70 ) mJ = 13.5 mJ = ( Energy stored )total Thus, the sums of the energies stored in the individual capacitors equals the total energy stored by the system.

16.57

In the absence of a dielectric, the capacitance of the parallel-plate capacitor is C0 = ∈0 A d.

1 d 3

1 d 3

k

c1

k

With the dielectric inserted, 2 d it fills one-third of the gap d between the plates as shown 3 in sketch (a) at the right. We 2 d c2 model this situation as con3 sisting of a pair of capacitors, C1 and C2, connected (a) in series as shown in sketch (b) at the right. In reality, the lower plate of C1 and the (b) upper plate of C2 are one and the same, consisting of the lower surface of the dielectric shown in sketch (a). The capacitances in the model of sketch (b) are given by C1 =

k ∈0 A 3k ∈0 A = d 3 d

and

C2 =

∈0 A 3 ∈0 A = 2d 3 2d

The equivalent capacitance of the series combination is d 1 2d ⎛1 ⎞ ⎛ d ⎞ ⎛ 2k + 1 ⎞ d ⎛ 2k + 1 ⎞ d ⎛ 2k + 1 ⎞ 1 = = + = ⎜ + 2⎟ ⎜ = = ⎠ ⎝ 3 ∈0 A ⎟⎠ ⎜⎝ k ⎟⎠ 3 ∈0 A ⎜⎝ 3k ⎟⎠ ∈0 A ⎜⎝ 3k ⎟⎠ C0 Ceq 3k ∈0 A 3 ∈0 A ⎝ k and Ceq = [ 3k (2k + 1) ]C0 . 16.58

For the parallel combination: C p = C1 + C2 which gives C2 = C p − C1 For the series combination:

Thus, we have C2 = C p − C1 =

1 1 1 = + Cs C1 C2

or

[1]

1 1 1 C1 − Cs = − = C2 Cs C1 Cs C1

Cs C1 and equating this to Equation [1] above gives C1 − Cs

Cs C1 C1 − Cs

We write this result as:

or

C p C1 − C p Cs − C12 + Cs C1 = Cs C1

C12 − C p C1 + C p Cs = 0 continued on next page

68719_16_ch16_p030-061.indd 57

1/7/11 2:33:42 PM

58

Chapter 16

C1 =

and use the quadratic formula to obtain C2 =

Then, Equation [1] gives 16.59

1 1 2 Cp ± C p − C p Cs 2 4

1 1 2 Cp ∓ C p − C p Cs 2 4

For a parallel-plate capacitor with plate separation d, ΔVm ax = Em ax ⋅ d

d=

or

ΔVm ax Em ax

The capacitance is then C=

⎛ E ⎞ k ∈0 A = k ∈0 A ⎜ m ax ⎟ d ⎝ ΔVm ax ⎠

and the needed area of the plates is A = C ⋅ ΔVm ax k ∈0 Em ax, or A= 16.60

(a)

( 0.250 × 10

( 3.00 )(8.85 × 10

F ) ( 4.00 × 10 3 V )

C N⋅m 2

2

) ( 2.00 × 10

8

V m)

= 0.188 m 2

The 1.0-mC is located 0.50 m from point P, so its contribution to the potential at P is V1 = ke

(b)

−12

−6

⎛ 1.0 × 10 −6 C ⎞ q1 = 1.8 × 10 4 V = (8.99 × 10 9 N ⋅ m 2 C2 ) ⎜ r1 ⎝ 0.50 m ⎟⎠

The potential at P due to the −2.0-mC charge located 0.50 m away is V2 = ke

⎛ − 2.0 × 10 −6 C ⎞ q2 4 = (8.99 × 10 9 N ⋅ m 2 C2 ) ⎜ ⎟⎠ = − 3.6 × 10 V r2 0.50 m ⎝

(c)

The total potential at point P is VP = V1 + V2 = ( +1.8 − 3.6 ) × 10 4 V = − 1.8 × 10 4 V

(d)

The work required to move a charge q = 3.0 mC to point P from infinity is W = qΔV = q (VP − V∞ ) = ( 3.0 × 10 −6 C ) ( −1.8 × 10 4 V − 0 ) = − 5.4 × 10 −2 J

16.61

The stages for the reduction of this circuit are shown below.

Thus, Ceq = 6.25 mF

68719_16_ch16_p030-061.indd 58

1/7/11 2:33:45 PM

Electrical Energy and Capacitance

16.62

(a)

59

Due to spherical symmetry, the charge on each of the concentric spherical shells will be uniformly distributed over that shell. Inside a spherical surface having a uniform charge distribution, the electric field due to the charge on that surface is zero. Thus, in this region, the potential due to the charge on that surface is constant and equal to the potential at the surface. Outside a spherical surface having a uniform charge distribution, the potential due to the charge on that surface is given by V = ke q r, where r is the distance from the center of that surface and q is the charge on that surface. In the region between a pair of concentric spherical shells, with the inner shell having charge + Q and the outer shell having radius b and charge − Q, the total electric potential at distance r from the center is given by V = Vdue to

+ Vdue to

inner shell

outer shell

=

ke Q ke ( −Q ) ⎛ 1 1⎞ + = ke Q ⎜ − ⎟ ⎝ r b⎠ r b

The potential difference between the two shells is therefore ⎛ 1 1⎞ ⎛ 1 1⎞ ⎛ b − a⎞ ΔV = V r = a − V r = b = ke Q ⎜ − ⎟ − ke Q ⎜ − ⎟ = ke Q ⎜ ⎝ a b⎠ ⎝ b b⎠ ⎝ ab ⎟⎠ The capacitance of this device is given by C= (b)

Q ab = ΔV ke ( b − a )

When b >> a, then b − a ≈ b. Thus, in the limit as b → ∞, the capacitance found above becomes C→

16.63

The energy stored in a charged capacitor is Estored = 12 C ( ΔV ) . Hence, 2

ΔV = 16.64

ab a = = 4p ∈0 a ke ( b ) ke

2 Estored = C

2 ( 300 J ) = 4.47 × 10 3 V = 4.47 kV 30.0 × 10 −6 F

From Q = C ( ΔV ), the capacitance of the capacitor with air between the plates is C0 =

Q0 150 mC = ΔV ΔV

After the dielectric is inserted, the potential difference is held to the original value, but the charge changes to Q = Q0 + 200 mC = 350 mC. Thus, the capacitance with the dielectric slab in place is C=

Q 350 mC = ΔV ΔV

The dielectric constant of the dielectric slab is therefore k =

68719_16_ch16_p030-061.indd 59

C ⎛ 350 mC ⎞ ⎛ ΔV ⎞ 350 = = = 2.33 C0 ⎜⎝ ΔV ⎟⎠ ⎜⎝ 150 mC ⎟⎠ 150

1/7/11 2:33:47 PM

60

16.65

Chapter 16

The charges initially stored on the capacitors are Q1 = C1 ( ΔV )i = ( 6.0 mF )( 250 V ) = 1.5 × 10 3 mC and

Q2 = C2 ( ΔV )i = ( 2.0 mF )( 250 V ) = 5.0 × 10 2 mC

When the capacitors are connected in parallel, with the negative plate of one connected to the positive pl ate of the other, the net stored charge is Q = Q1 − Q2 = 1.5 × 10 3 mC − 5.0 × 10 2 mC = 1.0 × 10 3 mC The equivalent capacitance of the parallel combination is Ceq = C1 + C2 = 8.0 mF. Thus, the final potential difference across each of the capacitors is Q 1.0 × 10 3 mC = = 125 V Ceq 8.0 mF

( ΔV )′ =

and the final charge on each capacitor is Q1′ = C1 ( ΔV )′ = ( 6.0 mF )(125 V ) = 750 mC = 0.75 mC

16.66

and

Q2′ = C2 ( ΔV )′ = ( 2.0 mF )(125 V ) = 250 mC = 0.25 mC

(a)

The distance from the charge 2q to either of the charges on the y-axis is r = d 2 + ( 2d ) = 5 d. Thus, 2

V =∑ i

ke qi kq kq 2ke q = e + e = ri 5d 5d 5d

ke q1 q3 ke q2 q3 ke q ( 2q ) ke q ( 2q ) 4ke q 2 + = + = r1 r2 5d 5d 5d

(b)

PE2q =

(c)

From conservation of energy with PE = 0 at r = ∞, KE f = KEi + PEi − PE f = 0 +

(d) 16.67

(

2 KE f

vf =

m

)=

4ke q 2 4ke q 2 −0= 5d 5d 1

2 ⎛ 4ke q 2 ⎞ ⎛ 8ke q 2 ⎞ 2 = m ⎜⎝ 5 d ⎟⎠ ⎜⎝ 5 md ⎟⎠

When excess charge resides on a spherical surface that is far removed from any other charge, this excess charge is uniformly distributed over the spherical surface, and the electric potential at the surface is the same as if all the excess charge were concentrated at the center of the spherical surface. In the given situation, we have two charged spheres, initially isolated from each other, with charges and potentials of QA = +6.00 mC and VA = ke QA RA, where RA = 12.0 cm, QB = − 4.00 mC, and VB = ke QB RB, with RB = 18.0 cm. When these spheres are then connected by a long conducting thread, the charges are redistributed

( yielding charges of Q ′ and Q ′ , respectively ) until the two surfaces come to a common potential (V ′ = kQ ′ R = V ′ = kQ ′ R ). When equilibrium is established, we have: A

A

A

A

B

B

B

B

From conservation of charge:

QA′ + QB′ = QA + QB

⇒

QA′ + QB′ = +2.00 mC

[1]

continued on next page

68719_16_ch16_p030-061.indd 60

1/7/11 2:33:49 PM

Electrical Energy and Capacitance

From equal potentials: kQA′ = kQB′ RA RB

16.68

⎛R ⎞ ⇒ QB′ = ⎜ B ⎟ QA′ ⎝ RB ⎠

QB′ = 1.50 QA′

or

61

[2]

+2.00 mC = 0.800 mC 2.50

Substituting Equation [2] into [1] gives

QA′ =

Then, Equation [2] gives

QB′ = 1.50 ( 0.800 mC ) = 1.20 mC

The electric field between the plates is directed downward with magnitude Ey =

ΔV 100 V = = 5.0 × 10 4 N m D 2.0 × 10 −3 m

Since the gravitational force experienced by the electron is negligible in comparison to the electrical force acting on it, the vertical acceleration is ay = (a)

Fy me

qE y

=

me

=

( −1.60 × 10

−19

C ) ( −5.0 × 10 4 N m )

9.11× 10 −31 kg

= + 8.8 × 1015 m s 2

At the closest approach to the bottom plate, v y = 0. Thus, the vertical displacement from point O is found from v y2 = v02 y + 2 a y ( Δy ) as Δy =

0 − ( v0 sinq 0 ) 2 ay

2

=

− ⎡⎣ − ( 5.6 × 10 6 m s ) sin 45° ⎤⎦ 2 (8.8 × 1015 m s 2 )

2

= − 8.9 × 10 −4 m = − 0.89 mm

The minimum distance above the bottom plate is then d= (b)

D + Δy = 1.0 mm − 0.89 mm = 0.1 mm 2

The time for the electron to go from point O to the upper plate (Δy = +1.0 mm) is found from Δy = v0 y t + 12 a y t 2 as m⎞ ⎡ ⎛ ⎤ 1⎛ 15 m ⎞ 2 +1.0 × 10 −3 m = ⎢ − ⎜ 5.6 × 10 6 ⎟⎠ sin 45° ⎥ t + ⎜⎝ 8.8 × 10 ⎟t ⎝ s s2 ⎠ 2 ⎣ ⎦ Solving for t gives a positive solution of t = 1.1× 10 −9 s. The horizontal displacement from point O at this time is Δx = v0 x t = ⎡⎣( 5.6 × 10 6 m s ) cos 45° ⎤⎦ (1.1× 10 −9 s ) = 4.4 mm

68719_16_ch16_p030-061.indd 61

1/7/11 2:33:52 PM

17 Current and Resistance QUICK QUIZZES 1.

Choice (d). Negative charges moving in one direction are equivalent to positive charges moving in the opposite direction. Thus, I a , I b , I c , and I d are equivalent to the movement of 5, 3, 4, and 2 charges respectively, giving I d < I b < I c < I a.

2.

Choice (b). Under steady-state conditions, the current is the same in all parts of the wire. Thus, the drift velocity, given by vd = I nqA, is inversely proportional to the cross-sectional area.

3.

Choices (c) and (d). Neither circuit (a) nor circuit (b) applies a difference in potential across the bulb. Circuit (a) has both lead wires connected to the same battery terminal. Circuit (b) has a low resistance path (a “short”) between the two battery terminals as well as between the bulb terminals.

4.

Choice (b). The slope of the line tangent to the curve at a point is the reciprocal of the resistance at that point. Note that as ΔV increases, the slope (and hence 1 R) increases. Thus, the resistance decreases.

5.

Choice (b). From Ohm’s Law, R = ΔV I = 120 V 6.00 A = 20.0 Ω.

6.

L L = r 2 . Doubling all linear A pr dimensions increases the numerator of this expression by a factor of 2 but increases the denominator by a factor of 4. Thus, the net result is that the resistance will be reduced to one-half of its original value. Choice (b). Consider the expression for resistance: R = r

7.

Choice (a). The resistance of the shorter wire is half that of the longer wire. The power dissipated, 2 P = ( ΔV ) R, (and hence the rate of heating) will be greater for the shorter wire. Consideration of the expression P = I 2 R might initially lead one to think that the reverse would be true. However, one must realize that the currents will not be the same in the two wires.

8.

Choice (b). I a = I b > I c = I d > I e = I f . Charges constituting the current I a leave the positive terminal of the battery and then split to flow through the two bulbs; thus, I a = I c + I e. Because the potential difference ΔV is the same across the two bulbs and because the power delivered to a device is P = I ( ΔV ), the 60-W bulb with the higher power rating must carry the greater current, meaning that I c > I e . Because charge does not accumulate in the bulbs, all the charge flowing into a bulb from the left has to flow out on the right; consequently I c = I d and I e = I f . The two currents leaving the bulbs recombine to form the current back into the battery, I f + I d = I b.

9.

Choice (a). The power dissipated by a resistor may be expressed as P = I 2 R, where I is the current carried by the resistor of resistance R. Since resistors connected in series carry the same current, the resistor having the largest resistance will dissipate the most power.

10.

Choice (c). Increasing the diameter of a wire increases the cross-sectional area. Thus, the crosssectional area of A is greater than that of B, and from R = rL A, we see that RA < RB . Since the power dissipated in a resistance may be expressed as P = (ΔV )2 R, the wire having the smallest resistance dissipates the most power for a given potential difference. 62

68719_17_ch17_p062-082.indd 62

1/7/11 2:39:04 PM

Current and Resistance

63

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

The average current in a conductor is the charge passing a given point per unit time, or I = ΔQ Δt = e ( Δn ) Δt, so the number of electrons passing this point per second is Δn I 1.6 C s = 1.0 × 1019 electron s = = Δt e 1.6 × 10 −19 C electron making (c) the correct choice.

2.

The drift velocity of charge carriers in a conductor is given by vd = I nqA. Wires A and B carry the same current I. Also, they are made of the same material, so the density and charge of the charge carriers (n and q) are the same for the two wires. Since the cross-sectional area is A = p r 2, and rA = 2rB, wire A has a cross-section 4 times that of B. This makes v A = vB 4, and the correct choice is (e).

3.

By cutting the wire of length L and cross-sectional area A into pieces and placing the pieces side by side, a new wire with length L ′ = L 3, and cross-sectional area A′ = 3A is created. The new resistance is R′ =

r L ′ rL 3 1 ⎛ rL ⎞ 1 = = ⎜ ⎟= R A′ 3A 9⎝ A ⎠ 9

so the correct choice is (a). 4.

The resistance of a conductor having length L ′ and cross-sectional area A is R = r L ′ A. Thus, the resistance of each wire is: R1 = rL / p r 2, R2 = rL / p (2r)2 = rL / 2p r 2 = 12 R1, and R3 = r(2L) / p (3r)2 = 4rL / 9p r 2 = 49 R1. We see that wire 3 has the smallest resistance, and choice (c) is the correct answer.

5.

The ampere is a unit of current (1 amp = 1coulomb second). Hence, the ampere-hour is a unit of charge 1 ampere-hour = 1 ampere ⋅1 hour = (1 C s ) ( 3 600 s ) = 3 600 C Thus, the ampere-hour rating is a measure of the charge the battery can supply, and choice (d) is the correct answer.

6.

When the potential across the device is 2 V, the current is 2 A, so the resistance is R = ΔV I = 2 V 2 A = 1 Ω, and (a) is the correct choice.

7.

The power consumption of the set is P = ( ΔV ) I = (120 V )( 2.5 A ) = 3.0 × 10 2 W = 0.30 kW. Thus, the energy used in 8.0 h of operation is E = P ⋅ t = ( 0.30 kW )(8.0 h ) = 2.4 kWh, at a cost of cost = ( 2.4 kWh )(8.0 cents kWh ) = 19 cents. The correct choice is (c).

8.

The temperature variation of resistance is given by R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦, where R0 is the resistance of the conductor at the reference temperature T0, usually 20.0°C. Using the given resistance at T = 90.0°C, the temperature coefficient of resistivity for this conducting material is found to be a=

68719_17_ch17_p062-082.indd 63

⎞ 1 ⎛ R 1 ⎛ 10.55 Ω ⎞ − 1⎟ = − 1⎟ = 7.86 × 10 −4 °C−1 ⎜ ⎜ T − T0 ⎝ R0 ⎠ 90.0°C − 20.0°C ⎝ 10.00 Ω ⎠

1/7/11 2:39:07 PM

64

Chapter 17

The resistance of this conductor at T = −20.0°C, where ΔT = −20.0°C − ( 20.0°C ) = −40.0°C, will be R = (10.00 Ω ) ⎡⎣1+ ( 7.86 × 10 −4 °C−1 )( −40.0°C )⎤⎦ = 9.69 Ω so (b) is the correct choice. 9.

When the potential difference across the device is 3.0 V, the current is 2.5 A, so the resistance is R = ΔV I = 3.0 V 2.5 A = 1.2 Ω, and (b) is the correct choice.

10.

Resistors in a parallel combination all have the same potential difference across them. Thus, from Ohm’s law, I = ΔV R, the resistor with the smallest resistance carries the largest current. Choice (a) is the correct response.

11.

The current through the resistor is I = ΔV R = 1.0 V 10.0 Ω = 0.10 A, and the charge passing through in a 20 s interval is ΔQ = I ⋅ Δt = ( 0.10 C s )( 20 s ) = 2.0 C. Thus, (c) is the correct choice.

12.

Resistors in a series combination all carry the same current. Thus, from Ohm’s law, ΔV = IR, the resistor with the highest resistance has the greatest voltage drop across it. Choice (c) is the correct response. 2 rLB (p d B2 4 ) ⎛ LB ⎞ ⎛ d A ⎞ PA ( ΔV ) RA RB rLB AB ⎛ 1⎞ 2 = = = = =⎜ = ⎜ ⎟ ( 2) = 2 2 2 ⎟ ⎜ ⎟ ⎝ 2⎠ PB ( ΔV ) RB RA rLA AA rLA (p d A 4 ) ⎝ LA ⎠ ⎝ d B ⎠ 2

13.

Thus, PA PB = 2, and the correct choice is (c). 14.

The resistance of a conductor may be expressed as R = rL A, so the cross-sectional area is A = rL R. Thus, R 1 AA rL RA RB = = = B = AB rL RB RA 3RB 3 and choice (e) is the correct answer.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The amplitude of atomic vibrations increases with temperature, thereby scattering electrons more efficiently and increasing the resistivity of the material.

4.

(a)

The number of cars would correspond to charge Q.

(b)

The rate of flow of cars past a point would correspond to current.

(a)

The 25 W bulb has the higher resistance. Because R = ( ΔV ) P, and both operate from 120 V, the bulb dissipating the least power has the higher resistance.

(b)

When the voltage is constant, the current and power are directly proportional to each other, P = ( ΔV ) I = ( constant ) I . Thus, the higher power bulb (100 W) carries more current.

6.

2

8.

An electrical shock occurs when your body serves as a conductor between two points having a difference in potential. The concept behind the admonition is to avoid simultaneously touching points that are at different potentials.

10.

The knob is connected to a variable resistor. As you increase the magnitude of the resistance in the circuit, the current is reduced, and the bulb dims.

68719_17_ch17_p062-082.indd 64

1/7/11 2:39:10 PM

Current and Resistance

65

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a) 5.57 × 10 −5 m s

4.

3.4 × 10 21 electrons

6.

159 mA

8.

0.130 mm s

10.

8.89 Ω

12.

(a) 1.8 m

14.

3.22 × 10 −8 Ω ⋅ m

16.

1.3

18.

(a)

2.8 × 108 A

20.

(a)

5.0 × 10 5 V

(b)

the drift speed is smaller

(b)

0.28 mm

(b) 1.8 × 10 7 A

(b) Rubber gloves and soles can increase resistance to current passing through the body. 22.

6.3 Ω

24.

(a) 1.6 × 10 2 °C (b) The expansion of the cross-sectional area contributes slightly more than the expansion of the length of the wire, so the answer would be slightly reduced.

26.

2.3 × 10 2 °C

28.

1.1× 10 −3 ( °C )

30.

(a) Yes, the design goal can be met with R0 ,C arbon = 4.4 Ω and R0 ,Nichrom e = 5.6 Ω.

−1

(b)

LC arbon = 0.89 m, LNichrom e = 26 m

32.

63.3°C

34.

(a) $0.29

(b)

$2.6

36.

(a) 50 MW

(b)

0.071 or 7.1%

38.

(a) 3.2 × 10 5 J

(b)

18 min

40.

(a)

184 W

(b) 461°C

42.

(a)

0.66 kWh

(b)

$0.079 = 7.9 cents

44.

(a) $1.61

(b)

0.582 cents

68719_17_ch17_p062-082.indd 65

(c)

$0.416 = 41.6 cents

1/7/11 2:39:11 PM

66

46.

48.

Chapter 17

(a) 3.5 × 10 2 W

(b)

41 Ω 0.38 mm

(d)

d = 4r0 L p R0

(e)

(a)

2.7 × 10 −3 Ω

(b) 1.45 × 10 3 K

(d) 1.1 V

(c)

4.0 × 101 Ω

(c) 1.7 × 10 −2 Ω

(e) They radiate only a small portion of the energy consumed as visible light.

50.

15.0 h

52.

1.5 × 10 2 °C

54.

90 mV

56.

(a) 9.3 m

(b)

0.93 mm

58.

(a) 18 C

(b)

3.6 A

60.

(a)

9.1 Ω

(b) We assume the temperature coefficient of resistivity for Nichrome remains constant over this temperature range. 62.

No, the fuse should limit the current to 3.87 A or less.

64.

(a)

66.

(a) 470 W

(b) 1.420 Ω versus 1.418 Ω

See Solution.

(b) 1.60 mm or more

(c)

2.93 mm or more

PROBLEM SOLUTIONS 17.1

(a)

The charge that moves past the cross section is ΔQ = I ( Δt ), and the number of electrons is n=

=

17.2

ΔQ I ( Δt ) = e e

(80.0 × 10

−3

C s ) ⎡⎣(10.0 min )( 60.0 s min ) ⎤⎦ = 3.00 × 10 20 electrons 1.60 × 10 −19 C

(b)

The negatively charged electrons move in the direction opposite to the conventional current flow.

(a)

From Example 17.2 in the textbook, the density of charge carriers (electrons) in a copper wire is n = 8.46 × 10 28 electrons m 3. With A = p r 2 and q = e, the drift speed of electrons in this wire is vd =

I 3.70 C s = 5.57 × 10 −5 m s = 2 n q A (8.46 × 10 28 m −3 ) (1.60 × 10 −19 C ) p (1.25 × 10 −3 m )

continued on next page

68719_17_ch17_p062-082.indd 66

1/7/11 2:39:13 PM

Current and Resistance

(b)

17.3

The drift speed is smaller because more electrons are being conducted. With a greater number of charge carriers in motion, they do not have to move as fast to have a specified number of them passing a given point each second.

The period of the electron in its orbit is T = 2p r v, and the current represented by the orbiting electron is I = ΔQ Δt = e T = v e 2p r, or I=

17.4

( 2.19 × 10

m s ) (1.60 × 10 −19 C )

= 1.05 × 10 −3 C s = 1.05 mA

Since I = ΔQ Δt, we have ΔQ = I ( Δt ), and the number of electrons passing through in time Δt is N = ΔQ e = I ( Δt ) e. Thus,

( 0.15

C s )(1 h )( 3 600 s 1 h ) = 3.4 × 10 21 electrons 1.60 × 10 −19 C electron

The resistance of a wire of length L and diameter d is R = rL A = rL (p d 2 4 ), giving d 2 = 4rL p R. Using Table 17.1, the needed diameter is found to be 4 ( 2.82 × 10 −8 Ω ⋅ m )( 32.0 m )

4rL = pR

d= 17.6

6

2p ( 5.29 × 10 −11 m )

N= 17.5

67

p ( 2.50 Ω )

= 6.78 × 10 −4 m = 0.678 mm

The mass of a single gold atom is matom = M N A , where M is the molecular weight of gold and N A is Avogadro’s number. Thus, the number of atoms deposited, and hence the number of ions moving to the negative electrode, is n=

−3 23 m mN A ( 3.25 × 10 kg ) ( 6.02 × 10 atoms mol ) = = = 9.93 × 10 21 matom M (197 g mol ) (10 −3 kg g)

The current in the cell is then I= 17.7

21 −19 ΔQ ne ( 9.93 × 10 ) (1.60 × 10 C ) = = = 0.159 A = 159 mA Δt Δt ( 2.78 h )( 3 600 s 1 h )

The drift speed of electrons in the line is vd = I nqA = I n e (p d 2 4 ). The time to travel the length of the 200-km line is then Δt = L vd = Ln e (p d 2 ) 4I, or

( 200 × 10 m ) ( 8.5 × 10 3

Δt = 17.8

28

m 3 ) (1.6 × 10 −19 C ) p ( 0.02 m )

4 (1 000 A ) ( 3.156 × 10 7 s yr )

2

= 27 yr

The mass of a single aluminum atom is matom = M N A, where the molecular weight (mass per mole) of aluminum is M = 27.0 g mol, and Avogadro’s number (number of atoms per mole) is N A = 6.02 × 10 23 mol. If each aluminum atom contributes one conduction electron, the number of conduction electrons per unit volume is n=

mass per unit volume density rN A = = mass per atom matom M

and the drift speed is vd = I nqA = MI rN A qA. Thus, for the given case, we find

continued on next page

68719_17_ch17_p062-082.indd 67

1/7/11 2:39:15 PM

68

Chapter 17

vd =

17.9

( 27.0

g mol ) ( 5.00 C s )

⎡⎛ g ⎞ ⎛ 10 cm ⎞ ⎤ ⎛ 6.02 × 10 23 ⎞ −19 −6 2 ⎢⎜⎝ 2.70 3 ⎟ ⎜ 3 ⎟⎠ ⎥ ⎜⎝ ⎟⎠ (1.60 × 10 C ) ( 4.00 × 10 m ) ⎠ 1 m mol cm ⎝ ⎣ ⎦ 6

3

or

vd = 1.30 × 10 −4 m s = 0.130 mm s .

(a)

Using the periodic table on the inside back cover of the textbook, we find M Fe = 55.85 g mol = ( 55.85 g mol ) (1 kg 10 3 g ) = 55.85 × 10 −3 kg mol

(b)

From Table 9.1, the density of iron is rFe = 7.86 × 10 3 kg m 3 , so the molar density is

( molar density ) (c)

=

Fe

rFe 7.86 × 10 3 kg m 3 = 1.41× 10 5 mol m 3 = M Fe 55.85 × 10 −3 kg mol

The density of iron atoms is density of atoms = N A ( molar density ) atoms ⎞ ⎛ ⎛ 5 mol ⎞ 28 atoms = ⎜ 6.02 × 10 23 ⎟ ⎜ 1.41× 10 ⎟ = 8.49 × 10 ⎝ mol ⎠ ⎝ m3 ⎠ m3

(d)

With two conduction electrons per iron atom, the density of charge carriers is n = ( charge carriers atom ) ( density of atoms ) ⎛ electrons ⎞ ⎛ 28 atoms ⎞ 29 3 = ⎜2 ⎟ ⎜ 8.49 × 10 ⎟ = 1.70 × 10 electrons m ⎝ atom ⎠ ⎝ m3 ⎠

(e)

With a current of I = 30.0 A and cross-sectional area A = 5.00 × 10 −6 m 2 , the drift speed of the conduction electrons in this wire is vd =

I 30.0 C s = = 2.21× 10 −4 m s 29 −3 nqA (1.70 × 10 m ) (1.60 × 10 −19 C ) ( 5.00 × 10 −6 m 3 ) R=

ΔV 1.20 × 10 2 V = 8.89 Ω = 13.5 A I

17.10

From Ohm’s law,

17.11

( ΔV )m ax = I m ax R = (80 × 10 −6 A ) R Thus, if R = 4.0 × 10 5 Ω, ( ΔV )m ax = 32 V and if R = 2 000 Ω, ( ΔV )m ax = 0.16 V

17.12

The volume of the copper is V=

m 1.00 × 10 −3 kg = = 1.12 × 10 −7 m 3 density 8.92 × 10 3 kg m 3

Since V = A ⋅ L, this gives A ⋅ L = 1.12 × 10 −7 m 3.

[1]

continued on next page

68719_17_ch17_p062-082.indd 68

1/7/11 2:39:18 PM

Current and Resistance

(a)

From R =

69

rL , where r is the resistivity of copper, we find that A

⎛ 1.7 × 10 −8 Ω ⋅ m ⎞ ⎛ r⎞ −8 A=⎜ ⎟L=⎜ ⎟⎠ L = ( 3.4 × 10 m ) L ⎝ R⎠ 0.500 Ω ⎝ Inserting this expression for A into Equation [1] gives

(3.4 × 10 (b)

−8

m ) L2 = 1.12 × 10 −7 m 3, which yields

From Equation [1], A =

d=

L = 1.8 m

p d 2 1.12 × 10 −7 m 3 = , or 4 L

4 (1.12 × 10 −7 m 3 ) pL

=

4 (1.12 × 10 −7 m 3 ) p (1.8 m )

= 2.8 × 10 −4 m = 0.28 mm 17.13

R=

ΔV 1.20 × 10 2 V = = 13.0 Ω I 9.25 A

(a)

From Ohm’s law,

(b)

Using R = rL A and data from Table 17.1, the required length is found to be 2 −3 RA R (p r ) (13.0 Ω )p ( 0.791× 10 m ) L= = = = 17.0 m r r 150 × 10 −8 Ω ⋅ m 2

17.14

From R = rL A = rL (p d 2 4 ), the resistivity is −3 p Rd 2 p (1.60 Ω )( 0.800 × 10 m ) r= = = 3.22 × 10 −8 Ω ⋅ m 4L 4 ( 25.0 m ) 2

17.15

ΔV 12 V = = 30 Ω I 0.40 A

(a)

R=

(b)

From R = rL A, 2 −2 ⎡ ⎤ R ⋅ A ( 30 Ω ) ⎣p ( 0.40 × 10 m ) ⎦ = 4.7 × 10 − 4 Ω ⋅ m r= = 3.2 m L

17.16

Using R = rL A and data from Table 17.1, we have rC u LC u p rC2u = rAl LAl p rAl2. This reduces to rAl2 rC2u = rAl rC u, and yields rAl = rC u

17.17

rAl = rC u

2.82 × 10 −8 Ω ⋅ m = 1.3 1.7 × 10 −8 Ω ⋅ m

From Ohm’s law, R = ΔV I , and R = rL A = rL (p d 2 4 ), the resistivity is −3 p Rd 2 p ( ΔV ) d 2 p ( 9.11 V )( 2.00 × 10 m ) r= = = = 1.59 × 10 −8 Ω ⋅ m 4L 4IL 4 ( 36.0 A )( 50.0 m ) 2

Then, from Table 17.1, we see that the wire is made of silver .

68719_17_ch17_p062-082.indd 69

1/7/11 2:39:20 PM

70

17.18

Chapter 17

With different orientations of the block, three different values of the ratio L A are possible. These are: 10 cm 1 1 ⎛ L⎞ = = , ⎜⎝ ⎟⎠ = A 1 ( 20 cm )( 40 cm ) 80 cm 0.80 m 20 cm 1 1 ⎛ L⎞ = = , ⎜⎝ ⎟⎠ = A 2 (10 cm )( 40 cm ) 20 cm 0.20 m

17.19

and

40 cm 1 1 ⎛ L⎞ = = ⎜⎝ ⎟⎠ = A 3 (10 cm )( 20 cm ) 5.0 cm 0.050 m

(a)

I m ax =

ΔV ΔV ( 6.0 V )( 0.80 m ) = = = 2.8 × 108 A Rm in r ( L A )m in 1.7 × 10 −8 Ω ⋅ m

(b)

I m in =

ΔV ΔV ( 6.0 V )( 0.050 m ) = = = 1.8 × 10 7 A Rm ax r ( L A )m ax 1.7 × 10 −8 Ω ⋅ m

The volume of material, V = AL0 = (p r02 ) L0 , in the wire is constant. Thus, as the wire is stretched to decrease its radius, the length increases such that p rf2 L f = (p r02 ) L0 giving 2 2 2 L f = r0 rf L0 = ( r0 0.25r0 ) L0 = ( 4.0 ) L0 = 16L0 . The new resistance is then

(

)

Rf = r 17.20

17.21

( )

Lf Af

=r

Lf pr

2 f

=r

16L0

p ( 0.25r0 )

⎛ L ⎞ = 256 ⎜ r 02 ⎟ = 256R0 = 256 (1.00 Ω ) = 256 Ω ⎝ p r0 ⎠

(a)

From Ohm’s law, ΔV = IR = ( 500 × 10 −3 A ) (1.0 × 10 6 Ω ) = 5 × 10 5 V

(b)

Rubber-soled shoes and rubber gloves can increase the resistance to current and help reduce the likelihood of a serious shock.

If a conductor of length L has a uniform electric field E maintained within it, the potential difference between the ends of the conductor is ΔV = EL. But, from Ohm’s law, the relation between the potential difference across a conductor and the current through it is ΔV = IR, where R = r L A. Combining these relations, we obtain ΔV = EL = IR = I ( r L A )

17.22

2

E = r ( I A ) = rJ

or

−1 Using R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦, with R0 = 6.00 Ω at T0 = 20.0°C and a silver = 3.8 × 10 −3 (°C ) (from Table 17.1 in the textbook), the resistance at T = 34.0°C is −1 R = ( 6.00 Ω ) ⎡⎣1+ 3.8 × 10 −3 (°C ) ( 34.0°C − 20.0°C )⎤⎦ = 6.3 Ω

17.23

From Ohm’s law, ΔV = I i Ri = I f R f , so the current in Antarctica is ⎛ R ⎡1+ a ( T − T ) ⎤ ⎞ ⎛ 1+ ⎡3.9 × 10 −3 ( °C )−1 ⎤ ( 58.0°C − 20.0°C ) ⎞ ⎛R ⎞ 0 ⎣ i 0 ⎦ ⎣ ⎦ ⎟ = (1.00 A ) ⎜ I f = Ii ⎜ i ⎟ = Ii ⎜ ⎟ ⎜⎝ 1+ ⎡3.9 × 10 −3 ( °C )−1 ⎤ ( −88.0°C − 20.0°C ) ⎟⎠ ⎜ R0 ⎡1+ a T f − T0 ⎤ ⎟ ⎝ Rf ⎠ ⎣ ⎦ ⎝ ⎣ ⎠ ⎦

(

)

or I f = 2.0 A

68719_17_ch17_p062-082.indd 70

1/7/11 2:39:23 PM

Current and Resistance

17.24

(a)

Given: Aluminum wire with a = 3.9 × 10 −3 (°C ) (see Table 17.1 in textbook), and R0 = 30.0 Ω at T0 = 20.0°C. If R = 46.2 Ω at temperature T, solving R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ gives the final temperature as −1

T = T0 + (b)

17.25

( R R ) − 1 = 20.0°C + ( 46.2 Ω 30.0 Ω) − 1 = 1.6 × 10 0

3.9 × 10 −3 (°C )

−1

a

2

°C

The expansion of the cross-sectional area contributes slightly more than the expansion of the length of the wire, so the answer would be slightly reduced.

The volume of the gold wire may be written as V = A ⋅ L = m rd , where rd is the density of gold. Thus, the cross-sectional area is A = m rd L. The resistance of the wire is R = re L A, where re is the electrical resistivity. Therefore, −8 3 3 3 rL r r L2 ( 2.44 × 10 Ω ⋅ m ) (19.3 × 10 kg m ) ( 2.40 × 10 m ) R= e = e d = m rd L m 1.00 × 10 −3 kg

2

R = 2.71× 10 6 Ω = 2.71 MΩ

giving 17.26

71

For aluminum, the resistivity at room temperature is r0 ,Al = 2.82 × 10 −8 Ω ⋅ m, and the temperature coefficient of resistivity is a Al = 3.9 × 10 −3 (°C)−1. Thus, if at some temperature, the aluminum has a resistivity of rAl, solving rAl = r0 ,Al ⎡⎣1+ a Al ( T − T0 ) ⎤⎦ for that temperature gives T = T0 + ⎡⎣ rAl r0 ,Al − 1⎤⎦ a Al , where T0 = 20°C.

{(

)

}

When rAl = 3r0 ,C u = 3 (1.7 × 10 −8 Ω ⋅ m ) = 5.1× 10 −8 Ω ⋅ m, the temperature must be ⎛ 5.1× 10 −8 Ω ⋅ m ⎞ ⎜⎝ 2.82 × 10 −8 Ω ⋅ m ⎟⎠ − 1 T = 20°C + = 2.3 × 10 2 °C −1 3.9 × 10 −3 (°C ) 17.27

At 80°C, I= or

17.28

I = 2.6 × 10 −2 A = 26 mA

If R = 41.0 Ω at T = 20.0°C and R = 41.4 Ω at T = 29.0°C, then R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ gives the temperature coefficient of resistivity of the material making up this wire as a=

17.29

ΔV ΔV 5.0 V = = 2 R R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ ( 2.0 × 10 Ω ) ⎡⎣1+ ( − 0.5 × 10 −3 °C −1 )(80°C − 20°C )⎤⎦

(a)

R − R0 41.4 Ω − 41.0 Ω −1 = = 1.1× 10 −3 (°C ) R0 ( T − T0 ) ( 41.0 Ω )( 29.0°C − 20.0°C )

The resistance at 20.0°C is R0 = r

−8 L (1.7 × 10 Ω ⋅ m )( 34.5 m ) = = 3.0 Ω 2 A p ( 0.25 × 10 −3 m )

and the current will be I =

ΔV 9.0 V = = 3.0 A R0 3.0 Ω

continued on next page

68719_17_ch17_p062-082.indd 71

1/7/11 2:39:26 PM

72

Chapter 17

(b)

At 30.0°C, R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦

(

= ( 3.0 Ω ) ⎡1+ 3.9 × 10 −3 ( °C ) ⎣ Thus, the current is I = 17.30

−1

)(30.0°C − 20.0°C)⎤⎦ = 3.1 Ω

ΔV 9.0 V = = 2.9 A . R 3.1 Ω

We call the carbon resistor 1 and the Nichrome resistor 2. Then, connecting the resistors end to end, the total resistance is R = R1 + R2 = R01 [1+ a 1 ⋅ ΔT ] + R02 [1+ a 2 ⋅ ΔT ] = R01 + R02 + ( R01a 1 + R02a 2 ) ΔT If this is not to vary with temperature, it is necessary that R01a 1 + R02a 2 = 0, or ⎛ 0.4 × 10 −3 ( °C )−1 ⎞ ⎛a ⎞ R01 = − ⎜ 2 ⎟ R02 = − ⎜ R02 = 0.8R02 −1 −3 ⎝ a1 ⎠ ⎝ −0.5 × 10 ( °C ) ⎟⎠ If the constant total resistance is to be R = 10.0 Ω, it is necessary that R01 + R02 = 0.8R02 + R02 = 10.0 Ω or (a) (b)

R02 =

10.0 Ω = 5.6 Ω 1.8

Yes it is possible for her to meet the design goal with this method. From R = rL A = rL p r 2, we have L = (p r 2 ) ⋅ R r. This yields L1 = (p r 2 )

2 R01 4.4 Ω = p (1.50 × 10 −3 m ) = 0.89 m (carbon) r01 3.5 × 10 −5 Ω ⋅ m 2 R02 5.6 Ω = p (1.50 × 10 −3 m ) = 26 m (Nichrome) r02 150 × 10 −8 Ω ⋅ m

and L2 = (p r 2 ) 17.31

(a)

From R = rL A, the initial resistance of the mercury is Ri =

(b)

R01 = 10.0 Ω − R02 = 4.4 Ω

and

−7 rLi ( 9.4 × 10 Ω ⋅ m ) (1.000 0 m ) = 1.2 Ω = 2 Ai p (1.00 × 10 −3 m ) 4

Since the volume of mercury is constant, V = A f ⋅ L f = Ai ⋅ Li gives the final cross-sectional area as A f = Ai ⋅ Li L f . Thus, the final resistance is given by R f = r L f A f = r L2f Ai ⋅ Li. The fractional change in the resistance is then

(

Δ=

R f − Ri Ri

)

Rf

=

Ri

−1 =

r L2f

(A ⋅ L ) −1 = ⎛ L i

r Li Ai

i

2

⎞ ⎜⎝ L ⎟⎠ − 1 i f

2

⎛ 100.04 ⎞ Δ=⎜ − 1 = 8.0 × 10 −4 or a 0.080% increase ⎝ 100.00 ⎟⎠

68719_17_ch17_p062-082.indd 72

1/7/11 2:39:29 PM

Current and Resistance

17.32

73

The resistance at the reference temperature of 20.0°C is R0 =

R 200.0 Ω = = 217 Ω 1+ a ( T − T0 ) 1+ ⎡3.92 × 10 −3 ( °C )−1 ⎤ ( 0°C − 20.0°C ) ⎣ ⎦

Solving R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ for T gives the temperature of the melting potassium as T = T0 + 17.33

(a)

R − R0 253.8 Ω − 217 Ω = 20.0°C + = 63.3°C a R0 ⎡⎣3.92 × 10 −3 ( °C )−1 ⎤⎦ ( 217 Ω )

The power consumed by the device is P = I ( ΔV ), so the current must be I=

17.34

P 1.00 × 10 3 W = 8.33 A = ΔV 1.20 × 10 2 V R=

ΔV 1.20 × 10 2 V = = 14.4 Ω I 8.33 A

(b)

From Ohm’s law, the resistance is

(a)

The energy used by a 100-W bulb in 24 h is E = P ⋅ Δt = (100 W )( 24 h ) = ( 0.100 kW )( 24 h ) = 2.4 kWh and the cost of this energy, at a rate of $0.12 per kilowatt-hour is cost = E ⋅rate = ( 2.4 kWh )( $0.12 kWh ) = $0.29

(b)

The energy used by the oven in 5.0 h is ⎡ ⎛ 1 kW ⎞ ⎤ E = P ⋅ Δt = [ I ( ΔV )] ⋅ Δt = ⎢( 20.0 C s ) ( 220 J C ) ⎜ ⎥ ( 5.0 h ) = 22 kWh ⎝ 10 3 J s ⎟⎠ ⎦ ⎣ and the cost of this energy, at a rate of $0.12 per kilowatt-hour is cost = E ⋅rate = ( 22 kWh )( $0.12 kWh ) = $2.6

17.35

(a)

RC u = and

(b)

RAl = and

17.36

−8 rC u L 4rC u L 4 (1.7 × 10 Ω ⋅ m )(1.00 m ) = = = 5.2 × 10 −3 Ω 2 A pd2 p ( 0.205 × 10 −2 m )

PC u = I 2 RC u = ( 20.0 A ) ( 5.2 × 10 −3 Ω ) = 2.1 W 2

−8 rAl L 4rAl L 4 ( 2.82 × 10 Ω ⋅ m )(1.00 m ) = = = 8.54 × 10 −3 Ω 2 −2 A pd2 p ( 0.205 × 10 m )

PAl = I 2 RAl = ( 20.0 A ) (8.54 × 10 −3 Ω ) = 3.42 W 2

(c)

No , the aluminum wire would not be as safe. If surrounded by thermal insulation, it would get much hotter than the copper wire.

(a)

The power loss in the line is Ploss = I 2 R = (1 000 A ) ⎡⎣( 0.31 Ω km )(160 km )⎤⎦, or Ploss = 5.0 × 10 7 W = 50 MW .

(b)

The total power transmitted is Pinput = ( ΔV ) I = ( 700 × 10 3 V ) (1 000 A ) , or Pinput = 7.0 × 108 W = 700 MW.

2

continued on next page

68719_17_ch17_p062-082.indd 73

1/7/11 2:39:32 PM

74

Chapter 17

Thus, the fraction of the total transmitted power represented by the line losses is fraction loss = 17.37

Ploss 50 MW = = 0.071 or 7.1% Pinput 700 MW

The energy required to bring the water to the boiling point is E = m c ( ΔT ) = ( 0.500 kg ) ( 4 186 J kg ⋅°C )(100°C − 23.0°C ) = 1.61× 10 5 J The power input by the heating element is Pinput = ( ΔV ) I = (120 V )( 2.00 A ) = 240 W = 240 J s Therefore, the time required is t=

17.38

E 1.61× 10 5 J ⎛ 1 min ⎞ = 671 s ⎜ = 11.2 min = ⎝ 60 s ⎟⎠ 240 J s Pinput

(a)

E = P ⋅ t = ( 90 W )(1 h ) = ( 90 J s ) ( 3 600 s ) = 3.2 × 10 5 J

(b)

The power consumption of the color set is P = ( ΔV ) I = (120 V )( 2.50 A ) = 300 W Therefore, the time required to consume the energy found in (a) is t=

17.39

(a)

RA =

( ΔV )2 PA

=

(120 V )2 25.0 W

( ΔV )2

= 576 Ω

(120 V )2

RB =

(b)

Δt A =

Q ⎛R ⎞ ⎛ 576 Ω ⎞ = Q ⎜ A ⎟ = (1.00 C ) ⎜ = 4.80 s ⎝ ΔV ⎠ ⎝ 120 V ⎟⎠ IA

(c)

The charge is the same. However, as it leaves the bulb, it is at a lower potential than when it entered the bulb.

(e) (f )

PB

=

⎛ 1 min ⎞ s⎜ = 18 min ⎝ 60 s ⎟⎠

and

(d)

100 W

= 144 Ω

W 1.00 J W, so Δt A = = = 0.040 0 s P 25.0 Js A Δt Energy enters the bulb by electrical transmission and leaves by heat and radiation. P=

⎡ ⎛ 24.0 h ⎞ ⎤ ⎛ 1 kW ⎞ ⎤ ⎡ E = PA ( Δt ) = ⎢( 25.0 W ) ⎜ 3 = 18.0 kWh 30.0 d ) ⎜ ( ⎟ ⎢ ⎥ ⎝ 1 d ⎟⎠ ⎥⎦ ⎝ 10 W ⎠ ⎦ ⎣ ⎣ and

68719_17_ch17_p062-082.indd 74

E 3.2 × 10 5 J = = 1.1× 10 3 P 300 J s

cost = E × rate = (18.0 kWh )( $0.110 kWh ) = $1.98

1/7/11 2:39:34 PM

Current and Resistance

17.40

(a)

75

At the operating temperature, P = ( ΔV ) I = (120 V )(1.53 A ) = 184 W

(b)

From R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ , the temperature T is given by T = T0 + (R − R0 ) a R0. The resistances are given by Ohm’s law as R=

( ΔV ) 120 V I

=

1.53 A

and

R0 =

( ΔV )0 I0

=

120 V 1.80 A

Therefore, the operating temperature is T = 20.0°C + 17.41

(120 1.53) − (120 1.80 ) = ( 0.400 × 10 (°C) )(120 1.80 ) −3

−1

461°C

The power loss per unit length of the cable is P L = (I 2 R) L = I 2 (R L). Thus, the resistance per unit length of the cable is R P L 2.00 W m = 2 = = 2.22 × 10 −5 Ω m L I ( 300 A )2 From R = rL A, the resistance per unit length is also given by R L = r A. Hence, the cross-sectional area is p r 2 = A = r ( R L ), and the required radius is r=

17.42

(a)

r 1.7 × 10 −8 Ω ⋅ m = = 0.016 m = 1.6 cm p (R L) p ( 2.22 × 10 −5 Ω m )

The rating of the 12-V battery is I ⋅ Δt = 55 A ⋅ h. Thus, the stored energy is Energy stored = P ⋅ Δt = ( ΔV ) I ⋅ Δt = (12 V )( 55 A ⋅ h ) = 660 W ⋅ h = 0.66 kWh

(b)

value = ( 0.66 kWh )( $0.12 kWh ) = $0.079 = 7.9 cents

17.43

P = ( ΔV ) I = ( 75.0 × 10 −3 V ) ( 0.200 × 10 −3 A ) = 1.50 × 10 −5 W = 15.0 × 10 −6 W = 15.0 mW

17.44

(a)

E = P ⋅ t = ( 40.0 W )(14.0 d )( 24.0 h d ) = 1.34 × 10 4 Wh = 13.4 kWh cost = E ⋅ ( rate ) = (13.4 kWh )( $0.120 kWh ) = $1.61

(b)

E = P ⋅ t = ( 0.970 kW )( 3.00 min )(1 h 60 min ) = 4.85 × 10 −2 kWh cost = ( 4.85 × 10 −2 kWh ) ( $0.120 kWh ) = $0.005 82 = 0.582 cents

(c)

E = P ⋅ t = ( 5.20 kW )( 40.0 min )(1 h 60 min ) = 3.47 kWh cost = E ⋅ ( rate ) = ( 3.47 kWh )( $0.120 kWh ) = $0.416 = 41.6 cents

17.45

Total length of transmission lines: L = 2 ( 50.0 m ) = 100 m. Thus, the resistance of these lines is R = ( 0.108 Ω 300 m )(100 m ) = 3.60 × 10 −2 Ω. (a)

The total potential drop along the transmission lines is ( ΔV )lines = IR, giving

( ΔV )house = ( ΔV )source − ( ΔV )lines = 120 V − (110 A )( 3.60 × 10 −2 Ω ) = 116 V (b)

68719_17_ch17_p062-082.indd 75

Pdelivered = I ( ΔV )house = (110 A )(116 V ) = 1.28 × 10 4 W = 12.8 kW

1/7/11 2:39:36 PM

76

17.46

Chapter 17

(a)

The thermal energy needed to raise the water temperature by ΔT is E = mc(ΔT ). If ΔT = 100°C − 20°C = 80°C and this energy is to be delivered in 4.00 min, the average power required is P=

(b)

The required resistance (at 100°C) of the heating element is then R=

(c)

E mc ( ΔT ) ( 0.250 kg ) ( 4 186 J kg ⋅°C )(100°C − 20°C ) = = = 3.5 × 10 2 W Δt 4.00 min 60 s 1 min Δt ( )( )

( ΔV )2 P

=

(120 V )2 3.5 × 10 2 W

= 41 Ω

The resistance at 20.0°C would then be R0 =

R 41 Ω = = 4.0 × 101 Ω 1+ a ( T − T0 ) 1+ ( 0.4 × 10 −3 °C−1 )(100°C − 20°C )

(d)

We find the needed dimensions of a Nichrome wire for this heating element from R0 = r0 L A = r0 L (p d 2 4 ) = 4r0 L p d 2, where L is the length of the wire and d is its diameter. This gives the diameter as d = 4r0 L p R0 .

(e)

If L = 3.00 m, the required diameter of the wire is ⎡ 4 (150 × 10 −8 Ω ⋅ m )( 3.00 m ) ⎤ ⎡ 4r L ⎤ d=⎢ 0 ⎥ =⎢ ⎥ = 3.8 × 10 −4 m = 0.38 mm 1 p R p 4.0 × 10 Ω ( ) ⎥⎦ 0 ⎦ ⎣ ⎣⎢ 1 2

1 2

17.47

The power dissipated in a conductor is P = ( ΔV ) R, so the resistance may be written as 2 R = ( ΔV ) P. Hence, 2

( ΔV ) RB PA P = ⋅ = A =3 2 RA PB ( ΔV ) PB 2

or

RB = 3RA

Since R = rL A = rL (p d 2 4 ), this result becomes ⎛ 4rL ⎞ = 3⎜ 2 2 ⎟ p dB ⎝ p dA ⎠

4rL

or

d A2 =3 d B2

and yields d A d B = 3 . 17.48

(a)

For tungsten, Table 17.1 from the textbook gives the resistivity at T0 = 20.0°C = 293 K as r0 = 5.6 × 10 −8 Ω ⋅ m and the temperature coefficient of resistivity as −1 a = 4.5 × 10 −3 (°C ) = 4.5 × 10 −3 K −1. Thus, for a tungsten wire having a radius of 1.00 mm and a length of 15.0 cm, the resistance at T0 = 293 K is R0 = r0

(b)

(15.0 × 10−2 m ) = 2.7 × 10−3 Ω L L −8 = r0 = 5.6 × 10 Ω ⋅ m ( ) 2 A (p r 2 ) p (1.00 × 10 −3 m )

From Stefan’s law (see Section 11.5 of the textbook), the radiated power is P = s AeT 4 , where A is the area of the radiating surface. Note that since we are computing the radiated power, not the net energy gained or lost as a result of radiation, the ambient temperature is not needed here. In the case of a wire, this is the cylindrical surface area A = 2p rL. The temperature of 1 4 the wire when it is radiating a power of P = 75.0 W is given by T = [ P s Ae ] as

continued on next page

68719_17_ch17_p062-082.indd 76

1/7/11 2:39:39 PM

Current and Resistance

77

1 4

⎡ ⎤ 75.0 W T=⎢ ⎥ −8 2 4 −3 ⎢⎣ ( 5.669 × 10 W m ⋅K ) 2p (1.00 × 10 m )( 0.150 m )( 0.320 ) ⎥⎦ (c)

= 1.45 × 10 3 K

Assuming a linear temperature variation of resistance, the resistance of the wire at this temperature is R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ = ( 2.7 × 10 −3 Ω ) ⎡⎣1+ ( 4.5 × 10 −3 K −1 ) (1.45 × 10 3 K − 293 K ) ⎤⎦ giving

(d)

R = 1.7 × 10 −2 Ω

The voltage drop across the wire when it is radiating 75.0 W and has the resistance found in 2 part (c) above is given by P = ( ΔV ) R as

(1.7 × 10

ΔV = R ⋅ P = (e)

17.49

−2

Ω )( 75.0 W ) = 1.1 V

Tungsten bulbs radiate little of the energy they consume in the form of visible light , making them inefficient sources of light.

The battery is rated to deliver the equivalent of 60.0 amperes of current (i.e., 60.0 C/s) for 1 hour. This is Q = I ⋅ Δt = ( 60.0 A )(1 h ) = ( 60.0 C s ) ( 3 600 s ) = 2.16 × 10 5 C

17.50

The energy available in the battery is Energy stored = P ⋅ t = ( ΔV ) I ⋅ t = (12.0 V )( 90.0 A ⋅ h ) = 1.08 × 10 3 W ⋅ h The two head lights together consume a total power of P = 2 ( 36.0 W ) = 72.0 W, so the time required to completely discharge the battery is Energy stored 1.08 × 10 3 W ⋅ h = 15.0 h = 72.0 W P

Δt =

17.51

17.52

(a)

R=

(b)

I=

−8 rL rL 4rL 4 ( 2.82 × 10 Ω ⋅ m )(15.0 m ) = = = = 1.50 Ω 2 A pd2 4 pd2 p ( 0.600 × 10 −3 m )

ΔV 9.00 V = = 6.00 A R 1.50 Ω

Using chemical symbols to denote the two different metals, the resistances are equal when R0 ⎡⎣1+ a C u ( ΔT )⎤⎦ = R0 ⎣⎡1+ a W ( ΔT )⎤⎦ Cu

or

R0

Cu

R0

W

Thus,

W

⎡ ⎤ ⎛ R0 ⎞ − 1 = ⎢a W − ⎜ ⎟ a C u ⎥ ( ΔT ) ⎢⎣ ⎥⎦ ⎝ R0 ⎠ Cu

W

(R R ) −1 a − ( R R )a (R R ) −1 − ( R R )a 0 Cu

ΔT = T − 20.0 °C =

W

or

T = 20.0 °C +

0 Cu

aW

0W

0 Cu

0W

Cu

0W

0 Cu

0W

Cu

continued on next page

68719_17_ch17_p062-082.indd 77

1/7/11 2:39:42 PM

78

Chapter 17

T = 20.0 °C +

(5.00 4.75) − 1

(

4.5 × 10 −3 ( °C ) − ( 5.00 4.75) 3.9 × 10 −3 ( °C ) −1

−1

)

= 20.0 °C + 1.3 × 10 2 °C = 1.5 × 10 2 °C 17.53

From P = ( ΔV ) R, the total resistance needed is R = (ΔV )2 P = (20 V)2 48 W = 8.3 Ω. 2

Thus, from R = rL A, the length of wire required is L= 17.54

−6 2 R ⋅ A (8.3 Ω )( 4.0 × 10 m ) = = 1.1× 10 3 m = 1.1 km r 3.0 × 10 −8 Ω ⋅ m

The resistance of the 4.0 cm length of wire between the feet is R=

−8 rL (1.7 × 10 Ω ⋅ m )( 0.040 m ) = 1.8 × 10 −6 Ω = 2 A p ( 0.022 m ) 4

so the potential difference is ΔV = IR = ( 50 A )(1.8 × 10 −6 Ω ) = 9.0 × 10 −5 V = 90 mV 17.55

Ohm’s law gives the resistance as R = ( ΔV ) I . From R = rL A, the resistivity is given by r = R ⋅ ( A L ). The results of these calculations for each of the three wires are summarized in the table below. L (m)

R (Ω)

r (Ω ⋅ m )

0.540

10.4

1.41× 10 −6

1.028

21.1

1.50 × 10 −6

1.543

31.8

1.50 × 10 −6

The average value found for the resistivity is rav =

Σri = 1.47 × 10 −6 Ω ⋅ m 3

which differs from the value of r = 150 × 10 −8 Ω ⋅ m = 1.50 × 10 −6 Ω ⋅ m given in Table 17.1 by 2.0% . 17.56

The volume of the material is V=

mass 50.0 g ⎛ 1 m 3 ⎞ = = 6.36 × 10 −6 m 3 density 7.86 g cm 3 ⎜⎝ 10 6 cm 3 ⎟⎠

Since V = A ⋅ L, the cross-sectional area of the wire is A = V L . (a)

From R = rL A = rL (V L ) = rL2 V , the length of the wire is given by L=

R ⋅V = r

(1.5 Ω )( 6.36 × 10 −6 m 3 ) 11× 10 −8 Ω ⋅ m

= 9.3 m

continued on next page

68719_17_ch17_p062-082.indd 78

1/7/11 2:39:45 PM

Current and Resistance

(b)

The cross-sectional area of the wire is A = p d 2 4 = V L. Thus, the diameter is

(a)

4 ( 6.36 × 10 −6 m 3 )

4V = pL

d= 17.57

79

p ( 9.3 m )

= 9.3 × 10 −4 m = 0.93 mm

The total power you now use while cooking breakfast is P = (1 200 + 500 ) W = 1.70 kW The cost to use this power for 0.500 h each day for 30.0 days is ⎡ ⎤ ⎛ h ⎞ cost = [ P × ( Δt )] × rate = ⎢(1.70 kW ) ⎜ 0.500 (30.0 days )⎥ ($0.120 kWh ) = $3.06 ⎟ day ⎠ ⎝ ⎣ ⎦

(b)

If you upgraded, the new power requirement would be P = ( 2 400 + 500 ) W = 2 900 W, and the required current would be I = P ΔV = 2 900 W 110 V = 26.4 A > 20 A. No , your present circuit breaker cannot handle the upgrade.

17.58

(a)

The charge passing through the conductor in the interval 0 ≤ t ≤ 5.0 s is represented by the area under the I vs. t graph given in Figure P17.58. This area consists of two rectangles and two triangles. Thus, ΔQ = Arectangle 1 + Arectangle 2 + Atriangle 1 + Atriangle 2 = ( 5.0 s − 0 )( 2.0 A − 0 ) + ( 4.0 s − 3.0 s )( 6.0 A − 2.0 A ) +

1 1 (3.0 s − 2.0 s ) (6.0 A − 2.0 A ) + (5.0 s − 4.0 s )(6.0 A − 2.0 A ) 2 2

ΔQ = 18 C (b)

The constant current that would pass the same charge in 5.0 s is I=

17.59

(a)

The power input to the motor is Pinput = ( ΔV ) I = Poutput efficiency, so the required current is I=

(b)

68719_17_ch17_p062-082.indd 79

Poutput

( ΔV )(efficiency)

E = Pinput ( Δt ) = yielding

(c)

ΔQ 18 C = = 3.6 A Δt 5.0 s

Poutput efficiency

=

( 2.50 hp) ( 746 W 1 hp) =

( Δt ) =

(120 V )( 0.900 )

17.3 A

( 2.50 hp) ( 0.746 kW 1 hp) (3.00 h ) 0.900

⎛ 3.60 MJ ⎞ E = 6.22 kWh = ( 6.22 kWh ) ⎜ = 22.4 MJ ⎝ 1 kWh ⎟⎠

cost = E × rate = ( 6.22 kWh )( $0.110 kWh ) = $0.684

1/7/11 2:39:47 PM

80

17.60

17.61

Chapter 17

(a)

R = R0 [1+ a ( ΔT )] = (8.0 Ω ) ⎡⎣1+ ( 0.4 × 10 −3 °C−1 )( 350°C − 20°C )⎤⎦ = 9.1 Ω

(b)

We assume that the temperature coefficient of resistivity for Nichrome remains constant over this temperature range.

The current in the wire is I = ΔV R = 15.0 V 0.100 Ω = 150 A. Then, from the expression for the drift velocity, vd = I nqA, the density of free electrons is n= or

17.62

I 150 A = 2 vd e (p r 2 ) ( 3.17 × 10 −4 m s ) (1.60 × 10 −19 C ) p ( 5.00 × 10 −3 m )

n = 3.77 × 10 28 m 3

Each speaker has a resistance of R = 4.00 Ω and can handle a maximum power of 60.0 W. From P = I 2 R, the maximum safe current is Pm ax = R

I m ax =

60.0 W = 3.87 A 4.00 Ω

Thus, the system is not adequately protected by a 4.00 A fuse. 17.63

2 2 − rinner The cross-sectional area of the conducting material is A = p ( router ).

Thus, R=

17.64

(a)

(3.5 × 105 Ω ⋅ m ) ( 4.0 × 10−2 m ) = 3.7 × 107 Ω = 37 MΩ rL = A p ⎡(1.2 × 10 −2 m )2 − ( 0.50 × 10 −2 m )2 ⎤ ⎣ ⎦

At temperature T, the resistance is R =

rL , where r = r0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ , A

L = L0 ⎡⎣1+ a ′ ( T − T0 ) ⎤⎦, and A = A0 ⎡⎣1+ 2a ′ ( T − T0 ) ⎤⎦ Thus, R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ ⋅ ⎡⎣1+ a ′ ( T − T0 ) ⎤⎦ ⎛ r L ⎞ ⎡1+ a ( T − T0 ) ⎤⎦ ⋅ ⎡⎣1+ a ′ ( T − T0 ) ⎤⎦ R=⎜ 0 0⎟ ⎣ = ⎡⎣1+ 2a ′ ( T − T0 ) ⎤⎦ ⎡⎣1+ 2a ′ ( T − T0 ) ⎤⎦ ⎝ A0 ⎠ (b)

R0 =

−8 r0 L0 (1.7 × 10 Ω ⋅ m ) ( 2.00 m ) = = 1.082 Ω 2 A0 p ( 0.100 × 10 −3 )

Then R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ gives R = (1.082 Ω ) ⎡⎣1+ ( 3.9 × 10 −3 °C−1 )(80.0°C )⎤⎦ = 1.420 Ω The more complex formula gives

(1.420 Ω ) ⋅ ⎡⎣1+ (17 × 10 −6 °C−1 )(80.0°C )⎤⎦ R= = 1.418 Ω ⎡1+ 2 (17 × 10 −6 °C−1 )(80.0°C )⎤ ⎣ ⎦ Note: Some rules for handing significant figures have been deliberately violated in this solution in order to illustrate the very small difference in the results obtained with these two expressions.

68719_17_ch17_p062-082.indd 80

1/7/11 2:39:49 PM

Current and Resistance

17.65

81

The power the beam delivers to the target is P = ( ΔV ) I = ( 4.0 × 10 6 V ) ( 25 × 10 −3 A ) = 1.0 × 10 5 W The mass of cooling water that must flow through the tube each second if the rise in the water temperature is not to exceed 50°C is found from P = ( Δm Δt ) c ( ΔT ) as Δm P 1.0 × 10 5 J s = = = 0.48 kg s Δt c ( ΔT ) ( 4 186 J kg ⋅°C )( 50°C )

17.66

Note: All potential differences in this solution have a value of ΔV = 120 V. First, we shall do a symbolic solution for many parts of the problem and then enter the specified numeric values for the cases of interest. From the marked specifications on the cleaner, its internal resistance (assumed constant) is Rvac =

( ΔV )2 P1

where P1 = 535 W

[1]

If each of the two conductors in the extension cord has resistance Rc, the total resistance in the path of the current (outside of the power source) is Rt = Rvac + 2Rc

[2]

so the current which will exist is I = ΔV Rt , and the power that is delivered to the cleaner is 2

Pdelivered = I Rvac 2

⎛ ΔV ⎞ ⎛ ΔV ⎞ =⎜ Rvac = ⎜ ⎟ ⎝ Rt ⎠ ⎝ Rt ⎟⎠

2

( ΔV )2 P1

=

( ΔV )4

[3]

Rt2 P1

The resistance of a copper conductor of length L and diameter d is Rc = rC u

4r L L L = rC u = C u2 2 A (p d 4 ) p d

Thus, if Rc , m ax is the maximum allowed value of Rc, the minimum acceptable diameter of the conductor is d m in = (a)

4rC u L p Rc , m ax

[4]

If Rc = 0.900 Ω, then from Equations [2] and [1], Rt = Rvac + 2 ( 0.900 Ω ) =

( ΔV )2 P1

+ 1.80 Ω =

(120 V )2 535 W

+ 1.80 Ω

and, from Equation [3], the power delivered to the cleaner is Pdelivered =

(120 V )4 2

⎡ (120 V )2 ⎤ + 1.80 Ω ⎥ ( 535 W ) ⎢ 535 W ⎣ ⎦

= 470 W

continued on next page

68719_17_ch17_p062-082.indd 81

1/7/11 2:39:52 PM

82

Chapter 17

If the minimum acceptable power delivered to the cleaner is Pmin, then the maximum allowable total resistance is given by Equations [2] and [3] as Rt , m ax = Rvac + 2Rc , m ax =

( ΔV )4 Pm in P1

=

( ΔV )2 Pm in P1

so Rc , m ax = (b)

2 ⎤ 1 ⎡ ( ΔV )2 1 ⎡ ( ΔV ) ( ΔV )2 − Rvac ⎥ = ⎢ − ⎢ 2 ⎢⎣ Pm in P1 P1 ⎥⎦ 2 ⎢⎣ Pm in P1

⎤ ( ΔV )2 ⎥= 2 ⎥⎦

⎡ 1 1⎤ − ⎥ ⎢ ⎢⎣ Pm in P1 P1 ⎥⎦

When Pm in = 525 W, then Rc , m ax =

(120 V )2 ⎡ 2

⎢ ⎢⎣

1 ⎤ 1 − ⎥ = 0.128 Ω ( 525 W )( 535 W ) 535 W ⎥⎦

and, from Equation [4], d m in = (c)

p ( 0.128 Ω )

= 1.60 mm

When Pm in = 532 W, then Rc , m ax =

and

68719_17_ch17_p062-082.indd 82

4 (1.7 × 10 −8 Ω ⋅ m )(15.0 m )

d m in =

(120 V )2 ⎡ 2

⎢ ⎢⎣

1 1 ⎤ − ⎥ = 0.037 9 Ω ( 532 W )( 535 W ) 535 W ⎥⎦

4 (1.7 × 10 −8 Ω ⋅ m )(15.0 m ) p ( 0.037 9 Ω )

= 2.93 mm

1/7/11 2:39:53 PM

18 Direct-Current Circuits QUICK QUIZZES 1.

True. When a battery is delivering a current, there is a voltage drop within the battery due to internal resistance, making the terminal voltage less than the emf.

2.

Because of internal resistance, power is dissipated into the battery material, raising its temperature.

3.

Choice (b). When the switch is opened, resistors R1 and R2 are in series, so that the total circuit resistance is larger than when the switch was closed. As a result, the current decreases.

4.

Choice (a). When the switch is opened, resistors R1 and R2 are in series, so the total circuit resistance is larger than and the current through R1 is less with the switch open than when it is closed. Since the power delivered to R1 is P = I 2 R1, Po < Pc.

5.

Choice (a). When the switch is closed, resistors R1 and R2 are in parallel, so that the total circuit resistance is smaller than when the switch was open. As a result, the current increases.

6.

Choice (b). Observe that the potential difference across R1 equals the terminal potential difference of the battery. If the battery has negligible internal resistance, the terminal potential difference is the same with the switch open or closed. Under these conditions, the power delivered to R1, equal 2 to P = ( ΔV ) R1, is unchanged when the switch is closed.

7.

The voltage drop across each bulb connected in parallel with each other and across the battery equals the terminal potential difference of the battery. As more bulbs are added, the current supplied by the battery increases. However, if the internal resistance is negligible, the terminal potential difference is constant and the current through each bulb is the same regardless of the number of bulbs connected. Under these conditions: (a) The brightness of a bulb, determined by the current flowing in the bulb, is unchanged as bulbs are added. (b) The individual currents in the bulbs, I = ΔV R, are constant as bulbs are added since ΔV does not change. (c) The total power 2 delivered by the battery increases by an amount ( ΔV ) R each time a bulb is added. (d) With the total delivered power increasing, energy is drawn from the battery at an increasing rate and the battery’s lifetime decreases.

8.

Adding bulbs in series with each other and the battery increases the total load resistance seen by the battery. This means that the current supplied by the battery decreases with each new bulb that is added. (a) The brightness of a bulb is determined by the power delivered to that bulb, Pbulb = I 2 R, which decreases as bulbs are added and the current decreases. (b) For a series connection, the individual currents in the bulbs are the same and equal to the total current supplied by the battery. This decreases as bulbs are added. (c) The total power delivered by the battery is given by Ptotal = ( ΔV ) I, where ΔV is the terminal potential difference of the battery and I is the total current supplied by the battery. With negligible internal resistance, ΔV is constant. Thus, with I decreasing as bulbs are added, the total delivered power decreases. (d) With the delivered power decreasing, energy is drawn from the battery at a decreasing rate, which increases the lifetime of the battery.

9.

Choice (c). After the capacitor is fully charged, current flows only around the outer loop of the circuit. This path has a total resistance of 3 Ω, so the 6-V battery will supply a current of 2 amperes. 83

68719_18_ch18_p083-119.indd 83

1/7/11 2:41:04 PM

84

Chapter 18

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

The same potential difference exists across all elements connected in parallel with each other, while the current through each element is inversely proportional to the resistance of that element (I = ΔV R). Thus, both (b) and (c) are true statements while the other choices are false.

2.

In a series connection, the same current exists in each element. The potential difference across a resistor in this series connection is directly proportional to the resistance of that resistor, ΔV = IR, and independent of its location within the series connection. The only true statement among the listed choices is (c).

3.

For these parallel resistors, 1 1 2 +1 2.00 Ω 1 = + = and Req = = 0.667 Ω 3 Req 1.00 Ω 2.00 Ω 2.00 Ω Choice (c) is the correct answer.

4.

The total power dissipated is Ptotal = P1 + P2 = 120 W + 60.0 W = 180 W, while the potential difference across this series combination is ΔV = 120 V. The current drawn through the series P 180 W = 1.5 A, and (b) is the correct choice. combination is then I = total = 120 V ΔV

5.

The equation of choice (b) is the result of a correct application of Kirchhoff’s junction rule at either of the two junctions in the circuit. The equation of choice (c) results from a correct application of Kirchhoff’s loop rule to the lower loop in the circuit, while the equation of choice (d) is obtained by correctly applying the loop rule to the loop forming the outer perimeter of the circuit. The equation of choice (a) is the result of an incorrect application (involving 2 sign errors) of the loop rule to the upper loop in the circuit. The correct answer is choice (a).

6.

The equivalent resistance of the parallel combination consisting of the 4.0-Ω, 6.0-Ω, and 10-Ω resistors is Rp = 1.9 Ω. This resistance is in series with a 2.0-Ω resistor, making the total resistance of the circuit Rtotal = 3.9 Ω. The total current supplied by the battery is I total = ΔV Rtotal = 12 V 3.9 Ω = 3.1 A. Thus, the potential difference across each resistor in the parallel combination is ΔVp = Rp I total = (1.9 Ω )( 3.1 A ) = 5.9 V and the current through the 10 Ω resistor is I10 = ΔVp 10 Ω = 5.9 V 10 Ω = 0.59 A. Choice (a) is the correct answer.

7.

The potential difference across each element of this parallel combination is ΔV = 120 V, and the total power dissipated is Ptotal = 1 200 W + 600 W = 1 800 W. The total current through the parallel combination, and hence, the current drawn from the external source is then P 1 800 W I = total = = 15 A. The correct choice is (e). 120 V ΔV

8.

When the two identical resistors are in series, the current supplied by the battery is I = ΔV 2R, 2 and the total power delivered is Ps = ( ΔV ) I = ( ΔV ) 2R. With the resistors connected in parallel, the potential difference across each resistor is ΔV and the power delivered to each resistor is 2 P1 = ( ΔV ) R. Thus, the total power delivered in this case is Pp = 2P1 = 2

( ΔV )2 R

⎡ ( ΔV )2 = 4⎢ ⎣ 2R

⎤ ⎥ = 4Ps = 4 (8.0 W ) = 32 W ⎦

and (b) is the correct choice.

68719_18_ch18_p083-119.indd 84

1/7/11 2:41:06 PM

Direct-Current Circuits

9.

85

The equivalent resistance for the series combination of five identical resistors is Req = 5R, and the equivalent capacitance of five identical capacitors in parallel is Ceq = 5C. The time constant for the circuit is therefore t = Req Ceq = ( 5R )( 5C ) = 25RC and (d) is the correct choice.

10.

When the switch is closed, the current has a large initial value but decreases exponentially in time. The bulb will glow brightly at first, but fade rapidly as the capacitor charges. After a time equal to many time constants of the circuit, the current is essentially zero and the bulb does not glow. The correct answer is choice (c).

11.

The equivalent resistance of a group of resistors connected in parallel is the reciprocal of the inverses of the individual resistances and is always less than the smallest resistance in the group. Therefore, both (b) and (e) are true statements while all other choices are false.

12.

When resistors are connected in series, Kirchhoff’s junction rule requires that the current be the same in all resistors at each instant in time. Thus, the charge entering each resistor in a given time interval must be the same for all resistors, and both choices (b) and (d) are correct. The other choices are false since the potential difference, ΔV = IR, and the power, P = I 2 R, must differ when the resistances are different.

13.

With the capacitor initially uncharged, the potential difference across the capacitor, ΔV = q C, starts at zero when the switch is first closed and rises exponentially toward the equilibrium value of ( ΔV )m ax = ΔVbattery = 6.00 V. The time constant of the circuit is the time required for the charge (and hence the potential difference) to increase from 0 to 63.2% of the maximum equilibrium value. Thus, after one time constant, the potential difference across the capacitor will be ΔV = 0.632 ( 6.00 V ) = 3.79 V. The correct answer is choice (d).

14.

The circuit breaker should be connected in series with the device so that an open circuit results when the circuit breaker trips, causing the current through the device to cease. Thus, choice (a) is the correct answer.

15.

The equivalent resistance of a series combination of resistors is the algebraic sum of the individual resistances and is always greater than any individual resistance. Therefore, choices (a) and (d) are true statements and all others are false.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

4.

68719_18_ch18_p083-119.indd 85

A short circuit can develop when the last bit of insulation frays away between the two conductors in a lamp cord. Then the two conductors touch each other, creating a low resistance path in parallel with the lamp. The lamp will immediately go out, carrying no current and presenting no danger. A very large current will be produced in the power source, the house wiring, and the wire in the lamp cord up to and through the short. The circuit breaker will interrupt the circuit quickly but not before considerable heating and sparking is produced in the short-circuit path.

1/7/11 2:41:09 PM

86

Chapter 18

6.

A wire or cable in a transmission line is thick and made of material with very low resistivity. Only when its length is very large does its resistance become significant. To transmit power over a long distance it is most efficient to use low current at high voltage. The power loss per unit length of the transmission line is Ploss L = I 2 ( R L ), where R L is the resistance per unit length of the line. Thus, a low current is clearly desirable, but to transmit a significant amount of power P = ( ΔV ) I with low current, a high voltage must be used.

8.

The bulbs of set A are wired in parallel. The bulbs of set B are wired in series, so removing one bulb produces an open circuit with infinite resistance and zero current.

10.

(a) ii. The power delivered may be expressed as P = I 2 R, and while resistors connected in series have the same current in each, they may have different values of resistance. (b) ii. The power delivered may also be expressed as P = (ΔV )2 / R, and while resistors connected in parallel have the same potential difference across them, they may have different values of resistance.

12.

Compare two runs in series to two resistors connected in series. Compare three runs in parallel to three resistors connected in parallel. Compare one chairlift followed by two runs in parallel to a battery followed immediately by two resistors in parallel. The junction rule for ski resorts says that the number of skiers coming into a junction must be equal to the number of skiers leaving. The loop rule would be stated as the total change in altitude must be zero for any skier completing a closed path.

14.

Because water is a good conductor, if you should become part of a short circuit when fumbling with any electrical circuit while in a bathtub, the current would follow a pathway through you, the water, and to ground. Electrocution would be the obvious result.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a)

27 Ω

(b)

0.44 A

4.

(a)

1.13 A

(b)

9.17 Ω

6.

(a)

35.0 Ω 3 = 11.7 Ω

(b)

1.00 A in the 12.0-Ω and 8.00-Ω resistors, 2.00 A in the 6.00-Ω and 4.00-Ω resistors, and 3.00 A in the 5.00-Ω resistor

(a)

3.33 Ω

(b)

7.33 Ω

(c)

2.13 Ω

(d)

4.13 Ω

(e)

1.94 A

(f )

3.88 V

(g)

4.12 V

(h)

1.37 A

(a)

ΔV1 = e 3, ΔV2 = 2e 9, ΔV3 = 4e 9, ΔV4 = 2e 3

(b)

I1 = I, I 2 = I 3 = I 3, I 4 = 2I 3

(a)

1.00 kΩ

(c)

3.00 kΩ

8.

10.

12.

68719_18_ch18_p083-119.indd 86

(b)

2.00 kΩ

(c)

3.0 Ω, 1.3 A

1/7/11 2:41:10 PM

Direct-Current Circuits

14.

(a) Yes, Req = 8.00 Ω

16.

(a)

(b)

2.25 A

(c)

87

40.5 W

I 6 = 3.0 A, I12 = 2.0 A, I 24 = 1.0 A

(b) Answers are the same as for part (a). (a)

20.

I 2 = 2.0 A flowing from b toward a, I 3 = 1.0 A in direction shown

22.

(a)

60.0 Ω

(d)

0.50 W

24.

(a)

26. 28.

0.909 A

(b)

ΔVab = 1.82 V, with point b at the lower potential.

18.

(b)

0.20 A

4.59 Ω

(b)

0.081 6 or 8.16%

(a)

1.7 × 10 2 A

(b)

0.20 A

(a)

18.0I12 + 13.0I18 = 30.0

(b)

5.00I 36 − 18.0I12 = 24.0 (c)

(d)

I 36 = I18 − I12

(e)

5.00I18 − 23.0I12 = 24.0

(f )

I12 = −0.416 A, I18 = 2.88 A

(g)

I 36 = 3.30 A

(h)

I12 flows opposite to the assumed direction and has magnitude 0.416 A.

30.

See Solution.

32.

(a)

34.

1.3 × 10 2 mC

36.

587 kΩ

38.

(a)

8.0 A

(d)

5.8 × 10 2 W

(a)

I coffee = 10 A, I toaster = 9.2 A, I waffle = 12 A

40.

2.00 ms

m aker

(c)

2.4 W

I18 = I12 + I 36

(b)

180 mC

(c)

114 mC

(b)

120 V

(c)

0.80 A

m aker

(b)

I total = 31 A

(c)

No, I total > 15 A.

42.

(a)

4.1× 10 −11 J

(b)

0.56 mA

44.

Sixteen distinct resistances are possible – See Solution for how these are produced.

46.

(a)

2.0 kΩ

(b)

15 V

(c)

9.0 V

(d) Assumed negligible current through the voltmeter and negligible resistance in the ammeter. 48.

(a)

40.0 W

(b)

ΔV1 = 80.0 V, ΔV2 = ΔV3 = 40.0 V

50.

(a)

0.708 A

(b)

2.51 W

68719_18_ch18_p083-119.indd 87

1/7/11 2:41:12 PM

88

Chapter 18

(c)

Only the circuit of Figure P18.50c. In the other circuits, the batteries can be combined into a single effective battery while the 5.00-Ω and 8.00-Ω resistors remain in parallel with each other.

(d) The power is lowest in Figure P18.50c. The circuits in Figures P18.50b and P18.50d have in effect 30-V batteries driving the current. 52.

14 Ω

54.

14 s

56.

R = 20 Ω or R = 98 Ω

58.

See Solution.

60.

q1 = ( 240 mC ) (1− e −1 000 t 6.0 s ) , q2 = ( 360 mC ) (1− e −1 000 t 6.0 s )

62.

(a)

Req = 0.099 9 Ω, I R = 50.0 A, I R = I R = I100 = 45.0 mA

(b)

Req = 1.09 Ω, I R = I R = 4.55 A, I R = I100 = 45.5 mA

(c)

Req = 9.99 Ω, I R = I R = I R = 0.450 A, I100 = 50.0 mA

(a)

In case 1, the two bulbs have the same current, power supplied, and brightness.

(b)

In case 2, the two bulbs have the same current, power supplied, and brightness.

64.

1

1

1

2

2

2

3

3

3

(c) The bulbs in case 2 are brighter, each having twice the current of a bulb in case 1.

66.

(d)

In case 1, both go out when one bulb fails. In case 2, the other bulb remains lit with unchanged brightness when one bulb fails.

(a)

61.6 mA

(b)

0.235 mC

(c)

1.96 A

PROBLEM SOLUTIONS 18.1

From ΔV = I ( R + r ), the internal resistance is r=

18.2

ΔV 9.00 V −R= − 72.0 Ω = 4.92 Ω I 0.117 A

(a)

When the three resistors are in series, the equivalent resistance of the circuit is Req = R1 + R2 + R3 = 3 (9.0 Ω) = 27 Ω .

(b)

The terminal potential difference of the battery is applied across the series combination of the three 9.0-Ω resistors, so the current supplied by the battery and the current through each resistor in the series combination is I=

ΔV 12 V = = 0.44 A Req 27 Ω

continued on next page

68719_18_ch18_p083-119.indd 88

1/7/11 2:41:16 PM

Direct-Current Circuits

(c)

89

If the three 9.0-Ω resistors are now connected in parallel with each other, the equivalent resistance is 1 1 1 3 1 = + + = Req 9.0 Ω 9.0 Ω 9.0 Ω 9.0 Ω

or

Req =

9.0 Ω = 3.0 Ω 3

When this parallel combination is connected to the battery, the potential difference across each resistor in the combination is ΔV = 12 V, so the current through each of the resistors is I= 18.3

ΔV 12 V = = 1.3 A R 9.0 Ω

(a)

The bulb acts as a 192-Ω resistor (see below), so the circuit diagram is:

(b)

For the bulb in use as intended, Rbulb = ( ΔV ) P = (120 V ) 75.0 W = 192 Ω . 2

2

Now, assuming the bulb resistance is unchanged, the current in the circuit shown is I=

ΔV 120 V = = 0.620 A Req 0.800 Ω + 192 Ω + 0.800 Ω

and the actual power dissipated in the bulb is P = I 2 Rbulb = ( 0.620 A ) (192 Ω ) = 73.8 W 2

18.4

(a)

When the 8.00-Ω resistor is connected across the 9.00-V terminal potential difference of the battery, the current through both the resistor and the battery is I=

(b)

ΔV 9.00 V = = 1.13 A R 8.00 Ω

The relation between the emf and the terminal potential difference of a battery supplying current I is ΔV = e − Ir, where r is the internal resistance of the battery. Thus, if the battery has r = 0.15 Ω and maintains a terminal potential difference of ΔV = 9.00 V while supplying the current found above, the emf of this battery must be e = ΔV + Ir = 9.00 V + (1.13 A )( 0.15 Ω ) = ( 9.00 + 0.17 ) Ω = 9.17 Ω

18.5

(a)

The equivalent resistance of the two parallel resistors is 1 ⎞ ⎛ 1 Rp = ⎜ + ⎝ 7.00 Ω 10.0 Ω ⎟⎠

−1

= 4.12 Ω

Thus, Rab = R4 + Rp + R9 = ( 4.00 + 4.12 + 9.00 ) Ω = 17.1 Ω continued on next page

68719_18_ch18_p083-119.indd 89

1/7/11 2:41:18 PM

90

Chapter 18

(b)

18.6

(a)

( ΔV )ab

I ab =

=

Rab

34.0 V = 1.99 A, so I 4 = I 9 = 1.99 A 17.1 Ω

Also,

( ΔV ) p = I ab Rp = (1.99 A )( 4.12 Ω ) = 8.20 V

Then,

I7 =

and

I10 =

( ΔV ) p R7

( ΔV ) p R10

=

8.20 V = 1.17 A 7.00 Ω

=

8.20 V = 0.820 A 10.0 Ω

The parallel combination of the 6.00-Ω and 12.0-Ω resistors has an equivalent resistance of 1 1 2 +1 1 = + = Rp1 6.00 Ω 12.0 Ω 12.0 Ω

or

Rp1 =

12.0 Ω = 4.00 Ω 3

Similarly, the equivalent resistance of the 4.00-Ω and 8.00-Ω parallel combination is 1 1 2 +1 1 = + = Rp 2 4.00 Ω 8.00 Ω 8.00 Ω

or

Rp 2 =

8.00 Ω 3

The total resistance of the series combination between points a and b is then Rab = Rp1 + 5.00 Ω + Rp 2 = 4.00 Ω + 5.00 Ω + (b)

8.00 35.0 Ω= Ω 3 3

If ΔVab = 35.0 V, the total current from a to b is I ab = ΔVab Rab = 35.0 V ( 35.0 Ω 3) = 3.00 A and the potential differences across the two parallel combinations are ΔVp1 = I ab Rp1 = ( 3.00 A )( 4.00 Ω ) = 12.0 V, and ⎛ 8.00 ⎞ ΔVp 2 = I ab Rp 2 = ( 3.00 A ) ⎜ Ω⎟ = 8.00 V ⎝ 3 ⎠ The individual currents through the various resistors are: I12 = ΔVp1 12.0 Ω = 1.00 A ;

I 6 = ΔVp1 6.00 Ω = 2.00 A ;

I 5 = I ab = 3.00 A ;

I 8 = ΔVp 2 8.00 Ω = 1.00 A ;

and 18.7

I 4 = ΔVp 2 4.00 Ω = 2.00 A

When connected in series, we have R1 + R2 = 690 Ω which we may rewrite as R2 = 690 Ω − R1 When in parallel,

1 1 1 + = R1 R2 150 Ω

[1] [1a]

or

R1 R2 = 150 Ω R1 + R2

[2]

continued on next page

68719_18_ch18_p083-119.indd 90

1/7/11 2:41:20 PM

Direct-Current Circuits

91

Substitute Equations [1] and [1a] into Equation [2] to obtain: R1 ( 690 Ω − R1 ) = 150 Ω 690 Ω

R12 − ( 690 Ω ) R1 + ( 690 Ω )(150 Ω ) = 0

or

[3]

Using the quadratic formula to solve Equation [3] gives R1 =

690 Ω ±

( 690 Ω )2 − 4 ( 690 Ω )(150 Ω ) 2

with two solutions of

R1 = 470 Ω

and R1 = 220 Ω

Then Equation [1a] yields

R2 = 220 Ω

or

R2 = 470 Ω

Thus, the two resistors have resistances of 220 Ω and 470 Ω . 18.8

(a)

The equivalent resistance of this first parallel combination is 1 1 1 = + Rp1 10.0 Ω 5.00 Ω

(b)

or

Rp1 = 3.33 Ω

For this series combination, Rupper = Rp1 + 4.00 Ω = 7.33 Ω

(c)

For the second parallel combination, 1 1 1 1 1 = + = + Rp 2 Rupper 3.00 Ω 7.33 Ω 3.00 Ω

(d)

Rp 2 = 2.13 Ω

or

For the second series combination (and hence the entire resistor network) Rtotal = 2.00 Ω + Rp 2 = 2.00 Ω + 2.13 Ω = 4.13 Ω

(e)

The total current supplied by the battery is I total =

(f )

ΔV 8.00 V = = 1.94 A Rtotal 4.13 Ω

The potential drop across the 2.00 Ω resistor is ΔV2 = R2 I total = ( 2.00 Ω )(1.94 A ) = 3.88 V

(g)

The potential drop across the second parallel combination must be ΔVp 2 = ΔV − ΔV2 = 8.00 V − 3.88 V = 4.12 V

(h)

68719_18_ch18_p083-119.indd 91

So the current through the 3.00 Ω resistor is

I total =

ΔVp 2 R3

=

4.12 V = 1.37 A 3.00 Ω

1/7/11 2:41:23 PM

92

18.9

Chapter 18

(a)

Using the rules for combining resistors in series and parallel, the circuit reduces as shown below: 25.0 V

10.0 Ω

10.0 Ω

− +

I b

a

25.0 V

− +

10.0 Ω

I b

a

25.0 V

− +

10.0 Ω

10.0 Ω

I 5.00 Ω

I20

c 20.0 Ω

5.00 Ω

I20

a

b 2.94 Ω

25.0 Ω Step 2

5.00 Ω Step 1

Step 3

From the figure of Step 3, observe that I=

18.10

25.0 V = 1.93 A and ΔVab = I ( 2.94 Ω ) = (1.93 A )( 2.94 Ω ) = 5.67 V 10.0 Ω + 2.94 Ω ΔVab 5.67 V = = 0.227 A 25.0 Ω 25.0 Ω

(b)

From the figure of Step 1, observe that I 20 =

(a)

The figures below show the simplification of the circuit in stages: a

I I4

R1 = R e

I2 = I3 = I23 R2 = 2R

R3 = 4R

e

I23

I4

R1

R4 = 3R

+ −

a

I

+ −

b Figure 1

R4

a

I I

R1 R23 e

b Figure 2

R234

+ −

b Figure 3

Note that R2 and R3 are in series with equivalent resistance R23 = R2 + R3 = 6R. Then, R4 and R23 are in parallel with equivalent resistance R234 =

R4 R23 ( 3R )( 6R ) = = 2R R4 + R23 3R + 6R

The total current supplied by the battery is then I=

e e e = = R1 + R234 R + 2R 3R

⎛ e ⎞ The potential difference across R1 is ΔV1 = IR1 = ⎜ R= e 3 ⎝ 3R ⎟⎠

continued on next page

68719_18_ch18_p083-119.indd 92

1/7/11 2:41:25 PM

93

Direct-Current Circuits

⎛ e ⎞ and that across R4 is ΔV4 = ΔVab = IR234 = ⎜ ( 2R ) = 2e 3 ⎝ 3R ⎟⎠ The current through R2 and R3 is I 23 = ΔVab R23 = ( 2e 3R ) 6R = e 9R e ⎞ so the potential difference across R2 is ΔV2 = I 23 R2 = ⎛⎜ ( 2R ) = 2e 9 ⎝ 9R ⎟⎠ ⎛ e ⎞ and that across R3 is ΔV3 = I 23 R3 = ⎜ ( 4R ) = 4e 9 ⎝ 9R ⎟⎠ From above, we have I1 = I and I 2 = I 3 = I 23 = e 9R = I 3

(b)

The current through R4 is I 4 = ΔV4 R4 = ( 2e 3) 3R = 2e 9R = 2I 3 18.11

The equivalent resistance is Req = R + Rp, where Rp is the total resistance of the three parallel branches: 1 1 ⎛ 1 ⎞ Rp = ⎜ + + ⎝ 120 Ω 40 Ω R + 5.0 Ω ⎟⎠ Thus,

75 Ω = R +

−1

1 ⎛ 1 ⎞ =⎜ + ⎝ 30 Ω R + 5.0 Ω ⎟⎠

−1

=

( 30 Ω )( R + 5.0 Ω ) R + 35 Ω

( 30 Ω )( R + 5.0 Ω ) R 2 + ( 65 Ω ) R + 150 Ω2 R + 35 Ω

=

R + 35 Ω

which reduces to R 2 − (10 Ω ) R − 2 475 Ω2 = 0 or ( R − 55 Ω )( R + 45 Ω ) = 0. Only the positive solution is physically acceptable, so R = 55 Ω . 18.12

The sketch at the right shows the equivalent circuit when the switch is in the open position. For this simple series circuit, R1 + R2 + R3 = or

e

[1] R2

e

R2 + R3 = 5.00 kΩ 2

− R1 +

R2

R3

Ia

R e 6.00 V R1 + 2 + R3 = = I a 1.20 × 10 −3 A 2 R1 +

R3

Figure 1

When the switch is closed in position a, the equivalent circuit is shown in Figure 2. The equivalent resistance of the two parallel resistors, R2, is Rp = R2 2 and the total resistance of the circuit is Ra = R1 + ( R2 2 ) + R3. Thus,

or

R2 Io

e 6.00 V = I o 1.00 × 10 −3 A

R1 + R2 + R3 = 6.00 kΩ

− R1 +

Figure 2

[2]

continued on next page

68719_18_ch18_p083-119.indd 93

1/7/11 2:41:27 PM

94

Chapter 18

When the switch is closed in position b, resistor R3 is shorted out, leaving R1 and R2 in series with the battery as shown in Figure 3. This gives

e

− R1 + Ib

e 6.00 V R1 + R2 = = I b 2.00 × 10 −3 A and

R2

Figure 3

R1 + R2 = 3.00 kΩ

[3]

Substitute Equation [3] into Equation [1] to obtain 3.00 kΩ + R3 = 6.00 kΩ

and

R3 = 3.00 kΩ

Now, Equation [1] minus Equation [2] gives R2 2 = 1.00 kΩ

or

R2 = 2.00 kΩ

Finally, Equation [3] tells that R1 + 2.00 kΩ = 3.00 kΩ,

or

R1 = 1.00 kΩ

In summary, we have (a) R1 = 1.00 kΩ , 18.13

(b) R2 = 2.00 kΩ ,

and

(c) R3 = 3.00 kΩ

The resistors in the circuit can be combined in the stages shown below to yield an equivalent resistance of Rad = ( 63 11) Ω.

Figure 1

Figure 2

Figure 4

Figure 3 From Figure 5,

I=

( ΔV )ad Rad

=

18 V

(63 11) Ω

Figure 5

= 3.1 A

continued on next page

68719_18_ch18_p083-119.indd 94

1/7/11 2:41:30 PM

Direct-Current Circuits

Then, from Figure 4,

95

( ΔV )bd = I Rbd = ( 3.1 A )( 30 11 Ω ) = 8.5 V

Now, look at Figure 2 and observe that I2 = so

( ΔV )bd 8.5 V = = 1.7 A 3.0 Ω + 2.0 Ω 5.0 Ω

( ΔV )be = I 2 Rbe = (1.7 A )( 3.0 Ω ) = 5.1 V

Finally, from Figure 1, I12 = 18.14

(a)

( ΔV )be R12

=

5.1 V = 0.43 A 12 Ω

The resistor network connected to the battery in Figure P18.14 can be reduced to a single equivalent resistance in the following steps. The equivalent resistance of the parallel combination of the 3.00 Ω and 6.00 Ω resistors is 1 1 3 1 = + = Rp 3.00 Ω 6.00 Ω 6.00 Ω

or

Rp = 2.00 Ω

2.00 Ω + 18.0 V − 4.00 Ω

6.00 Ω 3.00 Ω

Figure P18.14

This resistance is in series with the 4.00 Ω and the other 2.00 Ω resistor, giving a total equivalent resistance of Req = 2.00 Ω + Rp + 4.00 Ω = 8.00 Ω . (b)

The current in the 2.00 Ω resistor is the total current supplied by the battery and is equal to I total =

(c)

ΔV 18.0 V = = 2.25 A Req 8.00 Ω

The power the battery delivers to the circuit is P = ( ΔV ) I total = (18.0 V )( 2.25 A ) = 40.5 W

18.15

(a)

Connect two 50-Ω resistors in parallel to get 25 Ω. Then connect that parallel combination in series with a 20-Ω resistor for a total resistance of 45 Ω.

(b)

Connect two 50-Ω resistors in parallel to get 25 Ω. Also, connect two 20-Ω resistors in parallel to get 10 Ω. Then, connect these two parallel combinations in series to obtain 35 Ω.

18.16

(a)

The equivalent resistance of the parallel combination between points b and e is 1 1 1 = + Rbe 12 Ω 24 Ω

or

Rbe = 8.0 Ω

The total resistance between points a and e is then Rae = Rab + Rbe = 6.0 Ω + 8.0 Ω = 14 Ω

continued on next page

68719_18_ch18_p083-119.indd 95

1/7/11 2:41:32 PM

96

Chapter 18

The total current supplied by the battery (and also the current in the 6.0-Ω resistor) is I total = I 6 =

ΔVae 42 V = = 3.0 A 14 Ω Rae

The potential difference between points b and e is ΔVbe = Rbe I total = (8.0 Ω )( 3.0 A ) = 24 V so (b)

I12 =

ΔVbe 24 V = = 2.0 A 12 Ω Rbce

I 24 =

and

Applying the junction rule at point b yields

ΔVbe 24 V = = 1.0 A 24 Ω Rbde

I 6 − I12 − I 24 = 0

Using the loop rule on loop abdea gives +42 − 6I 6 − 24I 24 = 0

or

I 6 = 7.0 − 4I 24

[2]

and using the loop rule on loop bcedb gives −12I12 + 24I 24 = 0

or

I12 = 2I 24

[3]

Substituting Equations [2] and [3] into [1] yields 7I 24 = 7.0 Then, Equations [2] and [3] yield 18.17

[1]

I 6 = 3.0 A

and

or

I 24 = 1.0 A

I12 = 2.0 A

Going counterclockwise around the upper loop, applying Kirchhoff’s loop rule, gives +15.0 V − ( 7.00 ) I1 − ( 5.00 )( 2.00 A ) = 0 or

I1 =

15.0 V − 10.0 V = 0.714 A 7.00 Ω

e

From Kirchhoff’s junction rule, I1 + I 2 − 2.00 A = 0 so

I 2 = 2.00 A − I1 = 2.00 A − 0.714 A = 1.29 A

Going around the lower loop in a clockwise direction gives + e − ( 2.00 ) I 2 − ( 5.00 )( 2.00 A ) = 0 or 18.18

(a)

e = ( 2.00 Ω )(1.29 A ) + ( 5.00 Ω ) ( 2.00 A ) = 12.6 V Applying Kirchhoff’s junction rule at a gives I2 = I4 − I6 Going counterclockwise around the lower loop, and applying Kirchhoff’s loop rule, we obtain +8.00 V − ( 6.00 Ω ) I 6 + ( 2.00 Ω ) I 2 = 0 or

I6 =

4 1 + I2 3 3

12.0 V

+ −

[1]

4.00 Ω

I4 I6

b

I2 a

2.00 Ω

− + 8.00 V

6.00 Ω

[2]

continued on next page

68719_18_ch18_p083-119.indd 96

1/7/11 2:41:35 PM

97

Direct-Current Circuits

Applying the loop rule as we go counterclockwise around the upper loop: − ( 2.00 Ω ) I 2 − ( 4.00 Ω ) I 4 + 12.0 V = 0

I 4 = 3.00 −

or

1 I2 2

[3]

Substituting Equations [2] and [3] into Equation [1] yields 1 1⎞ 4 ⎛ ⎜⎝ 1+ + ⎟⎠ I 2 = 3.00 − 2 3 3 (b)

I 2 = 0.909 A

and

The potential difference between points a and b is ΔVab = −I 2 ( 2.00 Ω ) = − ( 0.909 A )( 2.00 Ω ) = −1.82 V or

18.19

ΔVab = 1.82 V with point b at the lower potential. R

Consider the circuit diagram at the right, in which Kirchhoff’s junction rule has already been applied at points a and e.

+ −

e

Applying the loop rule around loop abca gives I1 − I

e − R ( I1 − I ) − 4RI1 = 0 or

4R I1

a

2R

c

b

d

+ − 2e

3R I

I2 + I

I2

e

1⎛ e ⎞ I1 = ⎜ + I ⎟ 5⎝ R ⎠

[1]

Next, applying the loop rule around loop cedc gives −3RI 2 + 2e − 2R ( I 2 + I ) = 0

or

I2 =

2⎛ e ⎞ ⎜ − I⎟ 5⎝ R ⎠

[2]

Finally, applying the loop rule around loop caec gives −4RI1 + 3RI 2 = 0

or

4I1 = 3I 2

Substituting Equations [1] and [2] into Equation [3] yields

[3] I=

e 5R

Thus, if e = 250 V and R = 1.00 kΩ = 1.00 × 10 3 Ω, the current in the wire between a and e is I= 18.20

250 V = 50.0 × 10 −3 A = 50.0 mA flowing from a toward e. 5 (1.00 × 10 3 Ω )

Following the path of I1 from a to b and recording changes in potential gives Vb − Va = + 24 V − ( 6.0 Ω )( 3.0 A ) = + 6.0 V Now, following the path of I 2 from a to b and recording changes in potential gives Vb − Va = − ( 3.0 Ω ) I 2 = + 6.0 V, or I 2 = − 2.0 A

continued on next page

68719_18_ch18_p083-119.indd 97

1/7/11 2:41:37 PM

98

Chapter 18

Thus, I 2 is directed from b toward a and has magnitude of 2.0 A. Applying Kirchhoff’s junction rule at point a gives I 3 = I1 + I 2 = 3.0 A + ( −2.0 A ) = 1.0 A 18.21

(a)

a

The circuit diagram at the right shows the assumed directions of the current in each resistor. Note that the total current flowing out of the section of wire connecting points g and f must equal the current flowing into that section. Thus,

b

c

d

40.0 V

+ − 360 V

− 80.0 V +

200 Ω

80.0 Ω

20.0 Ω

I2

I3

I1 h

g

f

+ − 70.0 Ω

I4 e

I 3 = I1 + I 2 + I 4

[1]

Applying the loop rule around loop abgha gives −200I1 − 40.0 + 80.0I 2 = 0

or

I2 =

1 (5I1 + 1.00 ) 2

[2]

Next, applying the loop rule around loop bcfgb gives +360 − 20.0I 3 − 80.0I 2 + 40.0 = 0

I 3 = 20.0 − 4I 2

or

[3]

Finally, applying the loop rule around the outer loop abcdefgha yields −80.0 + 70I 4 − 200I1 = 0

or

I4 =

1 ( 20I1 + 8.00 ) 7

[4]

To solve this set of equations, we first substitute Equation [2] into Equation [3] to obtain I 3 = 20.0 − 2 ( 5I1 + 1.00 )

or

I 3 = 18.0 − 10I1

[3ⴕ]

Now, substitute Equations [2], [4], and [3⬘] into Equation [1] to find 5 20 ⎞ 1.00 8.00 ⎛ − ⎜⎝ 10 + + 1+ ⎟⎠ I1 = 18.0 − 2 7 2 7

and

I1 = 1.00 A

Substituting this result back into Equations [2], [4], and [3⬘] gives I 2 = 3.00 A (b)

I 4 = 4.00 A

and

I 3 = 8.00 A

The potential difference across the 200-Ω resistor is ΔV200 = I1 R200 = (1.00 A )( 200 Ω ) = 200 V with point a at a lower potential than point h.

68719_18_ch18_p083-119.indd 98

1/7/11 2:41:40 PM

99

Direct-Current Circuits

18.22

(a)

The 30.0-Ω and 50.0-Ω resistors in the upper branch are in series, and add to give a total resistance of Rupper = 80.0 Ω for this path. This 80.0-Ω resistance is in parallel with the 80.0-Ω resistance of the middle branch, and the rule for combining resistors in parallel yields a total resistance of Rab = 40.0 Ω between points a and b. This resistance is in series with the 20.0-Ω resistor, so the total equivalent resistance of the circuit is Req = 20.0 Ω + Rab = 20.0 Ω + 40.0 Ω = 60.0 Ω ΔV 12 V = = 0.20 A . Req 60.0 Ω

(b)

The current supplied to this circuit by the battery is I total =

(c)

2 = ( 60.0 Ω )( 0.20 A ) = 2.4 W . The power delivered by the battery is Ptotal = Req I total

(d)

The potential difference between points a and b is

2

ΔVab = Rab I total = ( 40.0 Ω )( 0.20 A ) = 8.0 V ΔVab 8.0 V and the current in the upper branch is I upper = = = 0.10 A, so the power Rupper 80.0 Ω delivered to the 50.0 Ω resistor is 2 P50 = R50 I upper = ( 50.0 Ω )( 0.10 A ) = 0.50 W 2

18.23

(a)

We name the currents I1 , I 2 , and I 3 as shown.

e

Applying Kirchhoff’s loop rule to loop abcfa gives + e 1 − e 2 − R2 I 2 − R1 I1 = 0 or and

e

e

3I 2 + 2I1 = 10.0 mA I1 = 5.00 mA − 1.50I 2

[1]

Applying the loop rule to loop edcfe yields + e 3 − R3 I 3 − e 2 − R2 I 2 = 0 and

or

3I 2 + 4I 3 = 20.0 mA

I 3 = 5.00 mA − 0.750I 2

[2]

Finally, applying Kirchhoff’s junction rule at junction c gives I 2 = I1 + I 3

[3]

Substituting Equations [1] and [2] into [3] yields I 2 = 5.00 mA − 1.50I 2 + 5.00 mA − 0.750I 2

or

3.25I 2 = 10.0 mA

This gives I 2 = (10.0 mA) 3.25 = 3.08 mA . Then, Equation [1] yields ⎛ 10.0 mA ⎞ I1 = 5.00 mA − 1.50 ⎜ = 0.385 mA ⎝ 3.25 ⎟⎠ continued on next page

68719_18_ch18_p083-119.indd 99

1/7/11 2:41:42 PM

100

Chapter 18

10.0 mA ⎞ and, from Equation [2], I 3 = 5.00 mA − 0.750 ⎛⎜ = 2.69 mA ⎝ 3.25 ⎟⎠ (b)

Start at point c and go to point f, recording changes in potential to obtain V f − Vc = − e 2 − R2 I 2 = −60.0 V − ( 3.00 × 10 3 Ω ) ( 3.08 × 10 −3 A ) = −69.2 V or

18.24

(a)

ΔV

cf

= 69.2 V and point c is at the higher potential .

Applying Kirchhoff’s loop rule to the circuit gives + 3.00 V − ( 0.255 Ω + 0.153 Ω + R )( 0.600 A ) = 0 or

(b)

R=

3.00 V − ( 0.255 Ω + 0.153 Ω ) = 4.59 Ω 0.600 A

The total power input to the circuit is Pinput = (e 1 + e 2 ) I = (1.50 V + 1.50 V )( 0.600 A ) = 1.80 W The power loss by heating within the batteries is Ploss = I 2 ( r1 + r2 ) = ( 0.600 A ) ( 0.255 Ω + 0.153 Ω ) = 0.147 W 2

Thus, the fraction of the power input that is dissipated internally is Ploss 0.147 W = = 0.081 6 or 8.16% 1.80 W Pinput 18.25

(a)

No. Some simplification could be made by recognizing that the 2.0-Ω and 4.0-Ω resistors are in series, adding to give a total of 6.0 Ω; and the 5.0-Ω and 1.0-Ω resistors form a series combination with a total resistance of 6.0 Ω. The circuit cannot be simplified any further, and Kirchhoff’s rules must be used to analyze the circuit.

(b)

Applying Kirchhoff’s junction rule at junction a gives I1 = I 2 + I 3

[1]

Using Kirchhoff’s loop rule on the upper loop yields + 24 V − ( 2.0 + 4.0 ) I1 − ( 3.0 ) I 3 = 0 or

I 3 = 8.0 A − 2 I1

For the lower loop:

[2] +12 V + ( 3.0 ) I 3 − (1.0 + 5.0 ) I 2 = 0

Using Equation [2], this reduces to I2 =

12 V + 3.0 (8.0 A − 2 I1 ) 6.0

or

I 2 = 6.0 A − I1

[3]

Substituting Equations [2] and [3] into [1] gives I1 = 3.5 A . Then, Equation [3] gives

68719_18_ch18_p083-119.indd 100

I 2 = 2.5 A , and Equation [2] yields I 3 = 1.0 A .

1/7/11 2:41:45 PM

Direct-Current Circuits

18.26

Using Kirchhoff’s loop rule on the outer perimeter of the circuit gives

I1 0.01 Ω

+12 V − ( 0.01) I1 − ( 0.06 ) I 3 = 0 or

I1 = 1.2 × 10 A − 6.0 I 3 3

+ − 12 V

[1]

For the rightmost loop, the loop rule gives

Live Battery

I2

101

I3

1.00 Ω

0.06 Ω Starter

+ − 10 V Dead Battery

+10 V + (1.00 ) I 2 − ( 0.06 ) I 3 = 0 or

I 2 = 0.06 I 3 − 10 A

[2]

Applying Kirchhoff’s junction rule at either junction gives I1 = I 2 + I 3 (a)

[3]

Substituting Equations [1] and [2] into [3] yields 7.1I 3 = 1.2 × 10 3 A and I 3 = 1.7 × 10 2 A ( in starter )

(b) 18.27

Then, Equation [2] gives I 2 = 0.20 A ( in dead battery ) .

(a)

No. This multi-loop circuit does not contain any resistors in series (i.e., connected so all the current in one must pass through the other) nor in parallel (connected so the voltage drop across one is always the same as that across the other). Thus, this circuit cannot be simplified any further, and Kirchhoff’s rules must be used to analyze it.

(b)

Assume currents I1 , I 2 , and I 3 in the directions shown. Then, using Kirchhoff’s junction rule at junction a gives I 3 = I1 + I 2

[1]

Applying Kirchhoff’s loop rule on the lower loop, +10.0 V − ( 5.00 ) I 2 − ( 20.0 ) I 3 = 0 or

I 2 = 2.00 A − 4 I 3

and for the loop around the perimeter of the circuit, or

[2] 20.0 V − 30.0I1 − 20.0I 3 = 0

I1 = 0.667 A − 0.667I 3

[3]

Substituting Equations [2] and [3] into [1]: I 3 = 0.667 A − 0.667I 3 + 2.00 A − 4 I 3 which reduces to 5.67I 3 = 2.67 A and gives I 3 = 0.471 A . Then, Equation [2] gives I 2 = 0.116 A , and from Equation [3], I1 = 0.353 A . All currents are in the directions indicated in the circuit diagram given above.

68719_18_ch18_p083-119.indd 101

1/7/11 2:41:48 PM

102

18.28

Chapter 18

(a)

Going counterclockwise around the upper loop, Kirchhoff’s loop rule gives

−11.0I12 + 12.0 − 7.00I12 − 5.00I18 + 18.0 − 8.00I18 = 0 or (b)

18.0I12 + 13.0I18 = 30.0

[1]

Going counterclockwise around the lower loop: −5.00I 36 + 36.0 + 7.00I12 − 12.0 + 11.0I12 = 0 or

(c)

5.00I 36 − 18.0I12 = 24.0

[2]

Applying the junction rule at the node in the left end of the circuit gives I18 = I12 + I 36

[3]

(d)

Solving Equation [3] for I 36 yields I 36 = I18 − I12 .

(e)

Substituting Equation [4] into [2] gives 5.00 ( I18 − I12 ) − 18.0I12 = 24.0, or

[4]

5.00I18 − 23.0I12 = 24.0 (f )

18.29

[5]

Solving Equation [5] for I18 yields I18 = ( 24.0 + 23.0I12 ) 5.00. Substituting this into Equation [1] and simplifying gives 389I12 = −162, and I12 = −0.416 A . Then, from Equation [1], I18 = ( 30.0 − 18.0I12 ) 13.0 which yields I18 = 2.88 A .

(g)

Equation [4] gives I 36 = 2.88 A − ( −0.416 A ), or

(h)

The negative sign in the answer for I12 means that this current flows in the opposite direction to that shown in the circuit diagram and assumed during this solution. That is, the actual current in the middle branch of the circuit flows from right to left and has a magnitude of 0.416 A.

I 36 = 3.30 A .

Applying Kirchhoff’s junction rule at junction a gives I 3 = I1 + I 2

[1]

Using Kirchhoff’s loop rule on the leftmost loop yields −3.00 V − ( 4.00 ) I 3 − ( 5.00 ) I1 + 12.0 V = 0 so

I1 = ( 9.00 A − 4.00I 3 ) 5.00

or

I1 = 1.80 A − 0.800 I 3

[2]

For the rightmost loop, −3.00 V − ( 4.00 ) I 3 − ( 3.00 + 2.00 ) I 2 + 18.0 V = 0 and

I 2 = (15.0 A − 4.00I 3 ) 5.00

or

I 2 = 3.00 A − 0.800 I 3

[3]

continued on next page

68719_18_ch18_p083-119.indd 102

1/7/11 2:41:50 PM

Direct-Current Circuits

103

Substituting Equations [2] and [3] into [1] and simplifying gives 2.60I 3 = 4.80 and I 3 = 1.846 A. Then Equations [2] and [3] yield I1 = 0.323 A and I 2 = 1.523 A. Therefore, the potential differences across the resistors are ΔV2 = I 2 ( 2.00 Ω ) = 3.05 V , ΔV3 = I 2 ( 3.00 Ω ) = 4.57 V ΔV4 = I 3 ( 4.00 Ω ) = 7.38 V , and ΔV5 = I1 ( 5.00 Ω ) = 1.62 V 18.30

The time constant is t = RC. Considering units, we find ⎛ Volts ⎞ ⎛ Coulombs ⎞ ⎛ Coulombs ⎞ RC → ( Ohms )( Farads ) = ⎜ ⎜ ⎟= ⎝ Amperes ⎟⎠ ⎝ Volts ⎠ ⎜⎝ Amperes ⎟⎠ ⎞ ⎛ Coulombs = Second =⎜ ⎝ Coulombs Second ⎟⎠ or

18.31

18.32

18.33

t = RC has units of time.

(a)

The time constant is: t = RC = ( 75.0 × 10 3 Ω ) ( 25.0 × 10 −6 F ) = 1.88 s .

(b)

At t = t , q = 0.632Qm ax = 0.632 (Ce ) = 0.632 ( 25.0 × 10 −6 F )(12.0 V ) = 1.90 × 10 −4 C .

(a)

t = RC = (100 Ω )( 20.0 × 10 −6 F ) = 2.00 × 10 −3 s = 2.00 ms

(b)

Qm ax = C e = ( 20.0 × 10 −6 F )( 9.00 V ) = 1.80 × 10 −4 C = 180 mC

(c)

⎛ 1⎞ Q = Qm ax (1− e − t t ) = Qm ax (1− e −t t ) = Qm ax ⎜ 1− ⎟ = 114 mC ⎝ e⎠

(a)

The time constant of an RC circuit is t = RC. Thus, t = (1.00 × 10 6 Ω ) ( 5.00 × 10 −6 F ) = 5.00 s

(b) (c)

Qm ax = Ce = ( 5.00 mF )( 30.0 V ) = 150 mC To obtain the current through the resistor at time t after the switch is closed, recall that the charge on the capacitor at that time is q = Ce (1− e − t t ) and the potential difference across a capacitor is VC = q C. Thus, VC =

Ce (1− e

−t t

C

) =e

(1− e )

e

R

+ −

C S

−t t

Then, considering switch S to have been closed at time t = 0, apply Kirchhoff’s loop rule around the circuit shown above to obtain +e − iR − VC = 0

or

i=

e − e (1− e − t t

)

R

The current in the circuit at time t after the switch is closed is then i = (e R ) e − t t , so the current in the resistor at t = 10.0 s is 10.0 s ⎛ 30.0 V ⎞ − 5.00 s i=⎜ = ( 30.0 mA ) e −2.00 = 4.06 mA e ⎟ ⎝ 1.00 × 10 6 Ω ⎠

68719_18_ch18_p083-119.indd 103

1/7/11 2:41:53 PM

104

18.34

Chapter 18

Assuming the capacitor is initially uncharged and the switch is closed at t = 0, the charge on the capacitor at time t > 0 is q = Qm ax (1− e − t t ), where Qm ax = Ce and t is the time constant of the circuit. When e = 30 V, C = 5.0 mF, and R = 1.0 MΩ, the time constant is t = RC = (1.0 × 10 6 Ω ) ( 5.0 × 10 −6 F ) = 5.0 s and we find the charge on the capacitor at t = 10 s to be q = ( 5.0 mF )( 30 V )(1− e −10 s 5.0 s ) = 1.3 × 10 2 mC

18.35

The charge remaining on the capacitor after time t is q = Qe − t t .

(a)

Thus, if q = 0.750Q, then e − t t = 0.750 and −t t = ln ( 0.750 ), t = −t ln ( 0.750 ) = − (1.50 s ) ln ( 0.750 ) = 0.432 s

or

t = RC , so

(b) 18.36

C=

t 1.50 s = = 6.00 × 10 −6 F = 6.00 mF R 250 × 10 3 Ω

At time t after the switch is closed, the potential difference between the plates of the initially uncharged capacitor is ΔV = qC = CQm ax (1− e − t t ) = e (1 − e − t RC ) e − t RC = 1−

Thus, giving

R=

ΔV e

and

−

t ΔV ⎞ ⎛ = ln ⎜ 1− ⎟ ⎝ RC e ⎠

− (t C ) ln (1− ΔV e )

If e = 10.0 V, C = 10.0 mF, and ΔV = 4.00 V at t = 3.00 s after closing the switch, the resistance must be R= 18.37

ln (1− 4.00 V 10.0 V )

=

−3.00 × 10 5 Ω = 5.87 × 10 5 Ω = 587 kΩ ln ( 0.600 )

The current drawn by a single 75-W bulb connected to a 120-V source is I1 = P ΔV = 75 W 120 V. Thus, the number of such bulbs that can be connected in parallel with this source before the total current drawn will equal 30.0 A is n=

18.38

− ( 3.00 s 10.0 × 10 −6 F )

(a)

30.0 A ⎛ 120 V ⎞ = ( 30.0 A ) ⎜ = 48 ⎝ 75 W ⎟⎠ I1

The equivalent resistance of the parallel combination is ⎛ 1 1 1⎞ Req = ⎜ + + ⎟ R R R ⎝ 1 2 3 ⎠

−1

1 1 ⎞ ⎛ 1 =⎜ + + ⎝ 150 Ω 25 Ω 50 Ω ⎟⎠

−1

= 15 Ω

so the total current supplied to the circuit is I total =

ΔV 120 V = = 8.0 A R 15 Ω

continued on next page

68719_18_ch18_p083-119.indd 104

1/7/11 2:41:56 PM

Direct-Current Circuits

18.39

105

(b)

Since the appliances are connected in parallel, the voltage across each one is ΔV = 120 V .

(c)

I lam p =

(d)

Pheater =

ΔV 120 V = = 0.80 A Rlam p 150 Ω

( ΔV )2 Rheater

=

(120 V )2 25 Ω

= 5.8 × 10 2 W

From P = ( ΔV ) R, the resistance of the element is 2

R=

( ΔV )2 P

=

( 240 V )2 3 000 W

=19.2 Ω

When the element is connected to a 120-V source, we find that

18.40

18.41

ΔV 120 V = = 6.25 A , and R 19.2 Ω

(a)

I=

(b)

P = ( ΔV ) I = (120 V )( 6.25 A ) = 750 W

(a)

The current drawn by each appliance operating separately is Coffee Maker:

I=

P 1 200 W = = 10 A ΔV 120 V

Toaster:

I=

P 1100 W = = 9.2 A ΔV 120 V

Waffle Maker:

I=

P 1 400 W = = 12 A ΔV 120 V

(b)

If the three appliances are operated simultaneously, they will draw a total current of I total = (10 + 9.2 + 12 ) A = 31 A .

(c)

No. The total current required exceeds the limit of the circuit breaker, so they cannot be operated simultaneously. In fact, with a 15 A limit, no two of these appliances could be operated at the same time without tripping the breaker.

(a)

The area of each surface of this axon membrane is A = L ( 2p r ) = ( 0.10 m ) ⎡⎣ 2p (10 × 10 −6 m ) ⎤⎦ = 2p × 10 −6 m 2 and the capacitance is C = k ∈0

⎛ 2p × 10 −6 m 2 ⎞ A = 1.67 × 10 −8 F = 3.0 (8.85 × 10 −12 C2 N ⋅ m 2 ) ⎜ d ⎝ 1.0 × 10 −8 m ⎟⎠

In the resting state, the charge on the outer surface of the membrane is Qi = C ( ΔV )i = (1.67 × 10 −8 F ) ( 70 × 10 −3 V ) = 1.17 × 10 −9 C → 1.2 × 10 −9 C The number of potassium ions required to produce this charge is NK = +

Qi 1.17 × 10 −9 C = = 7.3 × 10 9 K + ions e 1.6 × 10 −19 C continued on next page

68719_18_ch18_p083-119.indd 105

1/7/11 2:42:00 PM

106

Chapter 18

and the charge per unit area on this surface is −20 2 Qi 1.17 × 10 −9 C ⎛ 1e 1e ⎞ ⎛ 10 m ⎞ = = = ⎜ ⎟ −6 2 −19 2 ⎟ ⎜ ⎝ ⎠ A 2p × 10 m 1.6 × 10 C ⎝ 1 Å ⎠ 8.6 × 10 4 Å 2

s =

1e

( 290 Å )

2

This corresponds to a low charge density of one electronic charge per square of side 290 Å, compared to a normal atomic spacing of one atom every few Å. (b)

In the resting state, the net charge on the inner surface of the membrane is − Qi = −1.17 × 10 −9 C, and the net positive charge on this surface in the excited state is Q f = C ( ΔV ) f = (1.67 × 10 −8 F ) ( +30 × 10 −3 V ) = + 5.0 × 10 −10 C The total positive charge which must pass through the membrane to produce the excited state is therefore ΔQ = Q f − Qi = + 5.0 × 10 −10 C − ( −1.17 × 10 −9 C ) = 1.67 × 10 −9 C → 1.7 × 10 −9 C corresponding to N Na = +

(c)

If the sodium ions enter the axon in a time of Δt = 2.0 ms, the average current is I=

(d)

ΔQ 1.67 × 10 −9 C = = 8.3 × 10 −7 A = 0.83 mA Δt 2.0 × 10 −3 s

When the membrane becomes permeable to sodium ions, the initial influx of sodium ions neutralizes the capacitor with no required energy input. The energy input required to charge the now neutral capacitor to the potential difference of the excited state is W=

18.42

ΔQ 1.67 × 10 −9 C = = 1.0 × 1010 Na + ions e 1.6 × 10 −19 C Na + ion

2 1 1 2 C ( ΔV ) f = (1.67 × 10 −8 F ) ( 30 × 10 −3 V ) = 7.5 × 10 −12 J 2 2

The capacitance of the 10 cm length of axon was found to be C = 1.67 × 10 −8 F in the solution of Problem 18.41. (a)

When the membrane becomes permeable to potassium ions, these ions flow out of the axon with no energy input required until the capacitor is neutralized. To maintain this outflow of potassium ions and charge the now neutral capacitor to the resting action potential requires an energy input of W=

(b)

2 1 1 2 C ( ΔV ) = (1.67 × 10 −8 F ) ( 70 × 10 −3 V ) = 4.1× 10 −11 J 2 2

As found in the solution of Problem 18.41, the charge on the inner surface of the membrane in the resting state is −1.17 × 10 −9 C, and the charge on this surface in the excited state is + 5.0 × 10 −10 C. Thus, the positive charge which must flow out of the axon as it goes from the excited state to the resting state is ΔQ = 5.0 × 10 −10 C + 1.17 × 10 −9 C = 1.67 × 10 −9 C

continued on next page

68719_18_ch18_p083-119.indd 106

1/7/11 2:42:01 PM

Direct-Current Circuits

107

and the average current during the 3.0 ms required to return to the resting state is I= 18.43

From Figure 18.28, the duration of an action potential pulse is 4.5 ms. From the solution of Problem 18.41, the energy input required to reach the excited state is W1 = 7.5 × 10 −12 J. The energy input required during the return to the resting state is found in Problem 18.42 to be W2 = 4.1× 10 −11 J. Therefore, the average power input required during an action potential pulse is P=

18.44

ΔQ 1.67 × 10 −9 C = = 5.6 × 10 −7 A = 0.56 mA Δt 3.0 × 10 −3 s

Wtotal W1 + W2 7.5 × 10 −12 J + 4.1× 10 −11 J = = = 1.1× 10 −8 W = 11 nW Δt Δt 4.5 × 10 −3 s

Using a single resistor → 3 distinct values: R1 = 2.0 Ω, R2 = 4.0 Ω, R3 = 6.0 Ω 2 resistors in Series → 2 additional distinct values: R4 = 2.0 Ω + 6.0 Ω = 8.0 Ω, and R5 = 4.0 Ω + 6.0 Ω = 10 Ω. Note: 2.0 Ω and 4.0 Ω in series duplicates R3 above. 2 resistors in Parallel → 3 additional distinct values: R6 = 2.0 Ω and 4.0 Ω in parallel = 1.3 Ω R7 = 2.0 Ω and 6.0 Ω in parallel = 1.5 Ω R8 = 4.0 Ω and 6.0 Ω in parallel = 2.4 Ω 3 resistors in Series → 1 additional distinct value: R9 = 2.0 Ω + 4.0 Ω + 6.0 Ω = 12 Ω 3 resistors in Parallel → 1 additional distinct value: R10 = 2.0 Ω, 4.0 Ω, and 6.0 Ω in parallel = 1.1 Ω 1 resistor in Parallel with Series combination of the other 2: → 3 additional values:

( = (R = (R

) = 4.0 Ω; 2.0 Ω and 6.0 Ω in series ) = 2.7 Ω = 6.0 Ω; 2.0 Ω and 4.0 Ω in series ) = 3.0 Ω

R11 = Rp = 2.0 Ω; 4.0 Ω and 6.0 Ω in series = 1.7 Ω R12 R12

p

p

1 resistor in Series with Parallel combination of the other 2: → 3 additional values: R14 = ( Rs = 2.0 Ω; 4.0 Ω and 6.0 Ω in parallel ) = 4.4 Ω R15 = ( Rs = 4.0 Ω; 2.0 Ω and 6.0 Ω in parallel ) = 5.5 Ω R16 = ( Rs = 6.0 Ω; 2.0 Ω and 4.0 Ω in Parallel ) = 7.3 Ω Thus, 16 distinct values of resistance are possible using these three resistors. 18.45

Since the circuit is open at points a and b, no current flows through the 4.00-V battery or the 10.0-Ω resistor. A current I will flow around the closed path through the 2.00-Ω resistor, 4.00-Ω resistor, and the 12.0-V battery as shown in the sketch at the right. This current has magnitude

2.00 Ω

12.0 V

+ −

4.00 V

+ −

a

4.00 Ω I 10.0 Ω

b

continued on next page

68719_18_ch18_p083-119.indd 107

1/7/11 2:42:03 PM

108

Chapter 18

I=

ΔV 12.0 V = = 2.00 A Rpath 2.00 Ω + 4.00 Ω

Along the path from point a to point b, the change in potential that occurs is given by ΔVab = +e 4 − IR4 = + 4.00 V − ( 2.00 A )( 4.00 Ω ) = − 4.00 V (a)

The potential difference between points a and b has magnitude ΔVab = 4.00 V

18.46

(b)

Since the change in potential in going from a to b was negative, we conclude that point a is at the higher potential .

(a)

R=

(b)

The resistance in the circuit consists of a series combination with an equivalent resistance of Req = 2.0 kΩ + 3.0 kΩ = 5.0 kΩ. The emf of the battery is then

ΔV 6.0 V = = 2.0 × 10 3 Ω = 2.0 kΩ I 3.0 × 10 −3 A

e = IReq = ( 3.0 × 10 −3 A ) ( 5.0 × 10 3 Ω ) = 15 V (c)

18.47

ΔV3 = IR3 = ( 3.0 × 10 −3 A ) ( 3.0 × 10 3 Ω ) = 9.0 V

(d)

In this solution, we have assumed that we have ideal devices in the circuit. In particular, we have assumed that the battery has negligible internal resistance, the voltmeter has an extremely large resistance and draws negligible current, and the ammeter has an extremely low resistance and a negligible voltage drop across it.

(a)

The resistors combine to an equivalent resistance of Req = 15 Ω as shown.

Figure 1

Figure 3

Figure 2

Figure 4

Figure 5

continued on next page

68719_18_ch18_p083-119.indd 108

1/7/11 2:42:07 PM

Direct-Current Circuits

(b)

109

ΔVab 15 V = = 1.0 A 15 Ω Req

From Figure 5, I1 =

Then, from Figure 4, ΔVac = ΔVdb = I1 ( 6.0 Ω ) = 6.0 V and ΔVcd = I1 ( 3.0 Ω ) = 3.0 V I2 = I3 =

From Figure 2,

ΔVed = I 3 ( 3.6 Ω ) = 1.8 V

Then, from Figure 1,

and (c)

ΔVcd 3.0 V = = 0.50 A 6.0 Ω 6.0 Ω

From Figure 3,

I5 =

ΔV fd 9.0 Ω

=

I4 =

ΔVed 1.8 V = = 0.30 A 6.0 Ω 6.0 Ω

ΔVed 1.8 V = 0.20 A = 9.0 Ω 9.0 Ω

From Figure 2, ΔVce = I 3 ( 2.4 Ω ) = 1.2 V . All the other needed potential differences were calculated above in part (b). The results were ΔVac = ΔVdb = 6.0 V ; ΔVcd = 3.0 V ; and ΔV fd = ΔVed = 1.8 V

(d)

The power dissipated in each resistor is found from P = ( ΔV ) R with the following results: 2

Pac = Ped = Pcd = 18.48

(a)

( ΔV )2ac Rac

( ΔV )ed2 Red

( ΔV )cd2 Rcd

= = =

( 6.0 V )2 6.0 Ω

(1.8 V )2 6.0 Ω

( 3.0 V )2 6.0 Ω

= 6.0 W

Pce =

= 0.54 W

Pfd =

= 1.5 W

Pdb =

( ΔV )ce2 Rce

( ΔV )2fd R fd

( ΔV )2db Rdb

= = =

(1.2 V )2 2.4 Ω

(1.8 V )2 9.0 Ω

( 6.0 V )2 6.0 Ω

= 0.60 W = 0.36 W = 6.0 W

From P = ( ΔV ) R, the resistance of each of the three bulbs is given by 2

R=

( ΔV )2 P

=

(120 V )2 60.0 W

R1 120 V

= 240 Ω

R2 R3

As connected, the parallel combination of R2 and R3 is in series with R1. Thus, the equivalent resistance of the circuit is ⎛ 1 1⎞ Req = R1 + ⎜ + ⎟ R R ⎝ 2 3 ⎠

−1

1 ⎞ ⎛ 1 = 240 Ω + ⎜ + ⎝ 240 Ω 240 Ω ⎟⎠

−1

= 360 Ω

The total power delivered to the circuit is P=

( ΔV )2 Req

=

(120 V )2 360 Ω

= 40.0 W

continued on next page

68719_18_ch18_p083-119.indd 109

1/7/11 2:42:08 PM

110

Chapter 18

ΔV 120 V 1 The current supplied by the source is I = = = A. Thus, the potential Req 360 Ω 3 difference across R1 is

(b)

( ΔV )1 = I R1 = ⎛⎜⎝ A ⎞⎟⎠ ( 240 Ω ) = 80.0 V 3 1

The potential difference across the parallel combination of R2 and R3 is then

( ΔV )2 = ( ΔV )3 = ( ΔV )source − ( ΔV ) 1 = 120 V − 80.0 V = 40.0 V 18.49

When the two resistors are connected in series, the equivalent resistance is Rs = R1 + R2 and the power delivered when a current I = 5.00 A flows through the series combination is Ps = I 2 Rs = ( 5.00 A ) ( R1 + R2 ) = 225 W 2

R1 + R2 =

Thus,

225 W 25.0 A 2

giving

R1 + R2 = 9.00 Ω

When the resistors are connected in parallel, the equivalent resistance is Rp = R1 R2 the power delivered by the same current ( I = 5.00 A ) is

[1]

(R

1

+ R2 ) and

RR ⎞ 2 ⎛ Pp = I 2 Rp = ( 5.00 A ) ⎜ 1 2 ⎟ = 50.0 W ⎝ R1 + R2 ⎠ Rp =

giving

50.0 W 25.0 A 2

R1 R2 = 2.00 Ω R1 + R2

or

[2]

Substituting Equation [1] into Equation [2] yields R ( 9.00 Ω − R1 ) R1 R2 = 1 = 2.00 Ω R1 + R2 9.00 Ω or

R12 − ( 9.00 Ω ) R1 + 18.0 Ω2 = 0 . This quadratic equation factors as

(R

1

− 3.00 Ω ) ( R1 − 6.00 Ω ) = 0

Thus, either R1 = 3.00 Ω or R1 = 6.00 Ω, and from Equation [1], we find that either R2 = 6.00 Ω or R2 = 3.00 Ω. Therefore, the pair contains one 3.00 Ω resistor and one 6.00 Ω resistor. 18.50

(a)

Recognize that the 5.00-Ω and the 8.00-Ω resistors are connected in parallel and that the effective resistance of this parallel combination is Rp =

( 5.00 Ω )(8.00 Ω ) 5.00 Ω + 8.00 Ω

= 3.08 Ω

This resistance is in series with the 10.0-Ω resistor, giving a total resistance for the circuit of Req = 10.0 Ω + 3.08 = 13.1 Ω. Thus, the current supplied by the battery is I total =

e 15.0 V = = 1.15 A Req 13.1 Ω

and the potential difference across the parallel combination is ΔVp = I total Rp = (1.15 A )( 3.08 Ω ) = 3.54 V continued on next page

68719_18_ch18_p083-119.indd 110

1/7/11 2:42:11 PM

111

Direct-Current Circuits

The current through the 5.00-Ω in this parallel combination is then I5 = (b)

ΔVp R5

=

3.54 V = 0.708 A 5.00 Ω

The power delivered to the 5.00-Ω resistor is P5 = I 52 R5 = ( 0.708 A ) ( 5.00 Ω ) = 2.51 W 2

18.51

(c)

Only the circuit in Figure P18.50c requires the use of Kirchhoff’s rules for solution. In the other circuits, the batteries can be combined into a single effective battery while the 5.00-Ω and 8.00-Ω resistors remain in parallel with each other.

(d)

The power delivered is lowest in Figure 18.50c. The circuits in Figures P18.50b and P18.50d have in effect 30.0-V batteries driving current through the 10.0-Ω resistor, thus delivering more power than the circuit in Figure 18.50a. In Figure 18.50c, the two 15.0-V batteries tend to oppose each other’s efforts to drive current through the 10.0-Ω resistor, making them less effective than the single 15.0-V battery of Figure 18.50a.

(a)

When switch S is open, all three bulbs are in series and the equivalent resistance is Reqopen = R + R + R = 3R . When the switch is closed, bulb C is shorted across and no current will flow through that bulb. This leaves bulbs A and B in series with an equivalent resistance of Reqclosed = R + R = 2R .

(b)

(c)

18.52

A

e

B

C

S

e2 e2 With the switch open, the power delivered by the battery is Popen = open = , and with Req 3R the switch closed, Pclosed = e 2 Reqclosed = e 2 2R . When the switch is open, the three bulbs have equal brightness. When S is closed, bulb C goes out, while A and B remain equal at a greater brightness than they had when the switch was open.

With the switch open, the circuit may be reduced as follows:

With the switch closed, the circuit reduces as shown below:

Since the equivalent resistance with the switch closed is one-half that when the switch is open, we have R + 18 Ω =

68719_18_ch18_p083-119.indd 111

1 ( R + 50 Ω ), which yields R = 14 Ω 2

1/7/11 2:42:14 PM

112

18.53

Chapter 18

(a)

Note the assumed directions of the three distinct currents in the circuit diagram at the right. Applying the junction rule at point c gives I1 = I 3 − I 2 Applying Kirchhoff’s loop rule to loop gbcfg gives

a

− 3.00 V + k [1] + 6.00 mF − h

+ 8.00 V + 3.00I 2 − 5.00I1 = 0

or

b

c

+ I1 I2 −

8.00 V

I1

d 5.00 Ω

I3

3.00 Ω

− 4.00 V +

I1 I=0

g

5.00 Ω

I3 f

e

5.00I1 − 3.00I 2 = 8.00 V

[2]

Finally, applying Kirchhoff’s loop rule to loop fcdef yields −3.00I 2 − 5.00I 3 + 4.00 V = 0

or

5.00I 3 + 3.00I 2 = 4.00 V

[3]

Substituting Equation [1] into Equation [2] gives 5.00I 3 − 8.00I 2 = 8.00 V and subtracting this result from Equation [3] yields

I 2 = − 4.00 V 11.0

Equation [3] then gives the current in the 4.00-V battery as I3 = (b)

4.00 V − 3.00 ( −4.00 V 11.0 ) = +1.02 A 5.00

I 3 = 1.02 A down

From above, the current in the 3.00-Ω resistor is I2 = −

(c)

or

4.00 V = − 0.364 A 11.0

or

I 2 = 0.364 A down

Equation [1] now gives the current in the 8.00-V battery as I1 = 1.02 A − ( −0.364 A ) = +1.38 A

or

I1 = 1.38 A up

(d)

Once the capacitor is charged, the current in the 3.00-V battery is I = 0 because of the open circuit between the plates of the capacitor.

(e)

To obtain the potential difference between the plates of the capacitor, we start at the negative plate and go to the positive plate (noting the changes in potential) along path hgbak. The result is ΔVhk = +8.00 V + 3.00 V = 11.0 V so the charge on the capacitor is Qhk = Chk ( ΔVhk ) = ( 6.00 mF )(11.0 V ) = 66.0 mC

18.54

At time t, the charge on the capacitor will be q = Qm ax (1− e − t t ) , where t = RC = ( 2.0 × 10 6 Ω ) ( 3.0 × 10 −6 F ) = 6.0 s When q = 0.90Qm ax, this gives 0.90 = 1− e − t t or e − t t = 0.10. Thus, − t t = ln ( 0.10 ), giving t = − ( 6.0 s ) ln ( 0.10 ) = 14 s .

68719_18_ch18_p083-119.indd 112

1/7/11 2:42:16 PM

Direct-Current Circuits

18.55

(a)

113

For the first measurement, the equivalent circuit is as shown in Figure 1. From this, Rab = R1 = Ry + Ry = 2 Ry so

Ry =

1 R1 2

[1]

Figure 1

For the second measurement, the equivalent circuit is shown in Figure 2. This gives Rac = R2 =

1 Ry + Rx 2

[2] Figure 2

Substitute [1] into [2] to obtain R2 = (b)

1⎛1 ⎞ 1 ⎜⎝ R1 ⎟⎠ + Rx, or Rx = R2 − R1 2 2 4

If R1 = 13 Ω and R2 = 6.0 Ω, then Rx = 2.8 Ω . Since this exceeds the limit of 2.0 Ω, the antenna is inadequately grounded .

18.56

Assume a set of currents as shown in the circuit diagram at the right. Applying Kirchhoff’s loop rule to the leftmost loop gives + 75 − ( 5.0 ) I − ( 30 )( I − I1 ) = 0 or

7 I − 6 I1 = 15

[1]

For the rightmost loop, the loop rule gives − ( 40 + R ) I1 + ( 30 )( I − I1 ) = 0,

or

⎛7 R⎞ I = ⎜ + ⎟ I1 ⎝ 3 30 ⎠

[2]

Substituting Equation [2] into [1] and simplifying gives 310 I1 + 7 ( I1 R ) = 450

[3]

Also, it is known that PR = I12 R = 20 W, so

I1 R =

20 W I1

[4]

Substitution of Equation [4] into [3] yields 310 I1 +

140 = 450 I1

or

Using the quadratic formula: I1 =

310 I12 − 450 I1 + 140 = 0 − ( −450 ) ±

( −450 )2 − 4 ( 310 )(140 ) , 2 ( 310 )

20 W , we find two possible values for I12 R = 20 Ω or R = 98 Ω

yielding I1 = 1.0 A and I1 = 0.452 A. Then, from R = the resistance R. These are:

68719_18_ch18_p083-119.indd 113

1/7/11 2:42:19 PM

114

18.57

Chapter 18

When connected in series, the equivalent resistance is Req = R1 + R2 + ⋅⋅⋅ + Rn = n R. Thus, the current is I s = ( ΔV ) Req = ( ΔV ) n R, and the power consumed by the series configuration is Ps = I s2 Req =

( ΔV )2

(n R)

2

(n R) =

( ΔV )2 nR

For the parallel connection, the power consumed by each individual resistor is P1 = ( ΔV ) R, and the total power consumption is 2

n ( ΔV ) R

2

Pp = nP1 =

Ps ( ΔV ) 1 1 R = = 2 or Ps = 2 Pp ⋅ 2 n n Pp n R n ( ΔV ) 2

Therefore, 18.58

Consider a battery of emf e connected between points a and b as shown. Applying Kirchhoff’s loop rule to loop acbea gives

ε

− (1.0 ) I1 − (1.0 )( I1 − I 3 ) + e = 0 or

I 3 = 2 I1 − e

[1]

Applying the loop rule to loop adbea gives − ( 3.0 ) I 2 − ( 5.0 )( I 2 + I 3 ) + e = 0 or

8 I2 + 5 I3 = e

[2]

For loop adca, the loop rule yields I1 + I 3 3

[3]

I 2 = I1 − e 3

[4]

− ( 3.0 ) I 2 + (1.0 ) I 3 + (1.0 ) I1 = 0 or I 2 = Substituting Equation [1] into [3] gives

26 13 Now, substitute Equations [1] and [4] into [2] to obtain 18I1 = e , which reduces to I1 = e. 3 27 13 9 ⎤ 4 1 ⎡ Then, Equation [4] gives I 2 = ⎢ − ⎥ e = e , and [1] yields I 3 = − e . 27 27 ⎣ 27 27 ⎦ Then, applying Kirchhoff’s junction rule at junction a gives I = I1 + I 2 = 18.59

13 4 17 e e 27 e+ e = e . Therefore, Rab = = = Ω . 27 27 27 I (17e 27 ) 17

(a) and (b): With R the value of the load resistor, the current in a series circuit composed of a 12.0 V battery, an internal resistance of 10.0 Ω, and a load resistor is I=

12.0 V R + 10.0 Ω

and the power delivered to the load resistor is

(144 V ) R 2

PL = I 2 R =

( R + 10.0 Ω )2

continued on next page

68719_18_ch18_p083-119.indd 114

1/7/11 2:42:22 PM

Direct-Current Circuits

Some typical data values for the graph are

3.50 3.00

R (Ω) PL (W) 1.19

5.00

3.20

10.0

3.60

15.0

3.46

20.0

3.20

25.0

2.94

30.0

2.70

2.50 PL(W)

1.00

2.00 1.50 1.00

0.500

The curve peaks at PL = 3.60 W at a load resistance of R = 10.0 Ω. 18.60

115

0

5.00

10.0

15.0 R(Ω)

20.0

25.0

30.0

The total resistance in the circuit is ⎛ 1 1⎞ R=⎜ + ⎟ R R ⎝ 1 2 ⎠

−1

⎛ 1 1 ⎞ =⎜ + ⎝ 2.0 kΩ 3.0 kΩ ⎟⎠

−1

= 1.2 kΩ

and the total capacitance is C = C1 + C2 = 2.0 mF + 3.0 mF = 5.0 mF. Thus, Qm ax = Ce = ( 5.0 mF )(120 V ) = 600 mC and

t = RC = (1.2 × 10 3 Ω ) ( 5.0 × 10 −6 F ) = 6.0 × 10 −3 s =

6.0 s 1 000

The total stored charge at any time t is then q = q1 + q2 = Qm ax (1− e − t t

)

q1 + q2 = ( 600 mC ) (1− e −1 000 t 6.0 s )

or

[1]

Since the capacitors are in parallel with each other, the same potential difference exists across both at any time. Therefore,

( ΔV )C =

q1 q2 = , C1 C2

or

⎛C ⎞ q2 = ⎜ 2 ⎟ q1 = 1.5q1 ⎝ C1 ⎠

[2]

Substituting Equation [2] into [1] gives 2.5q1 = ( 600 mC ) (1− e −1 000 t 6.0 s )

and

q1 =

( 240 mC) (1− e

Then, Equation [2] yields q2 = 1.5 ( 240 mC ) (1− e −1 000 t 6.0 s ) = 18.61

(a)

−1 000 t 6.0 s

(360 mC) (1− e

)

−1 000 t 6.0 s

)

With 4.0 × 10 3 cells, each with an emf of 150 mV, connected in series, the total terminal potential difference is ΔV = ( 4.0 × 10 3 ) (150 × 10 −3 V ) = 6.0 × 10 2 V

continued on next page

68719_18_ch18_p083-119.indd 115

1/7/11 2:42:25 PM

116

Chapter 18

When delivering a current of I = 1.0 A, the power output is P = I ( ΔV ) = (1.0 A )( 6.0 × 10 2 V ) = 6.0 × 10 2 W (b)

The energy released in one shock is E1 = P ( Δt )1 = ( 6.0 × 10 2 W ) ( 2.0 × 10 −3 s ) = 1.2 J

(c)

The energy released in 300 such shocks is Etotal = 300E1 = 300 (1.2 J ) = 3.6 × 10 2 J. For a 1.0-kg object to be given a gravitational potential energy of this magnitude, the height the object must be lifted above the reference level is h=

18.62

(a)

PEg

=

mg

3.6 × 10 2 J = 37 m (1.0 kg) (9.80 m s2 )

When the power supply is connected to points A and B, the circuit reduces as shown below to an equivalent resistance of Req = 0.099 9 Ω .

From the center figure above, observe that I R = I1 = 1

and (b)

I R = I R = I100 = 2

3

5.00 V = 50.0 A 0.100 Ω

5.00 V = 0.045 0 A = 45.0 mA 111 Ω

When the power supply is connected to points A and C, the circuit reduces as shown below to an equivalent resistance of Req = 1.09 Ω . R3 + 100 Ω = 110 Ω 1.09 Ω

I100 R1 + R2 = 1.10 Ω

From the center figure above, observe that I R = I R = I1 = 1

and (c)

I R = I100 = 3

2

5.00 V = 4.55 A 1.10 Ω

5.00 V = 0.045 5 A = 45.5 mA 110 Ω

When the power supply is connected to points A and D, the circuit reduces as shown below to an equivalent resistance of Req = 9.99 Ω .

continued on next page

68719_18_ch18_p083-119.indd 116

1/7/11 2:42:27 PM

Direct-Current Circuits

From the center figure above, observe that I R = I R = I R = I1 = 1

and 18.63

I100 =

2

3

117

5.00 V = 0.450 A 11.1 Ω

5.00 V = 0.050 0 A = 50.0 mA 100 Ω

In the circuit diagram at the right, note that all points labeled a are at the same potential and equivalent to each other. Also, all points labeled c are equivalent. To determine the voltmeter reading, go from point e to point d along the path ecd, keeping track of all changes in potential to find ΔVed = Vd − Ve = −4.50 V + 6.00 V = + 1.50 V Apply Kirchhoff’s loop rule around loop abcfa to find − ( 6.00 Ω ) I + ( 6.00 Ω ) I 3 = 0

or

I3 = I

[1]

I 2 = 0.600 A − 0.600I

[2]

I1 = 0.900 A − 1.20I

[3]

Apply Kirchhoff’s loop rule around loop abcda to find − ( 6.00 Ω ) I + 6.00 V − (10.0 Ω ) I 2 = 0

or

Apply Kirchhoff’s loop rule around loop abcea to find − ( 6.00 Ω ) I + 4.50 V − ( 5.00 Ω ) I1 = 0

or

Finally, apply Kirchhoff’s junction rule at either point a or point c to obtain I + I 3 = I1 + I 2

[4]

Substitute Equations [1], [2], and [3] into Equation [4] to obtain the current through the ammeter. This gives I + I = 0.900 A − 1.20I + 0.600 A − 0.600I or

68719_18_ch18_p083-119.indd 117

3.80I = 1.50 A

and

I = 1.50 A 3.80 = 0.395 A

1/7/11 2:42:29 PM

118

18.64

18.65

Chapter 18

In the figure given below, note that all bulbs have the same resistance, R.

(a)

In the series situation, Case 1, the same current I1 flows through both bulbs. Thus, the same power, P1 = I12 R, is supplied to each bulb. Since the brightness of a bulb is proportional to the power supplied to it, they will have the same brightness. We conclude that the bulbs have the same current, power supplied, and brightness .

(b)

In the parallel case, Case 2, the same potential difference ΔV is maintained across each of the bulbs. Thus, the same current I 2 = ΔV R will flow in each branch of this parallel circuit. This means that, again, the same power P2 = I 22 R is supplied to each bulb, and the two bulbs will have equal brightness .

(c)

The total resistance of the single branch of the series circuit (Case 1) is 2R. Thus, the current in this case is I1 = ΔV 2R. Note that this is one half of the current I 2 that flows through each bulb in the parallel circuit (Case 2). Since the power supplied is proportional to the square of the current, the power supplied to each bulb in Case 2 is four times that supplied to each bulb in Case 1. Thus, the bulbs in Case 2 are much brighter than those in Case 1.

(d)

If either bulb goes out in Case 1, the only conducting path of the circuit is broken and all current ceases. Thus, in the series case, the other bulb must also go out . If one bulb goes out in Case 2, there is still a continuous conducting path through the other bulb. Neglecting any internal resistance of the battery, the battery continues to maintain the same potential difference ΔV across this bulb as was present when both bulbs were lit. Thus, in the parallel case, the second bulb remains lit with unchanged current and brightness when one bulb fails.

(a)

The equivalent capacitance of this parallel combination is Ceq = C1 + C2 = 3.00 mF + 2.00 mF = 5.00 mF When fully charged by a 12.0-V battery, the total stored charge before the switch is closed is Q0 = Ceq ( ΔV ) = ( 5.00 mF )(12.0 V ) = 60.0 mC Once the switch is closed, the time constant of the resulting RC circuit is t = RCeq = ( 5.00 × 10 2 Ω ) ( 5.00 mF ) = 2.50 × 10 −3 s = 2.50 ms Thus, at t = 1.00 ms after closing the switch, the remaining total stored charge is q = Q0 e − t t = ( 60.0 mC ) e −1.00 m s 2.50 m s = ( 60.0 mC ) e −0.400 = 40.2 mC

continued on next page

68719_18_ch18_p083-119.indd 118

1/7/11 2:42:30 PM

Direct-Current Circuits

119

The potential difference across the parallel combination of capacitors is then ΔV =

q 40.2 mC = = 8.04 V Ceq 5.00 mF

and the charge remaining on the 3.00-mF capacitor will be q3 = C3 ( ΔV ) = ( 3.00 mF )(8.04 V ) = 24.1 mC (b)

The charge remaining on the 2.00-mF capacitor at this time is q2 = q − q3 = 40.2 mC − 24.1 mC = 16.1 mC or alternately,

(c)

Since the resistor is in parallel with the capacitors, it has the same potential difference across it as do the capacitors at all times. Thus, Ohm’s law gives I=

18.66

(a)

q2 = C2 ( ΔV ) = ( 2.00 mF )(8.04 V ) = 16.1 mC

ΔV 8.04 V = = 1.61× 10 −2 A = 16.1 mA R 5.00 × 10 2 Ω

If the switch S in the circuit at the right is closed at t = 0, the charge remaining on the capacitor at time t is q = Q0 e − t t , where Q0 = 5.10 mC is the initial charge and t = RC = (1.30 × 10 Ω ) ( 2.00 × 10 3

−9

F ) = 2.60 ms

+q −q C i

S R

is the time constant of the circuit. The potential difference across the capacitor, ΔvC , at time t is ΔvC =

q Q0 e − t t = C C

Applying Kirchhoff’s loop rule to the above circuit gives ΔvC − iR = 0 , or i=

ΔvC Q0 e − t t ⎛ Q0 ⎞ − t t = =⎜ ⎟e ⎝ t ⎠ R RC

The current through the resistor at time t = 9.00 ms is then 9.00 m s

⎛ 5.10 mC ⎞ − 2.60 m s ⎛Q ⎞ i = ⎜ 0 ⎟ e− t t = ⎜ e = 6.16 × 10 −2 C s = 61.6 mA ⎝ t ⎠ ⎝ 2.60 ms ⎟⎠ (b)

The charge remaining on the capacitor at t = 8.00 ms is q = Q0 e − t t = ( 5.10 mC ) e

(c)

−

8.00 m s 2.60 m s

= 0.235 mC

The maximum current in the circuit occurs when the switch is first closed (at t = 0) and is given by 5.10 m C Q ⎛Q ⎞ i0 = ⎜ 0 ⎟ e −0 = 0 = = 1.96 A ⎝ t ⎠ t 2.60 m s

68719_18_ch18_p083-119.indd 119

1/7/11 2:42:33 PM

19 Magnetism QUICK QUIZZES 1.

Choice (b). The force that a magnetic field exerts on a charged particle moving through it is given by F = qvB sinq = qvB⊥, where B⊥ is the component of the field perpendicular to the particle’s velocity. Since the particle moves in a straight line, the magnetic force (and hence B⊥, since qv ≠ 0) must be zero.

2.

Choice (c). The magnetic force exerted on a charge by a magnetic field is proportional to the charge’s velocity relative to the field. If the charge is stationary, as in this situation, there is no magnetic force.

3.

Choice (c). The torque that a planar current loop will experience when it is in a magnetic field is given by t = BIAsinq . Note that this torque depends on the strength of the field, the current in the coil, the area enclosed by the coil, and the orientation of the plane of the coil relative to the direction of the field. However, it does not depend on the shape of the loop.

4.

Choice (a). The magnetic force acting on the particle is always perpendicular to the velocity of the particle and hence to the displacement the particle is undergoing. Under these conditions, the force does no work on the particle, and the particle’s kinetic energy remains constant.

5.

Choices (a) and (c). The magnitude of the force per unit length between two parallel current carrying wires is F = ( m 0 I1 I 2 ) ( 2p d ). The magnitude of this force can be doubled by doubling the magnitude of the current in either wire. It can also be doubled by decreasing the distance between them, d, by half. Thus, both choices (a) and (c) are correct.

6.

Choice (b). The two forces are an action-reaction pair. They act on different wires and have equal magnitudes but opposite directions.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

If the magnitude of the magnetic force on the wire equals the weight of the wire, then BI sinq = w, or B = w I sinq . The magnitude of the magnetic field is a minimum when q = 90° and sinq = 1. Thus, Bm in =

w 1.0 × 10 −2 N = = 0.20 T I ( 0.10 A )( 0.50 m )

and (a) is the correct answer for this question. 2.

The electron moves in a horizontal plane in a direction of 35° N of E, which is the same as 55° E of N. Since the magnetic field at this location is horizontal and directed due north, the angle between the direction of the electron’s velocity and the direction of the magnetic field is 55°. The magnitude of the magnetic force experienced by the electron is then F = q vB sinq = (1.6 × 10 −19 C ) ( 2.5 × 10 6 m s ) ( 0.10 × 10 −4 T ) sin 55° = 3.3 × 10 −18 N 120

68719_19_ch19_p120-147.indd 120

1/7/11 2:43:34 PM

Magnetism

121

The right-hand rule number 1 predicts a force directed upward, away from the Earth’s surface for a positively charged particle moving in the direction of the electron. However, the negatively charged electron will experience a force in the opposite direction, downward toward the Earth’s surface. Thus, the correct choice is (d). 3.

The z-axis is perpendicular to the plane of the loop, and the angle between the direction of this normal line and the direction of the magnetic field is q = 30.0°. Thus, the magnitude of the torque experienced by this coil containing N = 10 turns is t = BIAN sinq = ( 0.010 T )( 2.0 A )[( 0.20 m )( 0.30 m )](10 ) sin 30.0° = 6.0 × 10 −3 N ⋅ m meaning that (c) is the correct choice.

4.

The magnitude of the magnetic field at distance r from a long straight wire carrying current I is B = m 0 I 2p r. Thus, for the described situation, B=

( 4p × 10

−7

T⋅m A ) (1 A )

2p ( 2 m )

= 1× 10 −7 T

making (d) the correct response. 5.

Since the proton follows a semicircular path, not a helical path, it entered perpendicularly to the field. A charged particle moving perpendicular to a magnetic field experiences a centripetal force of magnitude Fc = m v 2 r = qvB and follows a circular path of radius r = mv qB. The speed of this proton must be −19 −3 qBr (1.6 × 10 C )( 0.050 T ) (1.0 × 10 m ) v= = = 4.8 × 10 3 m s m 1.67 × 10 −27 kg

and choice (e) is the correct answer. 6.

The force per unit length between this pair of wires is F

−7 m 0 I1 I 2 ( 4p × 10 T ⋅ m A )( 3 A ) = = 9 × 10 −7 N ∼ 1× 10 −6 N = 2p d 2p ( 2 m ) 2

and (d) is the best choice for this question. 7.

The magnitude of the magnetic field inside the specified solenoid is ⎛ 120 ⎞ ⎛ N⎞ B = m 0 nI = m 0 ⎜ ⎟ I = ( 4p × 10 −7 T ⋅ m A ) ⎜ ( 2.0 A ) = 6.0 × 10 −4 T ⎝ 0.50 m ⎟⎠ ⎝ ⎠ which is choice (e).

8.

68719_19_ch19_p120-147.indd 121

The force that a magnetic field exerts on a moving charge is always perpendicular to both the direction of the field and the direction of the particle’s motion. Since the force is perpendicular to the direction of motion, it does no work on the particle and hence does not alter its speed. Because the speed is unchanged, both the kinetic energy and the magnitude of the linear momentum will be constant. Correct answers among the list of choices are (d) and (e). All other choices are false.

1/7/11 2:43:36 PM

122

9.

Chapter 19

The magnitude of the magnetic force experienced by a charged particle in a magnetic field is given by F = qvB sinq , where v is the speed of the particle and q is the angle between the direction of the particle’s velocity and the direction of the magnetic field. If either v = 0 [choice (e)] or sinq = 0 [choice (c)], this force has zero magnitude. All other choices are false, so the correct answers are (c) and (e).

10.

By the right-hand rule number 1, when the proton first enters the field, it experiences a force directed upward, toward the top of the page. This will deflect the proton upward, and as the proton’s velocity changes direction, the force changes direction, always staying perpendicular to the velocity. The force, being perpendicular to the motion, causes the particle to follow a circular path, with no change in speed, as long as it is in the field. After completing a half circle, the proton will exit the field traveling toward the left. The correct answer is choice (d).

11.

The contribution made to the magnetic field at point P by the lower wire is directed out of the page, while the contribution due to the upper wire is directed into the page. Since point P is equidistant from the two wires, and the wires carry the same magnitude currents, these two oppositely directed contributions to the magnetic field have equal magnitudes and cancel each other. Therefore, the total magnetic field at point P is zero, making (a) the correct answer for this question.

12.

The magnetic field due to the current in the vertical wire is directed into the page on the right side of the wire and out of the page on the left side. The field due to the current in the horizontal wire is out of the page above this wire and into the page below the wire. Thus, the two contributions to the total magnetic field have the same directions at points B (both out of the page) and D (both contributions into the page), while the two contributions have opposite directions at points A and C. The magnitude of the total magnetic field will be greatest at points B and D where the two contributions are in the same direction, and smallest at points A and C where the two contributions are in opposite directions and tend to cancel. The correct choices for this question are (a) and (c).

13.

The torque exerted on a single turn coil carrying current I by a magnetic field B is t = BIAsinq . The line perpendicular to the plane of each coil is also perpendicular to the direction of the magnetic field (i.e., q = 90°). Since B and I are the same for all three coils, the torques exerted on them are proportional to the area A enclosed by each of the coils. Coil A is rectangular with area AA = (1 m )( 2 m ) = 2 m 2. Coil B is elliptical with semi-major axis a = 0.75 m and semi-minor axis 2 b = 0.5 m, giving an area AB = p ab or AB = p ( 0.75 m )( 0.5 m ) = 1.2 m . Coil C is triangular 1 1 with area AC = ( base )( height ), or AC = (1 m )( 3 m ) = 1.5 m 2. Thus, AA > AC > AB, meaning 2 2 that t A > t C > t B and choice (b) is the correct answer.

14.

Any point in region I is closer to the upper wire which carries the larger current. At all points in this region, the outward-directed field due the upper wire will have a greater magnitude than will the inward-directed field due to the lower wire. Thus, the resultant field in region I will be nonzero and out of the page, meaning that choice (d) is a true statement and choice (a) is false. In region II, the field due to each wire is directed into the page, so their magnitudes add and the resultant field cannot be zero at any point in this region. This means that choice (b) is false. In region III, the field due to the upper wire is directed into the page while that due to the lower wire is out of the page. Since points in this region are closer to the wire carrying the smaller current, there are points in this region where the magnitudes of the oppositely directed fields due to the two wires will possibly have equal magnitudes, canceling each other and producing a zero resultant field. Thus, choice (c) is true and choice (e) is false. The correct answers for this question are choices (c) and (d).

15.

According to right-hand rule number 2, the magnetic field at point P due to the current in the wire is directed out of the page, meaning that choices (c) and (e) are false. The magnitude of this field is given by B = m 0 I 2p r, so choices (b) and (d) are false. Choice (a) is correct about both the magnitude and direction of the field and is the correct answer for the question.

68719_19_ch19_p120-147.indd 122

1/7/11 2:43:37 PM

Magnetism

16.

123

The magnetic field inside a solenoid, carrying current I, with N turns and length L, is m ( 2N A ) I m N I m N I 1 ⎛ N⎞ B = m 0 nI = m 0 ⎜ ⎟ I. Thus, BA = 0 A , BB = 0 A = BA, and BC = 0 = 4BA . ⎝ L⎠ LA LA 2 2LA 2 Therefore, we see that BC > BA > BB, and choice (d) gives the correct rankings.

17.

Using right-hand rule number 1, we see that the magnetic force exerted on the positively charged proton in the situation described is in the positive y-direction, making choice (c) the correct answer.

18.

The correct answers to each of the parts of this question are as follows: (a) Yes. (b) No. (c) Yes. (d) Yes, unless the object moves parallel to the field. (e) No. (f) Yes, unless the current is parallel to the field. (g) Yes. (h) Yes, unless the electrons move parallel to the field.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

It should point straight down toward the surface of the Earth.

4.

No. The force that a constant magnetic field exerts on a charged particle is dependent on the velocity of that particle. If the particle has zero velocity, it will experience no magnetic force and cannot be set in motion by a constant magnetic field.

6.

Yes. Regardless of which pole is used, the magnetic field of the magnet induces magnetic poles in the nail, with each end of the nail taking on a magnetic polarity opposite to that of the pole of the magnet nearest to it. These opposite magnetic poles then attract each other.

8.

The magnet causes domain alignment in the iron, inducing magnetic poles such that the iron becomes magnetic and attracted to the original magnet. Now that the iron is magnetic, it can produce an identical effect in another piece of iron.

10.

No. The magnetic field created by a single current loop resembles that of a bar magnet—strongest inside the loop, and decreasing in strength as you move away from the loop. Neither is the field uniform in direction—the magnetic field lines form closed paths that curve around the conductor forming the current loop and pass through the area enclosed by that current loop.

12.

Near the poles the magnetic field of Earth points almost straight downward (or straight upward), in the direction (or opposite to the direction) the charges are moving. As a result, there is little or no magnetic force exerted on the charged particles at the pole to deflect them away from Earth.

14.

The loop can be mounted on an axle so it is free to rotate. The current loop will tend to rotate in a manner that brings the plane of the loop perpendicular to the direction of the field. As the current in the loop is increased, the torque causing it to rotate will increase in magnitude.

16.

(a)

The blue magnet experiences an upward magnetic force equal to its weight. The yellow magnet is repelled by the red magnets with a force whose magnitude equals the weight of the yellow magnet plus the magnitude of the reaction force exerted on this magnet by the blue magnet.

(b)

The rods prevent motion to the side and prevent the magnets from rotating under their mutual torques. Its constraint changes unstable equilibrium into stable.

68719_19_ch19_p120-147.indd 123

1/7/11 2:43:39 PM

124

Chapter 19

(c)

Most likely, the disks are magnetized perpendicular to their flat faces, making one face a north pole and the other a south pole. The yellow magnet has a pole on its lower face which is the same as the pole on the upper faces of the red magnets. The pole on the lower face of the blue magnet is the same as that on the upper face of the yellow magnet.

(d)

If the upper magnet were inverted, the yellow and blue magnets would attract each other and stick firmly together. The yellow magnet would continue to be repelled by and float above the red magnets.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a)

a⬘) toward the left ⬘)d toward top of page

b⬘) into the page e⬘) into the page

c⬘) out of the page f⬘) out of the page

(b) The answer for each subpart is opposite to that given in part (a) above. 4.

(a)

toward top of page

(d)

into the page

(b)

out of page (q < 0)

6.

B = 2.09 ×10 −2 T in the negative y-direction

8.

(a)

7.90 × 10 −12 N

(b)

0

(c)

zero force

10.

807 N

12.

(a)

toward the left

(b)

into the page

(c)

out of the page

(d)

toward top of page

(e)

into the page

(f)

out of the page

(a)

0.12 N

(b)

Both the direction of the field and the direction of the current must be known before the direction of the force can be determined.

14.

16.

Bm in = 0.245 T eastward

18.

(a)

9.0 × 10 −3 N at 15° above the horizontal in the northward direction

(b)

2.3 × 10 −3 N horizontal and due west

20.

(a)

0.109 A

22.

B = m k mg Id

24.

5.8 N into the page

26.

4.33 × 10 −3 N ⋅ m

28.

9.98 N·m, clockwise about the y-axis when viewed from above

30.

118 N·m

68719_19_ch19_p120-147.indd 124

(b)

toward the right

1/7/11 2:43:40 PM

Magnetism

125

(a)

+x-direction, zero torque about x-axis

(b)

−x-direction, zero torque about x-axis

(c)

No. The two forces are equal in magnitude and opposite in direction, canceling each other, and can have no effect on the motion of the loop.

(d)

in the yz-plane at 130° counterclockwise from + y-direction. Torque is counterclockwise about x-axis.

(e)

counterclockwise about x-axis

(f)

0.135 A ⋅ m 2

(g)

130°

34.

(a)

1.25 × 10 −13 N

(b)

7.49 × 1013 m s 2

36.

0.150 mm

38.

3.11 cm

40.

(a)

v = 2 ( KE ) qBR

(b)

m = q 2 B 2 R 2 2 ( KE )

42.

(a)

rd = 2 ⋅rp

(b)

ra = 2 ⋅rp

44.

(a)

toward the left

(b)

46.

675 A downward

48.

(a)

40.0 mT into the page

(b)

50.

(a)

Bnet = 4.00 mT toward the bottom of the page

(b)

Bnet = 6.67 mT upward at 77.0° to the left of vertical

(a)

in the –y-direction

32.

52.

(b)

(h)

0.155 N·m

out of the page

(c)

lower left to upper right

5.00 mT out of the page

(c)

1.67 mT out of the page

upward, in the positive z-direction

(c) The magnitude of the upward magnetic force must equal that of the downward gravitational force.

54.

56.

(d)

d = qvm 0 I 2p mg

(e)

5.40 cm

(a)

in the + y-direction

(b)

BP = m 0 Ix p ( x 2 + d 2 )

(c)

BP x = 0 = 0. This is as expected since the two field contributions have equal magnitudes and opposite directions at the point midway between the wires.

(a)

8.0 A

(c)

Reversing the direction of one current changes the force from repulsion to attraction. Doubling the magnitude of one current doubles the magnitude of the force.

58.

2.70 × 10 −5 N toward the left

60.

4.8 × 10 4 turns

68719_19_ch19_p120-147.indd 125

(b)

opposite directions

1/7/11 3:11:50 PM

126

Chapter 19

62.

207 W

64.

(a)

B = mg NIw directed out of the page

(b) The forces on the two sides of the loop are equal in magnitude and opposite in direction, canceling each other, and do not affect the state of balance of the system. (c)

B = 0.26 T

66.

(a)

5.00 cm

68.

3.92 × 10 −2 T

70.

(a)

6.2 m s 2

(b)

0.40 s

72.

(a)

opposite directions

(b)

68 A

74.

0.59 T

76.

(a) 1.00 × 10 −5 T

(b) 8.00 × 10 −5 N directed toward wire 1

(c) 1.60 × 10 −5 T

(d) 8.00 × 10 −5 N directed toward wire 2

(b) 8.78 × 10 6 m s

PROBLEM SOLUTIONS 19.1

19.2

Remember that the direction of the magnetic force exerted on the negatively charged electron is opposite to the direction predicted by right-hand rule number 1. The magnetic field near the Earth’s equator is horizontal and directed toward the north. The magnetic force experienced by a moving charged particle is always perpendicular to the plane formed by the vectors representing the magnetic field and the particle’s velocity. (a)

When the velocity of a positively charged particle is downward, right-hand rule number 1 predicts a magnetic force toward the east. Hence, the force experienced by the negatively charged electron (and also the deflection of its velocity) is directed toward the west .

(b)

When the particle moves northward, its velocity is parallel to the magnetic field, and it will experience zero force and zero deflection .

(c)

The direction of the force on the negatively charged electron (and the deflection of its velocity) will be vertically upward .

(d)

The direction of the force on the negatively charged electron (and the deflection of its velocity) will be vertically downward .

(a)

For a positively charged particle, the direction of the force is that predicted by the righthand rule number one. These are: (a⬘)

in plane of page and to left

(b⬘)

into the page

(c⬘)

out of the page

(d⬘)

in plane of page and toward the top

(e⬘)

into the page

(f⬘)

out of the page

continued on next page

68719_19_ch19_p120-147.indd 126

1/7/11 2:43:44 PM

Magnetism

(b)

19.3

For a negatively charged particle, the direction of the force is exactly opposite what the right-hand rule number 1 predicts for positive charges. Thus, the answers for part (b) are reversed from those given in part (a) .

Since the particle is positively charged, use the right-hand rule number 1. In this case, start with the fingers of the right hand in the direction of v and the thumb pointing in the direction of F. As you start closing the hand, the fingers point in the direction of B after they have moved 90°. The results are (a)

19.4

into the page

(b)

toward the right

(c)

toward bottom of page

Hold the right hand with the fingers in the direction of v so that as you close your hand, the fingers move toward the direction of B. The thumb will point in the direction of the force (and hence the deflection) if the particle has a positive charge. The results are (a)

toward top of page

(c) q = 180° ⇒ zero force 19.5

127

(a)

(b)

out of the page , since the charge is negative

(d)

into the page

The proton experiences maximum force when it moves perpendicular to the magnetic field, and the magnitude of this maximum force is Fm ax = qvB sin 90° = (1.60 × 10 −19 C ) ( 6.00 × 10 6 m s ) (1.50 T )(1) = 1.44 × 10 −12 N

19.6

Fm ax 1.44 × 10 −12 N = = 8.62 × 1014 m s 2 mp 1.67 × 10 −27 kg

(b)

am ax =

(c)

Since the magnitude of the charge of an electron is the same as that of a proton, the force experienced by the electron would have the same magnitude but would be in the opposite direction due to the negative charge of the electron.

(d)

The acceleration of the electron would have a much greater magnitude than that of the proton because of the mass of the electron is much smaller. +y

Since the acceleration (and hence the magnetic force) is in the positive x-direction, the magnetic field must be in the negative y-direction (see sketch at the right) according to right-hand rule number 1.

F

+x

v

The magnitude of the magnetic field is found from Fm = qvB sinq as B=

or

B=

yielding

68719_19_ch19_p120-147.indd 127

+z

B

Fm ma = qv sinq qv sinq

(1.67 × 10

(1.60 × 10

−19

−27

kg ) ( 2.00 × 1013 m s 2 )

C ) (1.00 × 10 7 m s ) sin 90.0°

= 2.09 × 10 −2 T

B = 2.09 ×10 −2 T in the negative y-direction .

1/7/11 2:43:46 PM

128

19.7

Chapter 19

The gravitational force is small enough to be ignored, so the magnetic force must supply the needed centripetal acceleration. Thus, m

v2 q Br = qvB sin 90°, or v = , where r = RE + 1000 km = 7.38 × 10 6 m m r

v=

(1.60 × 10

−19

C ) ( 4.00 × 10 −8 T ) ( 7.38 × 10 6 m ) 1.67 × 10 −27 kg

= 2.83 × 10 7 m s

If v is toward the west and B is northward, F will be directed downward as required. 19.8

The speed attained by the electron is found from

v= (a)

2e ( ΔV ) = m

1 2 mv = q ( ΔV ), or 2

2 (1.60 × 10 −19 C ) ( 2 400 V ) 9.11× 10 −31 kg

= 2.90 × 10 7 m s

Maximum force occurs when the electron moves perpendicular to the field. Fm ax = q vB sin 90° = (1.60 × 10 −19 C ) ( 2.90 × 10 7 m s )(1.70 T ) = 7.90 × 10 −12 N

(b)

Minimum force occurs when the electron moves parallel to the field. Fm in = q vB sin 0° = 0

19.9

The magnetic force experienced by a moving charged particle has magnitude Fm = qvB sinq , where q is the angle between the directions of the particle’s velocity and the magnetic field. Thus, sinq = and

19.10

Fm 8.20 × 10 −13 N = = 0.754 qvB (1.60 × 10 −19 C ) ( 4.00 × 10 6 m s )(1.70 T )

q = sin −1 ( 0.754 ) = 48.9°

or

q = 180° − 48.9° = 131°

The force on a single ion is F1 = qvB sinq = (1.60 × 10 −19 C ) ( 0.851 m s )( 0.254 T ) sin ( 51.0° ) = 2.69 × 10 −20 N The total number of ions present is ions ⎞ ⎛ 3 22 N = ⎜ 3.00 × 10 20 ⎟ (100 cm ) = 3.00 × 10 ⎝ cm 3 ⎠ Thus, assuming all ions move in the same direction through the field, the total force is F = N ⋅ F1 = ( 3.00 × 10 22 ) ( 2.69 × 10 −20 N ) = 807 N

19.11

Gravitational force: Fg = mg = ( 9.11× 10 −31 kg ) ( 9.80 m s 2 ) = 8.93 × 10 −30 N downward

continued on next page

68719_19_ch19_p120-147.indd 128

1/7/11 2:43:48 PM

Magnetism

129

Electric force: Fe = qE = ( −1.60 × 10 −19 C ) ( −100 N C ) = 1.60 × 10 −17 N upward Magnetic force: Fm = qvB sinq = ( −1.60 × 10 −19 C ) ( 6.00 × 10 6 m s ) ( 50.0 × 10 −6 T ) sin ( 90.0° ) = 4.80 × 10 −17 N in direction opposite right hand rule prediction Fm = 4.80 × 10 −17 N downward 19.12

19.13

Hold the right hand with the fingers in the direction of the current so, as you close the hand, the fingers move toward the direction of the magnetic field. The thumb then points in the direction of the force. The results are (a)

to the left

(b)

into the page

(c)

out of the page

(d)

toward top of page

(e)

into the page

(f )

out of the page

From F = BIL sinq , the magnetic field is B=

F L 0.12 N m = = 8.0 × 10 −3 T I sinq (15 A ) sin 90°

The direction of B must be the + z-direction to have F in the –y-direction when I is in the +x-direction. 19.14

19.15

(a)

F = BIL sinq = ( 0.28 T )( 3.0 A )( 0.14 m ) sin 90° = 0.12 N

(b)

Neither the direction of the magnetic field nor that of the current is given . Both must be known before the direction of the force can be determined. In this problem, you can only say that the force is perpendicular to both the wire and the field.

Use the right-hand rule number 1, holding your right hand with the fingers in the direction of the current and the thumb pointing in the direction of the force. As you close your hand, the fingers will move toward the direction of the magnetic field. The results are (a)

19.16

into the page

(b)

toward the right

(c)

toward the bottom of the page

In order to just lift the wire, the magnetic force exerted on a unit length of the wire must be directed upward and have a magnitude equal to the weight per unit length. That is, the magnitude is F

⎛ m⎞ = BI sinq = ⎜ ⎟ g ⎝ ⎠

giving

⎛ m⎞ g B=⎜ ⎟ ⎝ ⎠ I sinq

To find the minimum possible field, the magnetic field should be perpendicular to the current

(q = 90.0°). Then,

⎡ g g ⎛ 1 kg ⎞ ⎛ 10 2 cm ⎞ ⎤ 9.80 m s 2 ⎛ m⎞ Bm in = ⎜ ⎟ = 0.245 T = ⎢ 0.500 ⎥ ⎝ ⎠ I sin 90.0° ⎣ cm ⎜⎝ 10 3 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ ⎦ ( 2.00 A )(1) To find the direction of the field, hold the right hand with the thumb pointing upward (direction of the force) and the fingers pointing southward (direction of current). Then, as you close the hand, the fingers point eastward. The magnetic field should be directed eastward .

68719_19_ch19_p120-147.indd 129

1/7/11 2:43:50 PM

130

Chapter 19

19.17

F = BIL sinq = ( 0.300 T )(10.0 A )( 5.00 m ) sin ( 30.0° ) = 7.50 N

19.18

(a)

The magnitude is F = BIL sinq = ( 0.60 × 10 − 4 T )(15 A )(10.0 m ) sin ( 90° ) = 9.0 × 10 −3 N F is perpendicular to B. Using the right-hand rule number 1, the orientation of F is found to be 15° above the horizontal in the northward direction .

(b)

F = BIL sinq = ( 0.60 × 10 − 4 T )(15 A )(10.0 m ) sin (165° ) = 2.3 × 10 −3 N and, from the right-hand rule number 1, the direction is horizontal and due west .

19.19

For minimum field, B should be perpendicular to the wire. If the force is to be northward, the field must be directed downward . To keep the wire moving, the magnitude of the magnetic force must equal that of the kinetic friction force. Thus, BIL sin 90° = m k ( mg ), or B=

19.20

(a)

2 m k ( m L ) g ( 0.200 )(1.00 g cm ) ( 9.80 m s ) ⎛ 1 kg ⎞ ⎛ 10 2 cm ⎞ = ⎜⎝ 10 3 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ = 0.131 T I sin 90° (1.50 A )(1.00 )

To have zero tension in the wires, the magnetic force per unit length must be directed upward and equal to the weight per unit length of the conductor. Thus, Fm L I=

19.21

= BI =

mg , or L

( m L ) g = ( 0.040 B

kg m ) ( 9.80 m s 2 ) 3.60 T

= 0.109 A

(b)

From the right-hand rule number 1, the current must be to the right if the force is to be upward when the magnetic field is into the page.

(a)

The magnetic force must be directed upward and its magnitude must equal mg , the weight of the wire. Then, the net force acting on the wire will be zero, and it can move upward at constant speed.

(b)

The magnitude of the magnetic force must be BIL sinq = mg and for minimum field q = 90°. Thus, Bm in =

2 mg ( 0.015 kg ) ( 9.80 m s ) = = 0.20 T IL ( 5.0 A )( 0.15 m )

For the magnetic force to be directed upward when the current is toward the left, B must be directed out of the page . (c)

68719_19_ch19_p120-147.indd 130

If the field exceeds 0.20 T, the upward magnetic force exceeds the downward force of gravity, so the wire accelerates upward .

1/7/11 2:43:52 PM

Magnetism

19.22

As shown in end view in the sketch at the right, the rod (and hence, the current) is horizontal, while the magnetic field is vertical. For the rod to move with constant velocity (zero acceleration), it is necessary that

n = mg Fm

fk

ΣFx = Fm − fk = BI sin 90.0° − m k n = 0 B=

or

reducing to 19.23

mg

m ( mg ) mk n = k I sin 90.0° Id (1.00 ) B = m k mg Id

For each segment, the magnitude of the force is given by F = BIL sinq , and the direction is given by the right-hand rule number 1. The results of applying these to each of the four segments, with B = 0.020 0 T and I = 5.00 A, are summarized below. Segment

L (m)

q

F (N)

Direction

ab

0.400

180°

0

no direction

bc

0.400

90.0° 0.040 0 negative x

cd

0.400 2

45.0° 0.040 0 negative z

0.400 2

parallel to xzplane at 45° to 90.0° 0.056 6 both +x- and +z-directions

da

19.24

131

y B

a

d

z

I

c

b

x

B

The magnitude of the force is F = BIL sinq = ( 5.0 × 10 −5 N ) ( 2.2 × 10 3 A )( 58 m ) sin 65° = 5.8 N and the right-hand rule number 1 shows its direction to be into the page .

19.25

The torque exerted on a single turn current loop in a magnetic field is given by t = BIAsinq , where A is the area enclosed by the loop and q is the angle between the line normal to the plane of the loop and the direction of the magnetic field B. For maximum torque, q = 90.0°, so we have ⎛ pd2 ⎞ 1 2 sin 90.0° = ( 3.00 × 10 −3 T )( 5.00 A )p ( 0.100 m ) (1.00 ) t m ax = BI ⎜ 4 ⎝ 4 ⎟⎠ = 1.18 × 10 −4 N ⋅ m = 118 × 10 −6 N ⋅ m = 118 mN ⋅ m

19.26

The magnitude of the torque is t = NBIAsinq , where q is the angle between the field and the line perpendicular to the plane of the loop. The circumference of the loop is 2p r = 2.00 m, so the 1.00 m 1 2 radius is r = and the area is A = p r 2 = m. p p Thus,

68719_19_ch19_p120-147.indd 131

⎛1 ⎞ t = (1)( 0.800 T ) (17.0 × 10 −3 A ) ⎜ m 2 ⎟ sin 90.0° = 4.33 × 10 −3 N ⋅ m ⎝p ⎠

1/7/11 2:43:55 PM

132

19.27

Chapter 19

The area of the loop is A = p ab, where a = 0.200 m and b = 0.150 m. Since the field is parallel to the plane of the loop, q = 90.0° and the magnitude of the torque is t = NBIAsinq = 8 ( 2.00 × 10 −4 T )( 6.00 A )[p ( 0.200 m )( 0.150 m )] sin 90.0° = 9.05 × 10 −4 N ⋅ m The torque is directed to make the left-hand side of the loop move toward you and the right-hand side move away.

19.28

Note that the angle between the field and the line perpendicular to the plane of the loop is q = 90.0° − 30.0° = 60.0°. Then, the magnitude of the torque is t = NBIAsinq = 100 ( 0.800 T )(1.20 A )[( 0.400 m )( 0.300 m )] sin 60.0° = 9.98 N ⋅ m With current in the –y-direction, the outside edge of the loop will experience a force directed out of the page (+z-direction) according to the right-hand rule number 1. Thus, the loop will rotate clockwise as viewed from above .

19.29

(a)

The torque exerted on a coil by a uniform magnetic field is t = NBIAsinq , with maximum torque occurring when q = 90°. Thus, the current in the coil must be I=

(b)

t m ax 0.15 N ⋅ m = 0.56 A = NBA ( 200 )( 0.90 T ) ⎡( 3.0 × 10 −2 m ) ( 5.0 × 10 −2 m ) ⎤ ⎣ ⎦

If I has the value found above and q is now 25°, the torque on the coil is t = NBIAsinq = ( 200 )( 0.90 T )( 0.56 A )[( 0.030 m )( 0.050 m )] sin 25° = 0.064 N ⋅ m

19.30

The resistance of the loop is R=

−8 rL (1.70 × 10 Ω ⋅ m )(8.00 m ) = = 1.36 × 10 −3 Ω A 1.00 × 10 −4 m 2

and the current in the loop is I =

ΔV 0.100 V = = 73.5 A. R 1.36 × 10 −3 Ω

The magnetic field exerts torque t = NBIAsinq on the loop, and this is a maximum when sinq = 1. The wire forms a square loop with each side 2.00 m long. Thus, t m ax = NBIA = (1)( 0.400 T )( 73.5 A )( 2.00 m ) = 118 N ⋅ m 2

19.31

(a)

Let q be the angle the plane of the loop makes with the horizontal as shown in the sketch at the right. Then, the angle it makes with the vertical is f = 90.0° −q . The number of turns on the loop is N=

L 4.00 m = = 10.0 circumference 4 ( 0.100 m )

The torque about the z-axis due to gravity is ⎛s ⎞ t g = mg ⎜ cosq ⎟ , where s = 0.100 m is the length ⎝2 ⎠ of one side of the loop. This torque tends to rotate the loop clockwise. The torque due to the magnetic force tends to rotate the loop counterclockwise about

continued on next page

68719_19_ch19_p120-147.indd 132

1/7/11 2:43:57 PM

Magnetism

133

the z-axis and has magnitude t m = NBIAsinq . At equilibrium, t m = t g, or NBI ( s 2 ) sinq = mg ( s cosq ) 2. This reduces to tanq =

( 0.100 kg) (9.80 m s2 ) mg = = 14.4 2NBIs 2 (10.0 )( 0.010 0 T )( 3.40 A )( 0.100 m )

Since tanq = tan ( 90.0° − f ) = cotf , the angle the loop makes with the vertical at equilibrium is f = cot −1 (14.4 ) = 3.97° . (b)

At equilibrium, t m = NBI ( s 2 ) sinq = (10.0 )( 0.010 0 T )( 3.40 A )( 0.100 m ) sin ( 90.0° − 3.97° ) = 3.39 × 10 −3 N ⋅ m 2

19.32

(a)

The current in segment ab is in the + y-direction. Thus, by right-hand rule 1, the magnetic force on it is in the +x-direction . This force is parallel to the x-axis and therefore has zero torque about that axis.

(b)

The current in segment cd is in the − y-direction, and the right-hand rule 1 gives the direction of the magnetic force as the −x-direction . This force is parallel to the x-axis and can exert no torque about that axis.

(c)

No . These two forces are equal in magnitude and opposite in directions, so their sum is zero. Further, each force has zero torque about the axis the loop is hinged on. Since the two forces cancel each other and are both parallel to the rotation axis, they can have no effect on the motion of the loop.

(d)

The magnetic force is perpendicular to both the direction of the current in bc (the x-axis) and the magnetic field. As given by right-hand rule 1, this places it in the yz-plane at 130° counterclockwise from the +y-axis . The force acting on segment bc tends to rotate it counterclockwise about the x-axis, so the torque is in the +x direction .

(e)

The loop tends to rotate counterclockwise about the x-axis .

(f )

m = IAN = ( 0.900 A )[( 0.500 m )( 0.300 m )](1) = 0.135 A ⋅ m 2

(g)

19.33

The magnetic moment vector is perpendicular to the plane of the loop (the xy-plane) and is therefore parallel to the z-axis. Because the current flows clockwise around the loop, the magnetic moment vector is directed downward, in the negative z-direction. This means that the angle between it and the direction of the magnetic field is q = 90.0° + 40.0° = 130° .

(h)

t = mB sinq = ( 0.135 A ⋅ m 2 )(1.50 T ) sin (130° ) = 0.155 N ⋅ m

(a)

From KE = 12 me v 2, the speed of the electron is v=

2 ( KE ) = me

2 ( 3.30 × 10 −19 J ) 9.11× 10 −31 kg

= 8.51× 10 5 m

continued on next page

68719_19_ch19_p120-147.indd 133

1/7/11 2:43:59 PM

134

Chapter 19

(b)

The magnetic force acting on the electron must provide the necessary centripetal acceleration. Thus, me v 2 r = qvB sinq , which gives r=

(9.11 × 10−31 kg) (8.51× 105 m s) me v = qB sinq (1.60 × 10 −19 C )( 0.235 T ) sin 90.0°

= 2.06 × 10 −5 m = 20.6 × 10 −6 m = 20.6 mm 19.34

19.35

(a)

F = qvB sinq = (1.60 × 10 −19 C ) ( 5.02 × 10 6 m s )( 0.180 T ) sin ( 60.0° ) = 1.25 × 10 −13 N

(b)

a=

F 1.25 × 10 −13 N = = 7.49 × 1013 m s 2 m 1.67 × 10 −27 kg

For the particle to pass through with no deflection, the net force acting on it must be zero. Thus, the magnetic force and the electric force must be in opposite directions and have equal magnitudes. This gives Fm = Fe , or qvB = qE, which reduces to v = E B

19.36

The speed of the particles emerging from the velocity selector is v = E B (see Problem 35). In the deflection chamber, the magnetic force supplies the centripetal acceleration, so qvB =

mv 2 r

r=

or

mv m ( E B ) mE = = 2 qB qB qB

Using the given data, the radius of the path is found to be

( 2.18 × 10 (1.60 × 10

kg ) ( 950 V m )

−26

r= 19.37

−19

C )( 0.930 T )

2

= 1.50 × 10 −4 m = 0.150 mm

1 From conservation of energy, ( KE + PE ) f = ( KE + PE )i, we find that mv 2 + qV f = 0 + qVi , or the 2 speed of the particle is

(

2 q Vi − V f

v=

m

)=

2 q ( ΔV ) = m

2 (1.60 × 10 −19 C )( 250 V ) 2.50 × 10 −26 kg

= 5.66 × 10 4 m s

The magnetic force supplies the centripetal acceleration giving qvB = or 19.38

r=

−26 4 mv ( 2.50 × 10 kg ) ( 5.66 × 10 m s ) = = 1.77 × 10 −2 m = 1.77 cm −19 qB 1.60 × 10 C 0.500 T ( ) ( )

Since the centripetal acceleration is furnished by the magnetic force acting on the ions, mv 2 mv qvB = , or the radius of the path is r = . Thus, the distance between the impact points qB r (that is, the difference in the diameters of the paths followed by the U 238 and the U 235 isotopes) is Δd = 2 ( r238 − r235 ) = = or

68719_19_ch19_p120-147.indd 134

mv 2 r

2v ( m238 − m235 ) qB

2 ( 3.00 × 10 5 m s )

(1.60 × 10

−19

kg ⎞ ⎤ ⎡ ( 238 u − 235 u ) ⎛⎜⎝ 1.66 × 10 −27 ⎟ ⎢ u ⎠ ⎥⎦ C )( 0.600 T ) ⎣

Δd = 3.11× 10 −2 m = 3.11 cm

1/7/11 2:44:02 PM

Magnetism

19.39

135

In the perfectly elastic, head-on collision between the a -particle and the initially stationary proton, conservation of momentum requires that m p v p + ma va = ma v0 while conservation of kinetic energy also requires that v0 − 0 = − va − v p or v p = va + v0 . Using the fact that ma = 4m p and combining these equations gives

(

(

)

(

)

)

m p ( va + v0 ) + 4m p va = 4m p v0 and Thus,

va = 3v0 5

or

v p = ( 3v0 5) + v0 = 8v0 5 va =

3 3⎛ 5 ⎞ 3 v0 = ⎜ v p ⎟ = v p 5 5⎝8 ⎠ 8

After the collision, each particle follows a circular path in the horizontal plane with the magnetic force supplying the centripetal acceleration. If the radius of the proton’s trajectory is R, and that of the alpha particle is r, we have qp vp B = mp

19.40

R va2 r

or

R=

or

r=

mp vp qp B

=

mp vp eB

(

)(

)

4m p 3v p 8 ma va 3 3 ⎛ mp vp ⎞ = R = = ⎜ ⎟ 4 qa B 4 ⎝ eB ⎠ ( 2e ) B

and

qa va B = ma

(a)

A charged particle follows a circular path when it moves perpendicular to the magnetic field. The magnetic force acting on the particle provides the required centripetal acceleration. Therefore, F = qvB sin 90° = m v 2 R. Since the kinetic energy is KE = mv 2 2, we rewrite the force as F = qvB sin 90° = 2 ( KE ) R, and solving for the speed v gives v = 2 ( KE ) qBR .

(b)

From KE = mv 2 2, the mass of the particle is m=

19.41

v 2p

(a)

2 ( KE ) 2 ( KE ) q2 B2 R2 = = 2 2 v 2 ( KE ) 4 ( KE ) q 2 B 2 R 2

Within the velocity selector, the electric and magnetic fields exert forces in opposite directions on charged particles passing through. For particles having a particular speed, these forces have equal magnitudes, and the particles pass through without deflection. The selected speed is found from Fe = qE = qvB = Fm, giving v = E B. In the deflection chamber, the selected particles follow a circular path having a diameter of d = 2r = 2mv qB. Thus, the mass to charge ratio for these particles is Bd B 2 d ( 0.093 1 T ) ( 0.396 m ) m Bd = = = = 2.08 × 10 −7 kg C = q 2v 2 ( E B ) 2E 2 (8 250 V m ) 2

(b)

If the particle is doubly ionized (i.e., two electrons have been removed from the neutral atom), then q = 2e, and the mass of the ion is ⎛ m⎞ m = ( 2e ) ⎜ ⎟ = 2 (1.60 × 10 −19 C ) ( 2.08 × 10 −7 kg C ) = 6.66 × 10 −26 kg ⎝ q⎠

(c)

Assuming this is an element, the mass of the ion should be approximately equal to the atomic weight multiplied by the atomic mass unit (see Table C.5 in Appendix C of the textbook). This would give the atomic weight as At. wt. ≈

68719_19_ch19_p120-147.indd 135

m 6.66 × 10 −26 kg = = 40.1 , suggesting that the element is calcium . 1 u 1.66 × 10 −27 kg

1/7/11 2:44:05 PM

136

19.42

Chapter 19

Consider a particle of mass m and charge q accelerated from rest through a potential difference ΔV = Vi − V f , Applying conservation of energy gives KE f + PE f = 0 + PEi

KE f =

or

1 2 mv = PEi − PE f = qVi − qV f = q ( ΔV ) 2

or the speed given the particle is v = 2q ( ΔV ) m. If the particle now enters a magnetic field of strength B, moving perpendicular to the direction of the field, it will follow a circular path of radius r=

mv m = qB qB

2q ( ΔV ) m

which reduces to

r=

2m ( ΔV ) qB 2

[1]

For a proton (mass m p and charge q = e), Equation [1] gives rp = (a)

2m p ( ΔV ) eB 2

For the deuteron (mass md = 2m p and charge qd = e), Equation [1] gives rd =

(b)

2ma ( ΔV ) = qa B 2

(

)

2 4m p ( ΔV )

( 2e ) B

2

⎛ 2m p ( ΔV ) ⎞ = 2⎜ ⎟= eB 2 ⎝ ⎠

2 ⋅rp

−7 4 m 0 I ( 4p × 10 T ⋅ m A ) (1.00 × 10 A ) = = 2.00 × 10 −5 T = 20.0 mT 2p r 2p (100 m )

toward the left

(b)

out of page

(c)

lower left to upper right

The magnetic field at distance r from a long conducting wire is B = m 0 I 2p r. Thus, if B = 1.0 × 10 −15 T at r = 4.0 cm, the current must be I=

19.46

2 ⋅rp

Imagine grasping the conductor with the right hand so the fingers curl around the conductor in the direction of the magnetic field. The thumb then points along the conductor in the direction of the current. The results are (a)

19.45

eB 2

⎛ 2m p ( ΔV ) ⎞ = 2⎜ ⎟= eB 2 ⎝ ⎠

Treat the lightning bolt as a long, straight conductor. Then, the magnetic field is B=

19.44

)

For the alpha particle (mass ma = 4m p and charge qa = 2e), we find ra =

19.43

(

2 2m p ( ΔV )

2md ( ΔV ) = qd B 2

−15 2p rB 2p ( 0.040 m )(1.0 × 10 T ) = = 2.0 × 10 −10 A m0 4p × 10 −7 T ⋅ m A

Model the tornado as a long, straight, vertical conductor and imagine grasping it with the right hand so the fingers point northward on the western side of the tornado (that is, at the observatory’s location.) The thumb is directed downward, meaning that the conventional current is downward or negative charge flows upward .

continued on next page

68719_19_ch19_p120-147.indd 136

1/7/11 2:44:08 PM

Magnetism

137

The magnitude of the current is found from B = m 0 I 2p r as I= 19.47

3 −8 2p rB 2p ( 9.00 × 10 m ) (1.50 × 10 T ) = = 675 A m0 4p × 10 −7 T ⋅ m A

From B = m 0 I 2p r, the required distance is −7 m 0 I ( 4p × 10 T ⋅ m A )( 20 A ) r= = = 2.4 × 10 −3 m = 2.4 mm −3 2p B 2p (1.7 × 10 T )

19.48

Assume that the wire on the right is wire 1 and that on the left is wire 2. Also, choose the positive direction for the magnetic field to be out of the page and negative into the page. (a)

At the point half way between the two wires, ⎡m I m m I ⎤ Bnet = −B1 − B2 = − ⎢ 0 1 + 0 2 ⎥ = − 0 2p r 2p r 2p 1 2 ⎦ ⎣ =− or

(b)

At point P1, Bnet = +B1 − B2 =

m0 2p

⎡ I1 I 2 ⎤ m 0 ⎡ I I ⎤ m0 ⎡ I ⎤ ⎢+ − ⎥ = ⎢⎣ + d − 2d ⎥⎦ = 2p ⎢⎣ + 2d ⎥⎦ r r 2p 2 ⎦ ⎣ 1

4p × 10 −7 T ⋅ m A ⎡ 5.00 A ⎤ −6 ⎥ = 5.00 × 10 T = 5.00 mT out of page ⎢ 2p ⎣ 2 ( 0.100 m ) ⎦

At point P2, Bnet = −B1 + B2 = Bnet =

19.49

T ⋅ m A ) ⎡ 4 ( 5.00 A ) ⎤ −5 ⎢ 0.100 m ⎥ = − 4.00 × 10 T 2p ⎦ ⎣ −7

Bnet = 40.0 mT into the page

Bnet =

(c)

( 4p × 10

⎡ I m 0 ⎡ 4I ⎤ I ⎤ ⎢ d 2 + d 2 ⎥ = − 2p ⎢ d ⎥ ⎣ ⎦ ⎣ ⎦

m0 2p

⎡ I1 I 2 ⎤ m 0 ⎡ I I ⎤ m0 ⎡ I ⎤ − + ⎢− + ⎥ = ⎢ ⎥= ⎢ ⎥ ⎣ r1 r2 ⎦ 2p ⎣ 3d 2d ⎦ 2p ⎣ 6d ⎦

4p ×10 −7 T ⋅ m A ⎡ 5.00 A ⎤ −6 ⎢ ⎥ = 1.67 ×10 T = 1.67 mT out of page 2p ⎣ 6 ( 0.100 m ) ⎦

The distance from each wire to point P is given by r = 12 s 2 + s 2 = s 2, where s = 0.200 m. At point P, the magnitude of the magnetic field produced by each of the wires is −7 m 0 I ( 4p × 10 T ⋅ m A )( 5.00 A ) 2 B= = = 7.07 mT 2p r 2p ( 0.200 m )

Carrying currents into the page, the field A produces at P is directed to the left and down at –135°, while B creates a field to the right and down at – 45°. Carrying currents toward you, C produces a field downward and to the right at – 45°, while D’s contribution is down and to the left at –135°. The horizontal components of these equal magnitude contributions cancel in pairs, while the vertical components all add. The total field is then Bnet = 4 ( 7.07 mT ) sin 45.0° = 20.0 mT toward the bottom of the page

68719_19_ch19_p120-147.indd 137

1/7/11 2:44:11 PM

138

19.50

Chapter 19

Wire 1 has current I1 = 3.00 A and wire 2 has I 2 = 5.00 A, with both currents directed out of the page. Choose the line running from wire 1 to wire 2 as the positive x-direction. (a)

At the point midway between the wires, the field due to each wire is parallel to the y-axis, and the net field is Bnet = +B1 y − B2 y = m 0 ( I1 − I 2 ) ( 2p d 2 )

(b)

( 4p × 10

−7

T ⋅ m A)

( 3.00 A − 5.00 A ) = − 4.00 × 10 −6 T

Thus,

Bnet =

or

Bnet = 4.00 mT toward the bottom of the page

2p ( 0.100 m )

At point P, B1 is directed at q1 = +135°. The magnitude of B1 is 4p × 10 −7 T ⋅ m A )( 3.00 A ) ( m 0 I1 m 0 I1 B1 = = = = 2.12 mT 2p r1 2p d 2 2p 0.200 2 m

(

)

(

)

The contribution from wire 2 is in the –x-direction and has magnitude B2 =

−7 m 0 I 2 m 0 I 2 ( 4p × 10 T ⋅ m A )( 5.00 A ) = = 5.00 mT = 2p r2 2p d 2p ( 0.200 m )

Therefore, the components of the net field at point P are Bx = B1 cos135° + B2 cos180° = ( 2.12 mT ) cos135° + ( 5.00 mT ) cos180° = −6.50 mT and

By = B1 sin135° + B2 sin180° = ( 2.12 mT ) sin135° + 0 = +1.50 mT

Thus, Bnet = Bx2 + By2 = 6.67 mT at ⎛ Bx ⎞ ⎛ 6.50 mT ⎞ q = tan −1 ⎜ = tan −1 ⎜ = 77.0° ⎟ ⎝ 1.50 mT ⎟⎠ ⎝ By ⎠ or 19.51

Bnet = 6.67 mT upward at 77.0° to the left of vertical

Call the wire along the x-axis wire 1 and the other wire 2. Also, choose the positive direction for the magnetic fields at point P to be out of the page. At point P, Bnet = +B1 − B2 = or

Bnet

( 4p × 10 =

−7

2p

m 0 I1 m 0 I 2 m 0 ⎛ I1 I 2 ⎞ − = − 2p r1 2p r2 2p ⎜⎝ r1 r2 ⎟⎠

T ⋅ m A ) ⎛ 7.00 A 6.00 A ⎞ −7 − ⎜⎝ ⎟ = +1.67 × 10 T 3.00 m 4.00 m ⎠

Bnet = 0.167 mT out of the page

68719_19_ch19_p120-147.indd 138

1/7/11 2:44:13 PM

Magnetism

19.52

139

(a)

Imagine the horizontal xy-plane being perpendicular to the page, with the positive x-axis coming out of the page toward you and the positive y-axis toward the right edge of the page. Then, the vertically upward positive z-axis is directed toward the top of the page. With the current in the wire flowing in the positive x-direction, the right-hand rule 2 gives the direction of the magnetic field above the wire as being toward the left, or in the −y-direction .

(b)

With the positively charged proton moving in the –x-direction (into the page), right-hand rule 1 gives the direction of the magnetic force on the proton as being directed toward the top of the page or upward, in the positive z-direction .

(c)

Since the proton moves with constant velocity, a zero net force acts on it. Thus, the magnitude of the magnetic force must equal that of the gravitational force .

(d)

ΣFz = maz = 0 ⇒ Fm = Fg or qvB = mg, where B = m 0 I 2p d . This gives qvm 0 I 2p d = mg, or the distance the proton is above the wire must be d = qvm 0 I 2p mg .

(e)

d=

−19 4 −7 −6 qvm 0 I (1.60 × 10 C ) ( 2.30 × 10 m s ) ( 4p × 10 T ⋅ m A ) (1.20 × 10 A ) = 2p mg 2p (1.67 × 10 −27 kg ) ( 9.80 m s 2 )

d = 5.40 × 10 −2 m = 5.40 cm 19.53

(a)

From B = m 0 I 2p r, observe that the field is inversely proportional to the distance from the conductor. Thus, the field will have one-tenth its original value if the distance is increased by a factor of 10. The required distance is then r ′ = 10r = 10 ( 0.400 m ) = 4.00 m .

(b)

A point in the plane of the conductors and 40.0 cm from the center of the cord is located 39.85 cm from the nearer wire and 40.15 cm from the far wire. Since the currents are in opposite directions, so are their contributions to the net field. Therefore, Bnet = B1 − B2, or Bnet =

−7 ⎞ m 0 I ⎛ 1 1 ⎞ ( 4p × 10 T ⋅ m A ) ( 2.00 A ) ⎛ 1 1 − − = ⎜ ⎜ ⎟ 2p ⎝ r1 r2 ⎠ 2p ⎝ 0.398 5 m 0.401 5 m ⎟⎠

= 7.50 × 10 −9 T = 7.50 nT (c)

Call r the distance from cord center to field point P and 2d = 3.00 mm the distance between centers of the conductors. Bnet =

m0 I ⎛ 1 1 ⎞ m 0 I ⎛ 2d ⎞ − ⎜⎝ ⎟= ⎜ ⎟ 2p r − d r + d ⎠ 2p ⎝ r 2 − d 2 ⎠

7.50 × 10 −10 T = so

( 4p × 10

−7

T ⋅ m A ) ( 2.00 A ) ⎛ 3.00 × 10 −3 m ⎞ ⎜⎝ r 2 − 2.25 × 10 −6 m 2 ⎟⎠ 2p

r = 1.26 m

The field of the two-conductor cord is weak to start with and falls off rapidly with distance. (d)

68719_19_ch19_p120-147.indd 139

The cable creates zero field at exterior points, since a loop in Ampère’s law encloses zero total current.

1/7/11 2:44:15 PM

140

19.54

Chapter 19

(a)

Point P is equidistant from the two wires which carry identical currents. Thus, the contributions of the two wires, B upper and Blower, to the magnetic field at P will have equal magnitudes. The horizontal components of these contributions will cancel, while the vertical components add. The resultant field will be vertical in the + y-direction .

(b)

The distance of each wire from point P is r = x 2 + d 2 , and the cosine of the angle that B upper and Blower make with the vertical is cosq = x r . The magnitude of either B upper or Blower is Bwire = m 0 I 2p r, and the vertical components of either of these contributions have values of ⎛ m0 I ⎞

m 0 Ix

( B ) = ( B ) cosq = ⎜⎝ 2p r ⎟⎠ rx = 2p r wire

y

wire

2

The magnitude of the resultant field at point P is then BP = 2 ( Bwire ) y =

19.55

m 0 Ix m0 I x = 2 pr p (x2 + d 2 )

(c)

The point midway between the two wires is the origin (0, 0). From the above result for part (b), the resultant field at this midpoint is BP x = 0 = 0 . This is as expected, because right-hand rule 2 shows that at the midpoint the field due to the upper wire is toward the right, while that due to the lower wire is toward the left. Thus, the two fields cancel, yielding a zero resultant field.

(a)

The magnetic force per unit length on each of two parallel wires separated by the distance d and carrying currents I1 and I 2 has the magnitude F

=

m 0 I1 I 2 2p d

In this case, we have F

19.56

=

( 4p × 10

−7

T ⋅ m A )(1.25 A )( 3.50 A )

2p ( 2.50 × 10 −2 m )

= 3.50 × 10 −5 N m

(b)

The magnetic forces two parallel wires exert on each other are attractive if their currents are in the same direction and repulsive if the currents flow in opposite directions. In this case, the currents in the two wires are in opposite directions, so the forces are repulsive .

(a)

The force per unit length that parallel conductors exert on each other is F = m 0 I1 I 2 2p d. Thus, if F = 2.0 × 10 −4 N m, I1 = 5.0 A, and d = 4.0 cm, the current in the second wire must be I2 =

(b)

2p ( 4.0 × 10 −2 m ) 2p d ⎛ F ⎞ = ( 2.0 × 10 −4 N m ) = 8.0 A ⎜ ⎟ m 0 I1 ⎝ ⎠ ( 4p × 10 −7 T ⋅ m A )( 5.0 A )

Since parallel conductors carrying currents in the same direction attract each other (see Section 19.8 in the textbook), the currents in these conductors which repel each other must be in opposite directions.

continued on next page

68719_19_ch19_p120-147.indd 140

1/7/11 2:44:18 PM

Magnetism

(c)

19.57

141

The result of reversing the direction of either of the currents would be that the force of interaction would change from a force of repulsion to an attractive force . The expression for the force per unit length, F = m 0 I1 I 2 2p d, shows that doubling either of the currents would double the magnitude of the force of interaction .

In order for the system to be in equilibrium, the repulsive magnetic force per unit length on the top wire must equal the weight per unit length of this wire. Thus,

F

= m 0 I1 I 2 2p d = 0.080 N m, and the distance between the wires will be 4p × 10 −7 T ⋅ m A )( 60.0 A )( 30.0 A ) ( m 0 I1 I 2 d= = 2p ( 0.080 N m ) 2p ( 0.080 N m ) = 4.5 × 10 −3 m = 4.5 mm

19.58

The magnetic forces exerted on the top and bottom segments of the rectangular loop are equal in magnitude and opposite in direction. Thus, these forces cancel, and we only need consider the sum of the forces exerted on the right and left sides of the loop. Choosing to the left (toward the long, straight wire) as the positive direction, the sum of these two forces is Fnet = +

or

Fnet =

m 0 I1 I 2 m II m I I ⎛1 1 ⎞ − 0 1 2 = 0 1 2 ⎜ − ⎟ 2p c 2p ( c + a ) 2p ⎝ c c + a ⎠

( 4p × 10

−7

T ⋅ m A )( 5.00 A )(10.0 A )( 0.450 m ) ⎛ 1 1 ⎞ − ⎜⎝ ⎟ 2p 0.100 m 0.250 m ⎠

= + 2.70 × 10 −5 N = 2.70 × 10 −5 N to the left 19.59

The magnetic field inside a solenoid which carries current I is given by B = m 0 nI, where n = N is the number of turns of wire per unit length. Thus, the current in the windings of this solenoid must be 1.00 × 10 −4 T )( 0.400 m ) ( B B I= = = 3.18 × 10 −2 A = 31.8 mA = m 0 n m 0 N ( 4p × 10 −7 T ⋅ m A ) (10 3 )

19.60

The magnetic field inside of a solenoid is B = m 0 nI = m 0 ( N L ) I. Thus, the number of turns on this solenoid must be N=

19.61

(a)

BL ( 9.0 T )( 0.50 m ) = = 4.8 × 10 4 turns m 0 I ( 4p × 10 −7 T ⋅ m A )( 75 A )

From R = rL A, the required length of wire to be used is 2 −3 ⎡ ⎤ R ⋅ A ( 5.00 Ω ) ⎣p ( 0.500 × 10 m ) 4 ⎦ L= = = 58 m r 1.7 × 10 −8 Ω ⋅ m

The total number of turns on the solenoid (that is, the number of times this length of wire will go around a 1.00-cm radius cylinder) is N=

L 58 m = = 9.2 × 10 2 = 920 2p r 2p (1.00 × 10 −2 m ) continued on next page

68719_19_ch19_p120-147.indd 141

1/7/11 2:44:20 PM

142

Chapter 19

(b)

From B = m 0 nI, the number of turns per unit length on the solenoid is n=

B 4.00 × 10 −2 T = = 7.96 × 10 3 turns m m 0 I ( 4p × 10 −7 T ⋅ m A )( 4.00 A )

Thus, the required length of the solenoid is = 19.62

N 9.2 ×10 2 turns = 0.12 m = 12 cm = n 7.96 ×10 3 turns m

The wire used to wind the solenoid has diameter d wire = 0.100 cm = 1.00 × 10 −3 m. Thus, when wound in a single layer with adjacent turns touching each other, the number of turns per unit length on the solenoid is n = 1.00 × 10 3 per meter. The total number of turns on the solenoid, N = n , will then be N = (1.00 × 10 3 m )( 0.750 m ) = 7.50 × 10 2 If the magnetic field inside the solenoid is to be B = 8.00 mT, the required current in the windings is I=

B 8.00 × 10 −3 T = = 6.37 A m 0 n ( 4p × 10 −7 T ⋅ m A ) (1.00 × 10 3 m )

The length of wire required for the solenoid is L = N ⋅ ( solenoid circumference ), or L = N (p dsolenoid ) = ( 7.50 × 10 2 ) p ( 0.100 m ) = 75.0p meters and its resistance will be R=

4 (1.70 × 10 −8 Ω ⋅ m )( 75.0p m ) rC u L rC u L = = = 5.10 Ω 2 2 A p ( d wire 4) p (1.00 × 10 −3 m )

The power that must be delivered to the solenoid is then P = I 2 R = ( 6.37 A ) ( 5.10 Ω ) = 207 W 2

19.63

(a)

The magnetic force supplies the centripetal acceleration, so qvB = mv 2 r . The magnetic field inside the solenoid is then found to be B=

(b)

−31 4 mv ( 9.11× 10 kg ) (1.0 × 10 m s ) = = 2.8 × 10 −6 T = 2.8 mT qr (1.60 × 10 −19 C) ( 2.0 × 10−2 m )

From B = m 0 nI, the current is the solenoid is found to be I=

B 2.8 × 10 −6 T = −7 m 0 n ( 4p × 10 T ⋅ m A ) ⎡⎣( 25 turns cm ) (100 cm 1 m ) ⎤⎦

= 8.9 × 10 −4 A = 0.89 mA

68719_19_ch19_p120-147.indd 142

1/7/11 2:44:22 PM

Magnetism

19.64

(a)

143

When switch S is closed, a total current NI (current I in a total of N conductors) flows toward the right through the lower side of the coil. This results in a downward force of magnitude Fm = B ( NI ) w being exerted on the coil by the magnetic field, with the requirement that the balance exert a upward force F ′ = mg on the coil to bring the system back into balance. In order for the magnetic force to be downward, the right-hand rule number 1 shows that the magnetic field must be directed out of the page toward the reader. For the system to be restored to balance, it is necessary that Fm = F ′

19.65

or

B ( NI ) w = mg, giving B = mg NIw

(b)

The magnetic field exerts forces of equal magnitudes and opposite directions on the two sides of the coil. These forces cancel each other and do not affect the balance of the coil. Hence the dimension of the sides is not needed.

(c)

B=

(a)

The magnetic field at the center of a circular current loop of radius R and carrying current I is B = m 0 I 2R. The direction of the field at this center is given by right-hand rule number 2. Taking out of the page (toward the reader) as positive, the net magnetic field at the common center of these coplanar loops has magnitude

−3 2 mg ( 20.0 × 10 kg ) ( 9.80 m s ) = = 0.26 T NIw ( 50 )( 0.30 A ) ( 5.0 × 10 −2 m )

Bnet = B2 − B1 =

−7 m 0 I 2 m 0 I1 ( 4p × 10 T ⋅ m A ) 3.00 A 5.00 A − = − −2 2r2 2r1 2 9.00 × 10 m 12.0 × 10 −2 m

= −5.24 × 10 −6 T = 5.24 mT (b)

Since we chose out of the page as the positive direction, and now find that Bnet < 0, we conclude the net magnetic field at the center is into the page .

(c)

To have Bnet = 0, it is necessary that I 2 r2 = I1 r1, or ⎛I ⎞ ⎛ 3.00 A ⎞ r2 = ⎜ 2 ⎟ r1 = ⎜ (12.0 cm ) = 7.20 cm ⎝ 5.00 A ⎟⎠ ⎝ I1 ⎠

19.66

The angular momentum of a point mass moving in a circular path is ⎛ v⎞ L = Iw = ( mr 2 ) ⎜ ⎟ = mvr ⎝r⎠ where m is the mass of the particle, v is its speed, and r is the radius of its path. (a)

The magnetic force experienced by the moving electron supplies the needed centripetal acceleration, so ⎛ v2 ⎞ m ⎜ ⎟ = qvB sin 90.0° ⎝ r ⎠

or

mv = qBr

Thus, L = mvr = ( qBr ) r = qBr 2, and the radius of the path must be

continued on next page

68719_19_ch19_p120-147.indd 143

1/7/11 2:44:25 PM

144

Chapter 19

(b)

The speed of the electron may now be found from L = mvr as v=

19.67

4.00 × 10 −25 kg ⋅ m 2 s = 5.00 × 10 −2 m = 5.00 cm −19 −3 C 1.00 × 10 T 1.60 × 10 ( )( )

L = qB

r=

L 4.00 × 10 −25 kg ⋅ m 2 s = = 8.78 × 10 6 m s mr ( 9.11× 10 −31 kg ) ( 5.00 × 10 −2 m )

Assume wire 1 is along the x-axis and wire 2 is along the y-axis. (a)

Choosing out of the page as the positive field direction, the field at point P is B = B1 − B2 =

−7 m 0 ⎛ I1 I 2 ⎞ ( 4p × 10 T ⋅ m A ) ⎛ 5.00 A 3.00 A ⎞ − − = ⎜⎝ ⎟ ⎜ ⎟ 2p ⎝ r1 r2 ⎠ 2p 0.400 m 0.300 m ⎠

= 5.00 × 10 −7 T = 0.500 mT out of the page (b)

At 30.0 cm above the intersection of the wires, the field components are as shown at the right, where By = −B1 = −

m 0 I1 2p r

( 4p × 10 =−

−7

T ⋅ m A )( 5.00 A )

2p ( 0.300 m )

and

Bx = B2 =

= −3.33 × 10 −6 T

−7 m 0 I 2 ( 4p × 10 T ⋅ m A )( 3.00 A ) = = 2.00 × 10 −6 T 2p r 2p ( 0.300 m )

With Bz = 0, the resultant field is parallel to the xy-plane and B = Bx2 + By2 = 3.88 × 10 −6 T or 19.68

⎛ By ⎞ q = tan −1 ⎜ ⎟ = −59.0° ⎝ Bx ⎠

B = 3.88 mT parallel to the xy-plane and 59.0° clockwise from the +x-direction

For the rail to move at constant velocity, the net force acting on, it must be zero. Thus, the magnitude of the magnetic force must equal that of the friction force, giving BIL = m k ( mg ), or B=

19.69

at

(a)

2 m k ( mg ) ( 0.100 )( 0.200 kg ) ( 9.80 m s ) = = 3.92 × 10 −2 T IL (10.0 A )( 0.500 m )

Since the magnetic field is directed from N to S (that is, from left to right within the artery), positive ions with velocity in the direction of the blood flow experience a magnetic deflection toward electrode A. Negative ions will experience a force deflecting them toward electrode B. This separation of charges creates an electric field directed from A toward B. At equilibrium, the electric force caused by this field must balance the magnetic force, so qvB = qE = q ( ΔV d )

continued on next page

68719_19_ch19_p120-147.indd 144

1/7/11 2:44:27 PM

Magnetism

or

19.70

160 × 10 −6 V ΔV = = 1.33 m s Bd ( 0.040 0 T ) ( 3.00 × 10 −3 m )

v=

(b)

The magnetic field is directed from N to S. If the charge carriers are negative moving in the direction of v, the magnetic force is directed toward point B. Negative charges build up at point B, making the potential at A higher than that at B. If the charge carriers are positive moving in the direction of v, the magnetic force is directed toward A, so positive charges build up at A. This also makes the potential at A higher than that at B. Therefore the sign of the potential difference does not depend on the charge of the ions .

(a)

The magnetic force acting on the wire is directed upward and of magnitude Fm = BIL sin 90° = BI. ay =

Thus,

ay = (b)

ΣFy m

=

Fm − mg BIL − mg BI = = − g, or m m (m L )

( 4.0 × 10

−3

5.0 × 10

T ) ( 2.0 A )

−4

kg m

− 9.80 m s 2 = 6.2 m s 2

Using Δy = v0 y t + 12 a y t 2 , with v0 y = 0, gives t=

19.71

145

2 ( Δy ) = ay

2 ( 0.50 m ) = 0.40 s 6.2 m s 2

Label the wires 1, 2, and 3, as shown in Figure 1. Also, let B1 , B2 , and B3 , respectively, represent the magnitudes of the fields produced by the currents in those wires, and observe that q = 45°.

(

)

At point A, B1 = B2 = m 0 I 2p a 2 , or B1 = B2 =

( 4p × 10

−7

T ⋅ m A ) ( 2.0 A )

2p ( 0.010 m ) 2

= 28 mT FIGURE 1

( 4p × 10 T ⋅ m A ) ( 2.0 A ) = 13 mT m0 I = 2p ( 3a ) 2p ( 0.030 m ) −7

and

B3 =

These field contributions are oriented as shown in Figure 2. Observe that the horizontal components of B1 and B2 cancel while their vertical components add to B3. The resultant field at point A is then BA = ( B1 + B2 ) cos 45° + B3 = 53 mT, or

FIGURE 2

B A = 53 mT directed toward the bottom of the page At point B, B1 = B2 =

−7 m 0 I ( 4p × 10 T ⋅ m A ) ( 2.0 A ) = = 40 mT 2p a 2p ( 0.010 m )

m0 I = 20 mT. These contributions are oriented as 2p ( 2a ) shown in Figure 3. Thus, the resultant field at B is and B3 =

FIGURE 3

continued on next page

68719_19_ch19_p120-147.indd 145

1/7/11 2:44:29 PM

146

Chapter 19

B B = B3 = 20 mT directed toward the bottom of the page

(

)

At point C, B1 = B2 = m 0 I 2p a 2 , while B3 = m 0 I 2p a. These contributions are oriented as shown in Figure 4. Observe that the horizontal components of B1 and B2 cancel, while their vertical components add to oppose B3 . The magnitude of the resultant field at C is FIGURE 4

m I ⎛ 2sin 45° ⎞ −1⎟ = 0 BC = ( B1 + B2 ) sin 45° − B3 = 0 ⎜ 2p a ⎝ ⎠ 2 19.72

(a)

Since one wire repels the other, the currents must be in opposite directions .

(b)

Consider a free-body diagram of one of the wires as shown at the right. ΣFy = 0 ⇒ T cos8.0° = mg or

T=

mg cos8.0°

⎛ mg ⎞ ΣFx = 0 ⇒ Fm = T sin8.0° = ⎜ sin8.0° ⎝ cos8.0° ⎟⎠ or

Fm = ( mg ) tan8.0°. Thus, I=

m0 I 2 L = ( mg ) tan8.0°, which gives 2p d

d ⎡⎣( m L ) g ⎤⎦ tan8.0° m 0 2p

Observe that the distance between the two wires is d = 2 [( 6.0 cm ) sin8.0° ] = 1.7 cm, so I= 19.73

(1.7 × 10

−2

m ) ( 0.040 kg m ) ( 9.80 m s 2 ) tan8.0° 2.0 × 10 −7 T ⋅ m A

= 68 A

Note: We solve part (b) before part (a) for this problem. (b)

Since the magnetic force supplies the centripetal acceleration for this particle, qvB = mv 2 r, or the radius of the path is r = mv qB = p qB , where p = mv = 2m(KE) = 2 (1.67 × 10 −27 kg ) ( 5.00 × 10 6 eV ) (1.60 × 10 −19 J eV ) = 5.17 × 10 −20 kg ⋅ m s Consider the circular path shown at the right and observe that the desired angle is ⎡ (1.00 m ) qB ⎤ ⎛ 1.00 m ⎞ a = sin −1 ⎜ = sin −1 ⎢ ⎥ ⎝ r ⎟⎠ p ⎣ ⎦ or

⎡ (1.00 m )(1.60 × 10 −19 C ) ( 0.050 0 T ) ⎤ a = sin −1 ⎢ ⎥ = 8.90° 5.17 × 10 −20 kg ⋅ m s ⎢⎣ ⎥⎦

continued on next page

68719_19_ch19_p120-147.indd 146

1/7/11 2:44:31 PM

Magnetism

(a)

147

The linear momentum of the particle has constant magnitude p = mv, and its vertical component as the particle leaves the field is py = − psina , or py = − ( 5.17 × 10 −20 kg ⋅ m s ) sin (8.90° ) = − 8.00 × 10 −21 kg ⋅ m s

19.74

The force constant of the spring system is found from the elongation produced by the weight acting alone. F mg = x1 x1

k=

where

x1 = 0.50 cm

The total force stretching the springs when the field is turned on is ΣFy = Fm + mg = kx total

where

x total = x1 + 0.30 cm = 0.80 cm

Thus, the downward magnetic force acting on the wire is ⎛ mg ⎞ ⎛x ⎞ Fm = kx total − mg = ⎜ x total − mg = ⎜ total − 1⎟ mg ⎟ ⎝ x1 ⎠ ⎝ x1 ⎠ ⎛ 0.80 cm ⎞ =⎜ − 1 10.0 × 10 −3 kg ) ( 9.80 m s 2 ) = 5.9 × 10 −2 N ⎝ 0.50 cm ⎟⎠ ( Since the magnetic force is given by Fm = BIL sin 90°, the magnetic field is B= 19.75

(12 Ω )( 5.9 × 10 −2 N ) Fm Fm = = 0.59 T = IL ( ΔV R ) L ( 24 V ) ( 5.0 × 10 −2 m )

The magnetic force is very small in comparison to the weight of the ball, so we treat the motion as that of a freely falling body. Then, as the ball approaches the ground, it has velocity components with magnitudes of vx = v0 x = 20.0 m s, and v y = v02 y + 2a y ( Δy ) = 0 + 2 ( −9.80 m s 2 )( −20.0 m ) = 19.8 m s The velocity of the ball is perpendicular to the magnetic field and, just before it reaches the ground, has magnitude v = vx2 + v y2 = 28.1 m s. Thus, the magnitude of the magnetic force is Fm = qvB sinq = ( 5.00 × 10 −6 C ) ( 28.1 m s ) ( 0.010 0 T ) sin 90.0° = 1.41× 10 −6 N

19.76

(a)

(b)

(c)

(d)

68719_19_ch19_p120-147.indd 147

B1 = F21

= B1 I 2 = (1.00 × 10 −5 T )(8.00 A ) = 8.00 × 10 −5 N directed toward wire 1

B2 = F12

−7 m 0 I1 ( 4p × 10 T ⋅ m A )( 5.00 A ) = = 1.00 × 10 −5 T 2p d 2p ( 0.100 m )

−7 m 0 I 2 ( 4p × 10 T ⋅ m A )(8.00 A ) = = 1.60 × 10 −5 T 2p d 2p ( 0.100 m )

= B2 I1 = (1.60 × 10 −5 T )( 5.00 A ) = 8.00 × 10 −5 N directed toward wire 2

1/7/11 2:44:34 PM

20 Induced Voltages and Inductance QUICK QUIZZES 1.

b, c, a. At each instant, the magnitude of the induced emf is proportional to the magnitude of the rate of change of the magnetic field (hence, proportional to the absolute value of the slope of the curve shown on the graph).

2.

Choice (c). Taking downward as the normal direction, the north pole approaching from above produces an increasing positive flux through the area enclosed by the loop. To oppose this, the induced current must generate a negative flux through the interior of the loop, or the induced magnetic field must point upward through the enclosed area of the loop. Imagine gripping the wire of the loop with your right hand so the fingers curl upward through the area enclosed by the loop. You will find that your thumb, indicating the direction of the induced current, is directed counterclockwise around the loop as viewed from above.

3.

Choice (b). If the positive z-direction is chosen as the normal direction, the increasing counterclockwise current in the left-hand loop produces an increasing positive flux through the area enclosed by this loop. The magnetic field lines due to this current curl around and pass through the area of the xy-plane outside the left-hand loop in the negative direction. Thus, the right-hand loop has an increasing negative flux through it. To counteract this effect, the induced current must produce positive flux, or generate a magnetic field in the positive z-direction, through the area enclosed by the right-hand loop. Imagine gripping the wire of the right-hand loop with the right hand so the fingers point in the positive z-direction as they penetrate the area enclosed by this loop. You should find that the thumb is directed counterclockwise around the loop as viewed from above the xy-plane.

4.

Choice (a). All charged particles within the metal bar move straight downward with the bar. According to right-hand rule number 1, positive changes moving downward through a magnetic field that is directed northward will experience magnetic forces toward the east. This means that the free electrons (negative charges) within the metal will experience westward forces and will drift toward the west end of the bar, leaving the east end with a net positive charge.

5.

Choice (b). According to Equation 20.3, when B and v are constant, the emf depends only on the length of the wire cutting across the magnetic field lines. Thus, you want the long dimension of the rectangular loop perpendicular to the velocity vector. This means that the short dimension is parallel to the velocity vector, and (b) is the correct choice. From a more conceptual point of view, you want the rate of change of area in the magnetic field to be the largest, which you do by thrusting the long dimension into the field.

6.

Choice (b). When the iron rod is inserted into the solenoid, the inductance of the coil increases. As a result, more potential difference appears across the coil than before. Consequently, less potential difference appears across the bulb, and its brightness decreases.

148

68719_20_ch20_p148-170.indd 148

1/7/11 2:45:21 PM

Induced Voltages and Inductance

149

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

The flux through a flat coil in a uniform magnetic field is a maximum when the plane of the coil is perpendicular to the direction of the field. Thus, with the field parallel to the y-axis, the flux is a maximum when the plane of the coil is parallel to the xz-plane, and (c) is the correct choice.

2.

Choose the positive z-direction to be the reference direction (q = 0°) for the normal to the plane of the coil. Then, the change in flux through the coil is

(

)

ΔΦ B = B f cosq f − Bi cosq i A = [( 3.0 T ) cos 0° − (1.0 T ) cos180° ] ( 0.50 m ) = 1.0 Wb 2

and the magnitude of the induced emf is e =N

ΔΦ B Δt

⎛ 1.0 Wb ⎞ = 5.0 V = (10 ) ⎜ ⎝ 2.0 s ⎟⎠

The correct answer is choice (b). 3.

The magnitude of the voltage drop across the inductor is eL = L

ΔI = ( 5.00 H ) ( 2.00 A s ) = 10.0 V Δt

and (d) is the correct choice. 4.

The angular velocity of the rotating coil is w = 10.0 rev s = 20p rad s and the maximum emf induced in the coil is

(

)

e max = NBAw = (100 ) ( 0.050 0 T ) 0.100 m 2 ( 20p rad s ) = 31.4 V showing the correct choice to be (a). 5.

The motional emf induced in a conductor of length moving at speed v through a magnetic field of magnitude B is ΔV = B⊥ v = ( B sinq ) v, where B⊥ = B sinq is the component of the field perpendicular to the velocity of the conductor. In the described situation,

(

)

ΔV = B⊥ v = ( B sinq ) v = ⎡⎣ 60.0 × 10 −6 T sin 60.0° ⎤⎦ (12 m ) ( 60.0 m s ) = 3.7 × 10

−2

V = 37 mV

and the correct choice is (c). 6.

As the bar slides to the right along the rails, the magnetic flux through the conducting path formed by the bar, rails, and the resistive element at the left end is directed out of the page and increasing in magnitude. Thus, the induced current must generate a flux directed into the page through the area enclosed by the current path. This means the induced current must be in the clockwise direction and choice (b) is correct. Also, as the induced current flows, the rod will experience a magnetic force that tends to impede the motion of the rod. Therefore, an external force must be exerted on the bar to keep it moving at constant speed, and choice (d) is also correct.

7.

The amplitude of the induced emf in the coil of a generator is directly proportional to the angular velocity of the coil (e max = NBAw ). Therefore, when the rate of rotation is doubled, the amplitude of the induced emf is also doubled, and the correct choice is (b).

68719_20_ch20_p148-170.indd 149

1/7/11 2:45:21 PM

150

Chapter 20

8.

An emf is induced in the coil by any action which causes a change in the flux through the coil. The actions described in choices (a), (b), (d), and (e) all change the flux through the coil and induce an emf. However, moving the coil through the constant field without changing its orientation with respect to the field will not cause a change of flux. Thus, choice (c) is the correct answer.

9.

With the current in the long wire flowing in the direction shown in Figure MCQ20.9, the magnetic flux through the rectangular loop is directed into the page. If this current is decreasing in time, the change in the flux is directed opposite to the flux itself (or out of the page). The induced current will then flow clockwise around the loop, producing a flux directed into the page through the loop and opposing the change in flux due to the decreasing current in the long wire. The correct choice for this question is (b).

10.

A current flowing counterclockwise in the outer loop of Figure MCQ20.10 produces a magnetic flux through the inner loop that is directed out of the page. If this current is increasing in time, the change in the flux is in the same direction as the flux itself (or out of the page). The induced current in the inner loop will then flow clockwise around the loop, producing a flux through the loop directed into the page, opposing the change in flux due to the increasing current in the outer loop. The correct answer is choice (b).

11.

As the bar magnet approaches the loop from above, with its south end downward as shown in Figure MCQ20.11, magnetic flux through the area enclosed by the loop is directed upward and increasing in magnitude. To oppose this increasing upward flux, the induced current in the loop will flow clockwise, as seen from above, producing a flux directed downward through the area enclosed by the loop. After the bar magnet has passed through the plane of the loop and is departing with its north end upward, a decreasing flux is directed upward through the loop. To oppose this decreasing upward flux, the induced current in the loop flows counterclockwise as seen from above, producing flux directed upward through the area enclosed by the loop. From this analysis, we see that (a) is the only true statement among the listed choices.

12.

With the magnetic field perpendicular to the plane of the page in Figure MCQ20.12, the flux through the closed loop to the left of the bar is given by Φ B = BA, where B is the magnitude of the field and A is the area enclosed by the loop. Any action which produces a change in this product, BA, will induce a current in the loop and cause the bulb to light. Such actions include increasing or decreasing the magnitude of the field (B) and moving the bar to the right or left and changing the enclosed area A. Thus, the bulb will light during all of the actions in choices (a), (b), (c), and (d).

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

Consider the copper tube to be a large set of rings stacked one on top of the other. As the magnet falls toward or falls away from each ring, a current is induced in the ring. Thus, there is a current in the copper tube around its circumference.

4.

(a)

The flux is calculated as Φ B = BA cosq = B⊥ A. The flux is therefore maximum when the magnetic field vector is perpendicular to the plane of the loop.

(b)

The flux is zero when the magnetic field is parallel to the plane of the loop (and hence, q = 90°, making B⊥ = B cosq = 0).

6.

68719_20_ch20_p148-170.indd 150

As water falls, it gains velocity and kinetic energy. It then pushes against the blades of a turbine, transferring this energy to the rotor or coil of a large generator. The rotor moves in a strong external magnetic field and a voltage is induced in the coil. This induced emf is the voltage source for the current in our electric power lines.

1/7/11 2:45:23 PM

Induced Voltages and Inductance

8.

151

No. Once the bar is in motion and the charges are separated, no external force is necessary to maintain the motion. During the initial acceleration of the bar, an external applied force will be necessary to overcome both the inertia of the bar and a retarding magnetic force exerted on the bar.

10.

Let us assume the north pole of the magnet faces the ring. As the bar magnet falls toward the conducting ring, a magnetic field is induced in the ring pointing upward. This upward directed field will oppose the motion of the magnet, preventing it from moving as a freely falling body. Try it for yourself to show that an upward force also acts on the falling magnet if the south end faces the ring.

12.

The increasing counterclockwise current in the solenoid coil produces an upward magnetic field that increases rapidly. The increasing upward flux of this field through the ring induces an emf to produce a clockwise current in the ring. At each point on the ring, the field of the solenoid has a radially outward component as well as an upward component. This radial field component exerts an upward force on the current at each point in the ring. The resultant magnetic force on the ring is upward and exceeds the weight of the ring. Thus, the ring accelerates upward off of the solenoid.

14.

B I

Bsolenoid

F

F Bring

I

Iring Bsolenoid increasing Isolenoid

As the magnet moves at high speed past the fixed coil, the magnetic flux through the coil changes very rapidly, increasing as the magnet approaches the coil and decreasing as the magnet moves away. The rapid change in flux through the coil induces a large emf, large enough to cause a spark across the gap in the spark plug.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a) 1.00 × 10 −7 T ⋅ m 2

4.

zero

6.

(a) 6.98 mT

8.

0.10 mV

(b) 8.66 × 10 −8 T ⋅ m 2

0

(c)

right to left

(b) 1.97 × 10 −5 T ⋅ m 2

10.

34 mV

12.

2.26 mV

14.

into the page

16.

(a) left to right

(b)

18.

(a) 1.88 × 10 −7 T ⋅ m 2

(b) 6.27 × 10 −8 V

20.

8.8 A

68719_20_ch20_p148-170.indd 151

(c)

no induced current

1/7/11 2:45:24 PM

152

Chapter 20

e = NB0p r 2 t

(a)

24.

1.20 mV, west end is positive

26.

(a) counterclockwise when viewed from the right end

28.

(b)

clockwise

22.

(b)

clockwise when viewed from the right end

(c)

no induced current

(c)

I = B0p r 2 t R

(a) 6.0 m T (b) The magnitude and direction of the Earth’s field varies from one location to another, so the induced voltage in the wire will change.

30.

1.00 m s

32.

13 mV

34.

(a) 7.5 kV (b)

when the plane of the coil is parallel to the magnetic field

36.

(a) 8.0 A

(b)

3.2 A

38.

(a) 5.9 mH

(b)

−24 V

40.

See Solution.

42.

1.92 × 10 −5 T ⋅ m 2

44.

(a)

1.00 kΩ

(b)

3.00 ms

46.

(a)

0

(b)

3.8 V

(c) 6.0 V

(d)

2.2 V

(a) 2.00 ms

(b)

0.176 A

(d)

3.22 ms

48.

(c)

1.50 A

(c)

60 V

50.

(a) 4.44 mH

(b)

0.555 mJ

52.

(a) 1.3 Ω

(b)

4.8 × 10 2 turns

(c)

0.48 m

(f)

3.6 A

(d)

0.76 mH

(e)

0.46 ms

(g)

3.2 ms

(h)

4.9 mJ

54.

(a)

increasing

(b) 62.2 mT s

56.

(a) 0.73 m s , counterclockwise (c)

68719_20_ch20_p148-170.indd 152

(b)

0.65 mW

Work is being done on the bar by an external force to maintain constant speed.

1/7/11 2:45:24 PM

Induced Voltages and Inductance

58.

(a)

2.1 × 10 6 m s

(b)

153

(c) 1.7 × 1010 V

from side to side

(d) The very large induced emf would lead to powerful spontaneous electrical discharges. The strong electric and magnetic fields would disrupt the flow of ions in their bodies. 60.

0.158 mV

62.

0.120 A, clockwise

64.

1.60 A

66.

(a)

e = NB v

(b)

I = NB v R

(c)

P = N 2 B2 2 v2 R

(d)

F = N 2 B2 2 v R

(e)

clockwise

(f)

toward the left

PROBLEM SOLUTIONS 20.1

The angle between the direction of the constant field and the normal to the plane of the loop is q = 0°, so

(

)(

)

Φ B = BA cosq = ( 0.50 T ) ⎡⎣ 8.0 × 10 −2 m 12 × 10 −2 m ⎤⎦ cos 0° = 4.8 × 10 −3 T ⋅ m 2 20.2

The magnetic flux through the loop is given by Φ B = BA cosq , where B is the magnitude of the magnetic field, A is the area enclosed by the loop, and q is the angle the magnetic field makes with the normal to the plane of the loop. Thus, 2 ⎡ ⎛ 10 −2 m ⎞ ⎤ Φ B = BA cosq = 5.00 × 10 −5 T ⎢ 20.0 cm 2 ⎜ ⎥ cosq = 1.00 × 10 −7 T ⋅ m 2 cosq ⎝ 1 cm ⎟⎠ ⎥⎦ ⎢⎣

(

20.3

)

(

)

(a)

When B is perpendicular to the plane of the loop, q = 0° and Φ B = 1.00 × 10 −7 T ⋅ m 2 .

(b)

If q = 30.0°, then Φ B = 1.00 × 10 −7 T ⋅ m 2 cos 30.0° = 8.66 × 10 −8 T ⋅ m 2 .

(c)

If q = 90.0°, then Φ B

( = (1.00 × 10

−7

T ⋅ m2

) ) cos 90.0° =

0.

Φ B = BA cosq = B(p r 2 ) cosq , where q is the angle between the direction of the field and the normal to the plane of the loop. (a)

If the field is perpendicular to the plane of the loop, q = 0°, and B=

(b)

ΦB 8.00 × 10 −3 T ⋅ m 2 = = 0.177 T 2 p r cosq p ( 0.12 m ) cos 0°

( ) 2

If the field is directed parallel to the plane of the loop, q = 90°, and Φ B = BA cosq = BA cos 90° = 0

20.4

68719_20_ch20_p148-170.indd 153

The magnetic field lines are everywhere parallel to the surface of the cylinder, so no magnetic field lines penetrate the cylindrical surface. The total flux through the cylinder is zero .

1/7/11 2:45:26 PM

154

20.5

20.6

Chapter 20

(a)

Every field line that comes up through the area A on one side of the wire goes back down through area A on the other side of the wire. Thus, the net flux through the coil is zero .

(b)

The magnetic field is parallel to the plane of the coil, so q = 90.0°. Therefore, Φ B = BA cosq = BA cos 90.0° = 0 .

(a)

The magnitude of the field inside the solenoid is ⎛ N⎞ ⎛ 400 ⎞ B = m 0 nI = m 0 ⎜ ⎟ I = 4p × 10 −7 T ⋅ m A ⎜ ( 5.00 A ) ⎝ ⎠ ⎝ 0.360 m ⎟⎠

(

)

= 6.98 × 10 −3 T = 6.98 mT (b)

The field inside a solenoid is directed perpendicular to the cross-sectional area, so q = 0° and the flux through a loop of the solenoid is

( ) T ) p ( 3.00 × 10

Φ B = BA cosq = B p r 2 cosq

(

= 6.98 × 10 −3 20.7

(a)

−2

)

2

m cos 0° = 1.97 × 10 −5 T ⋅ m 2

The magnetic flux through an area A may be written as Φ B = ( B cosq ) A = ( component of B perpendicular to A ) ⋅ A Thus, the flux through the shaded side of the cube is

(

Φ B = Bx ⋅ A = ( 5.0 T ) ⋅ 2.5 × 10 −2 m (b)

20.8 20.9

20.10

68719_20_ch20_p148-170.indd 154

)

2

= 3.1 × 10 −3 T ⋅ m 2

Unlike electric field lines, magnetic field lines always form closed loops, without beginning or end. Therefore, no magnetic field lines originate or terminate within the cube and any line entering the cube at one point must emerge from the cube at some other point. The net flux through the cube, and indeed through any closed surface, is zero .

(

)

2 −3 ⎡ ⎤ ΔΦ B ( ΔB ) A cosq (1.5 T − 0 ) ⎣p 1.6 × 10 m ⎦ cos 0° e = = = = 1.0 × 10 −4 V = 0.10 mV Δt Δt 120 × 10 −3 s

(a)

As loop A moves parallel to the long straight wire, the magnetic flux through loop A does not change. Hence, there is no induced current in this loop.

(b)

As loop B moves to the left away from the straight wire, the magnetic flux through this loop is directed out of the page, and is decreasing in magnitude. To oppose this change in flux, the induced current flows counterclockwise around loop B producing a magnetic flux directed out of the page through the area enclosed by loop B.

(c)

As loop C moves to the right away from the straight wire, the magnetic flux through this loop is directed into the page and is decreasing in magnitude. In order to oppose this change in flux, the induced current flows clockwise around loop C producing a magnetic flux directed into the page through the area enclosed by loop C.

2 ΔΦ B B ΔA cosq ( 0.15 T ) ⎡⎣p ( 0.12 m ) − 0 ⎤⎦ cos 0° e = = = = 3.4 × 10 −2 V = 34 mV Δt Δt 0.20 s

1/7/11 2:45:28 PM

Induced Voltages and Inductance

20.11

155

The magnitude of the induced emf is e =

ΔΦ B Δ ( B cosq ) A = Δt Δt

If the normal to the plane of the loop is considered to point in the original direction of the magnetic field, then q i = 0° and q f = 180°. Thus, we find e = 20.12

1.5 s

= 9.4 × 10 −2 V = 94 mV

With the field directed perpendicular to the plane of the coil, the flux through the coil is Φ B = BA cos 0° = BA. As the magnitude of the field decreases, the magnitude of the induced emf in the coil is e =

20.13

( 0.20 T ) cos180° − ( 0.30 T ) cos 0° p ( 0.30 m )2

ΔΦ B ΔB 2 = A = ( 0.050 0 T s ) ⎡⎣p ( 0.120 m ) ⎤⎦ = 2.26 × 10 −3 V = 2.26 mV Δt Δt

The magnetic field changes from zero to a magnitude of 2.5 T, directed at 45° to the plane of the band, in a time of Δt = 0.18 s. If the band has a diameter of 6.5 cm, the magnitude of the average induced emf in the metal band is

(

−2 ΔΦ B ΔB ( 2.5 T − 0 ) 6.5 × 10 m e = = A cosq = p Δt Δt 0.18 s 4

)

2

cos 45° = 3.3 × 10 −2 V = 33 mV

20.14

When the switch is closed, the magnetic field due to the current from the battery will be directed toward the left along the axis of the cylinder. To oppose this increasing leftward flux, the induced current in the other loop must produce a field directed to the right through the area it encloses. Thus, the induced current is directed out of the page through the resistor.

20.15

(a)

When the magnet moves to the left, the flux through the interior of the coil is directed toward the right and is decreasing in magnitude. To oppose this change in flux, the magnetic field generated by the induced current should be directed to the right along the axis of the coil. The current must then be left to right through the resistor.

(b)

When the magnet moves to the right, the flux through the interior of the coil is directed toward the right and is increasing in magnitude. To oppose this increasing flux, the magnetic field generated by the induced current should be directed toward the left along the axis of the coil. The current must then be right to left through the resistor.

20.16

68719_20_ch20_p148-170.indd 155

When the switch is closed, the current from the battery produces a magnetic field directed toward the right along the axis of both coils. (a)

As the battery current is growing in magnitude, the induced current in the rightmost coil opposes the increasing rightward directed field by generating a field toward to the left along the axis. Thus, the induced current must be left to right through the resistor.

(b)

Once the battery current, and the field it produces, have stabilized, the flux through the rightmost coil is constant, and there is no induced current .

(c)

As the switch is opened, the battery current and the field it produces rapidly decrease in magnitude. To oppose this decrease in the rightward directed field, the induced current must produce a field toward the right along the axis, so the induced current is right to left through the resistor.

1/7/11 2:45:30 PM

156

20.17

20.18

Chapter 20

(a)

The current is zero . The magnetic flux the current produces through the right side of the loop is directed into the page, and is equal in magnitude to the outward-directed flux the current produces through the left side of the loop. Thus, the net flux through the loop has a constant value of zero and does not induce a current.

(b)

The flux through the loop due to the long wire is directed out of the page and is increasing in magnitude. To oppose this increasing outward flux, the induced current must generate a magnetic field that is directed into the page through the area enclosed by the loop. Thus, the induced current in the loop must be clockwise .

The initial magnetic field inside the solenoid is ⎛ N⎞ ⎛ 100 ⎞ B = m 0 nI = m 0 ⎜ ⎟ I = 4p × 10 −7 T ⋅ m A ⎜ ( 3.00 A ) = 1.88 × 10 −3 T ⎝ ⎠ ⎝ 0.200 m ⎟⎠

(

(

)(

)

2

(a)

Φ B = BAloop cosq = 1.88 × 10 −3 T 1.00 × 10 −2 m cos 0° = 1.88 × 10 −7 T ⋅ m 2

(b)

When the current is zero, the flux through the loop is Φ B = 0, and the average induced emf has been e =

20.19

)

(a)

ΔΦ B Δt

=

0 − 1.88 × 10 −7 T ⋅ m 2 3.00 s

= 6.27 × 10 −8 V

The initial field inside the solenoid is ⎛ 300 ⎞ Bi = m 0 nI i = 4p × 10 −7 T ⋅ m A ⎜ ( 2.00 A ) = 3.77 × 10 −3 T ⎝ 0.200 m ⎟⎠

(

(b)

)

The final field inside the solenoid is ⎛ 300 ⎞ B f = m 0 nI f = 4p × 10 −7 T ⋅ m A ⎜ ( 5.00 A ) = 9.42 × 10 −3 T ⎝ 0.200 m ⎟⎠

(

)

(

)

2

(c)

The 4-turn coil encloses an area A = p r 2 = p 1.50 × 10 −2 m

(d)

The change in flux through each turn of the 4-turn coil during the 0.900-s period is

(

)(

= 7.07 × 10 −4 m 2

)

ΔΦ B = ( ΔB ) A = 9.42 × 10 −3 T − 3.77 × 10 −3 T 7.07 × 10 −4 m 2 = 3.99 × 10 −6 Wb (e)

The average induced emf in the 4-turn coil is ⎛ 3.99 × 10 −6 Wb ⎞ ⎛ ΔΦ B ⎞ −5 e = N2 ⎜ = 4⎜ ⎟ ⎟⎠ = 1.77 × 10 V ⎝ Δt ⎠ 0.900 s ⎝ Since the current increases at a constant rate during this time interval, the induced emf at any instant during the interval is the same as the average value given above.

(f)

68719_20_ch20_p148-170.indd 156

The induced emf is small, so the current in the 4-turn coil will also be very small. This means that the magnetic field generated by this current will be negligibly small in comparison to the field generated by the solenoid.

1/7/11 2:45:32 PM

Induced Voltages and Inductance

20.20

The magnitude of the average emf is e =

=

N ΔΦ B NBA Δ ( cosq ) = Δt Δt

(

)

200 (1.1 T ) 100 × 10 −4 m 2 cos180° − cos 0° 0.10 s

Therefore, the average induced current is I = 20.21

157

= 44 V

e 44 V = = 8.8 A . R 5.0 Ω

If the magnetic field makes an angle of 28.0° with the plane of the coil, the angle it makes with the normal to the plane of the coil is q = 62.0°. Thus, e =

=

N ( ΔΦ B ) NB ( ΔA ) cosq = Δt Δt

(

)(

)(

)

200 50.0 × 10 –6 T ⎡⎣ 39.0 cm 2 1 m 2 10 4 cm 2 ⎤⎦ cos 62.0° 1.80 s

= 1.02 × 10 −5 V = 10.2 mV 20.22

With the magnetic field perpendicular to the plane of the coil, the flux through each turn of the coil is Φ B = BA = B p r 2 . Since the area remains constant, the change in flux due to the changing magnitude of the magnetic field is ΔΦ B = ( ΔB ) p r 2 .

( )

The induced emf is e = N

(b)

When looking down on the coil from a location on the positive z-axis, the magnetic field (in the positive z-direction) is directed up toward you and increasing in magnitude. This means the change in the flux through the coil is directed upward. In order to oppose this change in flux, the induced current must produce a magnetic field directed downward through the area enclosed by the coil. Thus, the current must flow clockwise as seen from your viewing location.

(c)

Since the turns of the coil are connected in series, the total resistance of the coil is Req = NR. Thus, the magnitude of the induced current is I=

20.23

⎡ B0 − 0 p r 2 ⎤ NB0p r 2 ΔΦ =N⎢ . ⎥= Δt t ⎥⎦ ⎣⎢ t − 0

(a)

B pr2 NB0p r 2 t e = = 0 NR Req tR

The motional emf induced in a metallic object of length moving through a magnetic field at speed v is given by e = B⊥ v , where B⊥ is the component of the magnetic field perpendicular to the velocity of the object. Thus,

(

)

e = 35.0 × 10 −6 T ( 25.0 m s ) (15.0 m ) = 1.31 × 10 −2 V = 13.1 mV 20.24

The vertical component of the Earth’s magnetic field is perpendicular to the horizontal velocity of the wire. Thus, the magnitude of the motional emf induced in the wire is

(

)

e = B⊥ v = 40.0 × 10 −6 T ( 2.00 m ) (15.0 m s ) = 1.20 × 10 −3 V = 1.20 mV continued on next page

68719_20_ch20_p148-170.indd 157

1/7/11 2:45:33 PM

158

Chapter 20

Imagine holding your right hand horizontal with the fingers pointing north (the direction of the wire’s velocity), such that when you close your hand the fingers curl downward (in the direction of B⊥). Your thumb will then be pointing westward. By right-hand rule number 1, the magnetic force on charges in the wire would tend to move positive charges westward. Thus, the west end of the wire will be positive relative to the east end . 20.25

The vertical component of the Earth’s magnetic field is perpendicular to the horizontal velocity of the metallic truck body. Thus, the motional emf induced across the width of the truck is ⎡ ⎛ 1m ⎞⎤ −3 e = B⊥ v = 35 × 10 −6 T ⎢( 79.8 in ) ⎜ ⎟⎠ ⎥ ( 37 m s ) = 2.6 × 10 V = 2.6 mV ⎝ 39.37 in ⎣ ⎦

(

20.26

20.27

)

(a)

As the loop passes position A, the flux through the area enclosed by the loop is directed right to left and is increasing in magnitude. The induced current must flow counterclockwise as seen from the right end in order to generate flux passing left to right through the loop, opposing the increase in flux due to the magnet.

(b)

When the loop reaches position B, the flux through the enclosed area is again directed right to left but is now decreasing in magnitude. The induced current must flow clockwise as seen from the right end to generate additional flux passing right to left through the loop, opposing the decrease in flux due to the magnet.

(c)

At position C, the flux through the loop is not changing so there is no induced emf, and hence no induced current , in the loop.

(a)

Observe that only the horizontal component, Bh , of Earth’s magnetic field is effective in exerting a vertical force on charged particles in the antenna. For the magnetic force, Fm = qvBh sinq , on positive charges in the antenna to be directed upward and have maximum magnitude (when q = 90°), the car should move eastward through the northward horizontal component of the magnetic field.

(b)

e = Bh v, where Bh is the horizontal component of the magnetic field. ⎡⎛ km ⎞ ⎛ 0.278 m s ⎞ ⎤ e = ⎡⎣ 50.0 × 10 −6 T cos 65.0° ⎤⎦ (1.20 m ) ⎢⎜ 65.0 ⎟ ⎥ ⎝ h ⎠ ⎜⎝ 1 km h ⎟⎠ ⎦ ⎣

(

)

= 4.58 × 10 − 4 V 20.28

(a)

Since e = B⊥ v, the magnitude of the vertical component of the Earth’s magnetic field at this location is Bvertical = B⊥ =

(b)

20.29

e 0.45 V = = 6.0 × 10 −6 T = 6.0 m T v ( 25 m ) 3.0 × 10 3 m s

(

)

Yes. The magnitude and direction of the Earth’s field varies from one location to the other, so the induced voltage in the wire changes. Further, the voltage will change if the tether cord changes its orientation relative to the Earth’s field.

The metal bar of length and moving at speed v through the magnetic field experiences an induced emf of magnitude e = B⊥ v, where B⊥ = B cosq is the component of the magnetic field perpendicular to the velocity of the bar as shown in figure (a) below.

continued on next page

68719_20_ch20_p148-170.indd 158

1/7/11 2:45:35 PM

Induced Voltages and Inductance

159

y end view of bar B

q

q

n x

Fm

v B

current into page

mg

B||

q

q (b)

(a)

As the bar slides down the rails, the magnetic flux through the conducting path formed by the bar, the rails, and the resistor R is directed downward and is increasing in magnitude. Thus, the induced current must flow counterclockwise around the conducting path to generate an upward flux, opposing the increase in flux due to the field B. This current flows into the page as indicated in figure (b) and has magnitude I = e R = B v cosq R. According to right-hand rule number 1, the bar will experience a magnetic force Fm directed horizontally toward the left as shown in figure (b). The magnitude of this force is B 2 2 v cosq ⎛ B v cosq ⎞ Fm = BI = B ⎜ = ⎟ ⎝ ⎠ R R Now, consider the free-body diagram of the bar in figure (b), where n is the normal force exerted on the bar by the rails. If the bar is to move with constant velocity (i.e., be in equilibrium), it is necessary that

and

mg cosq

ΣFy = 0 ⇒ n cosq = mg

or

n=

ΣFx = 0 ⇒ Fm = n sinq

or

B 2 2 v cosq ⎛ mg ⎞ =⎜ sinq = mg tanq ⎝ cosq ⎟⎠ R

Thus, the equilibrium speed of the bar is v= 20.30

B

2 2

cosq

)

m s2 tan 25.0° (1.00 Ω )

( 0.500 T ) (1.20 m )2 cos 25.0° 2

(a)

e IR ( 0.500 A ) ( 6.00 Ω ) = = = 1.00 m s B B ( 2.50 T )(1.20 m )

In the initial orientation of the coil, the magnitude of the flux passing through the loop is Φ B = NBA, where A is the area enclosed by the loop and N is the number of turns on the loop. After the loop has rotated 90°, the magnetic field is now parallel to the plane of the loop and the flux through the loop is zero. The average emf induced in the loop as it rotates is

(

−2 ΔΦ B NBA − 0 28 (1.25 T ) 2.80 × 10 m e= = = Δt Δt 0.335 s

(b)

68719_20_ch20_p148-170.indd 159

= 2.80 m s

From e = B v, the required speed is v=

20.31

( mg tanq ) R = ( 0.200 kg) (9.80

The average induced current is I =

)

2

= 8.19 × 10 −2 V = 81.9 mV

e 81.9 mV = = 105 mA . R 0.780 Ω

1/7/11 2:45:38 PM

160

20.32

Chapter 20

Note that the vertical component of the magnetic field is always parallel to the plane of the coil, and can never contribute to the flux through the coil. The maximum induced emf in the coil is then rev ⎞ ⎛ 2p rad ⎞ ⎛ 1 min ⎞ ⎤ 2 ⎡⎛ e max = NBhorizontal Aw = 100 2.0 × 10 −5 T ( 0.20 m ) ⎢⎜ 1500 ⎟⎠ ⎜⎝ ⎟⎜ ⎟ ⎝ min 1 rev ⎠ ⎝ 60 s ⎠ ⎥⎦ ⎣

(

)

= 1.3 × 10 −2 V = 13 mV 20.33

Note the similarity between the situation in this problem and a generator. In a generator, one normally has a loop rotating in a constant magnetic field so the flux through the loop varies sinusoidally in time. In this problem, we have a stationary loop in an oscillating magnetic field, and the flux through the loop varies sinusoidally in time. In both cases, a sinusoidal emf e = e max sinwt, where e max = NBAw , is induced in the loop. The loop in this case consists of a single band ( N = 1) around the perimeter of a red blood cell with diameter d = 8.0 × 10 −6 m. The angular frequency of the oscillating flux through the area of this loop is w = 2p f = 2p ( 60 Hz ) = 120p rad s. The maximum induced emf is then e max

20.34

(a)

(

) (

1.0 × 10 −3 T p 8.0 × 10 −6 m ⎛ p d2 ⎞ = NBAw = B ⎜ w = 4 ⎝ 4 ⎟⎠

) (120p s ) = 2

–1

1.9 × 10 −11 V

Using e max = NBAw , we find ⎡⎛ rev ⎞ ⎛ 2p rad ⎞⎤ e max = 1 000 ( 0.20 T) 0.10 m 2 ⎢⎜60 ⎟⎥ ⎟⎜ s ⎠ ⎝ 1 rev ⎠⎦ ⎣⎝

(

)

= 7.5×10 3 V = 7.5 kV (b)

20.35

The maximum induced emf occurs when the flux through the coil is changing the most rapidly. This is when the plane of the coil is parallel to the magnetic field .

rev ⎞ ⎛ 1 min ⎞ ⎛ 2p rad ⎞ ⎛ The angular frequency is w = ⎜ 120 ⎟⎜ ⎟⎜ ⎟ = 4p rad s. ⎝ min ⎠ ⎝ 60 s ⎠ ⎝ 1 rev ⎠ (a)

e max = NBAw = 500 ( 0.60 T ) [( 0.080 m ) ( 0.20 m )] ( 4p rad s ) = 60 V

(b)

Note that the calculator must be in radians mode for the next calculation. ⎡ ⎛p e = e max sin (w t ) = ( 60 V ) sin ⎢( 4p rad s ) ⎜ ⎝ 32 ⎣

(c)

The emf is first maximum when w t = p 2 radians, or when t=

20.36

(a)

⎞⎤ s⎟ ⎥ = 57 V ⎠⎦

p 2 rad p 2 rad = = 0.13 s w 4p rad s

Immediately after the switch is closed, the motor coils are still stationary and the back emf is zero. Thus, I=

e 240 V = = 8.0 A R 30 Ω

continued on next page

68719_20_ch20_p148-170.indd 160

1/7/11 2:45:40 PM

Induced Voltages and Inductance

(b)

At maximum speed, e back = 145 V and I=

20.37

161

e − e back 240 V − 145 V = = 3.2 A R 30 Ω

(c)

e back = e − IR = 240 V − ( 6.0 A ) ( 30 Ω ) = 60 V

(a)

When a coil having N turns and enclosing area A rotates at angular frequency w in a constant magnetic field, the emf induced in the coil is e = e max sinw t, where e max = NB⊥ Aw Here, B⊥ is the magnitude of the magnetic field perpendicular to the rotation axis of the coil. In the given case, B⊥ = 55.0 mT; A = p ab, where a = (10.0 cm ) 2 and b = ( 4.00 cm ) 2; and rev ⎞ ⎛ 1 min ⎞ ⎛ w = 2p f = 2p ⎜ 100 ⎟⎜ ⎟ = 10.5 rad s ⎝ min ⎠ ⎝ 60.0 s ⎠

20.38

Thus,

p e max = (10.0 ) 55.0 × 10 −6 T ⎡⎢ ( 0.100 m ) ( 0.040 0 m ) ⎤⎥ (10.5 rad s ) ⎣4 ⎦

or

e max = 1.81 × 10 −5 V = 18.1 mV

(

(b)

When the rotation axis is parallel to the field, then B⊥ = 0, giving e max = 0 . It is easily understood that the induced emf is always zero in this case if you recognize that the magnetic field lines are always parallel to the plane of the coil, and the flux through the coil has a constant value of zero.

(a)

In terms of its cross-sectional area A, length , and number of turns N, the self inductance of a solenoid is given as L = m 0 N 2 A . Thus, for the given solenoid, L=

(

m0 N 2 p d 2 4

= 5.9 × 10 (b) 20.39

−3

) = ( 4p × 10

−7

)

(

T ⋅ m A ( 580 ) p 8.0 × 10 −2 m 2

4 ( 0.36 m )

)

2

H = 5.9 mH

⎛ ΔI ⎞ e = −L ⎜ ⎟ = − 5.9 × 10 −3 H ( +4.0 A s ) = −2.4 × 10 −2 V = −24 mV ⎝ Δt ⎠

(

)

From e = L ΔI Δt , we have L=

20.40

)

(

)

−3 e e ( Δt ) 12 × 10 V ( 0.50 s ) = = = 4.0 × 10 −3 H = 4.0 mH ΔI Δt ΔI 2.0 A − 3.5 A

The units of NΦ B I are T ⋅ m 2 A . From the force on a moving charged particle, F = qvB, the magnetic field is B = F qv, and we find that 1T =1

N N ⋅s =1 C ⋅ ( m s) C⋅m

⎛ N ⋅ s ⎞ 2 (N ⋅ m) ⋅ s ⎛ J ⎞ Thus, T ⋅ m 2 = ⎜ = ⎜ ⎟ ⋅ s = V ⋅ s, and T ⋅ m 2 A = V ⋅ s A. ⋅m = ⎝ C⎠ ⎝ C ⋅ m ⎟⎠ C The units of e ( ΔI Δt ) are V ( A s ) = V ⋅ s A , the same as the units of NΦ B I.

68719_20_ch20_p148-170.indd 161

1/7/11 2:45:42 PM

162

20.41

20.42

Chapter 20

m0 N 2 A

( 4p × 10

(a)

L=

(b)

From e = L ( ΔI Δt ),

=

)

(

)

2 2 T ⋅ m A ( 400 ) ⎡p 2.5 × 10 −2 m ⎤ ⎣ ⎦ = 2.0 × 10 −3 H = 2.0 mH 0.20 m

−7

ΔI e 75 × 10 −3 V = = = 38 A s Δt L 2.0 × 10 −3 H

From e = L ( ΔI Δt ), the self-inductance is L=

e 24.0 × 10 −3 V = = 2.40 × 10 −3 H ΔI Δt 10.0 A s

Then, from L = NΦ B I, the magnetic flux through each turn is ΦB = 20.43

(a)

(

In the series circuit of Figure P20.43, maximum current occurs after the switch has been closed for a very long time, when current has stabilized and the back emf due to the inductance has decreased to zero. This maximum current is given by I max =

(b)

e 24.0 V = = 5.33 A R 4.50 Ω

The time constant of the RL circuit is t =

(c)

)

2.40 × 10 −3 H ( 4.00 A ) L⋅I = = 1.92 × 10 −5 T ⋅ m 2 N 500

L 12.0 H 12.0 Ω ⋅ s = = = 2.67 s R 4.50 Ω 4.50 Ω

If the switch in the RL circuit is closed at time t = 0, the current as a function of time is given by I = I max 1 − e −t t .

(

)

Thus, e −t t = 1 − I I max, or t = −t ln (1 − I I max ) . With t = 2.67 s, the current in this circuit will be I = 0.950I max at time t = − ( 2.67 s ) ln (1 − 0.950 ) = 8.00 s 20.44

(a)

The time constant of the RL circuit is t = L R, and that of the RC circuit is t = RC. If the two time constants have the same value, then RC = L R, or R=

(b)

3.00 H = 1.00 × 10 3 Ω = 1.00 kΩ 3.00 × 10 −6 F

The common value of the two time constants is t =

20.45

L = C

L 3.00 H = = 3.00 × 10 −3 s = 3.00 ms R 1.00 × 10 3 Ω

(a)

I max = e R , so e = I max R = (8.0 A ) ( 0.30 Ω ) = 2.4 V .

(b)

The time constant is t = L R, giving L = t R = ( 0.25 s ) ( 0.30 Ω ) = 7.5 × 10 −2 H = 75 mH

continued on next page

68719_20_ch20_p148-170.indd 162

1/7/11 2:45:45 PM

Induced Voltages and Inductance

(c)

(

163

)

The current as a function of time is I = I max 1 − e −t t , so at t = t ,

(

)

I = I max 1 − e −1 = 0.632I max = 0.632 (8.0 A ) = 5.1 A (d)

At t = t , I = 5.1 A, and the voltage drop across the resistor is ΔVR = −IR = − ( 5.1 A ) ( 0.30 Ω ) = −1.5 V

(e)

Applying Kirchhoff’s loop rule to the circuit shown in Figure P20.43 gives e + ΔVR + ΔVL = 0. Thus, at t = t , we have ΔVL = − (e + ΔVR ) = − ( 2.4 V − 1.5 V ) = −0.90 V

20.46

e 1 − e −t t . The potential difference across the R , and from Kirchhoff’s loop rule, the potential difference across

The current in the RL circuit at time t is I =

(

resistor is ΔVR = RI = e 1 − e −t t

)

(

)

the inductor is

(

)

ΔVL = e − ΔVR = e ⎡⎣1 − 1 − e −t t ⎤⎦ = ee −t t

20.47

(

)

(

)

(a)

At t = 0, ΔVR = e 1 − e −0 = e (1 − 1) = 0 .

(b)

At t = t , ΔVR = e 1 − e −1 = ( 6.0 V ) (1 − 0.368 ) = 3.8 V .

(c)

At t = 0, ΔVL = e e −0 = e = 6.0 V .

(d)

At t = t , ΔVL = e e −1 = ( 6.0 V ) ( 0.368 ) = 2.2 V .

(

)

From I = I max 1 − e −t t , we obtain e −t t = 1 − I I max . If I I max = 0.900 at t = 3.00 s, then e − (3.00 s) t = 0.100

or

t =

−3.00 s = 1.30 s ln ( 0.100 )

Since the time constant of an RL circuit is t = L R, the resistance is

20.48

R=

L 2.50 H = = 1.92 Ω t 1.30 s

(a)

t =

L 8.00 mH = = 2.00 ms R 4.00 Ω

(b)

I=

(c)

I max =

(d)

I = I max 1 − e −t t yields e −t t = 1 − I I max, and

V⎞ (1 − e ) = ⎛⎜⎝ 6.00 ⎟ (1 − e R 4.00 Ω ⎠

e

−t t

e R

(

=

−250×10 −6 s 2.00×10 −3 s

)=

0.176 A

6.00 V = 1.50 A 4.00 Ω

)

t = −t ln (1 − I I max ) = − ( 2.00 ms ) ln (1 − 0.800 ) = 3.22 ms

68719_20_ch20_p148-170.indd 163

1/7/11 2:45:49 PM

164

20.49

Chapter 20

(a)

The energy stored by an inductor is PE L = 12 LI 2 , so the self inductance is

(

(b)

If I = 3.0 A, the stored energy will be PE L =

20.50

(a)

)

−3 2 ( PE L ) 2 0.300 × 10 J = = 2.08 × 10 −4 H = 0.208 mH I2 (1.70 A )2

L=

1 2 1 2 LI = 2.08 ×10 −4 H (3.0 A ) = 9.36 ×10 −4 J = 0.936 mJ 2 2

(

)

The inductance of a solenoid is given by L = m 0 N 2 A , where N is the number of turns on the solenoid, A is its cross-sectional area, and is its length. For the given solenoid, L=

( ) = ( 4p × 10

m0 N 2 p r 2

−7

)

(

T ⋅ m A ( 300 ) p 5.00 × 10 −2 m 2

0.200 m

)

2

= 4.44 × 10 −3 H = 4.44 mH (b)

When the solenoid described above carries a current of I = 0.500 A, the stored energy is PE L =

20.51

1 2 1 2 LI = ( 4.44 mH ) ( 0.500 A ) = 0.555 mJ 2 2

(

(a)

As t → ∞, I → I max = PE L →

(b)

(a)

e 24 V = = 3.0 A, and R 8.0 Ω

1 2 1 2 LI max = ( 4.0 H ) (3.0 A ) = 18 J 2 2

(

)

At t = t , I = I max 1 − e −1 = ( 3.0 A ) (1 − 0.368 ) = 1.9 A, and PE L =

20.52

)

The current in the circuit at time t is I = I max 1 − e −t t , where I max = e R , and the energy stored in the inductor is PE L = 12 LI 2.

1 ( 4.0 H )(1.9 A )2 = 7.2 J 2

Use Table 17.1 to obtain the resistivity of the copper wire and find Rwire

(

)

1.7 × 10 −8 Ω ⋅ m ( 60.0 m ) rCu L rCu L = = 2 = = 1.3 Ω 2 Awire p rwire p 0.50 × 10 −3 m

(

)

L L 60.0 m = = = 4.8 ×10 2 turns circumference of a loop 2p rsolenoid 2p 2.0 ×10 −2 m

(b)

N=

(c)

The length of the solenoid is

(

)

(

) p ( 2.0 × 10

= N ( diameter of wire ) = N ( 2rwire ) = ( 480 ) 2 0.50 × 10 −3 m = 0.48 m (d)

L=

m 0 N 2 Asolenoid

giving

=

2 m 0 N 2p rsolenoid

( 4p × 10 =

−7

)

T ⋅ m A ( 480 ) 0.48 m

2

−2

m

)

2

L = 7.6 × 10 −4 H = 0.76 mH

continued on next page

68719_20_ch20_p148-170.indd 164

1/7/11 2:45:53 PM

Induced Voltages and Inductance

165

L L 7.6 × 10 −4 H = = = 4.6 × 10 −4 s = 0.46 ms Rtotal Rwire + rinternal 1.3 Ω + 0.350 Ω

(e)

t =

(f)

I max =

(g)

I = I max 1 − e −t t , so when I = 0.999I max , we have e −t t = 1 − 0.999 = 0.001. Thus, − t t = ln ( 0.001) , or t = −t ⋅ ln ( 0.001) = − ( 0.46 ms ) ⋅ ln ( 0.001) = 3.2 ms .

(h)

( PEL )max =

e 6.0 V = = 3.6 A Rtotal 1.3 Ω + 0.350 Ω

(

)

1 2 1 2 LI max = 7.6 ×10 −4 H (3.6 A ) = 4.9 ×10 −3 J = 4.9 mJ 2 2

(

)

20.53

The flux due to the current in loop 1 passes from left to right through the area enclosed by loop 2. As loop 1 moves closer to loop 2, the magnitude of this flux through loop 2 is increasing. The induced current in loop 2 generates a magnetic field directed toward the left through the area it encloses in order to oppose the increasing flux from loop 1. This means that the induced current in loop 2 must flow counterclockwise as viewed from the left end of the rod.

20.54

(a)

The clockwise induced current in the loop produces a flux directed into the page through the area enclosed by the loop. Since this flux opposes the change in flux due to the external field, the outward-directed flux due to the external field must be increasing in magnitude. This means that the magnitude of the external field itself must be increasing in time.

(b)

The induced emf in the loop must be e = IR = ( 2.50 mA ) ( 0.500 Ω ) = 1.25 mV Since e =

ΔΦ B Δ ( BA cos 0° ) ⎛ ΔB ⎞ = =⎜ A, the rate of change of the field is ⎝ Δt ⎟⎠ Δt Δt

ΔB e e 1.25 × 10 −3 V = = 2 = Δt A pr p 8.00 × 10 −2 m

(

20.55

(a)

)

2

= 6.22 × 10 −2 T s = 62.2 mT s

After the right end of the coil has entered the field, but the left end has not, the flux through the area enclosed by the coil is directed into the page and is increasing in magnitude. This increasing flux induces an emf of magnitude e =

ΔΦ B NB ( ΔA ) = = NBwv Δt Δt

in the loop. Note that in the above equation, ΔA = wv is the area enclosed by the coil that enters the field in time Δt. This emf produces a counterclockwise current in the loop to oppose the increasing inward flux. The magnitude of this current is I = e R = NBwv R. The right end of the loop is now a conductor, of length Nw, carrying a current toward the top of the page through a field directed into the page. The field exerts a magnetic force of magnitude N 2 B2 w 2 v ⎛ NBwv ⎞ directed toward the left F = BI ( Nw ) = B ⎜ Nw ) = ( ⎟ ⎝ R ⎠ R on this conductor, and hence, on the loop. continued on next page

68719_20_ch20_p148-170.indd 165

1/7/11 2:45:56 PM

166

Chapter 20

(b)

When the loop is entirely within the magnetic field, the flux through the area enclosed by the loop is constant. Hence, there is no induced emf or current in the loop, and the field exerts zero force on the loop.

(c)

After the right end of the loop emerges from the field, and before the left end emerges, the flux through the loop is directed into the page and is decreasing. This decreasing flux induces an emf of magnitude e = NBwv in the loop, which produces an induced current directed clockwise around the loop so as to oppose the decreasing flux. The current has magnitude I = e R = NBwv R. This current flowing upward, through conductors of total length Nw, in the left end of the loop, experiences a magnetic force given by N 2 B2 w 2 v ⎛ NBwv ⎞ F = BI ( Nw ) = B ⎜ Nw ) = ( ⎟ ⎝ R ⎠ R

20.56

(a)

directed toward the left

The motional emf induced in the bar must be e = IR, where I is the current in this series circuit. Since e = B⊥ v, the speed of the moving bar must be v=

(

)

8.5 × 10 −3 A ( 9.0 Ω ) e IR = = = 0.73 m s B⊥ B⊥ ( 0.30 T )( 0.35 m )

The flux through the closed loop formed by the rails, the bar, and the resistor is directed into the page and is increasing in magnitude. To oppose this increasing inward flux, the induced current must generate a magnetic field directed out of the page through the area enclosed by the loop. This means the current will flow countercloclockwise . (b)

The rate at which energy is delivered to the resistor is

(

P = I 2 R = 8.5 × 10 −3 A

20.57

)

2

( 9.0 Ω ) = 6.5 × 10 −4 W = 0.65 mW

(c)

An external force directed to the right acts on the bar to balance the magnetic force to the left. Hence, work is being done by the external force , which is transformed into internal energy within the resistor.

(a)

The current in the solenoid reaches I sol = 0.632I max in a time of t = t = L R, where L=

m0 N 2 A

=

( 4p × 10

−7

)

(

T ⋅ m A (12 500 ) 1.00 × 10 −4 m 2 7.00 × 10

2

−2

m

) = 0.280 H

Thus, t = ( 0.280 H ) (14.0 Ω ) = 2.00 × 10 −2 s = 20.0 ms . (b)

The change in the solenoid current during this time is ⎛ ΔV ⎞ ⎛ 60.0 V ⎞ ΔI sol = 0.632I max − 0 = 0.632 ⎜ = 0.632 ⎜ = 2.71 A ⎝ R ⎟⎠ ⎝ 14.0 Ω ⎟⎠ so the average back emf is ⎛ 2.71 A ⎞ ⎛ ΔI ⎞ = 37.9 V e back = L ⎜ sol ⎟ = ( 0.280 H ) ⎜ ⎝ Δt ⎠ ⎝ 2.00 × 10 −2 s ⎟⎠

(c)

The change in the magnitude of the magnetic field at the location of the coil is one-half the change in the magnitude of the field at the center of the solenoid. Thus, ΔBcoil = 12 ⎡⎣ m 0 nsol ( ΔI sol ) ⎤⎦ , and the average rate of change of flux through each turn of the coil is

continued on next page

68719_20_ch20_p148-170.indd 166

1/7/11 2:45:58 PM

Induced Voltages and Inductance

( ΔB )coil Acoil ⎛ ΔΦ B ⎞ = ⎜⎝ ⎟⎠ = Δt coil Δt =

(d)

I coil =

e coil Rcoil

=

( 4p × 10

1 2

⎡⎣ m 0 nsol ( ΔI sol ) ⎤⎦ Acoil m 0 N sol ( ΔI sol ) Acoil = Δt 2 sol ⋅ ( Δt )

)

(

T ⋅ m A (12 500 ) ( 2.71 A ) 1.00 × 10 −4 m 2

−7

(

)(

2 7.00 × 10 −2 m 2.00 × 10 −2 s

N coil ( ΔΦ B Δt )coil Rcoil

=

167

)

)=

1.52 × 10 −3 V

(820 ) (1.52 × 10 −3 V ) 24.0 Ω

= 0.051 9 A = 51.9 mA 20.58

(a)

The gravitational force exerted on the ship by the pulsar supplies the centripetal GM pulsar mship mship v 2 acceleration needed to hold the ship in orbit. Thus, Fg = , giving = 2 rorbit rorbit GM pulsar

v=

rorbit

=

(6.67 × 10

−11

)(

N ⋅ m kg2 2.0 × 10 30 kg 7

3.0 × 10 m

)=

2.1 × 10 6 m s

(b)

The magnetic force acting on charged particles moving through a magnetic field is perpendicular to both the magnetic field and the velocity of the particles (and therefore perpendicular to the ship’s length). Thus, the charged particles in the materials making up the spacecraft experience magnetic forces directed from one side of the ship to the other, meaning that the induced emf is directed from side to side within the ship.

(c)

e = B⊥ v, where = 2rship = 0.080 km = 80 m is the side-to-side dimension of the ship. This yields

(

)

(

)

e = 1.0 × 10 2 T (80 m ) 2.1 × 10 6 m s = 1.7 × 1010 V

20.59

(d)

The very large induced emf would lead to powerful spontaneous electric discharges. The strong electric and magnetic fields would disrupt the flow of ions in their bodies.

(a)

To move the bar at uniform speed, the magnitude of the applied force must equal that of the magnetic force retarding the motion of the bar. Therefore, Fapp = BI . The magnitude of the induced current is I=

( ΔΦ B Δt ) = B ( ΔA Δt ) = B v e = R R R R

so the field strength is B = IR v, giving Fapp = ( IR v ) I = I 2 R v, and the current is I=

20.60

Fapp ⋅ v R

=

(1.00 N ) ( 2.00 m s ) 8.00 Ω

= 0.500 A

(b)

Pdissipated = I 2 R = ( 0.500 A )2 (8.00 Ω ) = 2.00 W

(c)

Pinput = Fapp ⋅ v = (1.00 N ) ( 2.00 m s ) = 2.00 W

Since the magnetic field outside the solenoid is negligible in comparison to the field inside the solenoid, we shall assume that the flux through the single-turn square loop is the same as that through each turn of the solenoid. Then, the induced emf in the square loop is

continued on next page

68719_20_ch20_p148-170.indd 167

1/7/11 2:46:01 PM

168

Chapter 20

ΔΦ B e= = Δt or

⎛ ⎞ Δ ⎜ Binside Asolenoid ⎟ ⎝ solenoid ⎠ Δ ( m 0 nI solenoid ) Asolenoid ⎛ ΔI ⎞ = m 0 n ⎜ solenoid ⎟ Asolenoid = ⎝ Δt ⎠ Δt Δt

e = m 0 n ( ΔI solenoid Δt ) p r 2 , which gives

(

)(

)

(

e = 4p × 10 −7 T ⋅ m A 3.50 × 10 3 m ( 28.5 A s ) p 2.00 × 10 −2 m

)

2

= 1.58 × 10 −4 V = 0.158 × 10 −3 V = 0.158 mV 20.61

If d is the distance from the lightning bolt to the center of the coil, then e av =

N ( ΔΦ B ) N ( ΔB ) A N ⎡⎣ m 0 ( ΔI ) 2p d ⎤⎦ A N m 0 ( ΔI ) A = = = Δt Δt Δt 2p d ( Δt )

(

)(

) )

2 100 4p × 10 −7 T ⋅ m A 6.02 × 10 6 A − 0 ⎡⎣p ( 0.800 m ) ⎤⎦ = 2p ( 200 m ) 10.5 × 10 −6 s

(

= 1.15 × 10 5 V = 115 kV 20.62

When A and B are 3.00 m apart, the area enclosed by the loop consists of four triangular sections, each having hypotenuse of 3.00 m, altitude of 2 2 1.50 m, and base of ( 3.00 m ) − (1.50 m ) = 2.60 m. The decrease in the enclosed area has been 1 2 ΔA = Ai − A f = ( 3.00 m ) − 4 ⎡⎢ (1.50 m ) ( 2.60 m )⎤⎥ = 1.20 m 2 ⎣2 ⎦ The average induced current has been I av =

e av R

=

( ΔΦ B R

Δt )

=

(

)

2 B ( ΔA Δt ) ( 0.100 T ) 1.20 m 0.100 s = = 0.120 A R 10.0 Ω

As the enclosed area decreases, the flux due to the external field (directed into the page) through this area also decreases. Thus, the induced current will be directed clockwise around the loop to create additional flux directed into the page through the enclosed area. 20.63

(a)

(

=

)

2 ΔΦ B B ΔA B ⎡⎣ p d 4 − 0 ⎤⎦ = = Δt Δt Δt

e av =

( 25.0 mT ) p ( 2.00 × 10 –2 m )

(

4 50.0 × 10 −3 s

)

2

= 0.157 mV

As the inward-directed flux through the loop decreases, the induced current goes clockwise around the loop to create additional inward flux through the enclosed area. With positive charges accumulating at B, point B is positive relative to A .

continued on next page

68719_20_ch20_p148-170.indd 168

1/7/11 2:46:03 PM

Induced Voltages and Inductance

(b)

e av

(

–2 ΔΦ B ( ΔB ) A [(100 − 25.0 ) mT ] p 2.00 × 10 m = = = Δt Δt 4 4.00 × 10 −3 s

(

)

)

169

2

= 5.89 mV

As the inward-directed flux through the enclosed area increases, the induced current goes counterclockwise around the loop in to create flux directed outward through the enclosed area. With positive charges now accumulating at A, point A is positive relative to B . 20.64

The induced emf in the ring is e av =

=

ΔΦ B ( ΔB ) Asolenoid ( ΔBsolenoid 2 ) Asolenoid 1 ⎡ ⎛ ΔI ⎞⎤ = = = ⎢ m 0 n ⎜ solenoid ⎟ ⎥ Asolenoid ⎝ Δt Δt Δt 2⎣ Δt ⎠ ⎦

(

)

2 1⎡ 4p × 10 −7 T ⋅ m A (1 000 ) ( 270 A s ) p ⎡⎣3.00 × 10 −2 m ⎤⎦ ⎤ = 4.80 × 10 −4 V ⎢ ⎥⎦ 2⎣

(

)

Thus, the induced current in the ring is I ring = 20.65

(a)

e av R

4.80 × 10 −4 V = 1.60 A 3.00 × 10 –4 Ω

As the rolling axle (of length = 1.50 m) moves perpendicularly to the uniform magnetic field, an induced emf of magnitude e = B v will exist between its ends. The current produced in the closed-loop circuit by this induced emf has magnitude I=

(b)

=

e av R

=

( ΔΦ B R

Δt )

=

B ( Δ A Δt ) B v ( 0.800 T ) (1.50 m ) ( 3.00 m s ) = = = 9.00 A R R 0.400 Ω

The induced current through the axle will cause the magnetic field to exert a retarding force of magnitude Fr = BI on the axle. This force is directed opposite to the velocity v to oppose the motion of the axle. If the axle is to continue moving at constant speed, an applied force in the direction of v, and having magnitude Fapp = Fr , must be exerted on the axle. Fapp = BI = ( 0.800 T ) ( 9.00 A ) (1.50 m ) = 10.8 N

20.66

(c)

Using right-hand rule number 1, observe that positive charges within the moving axle experience a magnetic force toward the rail containing point b, and negative charges experience a force directed toward the rail containing point a. Thus, the rail containing b is positive relative to the other rail, so b is at the higher potential .

(d)

No . Both the velocity v of the rolling axle and the magnetic field B are unchanged. Thus, the polarity of the induced emf in the moving axle is unchanged, and the current continues to be directed from b to a through the resistor R.

(a)

The time required for the coil to move distance and exit the field is t = v , where v is the constant speed of the coil. Since the speed of the coil is constant, the flux through the area enclosed by the coil decreases at a constant rate. Thus, the instantaneous induced emf is the same as the average emf over the interval t seconds in duration, or e = −N

(b)

ΔΦ B 2 NB 2 ( 0 − BA) = −N =N = = NB v Δt t−0 t v

The current induced in the coil is I = e R = NB v R .

continued on next page

68719_20_ch20_p148-170.indd 169

1/7/11 2:46:04 PM

170

Chapter 20

(c)

Since the coil moves with constant velocity, the power delivered to the coil must equal the power being dissipated within the coil. This is given by P = I 2 R, or ⎛ N 2 B2 2 v2 ⎞ N 2 B2 2 v2 P=⎜ R= 2 ⎟ R R ⎝ ⎠

(d)

The rate that the applied force does work must equal the power delivered to the coil, so Fapp ⋅ v = P, or Fapp =

20.67

N 2 B2 2 v P N 2 B2 2 v2 R = = v R v

(e)

As the coil is emerging from the field, the flux through the area it encloses is directed into the page and is decreasing in magnitude. The induced current must flow clockwise around the coil to generate a magnetic field directed into the page through the area enclosed by the coil, opposing the decrease in the inward flux.

(f)

As the coil is emerging from the field, the left side of the coil carries an induced current directed toward the top of the page through a magnetic field that is directed into the page. Right-hand rule number 1 then shows that this side of the coil will experience a magnetic force directed to the left , opposing the motion of the coil.

(a)

As the bottom conductor of the loop falls, it cuts across the magnetic field lines coming out of the page. This induces an emf of magnitude e = Bwv in this conductor, with the left end at the higher potential. As a result, an induced current of magnitude I=

e Bwv = R R

flows clockwise around the loop. The field then exerts an upward force of magnitude B2 w 2 v ⎛ Bwv ⎞ Fm = BIw = B ⎜ w= ⎟ ⎝ R ⎠ R on this current-carrying conductor forming the bottom of the loop. If the loop is falling at terminal speed, the magnitude of this force must equal the downward gravitational force acting on the loop. That is, when v = vt, we must have Fm =

68719_20_ch20_p148-170.indd 170

B 2 w 2 vt = Mg R

or

vt =

MgR B2 w 2

(b)

A larger resistance would make the current smaller, so the loop must reach higher speed before the magnitude of the magnetic force will equal the gravitational force.

(c)

The magnetic force is proportional to the product of the field and the current, while the current itself is proportional to the field. If B is cut in half, the speed must become four times larger to compensate and yield a magnetic force with magnitude equal to the that of the gravitational force.

1/7/11 2:46:07 PM

21 Alternating Current Circuits and Electromagnetic Waves QUICK QUIZZES 1.

Choice (c). The average power is proportional to the rms current which is non-zero even though the average current is zero. (a) is only valid for an open circuit, for which R → ∞. (b) and (d) can never be true because iav = 0 for AC currents.

2.

Choice (b). Choices (a) and (c) are incorrect because the unaligned sine curves in Figure 21.9 mean the voltages are out of phase, and so we cannot simply add the maximum (or rms) voltages across the elements. (In other words, ΔV ≠ ΔVR + ΔVL + ΔVC even though Δv = ΔvR + ΔvL + ΔvC .)

3.

Choice (b). Closing switch A replaces a single resistor with a parallel combination of two resistors. Since the equivalent resistance of a parallel combination is always less than the lowest resistance in the combination, the total resistance of the circuit decreases, which causes the 2 + ( X L − XC ) to decrease. impedance Z = Rtotal 2

4.

Choice (a). Closing switch A replaces a single resistor with a parallel combination of two resistors. Since the equivalent resistance of a parallel combination is always less than the lowest resistance in the combination, the total resistance of the circuit decreases, which causes the phase angle, f = tan −1 ⎡⎣( X L − XC ) R ⎤⎦ , to increase.

5.

Choice (a). Closing switch B replaces a single capacitor with a parallel combination of two capacitors. Since the equivalent capacitance of a parallel combination is greater than that of either of the individual capacitors, the total capacitance increases, which causes the capacitive reactance XC = 1 2p fC to decrease. Thus, the net reactance, X L − XC , increases causing the phase angle, f = tan −1 ⎡⎣( X L − XC ) R ⎤⎦ , to increase.

6.

Choice (b). Inserting an iron core in the inductor increases both the self-inductance and the inductive reactance, X L = 2p f L. This means the net reactance, X L − XC , and hence the impedance, 2 Z = Rtotal + ( X L − XC ) , increases, causing the current (and therefore, the bulb’s brightness) to decrease. 2

7.

Choices (b) and (c). Since pressure is force per unit area, changing the size of the area without altering the intensity of the radiation striking that area will not cause a change in radiation pressure. In (b), the smaller disk absorbs less radiation, resulting in a smaller force. For the same reason, the momentum in (c) is reduced.

8.

Choices (b) and (d). The frequency and wavelength of light waves are related by the equation l f = v or f = v l , where the speed of light v is a constant within a given medium. Thus, the frequency and wavelength are inversely proportional to each other; when one increases the other must decrease.

171

68719_21_ch21_p171-197.indd 171

1/7/11 2:46:50 PM

172

Chapter 21

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

In an electromagnetic wave, the electric ﬁeld E, the magnetic ﬁeld B, and the direction of propagation of the wave are always mutually perpendicular to each other. Thus, with B(in the −x-direction) and the direction of propagation (+ y-direction) both in the xy-plane, E must be parallel to the z-axis, meaning that either (c) or (d) must be the correct answer. To choose between these possible answers, recall the right-hand rule for electromagnetic waves (see Section 21.11 in the textbook). Hold your right hand, with the ﬁngers extended, so the thumb is in the direction of propagation ( + y) and your palm is facing the direction of B (−x-direction). Then the orientation of the extended ﬁngers is the direction of the electric ﬁeld E. You should ﬁnd this to be the negative z-direction, so the correct choice is (d).

2.

When the frequency doubles, the rms current I rm s = ΔVL , rm s X L = ΔVL , rm s 2p fL is cut in half. Thus, the new current is I rm s = 3.0 A 2 = 1.5 A, and (e) is the correct answer.

3.

At the resonance frequency, X L = XC and the impedance is Z = R 2 + ( X L − XC ) = R. Thus, the rms current is I rm s = ΔVrm s Z = (120 V ) ( 20 Ω ) = 6.0 A , and (b) is the correct choice. 2

4.

ΔVL , rm s = I rm s X L = I rm s ( 2p fL) = ( 2.0 A ) 2p ( 60.0 Hz ) ⎡⎣ (1.0 2p ) H ⎤⎦ = 120 V, and the correct answer is choice (c).

5.

ΔVC , rm s = I rm s XC =

6.

The battery produces a constant current in the primary coil, which generates a constant ﬂux through the secondary coil. With no change in ﬂux through the secondary coil, there is no induced voltage across the secondary coil, and choice (e) is the correct answer.

7.

The speed c, frequency f, and wavelength l of an electromagnetic wave are related by c = fl. The wavelength of the waves in the oven is then l=

I rm s 1.00 × 10 −3 A = = 16.7 V, so choice (a) is correct. 2p fC 2p ( 60.0 Hz ) ⎡⎣ (1.00 2p ) × 10 −6 F ⎤⎦

c 3.00 × 108 m s = = 0.122 m = 12.2 cm f 2.45 × 10 9 Hz

and (b) is the correct choice. 8.

When a power source, AC or DC, is ﬁrst connected to a RL combination, the presence of the inductor impedes the buildup of a current in the circuit. The value of the current starts at zero and increases as the back emf induced across the inductor decreases somewhat in magnitude. Thus, the correct choice is (c).

9.

The voltage across the capacitor is proportional to the stored charge. This charge, and hence the voltage ΔvC , is a maximum when the current has zero value and is in the process of reversing direction after having been ﬂowing in one direction for a half cycle. Thus, the voltage across the capacitor lags behind the current by 90°, and (a) is the correct choice.

10.

68719_21_ch21_p171-197.indd 172

In an AC circuit, both an inductor and a capacitor store energy for one half of the cycle of the current and return that energy to the circuit during the other half of the cycle. On the other hand, a resistor converts electrical potential energy into thermal energy during all parts of the cycle of the current. Thus, only the resistor has a non-zero average power loss. The power delivered to a circuit by a generator is given by Pav = I rm s ΔVrm s cosf , where f is the phase difference between the generator voltage and the current. Among the listed choices, the only true statement is choice (c).

1/7/11 2:46:52 PM

Alternating Current Circuits and Electromagnetic Waves

173

11.

In an RLC circuit, the instantaneous voltages ΔvR , ΔvL , and ΔvC (across the resistance, inductance, and capacitance respectively) are not in phase with each other. The instantaneous voltage ΔvR is in phase with the current, ΔvL leads the current by 90°, while ΔvC lags behind the current by 90°. The instantaneous values of these three voltages do add algebraically to give the instantaneous voltage across the RLC combination, but the rms voltages across these components do not add algebraically. The rms voltages across the three components must be added as vectors (or phasors) to obtain the correct rms voltage across the RLC combination. Among the listed choices, choice (e) is the false statement.

12.

If the voltage is a maximum when the current is zero, the voltage is either leading or lagging the current by 90° (or a quarter cycle) in phase. Thus, the element could be either an inductor or a capacitor. It could not be a resistor since the voltage across a resistor is always in phase with the current. If the current and voltage were out of phase by 180°, one would be a maximum in one direction when the other was a maximum value in the opposite direction. The correct choice for this question is (d).

13.

At resonance of the RLC series circuit, X L = XC , and the impedance becomes Z = R 2 + ( X L − XC ) = R 2 + 0 = R 2

so choice (c) is correct. 14.

At a frequency of f = 5.0 × 10 2 Hz, the inductive reactance, capacitive reactance, and impedance are X L = 2p f L, XC =

1 2 , and Z = R 2 + ( X L − XC ) , respectively. This yields 2p f C 2

⎡ ⎤ 1 Z = ( 20.0 Ω ) + ⎢ 2p ( 5.0 × 10 2 Hz ) ( 0.120 H ) − ⎥ = 51 Ω , 2 −6 2p ( 5.0 × 10 Hz ) ( 0.75 × 10 F ) ⎥⎦ ⎢⎣ ΔVrm s 120 V and I rm s = = = 2.3 A. Choice (a) is the correct answer. 51 Ω Z 2

15.

Choices (c) and (d) are the only true statements. Electromagnetic waves consist of oscillating electric and magnetic ﬁelds that are perpendicular to each other and to the direction of propagation. In a vacuum, all electromagnetic waves travel at the same speed, c. The electromagnetic spectrum consists of waves having broad ranges of frequencies and wavelengths, which are related by l f = c.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The phase angle in an RLC series circuit is given by ⎛ X − XC ⎞ −1 ⎛ 2p f L − 1 2p fC ⎞ f = tan −1 ⎜ L ⎟⎠ ⎟⎠ = tan ⎜⎝ ⎝ R R which is clearly frequency dependent. As f → 0, f → tan −1 ( − ∞ ) = − 90° and as f → ∞, f → tan −1 ( + ∞ ) = + 90°. At resonance, X L = XC , and f = tan −1 ( 0 ) = 0°.

4.

68719_21_ch21_p171-197.indd 173

An antenna that is a conducting line responds to the electric ﬁeld of the electromagnetic wave—the oscillating electric ﬁeld causes an electric force on electrons in the wire along its length. The movement of electrons along the wire is detected as a current by the radio and is ampliﬁed. Thus, a line antenna must have the same orientation as the broadcast antenna. A loop antenna responds to the magnetic ﬁeld in the radio wave. The varying magnetic ﬁeld induces a varying current in the loop (by Faraday’s law), and this signal is ampliﬁed. The loop should be in the

1/7/11 2:46:55 PM

174

Chapter 21

vertical plane containing the line of sight to the broadcast antenna, so the magnetic ﬁeld lines go through the area of the loop. 6.

(a)

In an electromagnetic wave, electric and magnetic ﬁelds oscillate at right angles to each other and perpendicular to the direction of propagation of the wave.

(b)

Energy is the quantity transported by the oscillating electric and magnetic ﬁelds of the electromagnetic wave.

8.

Consider a typical metal rod antenna for a car radio. Charges in the rod respond to the electric ﬁeld portion of the carrier wave. Variations in the amplitude of the incoming radio wave cause the electrons in the rod to vibrate with amplitudes emulating those of the carrier wave. Likewise, for frequency modulation, the variations of the frequency of the carrier wave cause constantamplitude vibrations of the electrons in the rod but at frequencies that imitate those of the carrier.

10.

The brightest portion of your face shows where you radiate the most. Your nostrils and the openings of your ear canals are particularly bright. Brighter still are the pupils of your eyes.

12.

The changing magnetic ﬁeld of the solenoid induces eddy currents in the conducting core. This is accompanied by I 2 R conversion of electrically-transmitted energy into internal energy in the conductor.

14.

The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams throughout the chapter. Kirchhoff’s loop rule is true at any instant, but the voltages across different circuit elements are not simultaneously at their maximum values. Do not forget that an inductor can induce an emf in itself and that the voltage across it is 90° in phase ahead of the current in the circuit.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a)

1.70 × 10 2 V

(b)

2.40 × 10 2 Ω

(c) The 100-W bulb will have the lower resistance. 4.

6.

(a)

I1, rm s = I 2 , rm s = 1.25 A, I 3 , rm s = 0.833 A

(b)

R1 = R2 = 96.0 Ω, R3 = 144 Ω

(a) 170 V

(b)

30.0 Hz

(c)

120 V

(d)

6.00 A

(e)

8.49 A

(f)

720 W

(g)

6.9 A

(h)

radians

8.

(a)

141 mA

(b)

235 mA

10.

(a)

221 Ω

(b)

0.163 A

(c)

0.231 A

(d) No. The charge is a maximum when the voltage is a maximum, but the voltage across a capacitor is 90° out of phase with the current. 12.

68719_21_ch21_p171-197.indd 174

(a) 69.3 V

(b)

40 Hz

(d) 196 Ω

(e)

20.3 mF

(c)

0.354 A

1/7/11 2:46:58 PM

Alternating Current Circuits and Electromagnetic Waves

14.

(a)

12.6 Ω

(b)

6.19 A

(c)

8.75 A

16.

(a)

15.0 Hz

(b)

84.9 V

(c)

47.1 Ω

(e)

2.55 A

(f )

0

(d) 1.80 A (g)

175

i = ( 2.55 A ) sin ( 30p t − p 2 )

(h) 20.9 ms 57.5 Ω

18.

(a)

20.

(a) 194 V

(b) The current leads the voltage by 49.9°.

22.

(a) 138 V

(b)

104 V

(d) 641 V

(e)

See Solution.

(a) 0.11 A

(b)

ΔVR, m ax = 1.3×10 2 V, ΔVL , m ax = 1.2 ×10 2 V

24.

26.

(b)

1.39 A

(c)

ΔvR = 1.3 × 10 2 V, ΔvL = 0, Δvsource = 1.3 × 10 2 V

(d)

ΔvR = 0, ΔvL = 1.2 × 10 2 V, Δvsource = 1.2 × 10 2 V

(a)

88.4 Ω

(b)

107 Ω

(c)

(c)

729 V

1.12 A

(d) The voltage lags the current by 55.8°. (e) It changes the impedance, and therefore the current, in the circuit. 28.

2.79 kHz

30.

(a) 0.11 A

(b)

ΔVR, m ax = 1.3×10 2 V, ΔVC , m ax = 1.2 ×10 2 V

(c)

ΔvR = 0, ΔvC = 1.2 × 10 2 V, Δvsource = 1.2 × 10 2 V, qC = 300 mC

(d)

ΔvR = 1.3 × 10 2 V, ΔvC = 0, Δvsource = 1.3 × 10 2 V, qC = 0

32.

(a)

66.8 Ω

34.

88.0 W

36.

(a) No, the algebraic sum of the rms voltage drops is 20.9 V. (b) to the resistor

38.

2.31 kHz

40.

(a)

Z = R = 15 Ω

(c) at resonance

68719_21_ch21_p171-197.indd 175

(b)

0.953 A

(c)

3.2 W

(b)

41 Hz

(d)

2.5 A

(c)

45.4 W

1/7/11 2:46:59 PM

176

42.

Chapter 21

(a) 480 W

(b)

0.192 W

(c) 30.7 mW

(d)

0.192 W

(e)

30.7 mW; Maximum power is delivered at the resonance frequency.

44.

(a) 9.23 V

(b)

30.0 W

46.

(a) fewer turns

(b)

25 mA

(c)

20 turns

48.

(a) 29.0 kW

(b)

0.577%

(c)

See Solution.

50.

(a)

(b)

8.31 min

(c)

2.56 s

52.

2.998 × 108 m s

54.

0.80 or 80%

56.

(a)

I = mgc 2A

(b)

1.46 × 10 9 W m 2

(c)

Propulsion by light pressure in a signiﬁcant gravity ﬁeld is impractical because of the enormous power requirements. In addition, no material is perfectly reﬂecting, so the absorbed energy would melt the reﬂecting surface.

58.

(a)

8.89 × 10 −8 W m 2

60.

11.0 m

62.

The radio listeners hear the news 8.4 ms before the studio audience because radio waves travel so much faster than sound waves.

64.

(a)

66.

1.1× 10 7 m s

68.

∼ 10 6 J

70.

XC , i = 3R

72.

(a) 0.63 pF

74.

(a)

6.0 Ω

76.

(a)

6.2 mW cm 2, 24% higher than the maximum allowed leakage from a microwave

(b)

0.024 mW cm 2

6.80 × 10 2 y

(b)

6.003 6 × 1014 Hz

11.4 MW

(b)

3.6 × 1011 Hz

(b)

8.4 mm

(b)

12 mH

(c)

25 Ω

PROBLEM SOLUTIONS 21.1

For an AC circuit containing only resistance (the ﬁlament of the lightbulb), the power dissipated 2 2 is P = I rm2 s R = ( ΔVrm s R ) R = ΔVrm2 s R = ΔVm ax 2 R.

(

(a)

(

170 V 2 ΔVrm2 s R= = P 75.0 W

)

)

2

= 193 Ω

continued on next page

68719_21_ch21_p171-197.indd 176

1/7/11 2:47:01 PM

Alternating Current Circuits and Electromagnetic Waves

21.2

(

)

2

(b)

170 V 2 ΔVrm2 s R= = P 100.0 W

(a)

ΔVR , m ax = 2 ΔVR , rm s = 2 (1.20 × 10 2 V ) = 1.70 × 10 2 V

(b)

Pav = I

(c)

Because R =

(

177

= 145 Ω

)

2 ΔVrm2 s ΔVrm2 s (1.20 × 10 V ) R= ⇒ R= = = 2.40 × 10 2 Ω R Pav 60.0 W 2

2 rm s

ΔVrm2 s (see above), if the bulbs are designed to operate at the same voltage, Pav

the 100 W will have the lower resistance . 21.3

For a simple resistance, i ( t ) = v ( t ) R = ( ΔVm ax sinwt ) R = I m ax sinwt. Thus, if i = 0.600I m ax at t = 7.00 ms, we have wt = 2p ft = sin −1 ( i I m ax ) giving f =

21.4

sin −1 ( i I m ax ) 2p t

0.644 rad ( 0.600 ) = = 14.6 s −1 = 14.6 Hz . 2p t ( 2p rad )( 7.00 × 10 −3 s )

−1

All lamps are connected in parallel with the voltage source, so ΔVrm s = 120 V for each lamp. Also, for each bulb, the current is I rm s = Pav ΔVrm s and the resistance is R = ΔVrm s I rm s . (a)

(b) 21.5

sin

f=

and

For bulbs 1 and 2:

I1, rm s = I 2 , rm s =

For bulb 3:

I 3 , rm s =

R1 = R2 =

150 W = 1.25 A 120 V

100 W = 0.833 A 120 V

120 V = 96.0 Ω 1.25 A

R3 =

and

120 V = 144 Ω 0.833 A

The total resistance (series connection) is Req = R1 + R2 = 8.20 Ω +10.4 Ω = 18.6 Ω, so the current in the circuit is I rm s =

ΔVrm s 15.0 V = = 0.806 A 18.6 Ω Req

The power to the speaker is then Pav = I rm2 s Rspeaker = ( 0.806 A ) (10.4 Ω ) = 6.76 W . 2

21.6

The general form of the generator voltage is Δv = ( ΔVm ax ) sin (wt ), so by inspection (a)

ΔVR , m ax = 170 V

(c)

ΔVR , rm s =

(d)

I rm s =

ΔVR , m ax 2

ΔVR , rm s R

=

and

=

(b)

f=

w 60p rad s = = 30.0 Hz 2p 2p rad

170 V = 120 V 2

120 V = 6.00 A 20.0 Ω

continued on next page

68719_21_ch21_p171-197.indd 177

1/7/11 2:47:03 PM

178

Chapter 21

(e)

I m ax = 2I rm s = 2 ( 6.00 A ) = 8.49 A

(f )

Pav = I rm2 s R = ( 6.00 A ) ( 20.0 Ω ) = 720 W

(g)

At t = 0.005 0 s, the instantaneous current is

2

i=

21.7

Δv (170 V ) (170 V ) = sin ⎡⎣( 60p rad s ) ( 0.005 0 s ) ⎤⎦ = sin ( 0.94 rad ) = 6.9 A R 20.0 Ω 20.0 Ω

(h)

The argument of the sine function has units of [wt ] = ( rad s )( s ) = radians .

(a)

The expression for capacitive reactance is XC = 1 2p fC. Thus, if XC < 175 Ω, it is necessary that f=

1 1 > 2p C XC 2p ( 22.0 ×10 −6 F ) (175 Ω)

or

f > 41.3 Hz

For C1 , the reactance is XC,1 = 1 2p fC1, while for C2 , XC, 2 = 1 2p fC 2. Thus, for the same frequencies, the ratio of the reactance for the two capacitors is

(b)

XC, 2 XC,1

⎛ 1 ⎞ ⎛ 2p f C ⎞ C 1 ⎟⎟ ⎜⎜ ⎟⎟ = 1 = ⎜⎜ 2p f C 1 ⎝ ⎠ C2 2 ⎝ ⎠

or

⎛C ⎞ XC, 2 = ⎜ 1 ⎟ XC,1 ⎝ C2 ⎠

If C1 = 22.0 mF, C2 = 44.0 mF, and XC,1 < 175 Ω, we have ⎛ 22.0 mF ⎞ XC, 2 < ⎜ ⎟ (175 Ω) ⎝ 44.0 mF ⎠ 21.8

21.9

I m ax = 2 I rm s =

68719_21_ch21_p171-197.indd 178

XC

XC, 2 < 87.5 Ω

= 2 ( ΔVrm s ) 2p f C

(a)

I m ax = 2 (120 V ) 2p ( 60.0 Hz )( 2.20 × 10 − 6 C V ) = 0.141 A = 141 mA

(b)

I m ax = 2 ( 240 V ) 2p ( 50.0 Hz )( 2.20 × 10 − 6 C V ) = 0.235 A = 235 mA

I rm s =

ΔVrm s =2p f C ( ΔVrm s ) , so XC

f=

21.10

2 ( ΔVrm s )

or

I rm s 0.30 A = = 4.0 × 10 2 Hz 2p C ( ΔVrm s ) 2p ( 4.0 × 10 −6 F )( 30 V ) 1 1 = = 221 Ω 2p fC 2p ( 60.0 Hz )(12.0 × 10 −6 F )

(a)

XC =

(b)

I rm s =

(c)

I m ax = 2 I rm s = 2 ( 0.163 A ) = 0.231 A

(d)

No. The charge is a maximum when the voltage is a maximum, but the voltage across a capacitor is 90° out of phase with the current.

ΔVC , rm s XC

=

36.0 V = 0.163 A 221 Ω

1/7/11 2:47:06 PM

Alternating Current Circuits and Electromagnetic Waves

21.11

21.12

21.13

I m ax =

ΔVm ax = 2p fC ( ΔVm ax ) = 2p ( 90.0 Hz )( 3.70 × 10 −6 F )( 48.0 V ) XC

or

I m ax = 1.00 × 10 −1 A = 100 mA

(a)

By inspection, ΔVC, m ax = 98.0 V, so ΔVC , rm s =

(b)

Also by inspection, w = 80p rad s, so f =

(c)

I rm s =

(d)

XC =

(e)

XC =

ΔVC, m ax 2

=

179

98.0 V = 69.3 V 2

w 80p rad s = = 40 Hz 2p 2p rad

I m ax 0.500 A = = 0.354 A 2 2 ΔVC , m ax I m ax

=

98.0 V = 196 Ω 0.500 A

1 1 = , 2p fC wC

C=

so

X L = 2p ( 60.0 Hz ) L = 54.0 Ω

⇒

1 1 = = 2.03 × 10 −5 F = 20.3 mF w XC (80π rad s )(196 Ω )

2p L =

54.0 Ω = 0.900 Ω ⋅s 60.0 s −1

Then, when ΔVrm s = 100 V and f = 50.0 Hz, the maximum current will be I m ax = 21.14

21.15

ΔVm ax = XL

2 ( ΔVrm s )

( 2p L ) f

=

2 (100 V ) = 3.14 A ( 0.900 Ω ⋅s )( 50.0 Hz )

(a)

X L = 2p f L = 2p (80.0 Hz )( 25.0 × 10 −3 H ) = 12.6 Ω

(b)

I rm s =

(c)

I m ax = 2 I rm s = 2 ( 6.19 A ) = 8.75 A

(a)

I m ax = L=

(b)

ΔVL , rm s XL

=

78.0 V = 6.19 A 12.6 Ω

ΔVm ax ΔVm ax , so = XL 2p fL ΔVm ax 100 V = = 4.24 × 10 −2 H = 42.4 mH 2p f I m ax 2p ( 50.0 Hz )( 7.50 A )

I m ax = ΔVm ax X L = ΔVm ax w L , or I m ax is inversely proportional to w . Thus, I m ax,1 I m ax, 2 = w 2 w 1 , or ⎛I ⎞ ⎛ 7.50 A ⎞ w 2 = ⎜ m ax,1 ⎟ w 1 = ⎜ 2p ( 50.0 Hz )] = 942 rad s ⎝ 2.50 A ⎟⎠ [ ⎝ I m ax,2 ⎠

21.16

Given: vL = (1.20 × 10 2 V ) sin ( 30p t ) and L = 0.500 H (a)

By inspection, w = 30p rad s, so f =

w 30p rad s = = 15.0 Hz 2p 2p

continued on next page

68719_21_ch21_p171-197.indd 179

1/7/11 2:47:10 PM

180

Chapter 21

(b)

Also by inspection, ΔVL , m ax = 1.20 × 10 2 V, so ΔVL, rm s =

ΔVL, m ax 2

=

1.20 ×10 2 V = 84.9 V 2

(c)

X L = 2p fL = w L = ( 30p rad s )( 0.500 H ) = 47.1 Ω

(d)

I rm s =

(e)

I m ax = 2 I rm s = 2 (1.80 A ) = 2.55 A

(f )

The phase difference between the voltage across an inductor and the current through the inductor is fL = 90°, so the average power delivered to the inductor is

ΔVL, rm s XL

=

84.9 V = 1.80 A 47.1 Ω

PL, av = I rm s ΔVL, rm s cosfL = I rm s ΔVL, rm s cos (90°) = 0 (g)

When a sinusoidal voltage with a peak value ΔVL, m ax is applied to an inductor, the current through the inductor also varies sinusoidally in time, with the same frequency as the applied voltage, and has a maximum value of I m ax = ΔVL, m ax X L . However, the current lags behind the voltage in phase by a quarter-cycle, or p 2 radians. Thus, if the voltage is given by ΔvL = ΔVL, m ax sin (wt ) , the current as a function of time is i = I m ax sin (wt − p 2 ) . In the case of the given inductor, the current through it will be i = ( 2.55 A ) sin ( 30p t − p 2 ) .

(h)

When i = +1.00 A, we have sin ( 30p t − p 2 ) = (1.00 A 2.55 A ) , or 30p t − p 2 = sin −1 (1.00 A 2.55 A ) = sin −1 ( 0.392 ) = 0.403 rad t=

and 21.17

p 2 rad + 0.403 rad = 2.09 × 10 −2 s = 20.9 ms 30p rad s

From L = N Φ B I (see Section 20.5 in the textbook), the total ﬂux through the coil is Φ B, total = N Φ B = L ⋅ I, where Φ B is the ﬂux through a single turn on the coil. Thus,

(Φ

B, total

)

m ax

⎡ ΔV ⎤ = L ⋅ I m ax = L ⋅ ⎢ m ax ⎥ ⎣ XL ⎦ =L

21.18

(a)

2 ( ΔVrm s ) 2 (120 V ) = = 0.450 T ⋅ m 2 2p fL 2p (60.0 Hz )

The applied voltage is Δv = ΔVm ax sin (wt ) = (80.0 V ) sin (150t ) , so ΔVm ax = 80.0 V and w = 2p f = 150 rad s. The impedance for the circuit is Z = R 2 + ( X L − XC ) = R 2 + ( 2p fL − 1 2p fC ) = R 2 + (w L − 1 w C ) 2

or

2

2

⎡ ⎤ 1 Z = ( 40.0 Ω ) + ⎢(150 rad s ) (80.0 × 10 −3 H ) − ⎥ −6 (150 rad s ) (125 × 10 F ) ⎦⎥ ⎢⎣

2

2

= 57.5 Ω (b)

68719_21_ch21_p171-197.indd 180

I m ax =

ΔVm ax 80.0 V = = 1.39 A Z 57.5 Ω

1/7/11 2:47:13 PM

Alternating Current Circuits and Electromagnetic Waves

21.19

XC =

181

1 1 = = 66.3 Ω 2p f C 2p ( 60.0 Hz )( 40.0 × 10 −6 F )

Z = R 2 + ( X L − XC ) = 2

( 50.0 Ω )2 + ( 0 − 66.3 Ω )2 = 83.0 Ω

ΔVrm s 30.0 V = = 0.361 A Z 83.0 Ω

(a)

I rm s =

(b)

ΔVR , rm s = I rm s R = ( 0.361 A )( 50.0 Ω ) = 18.1 V

(c)

ΔVC , rm s = I rm s XC = ( 0.361 A )( 66.3 Ω ) = 23.9 V

(d)

⎛ X − XC ⎞ −1 ⎛ 0 − 66.3 Ω ⎞ f = tan −1 ⎜ L ⎟ = −53.0° ⎟ = tan ⎜⎝ ⎝ R ⎠ 50.0 Ω ⎠ so the voltage lags behind the current by 53° .

21.20

(a)

X L = 2p f L = 2p ( 50.0 Hz )( 400 × 10 −3 H ) = 126 Ω XC =

1 1 = = 719 Ω 2p f C 2p ( 50.0 Hz )( 4.43 × 10 −6 F )

Z = R 2 + ( X L − XC ) = 2

m

( 500 Ω )2 + (126 Ω − 719 Ω )2 = 776 Ω

ΔVm ax = I m ax Z = ( 0.250 A ) ( 776 Ω ) = 194 V (b)

⎛ X − XC ⎞ −1 ⎛ 126 Ω − 719 Ω ⎞ f = tan −1 ⎜ L ⎟⎠ = − 49.9° ⎟ = tan ⎜⎝ ⎝ R ⎠ 500 Ω Thus, the current leads the voltage by 49.9° .

21.21

(a)

X L = 2p f L = 2p ( 240 Hz )( 2.50 H ) = 3.77 × 10 3 Ω XC =

1 1 = = 2.65 × 10 3 Ω 2p f C 2p ( 240 Hz )( 0.250 × 10 −6 F )

Z = R 2 + ( X L − XC ) = 2

21.22

( 900 Ω )2 + ⎡⎣( 3.77 − 2.65) × 103 Ω ⎤⎦ = 1.44 kΩ 2

ΔVm ax 140 V = = 0.097 2 A 1.44 × 10 3 Ω Z

(b)

I m ax =

(c)

⎡ ( 3.77 − 2.65) × 10 3 Ω ⎤ ⎛ X − XC ⎞ f = tan −1 ⎜ L = tan −1 ⎢ ⎟ ⎥ = 51.2° ⎝ 900 Ω R ⎠ ⎣ ⎦

(d)

f > 0 , so the voltage leads the current

X L = 2p fL = 2p ( 60.0 Hz )( 0.100 H ) = 37.7 Ω XC =

1 1 = = 265 Ω 2p fC 2p ( 60.0 Hz )(10.0 × 10 −6 F )

Z = R 2 + ( X L − XC ) = 2

( 50.0 Ω )2 + ( 37.7 Ω − 265 Ω )2 = 233 Ω

(a)

ΔVR, rm s = I rm s R = ( 2.75 A ) (50.0 Ω) = 138 V

(b)

ΔVL, rm s = I rm s X L = ( 2.75 A ) (37.7 Ω) = 104 V continued on next page

68719_21_ch21_p171-197.indd 181

1/7/11 2:47:16 PM

182

Chapter 21

(c)

ΔVC = I rm s XC = ( 2.75 A )( 265 Ω ) = 729 V

(d)

ΔVrm s = I rm s Z = ( 2.75 A )( 233 Ω ) = 641 V

(e)

21.23

(a)

X L = 2p f L = 2p ( 60.0 Hz )( 460 × 10 −3 H ) = 173 Ω XC =

1 1 = = 126 Ω 2p f C 2p ( 60.0 Hz )( 21.0 × 10 −6 F )

⎛ X − XC ⎞ −1 ⎛ 173 Ω − 126 Ω ⎞ Thus, f = tan −1 ⎜ L ⎟⎠ = + 17.4° ⎟ = tan ⎜⎝ ⎝ R ⎠ 150 Ω (b)

21.24

X L > XC and f > 0, so the voltage leads the current. Hence, the voltage reaches its maximum ﬁrst.

X L = 2p fL = 2p ( 60 Hz ) ( 2.8 H ) = 1.1 × 10 3 Ω Z = R 2 + ( X L − XC ) = 2

(1.2 × 10 Ω) + (1.1× 10 3

2

3

Ω − 0 ) = 1.6 × 10 3 Ω 2

ΔVm ax 170 V = = 0.11 A 1.6 × 10 3 Ω Z

(a)

I m ax =

(b)

ΔVR, m ax = I m ax R = ( 0.11 A ) (1.2 ×10 3 Ω) = 1.3×10 2 V ΔVL , m ax = I m ax X L = ( 0.11 A ) (1.1×10 3 Ω) = 1.2 ×10 2 V

(c)

When the instantaneous current is a maximum ( i = I m ax ) , the instantaneous voltage across the resistor is ΔvR = iR = I m ax R = ΔVR, m ax = 1.3×10 2 V . The instantaneous voltage across an inductor is always 90° or a quarter cycle out of phase with the instantaneous current. Thus, when i = I m ax , ΔvL = 0 . Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed path. Thus, for this series circuit Δvsource = ΔvR + ΔvL , and at this instant when i = I m ax , we have Δvsource = I m ax R + 0 = 1.3 × 10 2 V .

(d)

68719_21_ch21_p171-197.indd 182

When the instantaneous current i is zero, the instantaneous voltage across the resistor is ΔvR = iR = 0 . Again, the instantaneous voltage across an inductor is a quarter cycle out of phase with the current. Thus, when i = 0, we must have ΔvL = ΔVL, m ax = 1.2 ×10 2 V . Then, applying Kirchhoff’s loop rule to the instantaneous voltages around the series circuit at the instant when i = 0 gives Δvsource = ΔvR + ΔvL = 0 + ΔVL, m ax = 1.2 ×10 2 V .

1/7/11 2:47:20 PM

Alternating Current Circuits and Electromagnetic Waves

21.25

XC =

1 1 = = 1.33 × 108 Ω 2p f C 2p ( 60.0 Hz )( 20.0 × 10 −12 F )

(50.0 × 10 Ω) + (1.33 × 10 Ω)

2 Z RC = Rbody + XC2 =

and

183

I rm s =

3

( ΔV

secondary

)

rm s

Z RC

=

2

8

2

= 1.33 × 108 Ω

5 000 V = 3.76 × 10 −5 A 1.33 × 108 Ω

Therefore, ΔVbody, rm s = I rm s Rbody = ( 3.76 × 10 −5 A ) ( 50.0 × 10 3 Ω ) = 1.88 V . 21.26

1 1 = = 88.4 Ω 2p f C 2p ( 60.0 Hz )( 30.0 × 10 −6 F )

(a)

XC =

(b)

Z = R 2 + ( 0 − XC ) = R 2 + XC2 =

(c)

I m ax =

(d)

The phase angle in this RC circuit is

( 60.0 Ω )2 + (88.4 Ω )2 = 107 Ω

2

ΔVm ax 1.20 × 10 2 V = = 1.12 A Z 107 Ω

⎛ X − XC ⎞ −1 ⎛ 0 − 88.4 Ω ⎞ f = tan −1 ⎜ L ⎟ = −55.8° ⎟ = tan ⎜⎝ ⎝ R ⎠ 60.0 Ω ⎠ Since f < 0,

21.27

(e)

Adding an inductor will change the impedance and the current in the circuit. If the added inductive reactance is X L < 2XC , the impedance will be decreased and the current will increase. However, if X L > 2XC , the impedance will be increased and the current will decrease.

(a)

X L = 2p fL = 2p ( 50.0 Hz )( 0.150 H ) = 47.1 Ω

(b)

XC =

(c)

Z=

(d)

Z = R 2 + ( X L − XC ) R=

(e) 21.28

the voltage lags behind the current by 55.8° .

1 1 = = 637 Ω 2p f C 2p ( 50.0 Hz )( 5.00 × 10 −6 F )

ΔVm ax 240 V = = 2.40 × 10 3 Ω = 2.40 kΩ 1.00 × 10 −1 A I m ax

( 2.40 × 10 Ω) 3

2

2

so

R = Z 2 − ( X L − XC )

2

− ( 47.1 Ω − 637 Ω ) = 2.33 × 10 3 Ω = 2.33 kΩ 2

⎛ X − XC ⎞ −1 ⎛ 47.1 Ω − 637 Ω ⎞ f = tan −1 ⎜ L ⎟ = − 14.2° ⎟⎠ = tan ⎜⎝ ⎝ R 2.33 × 10 3 Ω ⎠

X L = w L and XC =1 wC , where w = 2p f . At resonance (i.e., when X L = XC ), we have w 0 = 2p f0 =

1 LC

or

f0 =

1 2p LC

With L = 57.0 mH and C = 57.0 mF, this resonance frequency is f0 =

68719_21_ch21_p171-197.indd 183

1 2p

(57.0 × 10

−6

H ) ( 57.0 × 10

−6

F)

= 2.79 × 10 3 Hz = 2.79 kHz

1/7/11 2:47:23 PM

184

21.29

Chapter 21

X L = 2p fL = 2p ( 50.0 Hz )( 0.185 H ) = 58.1 Ω XC =

1 1 = = 49.0 Ω 2p f C 2p ( 50.0 Hz )( 65.0 × 10 −6 F )

Z ad = R 2 + ( X L − XC ) = 2

21.30

(

( 40.0 Ω )2 + ( 58.1 Ω − 49.0 Ω )2 = 41.0 Ω

)

ΔVm ax 2 ΔVrm s 150 V = = = 2.59 A Z ad Z ad ( 41.0 Ω ) 2

and

I rm s =

(a)

Z ab = R = 40.0 Ω, so ( ΔVrm s )ab = I rm s Z ab = ( 2.59 A )( 40.0 Ω ) = 104 V

(b)

Z bc = X L = 58.1 Ω, and ( ΔVrm s )bc = I rm s Z bc = ( 2.59 A )( 58.1 Ω ) = 150 V

(c)

Z cd = XC = 49.0 Ω, and ( ΔVrm s )cd = I rm s Z cd = ( 2.59 A )( 49.0 Ω ) = 127 V

(d)

Z bd = X L − XC = 9.10 Ω, so ( ΔVrm s )bd = I rm s Z bd = ( 2.59 A )( 9.10 Ω ) = 23.6 V

XC =

1 1 = = 1.1× 10 3 Ω 2p fC 2p ( 60 Hz )( 2.5 × 10 −6 F )

Z = R 2 + ( X L − XC ) = 2

(1.2 × 10 Ω) + ( 0 − 1.1× 10 Ω) 3

2

3

2

= 1.6 × 10 3 Ω

ΔVm ax 170 V = = 0.11 A 1.6 × 10 3 Ω Z

(a)

I m ax =

(b)

ΔVR , m ax = I m ax R = ( 0.11 A )(1.2 × 10 3 Ω ) = 1.3 × 10 2 V ΔVC , m ax = I m ax XC = ( 0.11 A )(1.1× 10 3 Ω ) = 1.2 × 10 2 V

(c)

When the instantaneous current i is zero, the instantaneous voltage across the resistor is ΔvR = iR = 0 . The instantaneous voltage across a capacitor is always 90°, or a quarter cycle, out of phase with the instantaneous current. Thus, when i = 0, ΔvC = ΔVC, m ax = 1.2 ×10 2 V , and qC = C ( ΔvC ) = ( 2.5 × 10 −6 F ) (1.2 × 10 2 V ) = 3.0 × 10 −4 C = 300 mC Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed path. Thus, for this series circuit, Δvsource + ΔvR + ΔvC = 0, and at this instant when i = 0, we have Δvsource = 0 + ΔvC = ΔVC , m ax = 1.2 ×10 2 V .

(d)

When the instantaneous current is a maximum (i = I m ax ), the instantaneous voltage across the resistor is ΔvR = iR = I m ax R = ΔVR, m ax = 1.3×10 2 V . Again, the instantaneous voltage across a capacitor is a quarter cycle out of phase with the current. Thus, when i = I m ax , we must have ΔvC = 0 and qC = C ΔvC = 0 . Then, applying Kirchhoff’s loop rule to the instantaneous voltages around the series circuit at the instant when i = I m ax and ΔvC = 0 gives Δvsource + ΔvR + ΔvC = 0

68719_21_ch21_p171-197.indd 184

⇒

Δvsource = ΔvR = ΔVR, m ax = 1.3×10 2 V

1/7/11 2:47:26 PM

Alternating Current Circuits and Electromagnetic Waves

21.31

ΔVrm s 104 V = = 208 Ω I rm s 0.500 A

(a)

Z=

(b)

Pav = I rm2 s R gives R =

(c)

Z = R 2 + X L2 , so X L = Z 2 − R 2 = L=

and 21.32

185

Pav 10.0 W = = 40.0 Ω 2 I rm2 s ( 0.500 A )

( 208 Ω )2 − ( 40.0 Ω )2 = 204 Ω

XL 204 Ω = = 0.541 H 2p f 2p ( 60.0 Hz )

Given v = ΔVm ax sin (wt ) = ( 90.0 V ) sin ( 350t ) , observe that ΔVm ax = 90.0 V and w = 350 rad s. Also, the net reactance is X L − XC = 2p fL − 1 2p fC = w L − 1 w C. (a)

X L − XC = w L −

1 1 = ( 350 rad s )( 0.200 H ) − = − 44.3 Ω wC (350 rad s ) ( 25.0 × 10 −6 F ) Z = R 2 ( X L − XC ) = 2

so the impedance is ΔVrm s ΔVm ax = Z Z

2

( 50.0 Ω )2 + ( − 44.3 Ω )2 = 66.8 Ω

90.0 V = 0.953 A 2 ( 66.8 Ω )

(b)

I rm s =

=

(c)

The phase difference between the applied voltage and the current is ⎛ X − XC ⎞ −1 ⎛ − 44.3 Ω ⎞ f = tan −1 ⎜ L ⎟ = −41.5° ⎟⎠ = tan ⎜⎝ ⎝ R 50.0 Ω ⎠ so the average power delivered to the circuit is ⎛ ΔV ⎞ ⎛ 90.0 V ⎞ Pav = I rm s ΔVrm s cosf = I rm s ⎜ m ax ⎟ cosf = ( 0.953 A ) ⎜ ⎟ cos ( −41.5° ) = 45.4 W ⎝ 2 ⎠ ⎝ 2 ⎠

21.33

If Δv = (100 V ) sin ⎡⎣(1 000 rad s ) t ⎤⎦ , then ΔVm ax = 100 V and w = 1 000 rad s. Thus,

X L = w L = (1 000 rad s )( 0.500 H ) = 500 Ω

and

XC =

Therefore, and

I rm s =

1 1 = = 200 Ω w C (1 000 rad s ) ( 5.00 × 10 −6 C ) Z = R 2 + ( X L − XC ) = 2

ΔVrm s ΔVm ax = Z Z

2

=

( 400 Ω )2 + ( 300 Ω )2 = 500 Ω

100 V 0.200 A = 2 ( 500 Ω ) 2

The power delivered to the circuit equals the power dissipated in the resistor, or 2

⎛ 0.200 A ⎞ P = I rm2 s R = ⎜ ⎟ ( 400 Ω ) = 8.00 W ⎝ 2 ⎠ 21.34

The rms current in the circuit is I rm s =

ΔVrm s 160 V = = 2.00 A 80.0 Ω Z continued on next page

68719_21_ch21_p171-197.indd 185

1/7/11 2:47:30 PM

186

Chapter 21

and the average power delivered to the circuit is Pav = I rm s ( ΔVrm s cosf ) = I rm s ΔVR , rm s = I rm s ( I rm s R ) = I rm2 s R = ( 2.00 A ) ( 22.0 Ω ) = 88.0 W 2

21.35

Pav = I rm2 s R = I rm s ( I rm s R) = I rm s ( ΔVR, rm s ), so I rm s =

(a)

R=

Thus,

ΔVR, rm s

50 V = 1.8 ×10 2 Ω 0.28 A

=

I rm s

Pav 14 W = = 0.28 A ΔVR , rm s 50 V

Z = R 2 + X L2 , which yields

(b)

2

2 ⎛ ΔV ⎞ 2 ⎛ 90 V ⎞ X L = Z − R = ⎜ rm s ⎟ − R 2 = ⎜ − (1.8 × 10 2 Ω ) = 2.7 × 10 2 Ω ⎟ ⎝ ⎠ I 0.28 A ⎝ rm s ⎠ 2

and 21.36

L=

2

XL 2.7 × 10 2 Ω = = 0.72 H 2p f 2p ( 60 Hz )

X L = 2p fL = 2p ( 600 Hz )( 6.0 × 10 −3 H ) = 23 Ω XC =

1 1 = = 11 Ω 2p fC 2p ( 600 Hz )( 25 × 10 −6 F )

Z = R 2 + ( X L − XC ) = 2

(a)

( 25 Ω )2 + ( 23 Ω − 11 Ω )2 = 28 Ω

⎛ ΔV ⎞ ⎛ 10 V ⎞ ΔVR, rm s = I rm s R = ⎜ rm s ⎟ R = ⎜ ⎟ ( 25 Ω) = 8.9 V ⎝ 28 Ω ⎠ ⎝ Z ⎠ ⎛ ΔV ⎞ ⎛ 10 V ⎞ ΔVL, rm s = I rm s X L = ⎜ rm s ⎟ X L = ⎜ ⎟ ( 23 Ω) = 8.2 V ⎝ 28 Ω ⎠ ⎝ Z ⎠ ⎛ ΔV ⎞ ⎛ 10 V ⎞ ΔVC, rm s = I rm s XC = ⎜ rm s ⎟ XC = ⎜ ⎟ (11 Ω) = 3.9 V ⎝ 28 Ω ⎠ ⎝ Z ⎠ No , ΔVR, rm s + ΔVL, rm s + ΔVC , rm s = 8.9 V + 8.2 V + 3.8 V = 20.9 V ≠ 10 V

(b)

The power delivered to the resistor is the greatest. No power losses occur in an ideal capacitor or inductor. ⎛ ΔV ⎞ ⎛ 10 V ⎞ Pav = I rm2 s R = ⎜ rm s ⎟ R = ⎜ ( 25 Ω ) = 3.2 W ⎝ 28 Ω ⎟⎠ ⎝ Z ⎠ 2

(c) 21.37

(

)

The resonance frequency of a series RLC circuit is f0 = 1 2p LC . Thus, if L = 1.40 mH and the desired resonance frequency is f0 = 99.7 MHz, the needed capacitance is C=

21.38

2

1 1 = = 1.82 × 10 −12 F = 1.82 pF 2 4p 2 f02 L 4p 2 ( 99.7 × 10 6 Hz ) (1.40 × 10 −6 H )

(

)

The resonance frequency of a series RLC circuit is f0 = 1 2p LC . Thus, the ratio of the resonance frequencies when the same inductance is used with two different capacitances in the circuit is f0, 2 f0,1

⎛ 1 = ⎜⎜ ⎝ 2p LC2

⎞ ⎛ 2p LC ⎞ C1 1 ⎟⎟ ⎜⎜ ⎟⎟ = 1 C2 ⎠⎝ ⎠

continued on next page

68719_21_ch21_p171-197.indd 186

1/7/11 2:47:33 PM

Alternating Current Circuits and Electromagnetic Waves

187

If f0,1 = 2.84 kHz when C1 = 6.50 mF, the resonance frequency when the capacitance is C2 = 9.80 mF will be f0, 2 = f0,1

21.39

f0 =

C1 6.50 mF = ( 2.84 kHz ) = 2.31 kHz C2 9.80 mF

1 1 , so C = 4p 2 f02 L 2p LC

For f0 = ( f0 )m in = 500 kHz = 5.00 × 10 5 Hz C = Cm ax =

1

4p

2

= 5.1× 10 −8 F = 51 nF

(5.00 × 10 Hz ) ( 2.0 × 10−6 H ) 2

5

For f0 = ( f0 )m ax = 1600 kHz = 1.60 × 10 6 Hz C = Cm in = 21.40

(a)

4p

2

(1.60 × 10

1

Hz ) ( 2.0 × 10 −6 H ) 2

6

= 4.9 × 10 −9 F = 4.9 nF

At resonance, X L = XC , so the impedance will be Z = R 2 + ( X L − XC ) = R 2 + 0 = R = 15 Ω 2

(b)

When X L = XC , we have 2p fL = f=

1 2p LC

=

1 , which yields 2p fC 1

2p

( 0.20 H )( 75 × 10 −6 F )

= 41 Hz

(c)

The current is a maximum at resonance , where the impedance has its minimum value of Z = R.

(d)

At f = 60 Hz, X L = 2p ( 60 Hz )( 0.20 H ) = 75 Ω, XC =

1 = 35 Ω, 2p ( 60 Hz )( 75 × 10 −6 F )

and Z=

(15 Ω )2 + ( 75 Ω − 35 Ω )2 = 43 Ω

Thus, I rm s = 21.41

(

ΔVm ax ΔVrm s = Z Z

2

)=

150 V = 2.5 A . 2 ( 43 Ω )

(

)

The resonance frequency of an LC circuit is f0 = 1 2p LC . If the capacitance remains constant while the inductance increases by 1.000%, the new resonance frequency will be f0′ =

f0 1 1 ⎛ 1 ⎞ =⎜ ⎟⎠ 2p LC = 1.010 ⎝ 1.010 2p (1.010L )C

The beat frequency detected in this case will be fbeat = f0 − f0′, or 1 ⎞ 1 ⎞ ⎛ ⎛ fbeat = ⎜ 1− f0 = ⎜ 1− ( 725 kHz ) = 3.60 kHz ⎟ ⎝ ⎝ 1.010 ⎠ 1.010 ⎟⎠

68719_21_ch21_p171-197.indd 187

1/7/11 2:47:36 PM

188

21.42

Chapter 21

The resonance frequency is w 0 = 2p f0 = 1 (a)

LC . Also, X L = w L and XC = 1 w C .

L ⎛ 1 ⎞ At resonance, XC = X L = w 0 L = ⎜ L= = ⎝ LC ⎟⎠ C

3.00 H = 1 000 Ω. 3.00 × 10 −6 F

Thus, Z = R 2 + 0 2 = R, I rm s = ΔVrm s Z = 120 V 30.0 Ω = 4.00 A, and 2 Pav = I rm2 s R = ( 4.00 A ) ( 30.0 Ω ) = 480 W . (b)

(

1 1 At w = w 0 ; X L = X L 2 2

w0

) = 500 Ω, X = 2 ( X ) = 2 000 Ω, C

Z = R 2 + ( X L − XC ) =

( 30.0 Ω )2 + ( 500 Ω − 2 000 Ω )2 = 1 500 Ω

2

and I rm s = (c)

C w 0

120 V 2 = 0.080 0 A. Thus, Pav = I rm2 s R = ( 0.080 0 A ) ( 30.0 Ω ) = 0.192 W . 1 500 Ω

(

(

)

1 1 At w = w 0 ; X L = X L w = 250 Ω, XC = 4 XC 4 4 120 V and I rm s = = 0.032 0 A. 3 750 Ω 0

w0

) = 4 000 Ω,

Z = 3 750 Ω,

Therefore, Pav = I rm2 s R = ( 0.032 0 A ) ( 30.0 Ω ) = 3.07 × 10 −2 W = 30.7 mW . 2

(d)

(

At w = 2w 0 ; X L = 2 X L I rm s =

(e)

w0

) = 2 000 Ω, X

C

=

(

1 XC 2

w0

) = 500 Ω,

Z = 1 500 Ω, and

120 V 2 = 0.080 0 A. Then, Pav = I rm2 s R = ( 0.080 0 A ) ( 30.0 Ω ) = 0.192 W . 1 500 Ω

(

At w = 4w 0 ; X L = 4 X L and I rm s =

w0

) = 4 000 Ω, X

C

=

(

1 XC 4

w0

) = 250 Ω,

Z = 3 750 Ω,

120 V = 0.032 0 A. Hence, 3 750 Ω

Pav = I rm2 s R = ( 0.032 0 A ) ( 30.0 Ω ) = 3.07 × 10 −2 W = 30.7 mW 2

The power delivered to the circuit is a maximum when the rms current is a maximum. This occurs when the frequency of the source is equal to the resonance frequency of the circuit. 21.43

The maximum output voltage ( ΔVm ax )2 is related to the maximum input voltage ( ΔVm ax )1 by the N expression ( ΔVm ax )2 = 2 ( ΔVm ax )1, where N1 and N 2 are the number of turns on the primary coil N1 and the secondary coil, respectively. Thus, for the given transformer,

( ΔV ) m ax

2

=

1 500 (170 V ) = 1.02 × 103 V 250

and the rms voltage across the secondary is ( ΔVrm s )2 = 21.44

(a)

( ΔV ) m ax

2

2

=

1.02 × 10 3 V = 721 V . 2

The output voltage of the transformer is ⎛N ⎞ ⎛ 1⎞ ΔV2 , rm s = ⎜ 2 ⎟ ΔV1, rm s = ⎜ ⎟ (120 V ) = 9.23 V ⎝ 13 ⎠ ⎝ N1 ⎠

continued on next page

68719_21_ch21_p171-197.indd 188

1/7/11 2:47:40 PM

Alternating Current Circuits and Electromagnetic Waves

(b)

Assuming an ideal transformer, Poutput = Pinput , and the power delivered to the CD player is

(

(P ) = (P ) av

21.45

189

)

= I1, rm s ΔV1, rm s = ( 0.250 A )(120 V ) = 30.0 W

av 1

2

The power input to the transformer is

(P ) av

(

)

= ΔV1, rm s I1, rm s = ( 3 600 V )( 50 A ) = 1.8 × 10 5 W

input

For an ideal transformer, ( Pav )ouput = ( ΔV2, rm s ) I 2, rm s = ( Pav )input, so the current in the long-distance power line is I 2 , rm s =

(P ) av

(ΔV

=

input

2, rm s

)

1.8 ×10 5 W = 1.8 A 100 000 V

The power dissipated as heat in the line is then Plost = I 22, rm s Rline = (1.8 A ) (100 Ω ) = 3.2 × 10 2 W 2

The percentage of the power delivered by the generator that is lost in the line is % Lost = 21.46

(a)

(b)

⎛ 3.2 × 10 2 W ⎞ Plost × 100% = ⎜ × 100% = 0.18% Pinput ⎝ 1.8 × 10 5 W ⎟⎠

Since the transformer is to step the voltage down from 120 volts to 6.0 volts, the secondary must have fewer turns than the primary. For an ideal transformer, ( Pav )input = ( Pav )ouput , or ( ΔV1, rm s ) I1, rm s = ( ΔV2, rm s ) I 2, rm s , so the current in the primary will be

(ΔV

I1, rm s = (c)

)I

2, rm s

2, rm s

ΔV1, rm s

=

(6.0 V ) (500 mA ) = 25 mA 120 V

The ratio of the secondary to primary voltages is the same as the ratio of the number of turns on the secondary and primary coils, ΔV2 ΔV1 = N 2 N1. Thus, the number of turns needed on the secondary coil of this step down transformer is ⎛ ΔV ⎞ ⎛ 6.0 V ⎞ N 2 = N1 ⎜ 2 ⎟ = ( 400 ) ⎜ = 20 turns ⎝ 120 V ⎟⎠ ⎝ ΔV1 ⎠

21.47

(a)

At 90% efﬁciency, ( Pav )output = 0.90 ( Pav )input. Thus, if ( Pav )output = 1 000 kW, the input power to the primary is

(P ) av

68719_21_ch21_p171-197.indd 189

(b)

I1, rm s =

(c)

I 2, rm s =

input

(P ) av

=

(P )

input

ΔV1, rm s

(P ) av

output

ΔV2, rm s

av

output

0.90 =

=

1 000 kW = 1.1× 10 3 kW 0.90

1.1×10 3 kW 1.1×10 6 W = 3.1×10 2 A = ΔV1, rm s 3 600 V

=

1 000 kW 1.0 ×10 6 W = = 8.3×10 3 A ΔV2, rm s 120 V

1/7/11 2:47:44 PM

190

21.48

Chapter 21

Rline = ( 4.50 × 10 −4 Ω m ) ( 6.44 × 10 5 m ) = 290 Ω (a)

The power transmitted is ( Pav )transm itted = ( ΔVrm s ) I rm s, so I rm s =

(P ) av

transm itted

ΔVrm s

=

5.00 × 10 6 W = 10.0 A 500 × 10 3 V

Thus, ( Pav )loss = I rm2 s Rline = (10.0 A ) ( 290 Ω ) = 2.90 × 10 4 W = 29.0 kW . 2

(b)

The power input to the line is

(P ) av

input

= ( Pav )transm itted + ( Pav )loss = 5.00 × 10 6 W + 2.90 × 10 4 W = 5.03 × 10 6 W

and the fraction of input power lost during transmission is fraction =

(P ) (P ) av

av

(c)

loss

=

input

2.90 × 10 4 W = 0.005 77 or 0.577% 5.03 × 10 6 W

It is impossible to deliver the needed power at the generator voltage of 4.50 kV. The maximum line current with an input voltage of 4.50 kV to the line occurs when the line is shorted out at the customer’s end, and this current is

(I ) rm s

m ax

=

ΔVrm s 4 500 V = = 15.5 A 290 Ω Rline

The maximum input power is then

(P ) input

m ax

= ( ΔVrm s ) ( I rm s )m ax = ( 4.50 × 10 3 V )(15.5 A ) = 6.98 × 10 4 W = 69.8 kW

This is far short of meeting the customer’s request, and all of this power is lost in the transmission line. 21.49

From v = l f , the wavelength is l=

v 3.00 × 108 m s = = 4.00 × 10 6 m = 4 000 km f 75 Hz

The required length of the antenna is then L = l 4 = 1 000 km , or about 621 miles. Not very practical at all. 21.50

d 6.44 × 1018 m ⎛ 1y ⎞ 2 = ⎜ ⎟ = 6.80 × 10 y c 3.00 × 108 m s ⎝ 3.156 × 10 7 s ⎠

(a)

t=

(b)

From Table C.4 (in Appendix C of the textbook), the average Earth-Sun distance is d = 1.496 × 1011 m, giving the transit time as t=

d 1.496 × 1011 m ⎛ 1 min ⎞ = ⎜ ⎟ = 8.31 min c 3.00 × 108 m s ⎝ 60 s ⎠

continued on next page

68719_21_ch21_p171-197.indd 190

1/7/11 2:47:46 PM

Alternating Current Circuits and Electromagnetic Waves

(c)

Also from Table C.4, the average Earth-Moon distance is d = 3.84 × 108 m, giving the time for the round trip as 8 2d 2 ( 3.84 × 10 m ) = = 2.56 s 8 c 3.00 × 10 m s

t= 21.51

(a)

191

The solar energy incident each second on 1.00 m 2 of the surface of Earth’s atmosphere is W Js N⋅m N = 1370 2 = 1370 2 = 1370 m2 m m ⋅s m ⋅s

U total = 1 370

Of this, 38.0% is reﬂected and 62.0% is absorbed. From Equation 21.29 and Equation 21.30 in the textbook, the radiation pressures P1 due to the reﬂected radiation and P2 due to the absorbed radiation are given by P1 =

2 U reflected 2 ( 0.380 U total ) 0.760 U total = = c c c

P2 =

and

U absorbed 0.620 U total = c c

The total radiation pressure is then

( 0.760 + 0.620 ) U total

Prad = P1 + P2 = or (b)

Prad =

c

1.38 (1 370 N m ⋅s ) = 6.30 × 10 −6 N m 2 = 6.30 × 10 −6 Pa 3.00 × 108 m s

In comparison, atmospheric pressure at the surface of the Earth is Patm 101× 10 3 Pa = = 1.60 × 1010 times greater than the radiation pressure Prad 6.30 × 10 −6 Pa

21.52

21.53

c=

1 = m 0 ∈0

or

c = 2.998 × 108 m s

(a)

The frequency of an electromagnetic wave is f = c l, where c is the speed of light and l is the wavelength of the wave. The frequencies of the two light sources are then Red:

1

( 4p × 10

fred =

−7

N ⋅s C 2

2

) (8.854 × 10

−12

C2 N ⋅ m 2 )

c 3.00 × 108 m s = = 4.55 × 1014 Hz lred 660 × 10 −9 m

and Infrared: (b)

fIR =

c 3.00 × 108 m s = = 3.19 × 1014 Hz lIR 940 × 10 −9 m

The intensity of an electromagnetic wave is proportional to the square of its amplitude. If 67% of the incident intensity of the red light is absorbed, then the intensity of the

continued on next page

68719_21_ch21_p171-197.indd 191

1/7/11 2:47:48 PM

192

Chapter 21

emerging wave is (100% − 67%) = 33% of the incident intensity, or I f = 0.33I i . Hence, we must have Em ax, f Em ax, i 21.54

=

If Ii

= 0.33 = 0.57

If I 0 is the incident intensity of a light beam, and I is the intensity of the beam after passing through length L of a ﬂuid having concentration C of absorbing molecules, the Beer-Lambert law states that log10 ( I I 0 ) = − ∈CL, where ∈ is a constant. For 660-nm light, the absorbing molecules are oxygenated hemoglobin. Thus, if 33% of this wavelength light is transmitted through blood, the concentration of oxygenated hemoglobin in the blood is CHB O2 =

− log10 ( 0.33) ∈L

[1]

The absorbing molecules for 940-nm light are deoxygenated hemoglobin, so if 76% of this light is transmitted through the blood, the concentration of these molecules in the blood is CHB =

− log10 ( 0.76) ∈L

[2]

Dividing Equation [2] by Equation [1] gives the ratio of deoxygenated hemoglobin to oxygenated hemoglobin in the blood as log10 ( 0.76 ) CHB = = 0.25 CHB O2 log10 ( 0.33)

or

CHB = 0.25CHBO2

Since all the hemoglobin in the blood is either oxygenated or deoxygenated, it is necessary that CHB + CHB O2 = 1.00, and we now have 0.25CHBO2 + CHB O2 = 1.0. The fraction of hemoglobin that is oxygenated in this blood is then CHB O2 =

1.0 = 0.80 1.0 + 0.25

or

80%

Someone with only 80% oxygenated hemoglobin in the blood is probably in serious trouble, needing supplemental oxygen immediately. 21.55

From Intensity =

Em ax Bm ax c Bm2 ax E and m ax = c, we ﬁnd Intensity = . 2 m0 Bm ax 2 m0

Thus, Bm ax = and 21.56

(a)

2 m0 ( Intensity ) = c

2 ( 4p × 10 −7 T ⋅ m A ) 3.00 × 108 m s

(1 370

W m 2 ) = 3.39 × 10 −6 T

Em ax = Bm ax c = ( 3.39 × 10 −6 T ) ( 3.00 × 108 m s ) = 1.02 × 10 3 V m To exert an upward force on the disk, the laser beam should be aimed vertically upward, striking the lower surface of the disk. To just levitate the disk, the upward force exerted on the disk by the beam should equal the weight of the disk.

continued on next page

68719_21_ch21_p171-197.indd 192

1/7/11 2:47:50 PM

Alternating Current Circuits and Electromagnetic Waves

193

The momentum that electromagnetic radiation of intensity I, incident normally on a perfectly reﬂecting surface of area A, delivers to that surface in time Δt is given by Equation 21.30 in the textbook as Δp = 2U c = 2 ( IAΔt ) c. Thus, from the impulse-momentum theorem, the average force exerted on the reﬂecting surface is F = Δp Δt = 2IA c. To just levitate the surface, F = 2IA c = mg, and the required intensity of the incident radiation is I = mgc 2A .

21.57

−3 2 8 mgc mgc ( 5.00 × 10 kg ) ( 9.80 m s ) ( 3.00 × 10 m s ) = = = 1.46 × 10 9 W m 2 2 −2 2A 2p r 2 2p ( 4.00 × 10 m )

(b)

I=

(c)

Propulsion by light pressure in a signiﬁcant gravity ﬁeld is impractical because of the enormous power requirements. In addition, no material is perfectly reﬂecting, so the absorbed energy would melt the reﬂecting surface.

The distance between adjacent antinodes in a standing wave is l 2. Thus, l = 2 ( 6.00 cm ) = 12.0 cm = 0.120 m, and c = lf = ( 0.120 m )( 2.45 × 10 9 Hz ) = 2.94 × 108 m s

21.58

(a)

The intensity of the radiation at distance r = 20.0 ly from the star is I=

(b)

P 4.00 × 10 28 W = = 8.89 × 10 −8 W m 2 2 4p r 2 4p ⎡( 20.0 ly ) ( 9.461× 1015 m ly ) ⎤ ⎣ ⎦

The power of the starlight intercepted by Earth, with cross-sectional area of Acs = p RE2 , is P = IAcs = I ⋅p RE2 = (8.89 × 10 −8 W m 2 ) p ( 6.38 × 10 6 m )

21.59

or

P = 1.14 × 10 7 W = 11.4 MW

(a)

l=

c 3.00 × 108 m s = = 6.00 × 10 −12 m = 6.00 pm f 5.00 × 1019 Hz

(b)

l=

c 3.00 × 108 m s = = 7.50 × 10 −2 m = 7.50 cm f 4.00 × 10 9 Hz

21.60

l=

21.61

(a)

c 3.00 × 108 m s = = 11.0 m f 27.33 × 10 6 Hz For the AM band, lm in = lm ax =

(b)

c fm ax c fm in

=

3.00 × 108 m s = 188 m 1 600 × 10 3 Hz

=

3.00 × 108 m s = 556 m 540 × 10 3 Hz

For the FM band, lm in = lm ax =

68719_21_ch21_p171-197.indd 193

2

c fm ax c fm in

=

3.00 × 108 m s = 2.78 m 108 × 10 6 Hz

=

3.00 × 108 m s = 3.4 m 88 × 10 6 Hz

1/7/11 2:47:53 PM

194

21.62

Chapter 21

The transit time for the radio wave is tR =

dR 100 × 10 3 m = = 3.33 × 10 −4 s = 0.333 ms c 3.00 × 108 m s

and that for the sound wave is ds 3.0 m = = 8.7 × 10 −3 s = 8.7 ms vsound 343 m s

ts =

Thus, the radio listeners hear the news 8.4 ms before the studio audience because radio waves travel so much faster than sound waves. 21.63

If an object of mass m is attached to a spring of spring constant k, the natural frequency of vibration of that system is f = k m 2p. Thus, the resonance frequency of the C=O double bond will be f=

1 2p

k 1 = moxygen 2p

2 800 N m = 5.2 × 1013 Hz 2.66 × 10 −26 kg

atom

and the light with this frequency has wavelength l=

c 3.00 × 108 m s = = 5.8 × 10 −6 m = 5.8 mm f 5.2 × 1013 Hz

The infrared region of the electromagnetic spectrum ranges from lm ax ≈ 1 mm down to lm in = 700 nm = 0.7 mm. Thus, the computed wavelength falls within the infrared region . 21.64

(a)

Since the space station and the ship are moving toward one another, the frequency after being Doppler shifted is fO = fS (1+ u c ) , where u is the relative speed between the observer and source. Thus, ⎡ 1.800 0 × 10 5 m s ⎤ 14 fO = ( 6.000 0 × 1014 Hz ) ⎢1+ ⎥ = 6.003 6 × 10 Hz 8 3.000 0 × 10 m s ⎣ ⎦

(b)

The change in frequency is Δf = fO − fS = 6.003 6 × 1014 Hz − 6.000 0 × 1014 = 3.6 × 1011 Hz

21.65

Since you and the car ahead of you are moving away from each other (getting farther apart) at a rate of u = 120 km h − 80 km h = 40 km h, the Doppler-shifted frequency you will detect is fO = fS (1− u c ), and the change in frequency is ⎛ 40 km h ⎞ ⎛ 0.278 m s ⎞ ⎛ u⎞ Δf = fO − fS = − fS ⎜ ⎟ = − ( 4.3 × 1014 Hz ) ⎜ = − 1.6 × 10 7 Hz ⎝ c⎠ ⎝ 3.0 × 108 m s ⎟⎠ ⎜⎝ 1 km h ⎟⎠ The frequency you will detect will be fO = fS + Δf = 4.3 × 1014 Hz − 1.6 × 10 7 Hz = 4.299 999 84 × 1014 Hz

21.66

The driver was driving toward the warning lights, so the correct form of the Doppler shift equation is fO = fS (1+ u c ) . The frequency emitted by the yellow warning light is fS =

c 3.00 × 108 m s = = 5.17 × 1014 Hz lS 580 × 10 −9 m

continued on next page

68719_21_ch21_p171-197.indd 194

1/7/11 2:47:55 PM

Alternating Current Circuits and Electromagnetic Waves

195

and the frequency the driver claims that she observed is fO =

c 3.00 × 108 m s = 5.36 × 1014 Hz = lO 560 × 10 −9 m

The speed with which she would have to approach the light for the Doppler effect to yield this claimed shift is ⎛ f ⎞ ⎛ 5.36 × 1014 Hz ⎞ − 1 = 1.1× 10 7 m s u = c ⎜ O − 1⎟ = ( 3.00 × 108 m s ) ⎜ ⎝ 5.17 × 1014 Hz ⎟⎠ ⎝ fS ⎠ 21.67

The energy incident on the mirror in time Δt is U = Plaser ⋅ Δt, where Plaser is the power transmitted by the laser beam Plaser = 25.0 × 10 −3 W = 25.0 × 10 −3 J s = 25.0 × 10 −3 N ⋅ m s From Equation 21.30 in the textbook, the rate of change in the momentum of the mirror as the beam reﬂects from it is Δp 2U c 2Plaser = = Δt c Δt The impulse-momentum theorem then gives the force exerted on the mirror as F = Δp Δt = 2Plaser c , and the radiation pressure on the mirror is Prad =

F 2Plaser c 2Plaser = = A A cA

where A = p r 2 = p d 2 4 is the area of the mirror illuminated (i.e., the cross-sectional area of the laser beam). Thus, Prad =

Prad = 5.31× 10 −5 N m 2 = 5.31× 10 −5 Pa

or 21.68

8 ( 25.0 × 10 −3 N ⋅ m s ) 2Plaser 8Plaser = = 2 c (p d 2 4 ) p cd 2 p ( 3.00 × 108 m s ) ( 2.00 × 10 −3 m )

Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle of the Sun is 60°. Then the target area you ﬁll in the Sun’s ﬁeld of view is (1.7 m )( 0.3 m ) cos30° = 0.4 m 2 . The intensity the radiation at Earth’s surface is I surface = 0.6 I incom ing , and only 50% of this is absorbed. Since I = Pav A = ( ΔE Δt ) A, the absorbed energy is

(

)

ΔE = ( 0.5 I surface ) A ( Δt ) = ⎡⎣ 0.5 0.6 I incom ing ⎤⎦ A ( Δt ) = ( 0.5)( 0.6 )(1 370 W m 2 ) ( 0.4 m 2 ) ( 3 600 s ) = 6 × 10 5 J or 21.69

Z = R 2 + ( XC ) = R 2 + ( 2p f C ) 2

=

−2

( 200 Ω )2 + ⎡⎣ 2p ( 60 Hz )( 5.0 × 10 −6 F )⎤⎦ = 5.7 × 10 2 Ω −2

⎛ ΔV ⎞ ⎛ 120 V ⎞ Pav = I rm2 s R = ⎜ rm s ⎟ R = ⎜ ( 200 Ω ) = 8.9 W = 8.9 × 10 −3 kW ⎝ 5.7 × 10 2 Ω ⎟⎠ ⎝ Z ⎠ 2

Thus,

~10 6 J

2

continued on next page

68719_21_ch21_p171-197.indd 195

1/7/11 2:47:58 PM

196

Chapter 21

and

cost = ΔE ⋅ ( rate ) = Pav ⋅ Δt ⋅ ( rate ) = (8.9 × 10 −3 kW ) ( 24 h ) (8.0 cents kWh ) = 1.7 cents

21.70

For a parallel-plate capacitor, C = ∈ A d : cutting the plate separation d in half will double the capacitance C f = 2Ci . Since the capacitive reactance is XC = 1 (2p f C), doubling the capacitance reduces the capacitive reactance by a factor of 2, giving XC, f = XC, i 2.

(

)

(

)

The current in the circuit is I rm s = ΔVrm s Z , so the impedance must be cut in half Z f = Z i 2 if the current doubles while the applied voltage remains constant. With an inductive reactance equal to the resistance ( X L = R ) , we then have

(

)

R 2 + R − XC, f

2

=

(

1 R 2 + R − XC, i 2

)

2

or

(

)

(

2 4 ⎡⎢ R 2 + R − XC, i 2 ⎤⎥ = R 2 + R − XC, i ⎣ ⎦

)

2

Simplifying gives 6R 2 = 2RXC, i , and XC, i = 6R 2 2R = 3R . 21.71

R=

( ΔV )DC I DC

=

12.0 V = 19.0 Ω 0.630 A ΔVrm s 24.0 V = = 42.1 Ω I rm s 0.570 A

Z = R 2 + ( 2p f L ) = 2

Thus, L = 21.72

(a)

Z 2 − R2 = 2p f

( 42.1 Ω )2 − (19.0 Ω )2 = 9.97 × 10 −2 H = 99.7 mH 2p ( 60.0 Hz )

The frequency of a 3.0-cm radar wave, and hence the desired resonance frequency of the tuning circuit, f0 = 1 2p LC , is f0 = f =

c 3.00 × 108 m s = = 1.0 × 1010 Hz l 3.00 × 10 −2 m

Therefore, the required capacitance is C=

1

( 2p f )

2

0

L

=

( 2p × 10

10

1

Hz ) ( 400 × 10 2

−12

H)

= 6.3 × 10 −13 F = 0.63 pF

∈0 A ∈0 = , so d d 2

(b)

C= =

21.73

C ⋅d = ∈0

(6.3 × 10

−13

F ) (1.0 × 10 −3 m )

8.85 × 10 −12 C2 N ⋅ m

= 8.4 × 10 −3 m = 8.4 mm

(c)

XC = X L = ( 2p f0 ) L = 2p (1.0 × 1010 Hz ) ( 400 × 10 −12 H ) = 25 Ω

(a)

Em ax E 0.20 × 10 −6 V m = c, so Bm ax = m ax = = 6.7 × 10 −16 T 3.00 × 108 m s Bm ax c

(b)

Intensity =

(c)

⎡pd2 ⎤ Pav = (Intensity) ⋅A = (Intensity) ⎢ ⎥ ⎣ 4 ⎦

−6 −16 Em ax Bm ax ( 0.20 × 10 V m ) ( 6.7 × 10 T ) = = 5.3 × 10 −17 W m 2 2m 0 2 ( 4p × 10 −7 T ⋅ m A )

⎡ p ( 20.0 m )2 ⎤ −14 = ( 5.3 × 10 −17 W m 2 ) ⎢ ⎥ = 1.7 × 10 W 4 ⎣ ⎦

68719_21_ch21_p171-197.indd 196

1/7/11 2:48:00 PM

Alternating Current Circuits and Electromagnetic Waves

21.74

(a)

Z=

ΔVrm s 12 V = = 6.0 Ω I rm s 2.0 A

(b)

R=

ΔVDC 12 V = = 4.0 Ω I DC 3.0 A

197

From Z = R 2 + X L2 = R 2 + ( 2p f L ) , we ﬁnd 2

Z 2 − R2 = 2p f

L= 21.75

(a)

( 6.0 Ω )2 − ( 4.0 Ω )2 = 1.2 × 10 −2 H = 12 mH 2p ( 60 Hz )

From Equation 21.30 in the textbook, the momentum imparted in time Δt to a perfectly reﬂecting sail of area A by normally incident radiation of intensity I is Δp = 2U c = 2 ( IAΔt ) c. From the impulse-momentum theorem, the average force exerted on the sail is then Fav =

Fav 0.536 N = = 8.93 × 10 −5 m s 2 m 6 000 kg

(b)

aav =

(c)

From Δx = v0 t +

t= 21.76

2 4 2 Δp 2 ( IAΔt ) c 2IA 2 (1 340 W m ) ( 6.00 × 10 m ) = = = = 0.536 N 8 Δt Δt c 3.00 × 10 m s

(a)

1 2 at , with v0 = 0, the time is 2

2 ( Δx ) = aav

2 ( 3.84 × 108 m )

8.93 × 10

−5

1d ⎛ = ( 2.93 × 10 6 s ) ⎜ ⎝ m s 8.64 × 10 4 2

⎞ ⎟ = 33.9 d s⎠

The intensity of radiation at distance r from a point source, which radiates total power P, is I = P A = P 4p r 2. Thus, at distance r = 2.0 in from a cell phone radiating a total power of P = 2.0 W = 2.0 × 10 3 mW, the intensity is I=

2.0 × 10 3 mW 4p ⎡⎣( 2.0 in )( 2.54 cm 1 in ) ⎤⎦

2

= 6.2 mW cm 2

This intensity is 24% higher than the maximum allowed leakage from a microwave at this distance of 2.0 inches. (b)

If when using a Bluetooth headset (emitting 2.5 mW of power) in the ear at distance rh = 2.0 in = 5.1 cm from the brain, the cell phone (emitting 2.0 W of power) is located in the pocket at distance rp = 1.0 m = 1.0 × 10 2 cm from the brain, the total radiation intensity at the brain is I total = I phone + I headset =

or

68719_21_ch21_p171-197.indd 197

I total = 1.6 × 10 −2

2.0 × 10 3 mW

4p (1.0 × 10 cm ) 2

2

+

2.5 mW 2 4p ( 5.1 cm )

mW mW mW + 7.6 × 10 −3 = 2.4 × 10 −2 = 0.024 mW cm 2 2 2 cm cm cm 2

1/7/11 2:48:03 PM

22 Reflection and Refraction of Light QUICK QUIZZES 1.

Choice (a). In part (a), you can see clear reflections of the headlights and the lights on the top of the truck. The reflection is specular. In part (b), although bright areas appear on the roadway in front of the headlights, the reflection is not as clear, and no separate reflection of the lights from the top of the truck is visible. The reflection in part (b) is mostly diffuse.

2.

Beams 2 and 4 are reflected; beams 3 and 5 are refracted.

3.

Choice (b). When light goes from one material into one having a higher index of refraction, it refracts toward the normal line of the boundary between the two materials. If, as the light travels through the new material, the index of refraction continues to increase, the light ray will refract more and more toward the normal line.

4.

Choice (c). Both the wave speed and the wavelength decrease as the index of refraction increases. The frequency is unchanged.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

The index of refraction of a material in which light has speed v is n = c v = c l f. Thus, nliquid nair

=

c (lliquid f ) c (lair f )

=

lair lliquid

⎛ l ⎞ ⎛ 495 nm ⎞ nliquid = ⎜ air ⎟ nair = ⎜ (1.00 ) = 1.14 ⎝ 434 nm ⎟⎠ ⎝ lliquid ⎠

and

so the correct choice is (c). 2.

The critical angle as light passes from medium 1 into medium 2 is q c = sin −1 (n2 n1 ), so as light is incident on crown glass (n2 = 1.52) from carbon disulfide (n1 = 1.63), ⎛ 1.52 ⎞ q c = sin −1 ⎜ = 68.8° ⎝ 1.63 ⎟⎠ and choice (b) is the correct response.

3.

The energy of a photon is E = hf = hc l. Thus, if n 800-nm photons have the same energy as four 200-nm photons, it is necessary that n(hc 800 nm) = 4(hc 200 nm)

or

n = 4(800 nm 200 nm) = 16

Therefore, the correct answer is (e). 4.

Observe that the angle of refraction is greater than the angle of incidence as the light ray passes from medium 1 into medium 2. Thus, the speed of light increases as the light crosses the boundary between these materials. Since n = c v , the index of refraction of medium 2 is less than that of medium 1, or n1 > n2 . The correct choice is (d). 198

68719_22_ch22_p198-222.indd 198

1/7/11 2:48:49 PM

Reflection and Refraction of Light

5.

6.

7.

199

Total internal reflection will occur when light, in attempting to go from a medium with one index of refraction n1 into a second medium where it travels faster than in the first medium (or where n2 < n1), strikes the surface at an angle of incidence greater than or equal to the critical angle. The correct choice is (b). Water and air have different indices of refraction, with nwater ≈ 4nair 3. In passing from one of these media into the other, light will be refracted (deviated in direction) unless the angle of incidence is zero (in which case, the angle of refraction is also zero). Thus, rays B and D cannot be correct. In refraction, the incident ray and the refracted ray are never on the same side of the line normal to the surface at the point of contact, so ray A cannot be correct. Also in refraction, the ray makes a smaller angle with the normal in the medium having the highest index of refraction. Therefore, ray E cannot be correct, leaving only ray C as a likely path. Choice (c) is the correct answer. When light is in water, the relationships between the values of its frequency, speed, and wavelength to the values of the same quantities in air are fwater = fair ,

⎛ n ⎞ 3 lwater = ⎜ air ⎟ lair ≈ lair , 4 ⎝ nwater ⎠

and

⎛ n ⎞ ⎛ 3⎞ vwater = ⎜ air ⎟ vair = ⎜ ⎟ c ⎝ 4⎠ ⎝ nwater ⎠

Therefore, only choice (b) is a completely true statement. 8.

In a dispersive medium, the index of refraction is largest for the shortest wavelength. Thus, the violet light will be refracted (or bent) the most as it passes through a surface of the crown glass, making (a) the correct choice.

9.

For any medium, other than vacuum, the index of refraction for red light is slightly lower than that for blue light. This means that when light goes from vacuum (or air) into glass, the red light deviates from its original direction less than does the blue light. Also, as the light reemerges from the glass into vacuum (or air), the red light again deviates less than the blue light. If the two surfaces of the glass are parallel to each other, the red and blue rays will emerge traveling parallel to each other but displaced laterally from one another. The sketch that best illustrates this process is C, so choice (c) is the best answer.

10.

Consider the sketch at the right and apply Snell’s law to the refraction at each of the three surfaces. The resulting equations are

and

(1.00)sinq = n1 sina

(1st surface)

n1 sina = n2 sin b

(2nd surface)

n2 sin b = (1.00 ) sinf

(3rd surface)

Combining these three equations yields (1.00)sinf = (1.00)sinq , and f = q . Hence, choice (c) is the correct answer.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

68719_22_ch22_p198-222.indd 199

(a)

From Snell’s law, sinq 2 = (n1 n2 )sinq1 . Thus, if n2 < n1, then sinq 2 > sinq1 and q 2 > q1, so the ray refracts away from the normal.

1/7/11 2:48:51 PM

200

Chapter 22

(b)

The index of refraction is defined as n = c v = c l f . The wavelength may then be written as l = c nf = (c f ) n = l0 n , where l0 is the wavelength of the light in vacuum. Thus, as the ray moves into a medium of lower index of refraction, the wavelength will increase.

(c)

The frequency at which wavefronts move away from a boundary equals the frequency with which they arrive at the boundary. That is, the frequency of the light stays the same as it moves between the two materials.

4.

A mirage occurs when light changes direction as it moves between batches of air having different indices of refraction. The different indices of refraction occur because the air has different densities at different temperatures. Two images are seen: one from a direct path from the object to you, and the second arriving by rays originally heading toward Earth but refracted to your eye. On a hot day, the Sun makes the surface of asphalt hot, so the air is hot directly above it, becoming cooler as one moves higher into the sky. The “water” we see far in front of us is an image of the blue sky. Adding to the effect is the fact that the image shimmers as the air changes in temperature, giving the appearance of moving water.

6.

The upright image of the hill is formed by light that has followed a direct path from the hill to the eye of the observer. The second image is a result of refraction in the atmosphere. Some light is reflected from the hill toward the water. As this light passes through warmer layers of air directly above the water, it is refracted back up toward the eye of the observer, resulting in the observation of an inverted image of the hill directly below the upright image.

8.

The index of refraction of water is 1.333, quite different from that of air, which has an index of refraction of about 1. The boundary between the air and water is therefore easy to detect, because of the differing refraction effects above and below the boundary. (Try looking at a glass half full of water.) The index of refraction of liquid helium, however, happens to be much closer to that of air. Consequently, the refractive differences above and below the helium-air boundary are harder to see.

10.

Total internal reflection can only occur when light attempts to move from a medium of high index of refraction to a medium of lower index of refraction. Thus, light moving from air (n = 1) to water (n = 1.333) cannot undergo total internal reflection.

12.

With no water in the cup, light rays from the coin do not reach the eye because they are blocked by the side of the cup. With water in the cup, light rays are bent away from the normal as they leave the water so that some reach the eye. In figure (a) below, ray a is blocked by the side of the cup so it cannot enter the eye, and ray b misses the eye. In figure (b), ray a is still blocked by the side of the cup, but ray b refracts at the water’s surface so that it reaches the eye. Ray b seems to come from position B, directly above the coin at position A.

eye b

b a

a

B

A

(a)

68719_22_ch22_p198-222.indd 200

(b)

1/7/11 2:48:53 PM

Reflection and Refraction of Light

201

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

4.

(a) 1.99 × 10 −15 J

(b) 12.4 keV

(c)

decreased

(d)

increased

(a)

550 nm

(b)

366 nm

(c)

2.26 eV

(d) The energy does not change.

6.

(a) l = hc E

(b)

Higher energy photons have shorter wavelengths.

8.

(a)

(b)

50.0° above the horizontal

1.94 m

10.

See Solution.

12.

(a) The angle of refraction increases with wavelength, so the longest wavelength deviates the least from the original path. (b) l = 400 nm, q r = 16.0°; l = 500 nm, q r = 16.1°; l = 650 nm, q r = 16.3°

14.

80.0°

16.

(a) B

(b)

18.

(a) q r, top = 19.5°

(b) q i, bottom = 19.5°, q r , bottom = 30.0°

(c)

(d)

0.386 cm

A, B, and C

2.00 × 108 m s

(e) 1.06 × 10 −10 s

(f) The angle of incidence affects the distance the ray travels in the glass and hence affects the travel time. 20.

1.22

22.

6.30 cm

24.

(a)

See Solution.

(b)

42.0°

(d)

26.9°

(e)

107 m

(b)

42.2°

26.

See Solution.

28.

0.39°

30.

q red = 34.9°, q violet = 34.5°

32.

4.5°

34.

(a)

36.

48.5°

38.

67.3°

68719_22_ch22_p198-222.indd 201

43.3°

(c)

63.1°

(c)

40.4°

1/7/11 2:48:54 PM

202

40.

Chapter 22

(a)

(b)

34.2°

(c) and (d) 42.

34.2°

Neither thickness nor index of refraction affects the result.

(a)

24.42°

(b)

(c)

33.44°

(d) Total internal reflection still occurs.

(e)

rotate clockwise

(f)

2.9°

(b)

q r → q i = 30.0°

(b)

q incidence ≤ 29.9°

44.

4.54 m

46.

(a)

48.

(a) q incidence ≤ 90.0° (c)

23.7°

See Solution.

(c)

q r > q i = 30.0°

Total internal reflection is not possible since npolystyrene < ncarbon disulfide .

50.

77.5°

52.

(a)

Rm in = nd ( n − 1)

(b)

Rm in → 0 as d → 0; Rm in decreases as n increases; Rm in increases as n → 1. All are reasonable behaviors.

(c)

Rm in = 350 mm

54.

f = 8.0°

56.

82 complete reflections

58.

1.33

60.

23.1°

PROBLEM SOLUTIONS 22.1

The total distance the light travels is ⎛ ⎞ Δd = 2 ⎜ Dcenter to − REarth − RM oon ⎟ ⎝ center ⎠ = 2 ( 3.84 × 108 − 6.38 × 10 6 − 1.76 × 10 6 ) m = 7.52 × 108 m v=

Therefore, 22.2

(a)

Δd 7.52 × 108 m = = 3.00 × 108 m s Δt 2.51 s

The energy of a photon is E = hf = hc l, where Planck’s constant is h = 6.63×10 −34 J ⋅ s, and the speed of light in vacuum is c = 3.00 × 108 m s. If l = 1.00 × 10 −10 m,

(6.63 × 10 E=

−34

J ⋅ s ) ( 3.00 × 108 m s )

1.00 × 10 −10 m

= 1.99 × 10 −15 J

continued on next page

68719_22_ch22_p198-222.indd 202

1/7/11 2:48:55 PM

Reflection and Refraction of Light

(b)

203

1 eV ⎛ ⎞ E = (1.99 × 10 −15 J ) ⎜ = 1.24 × 10 4 eV = 12.4 keV ⎝ 1.602 × 10 −19 J ⎟⎠

(c) and (d)

For the x-rays to be more penetrating, the photons should be more energetic. Since the energy of a photon is directly proportional to the frequency and inversely proportional to the wavelength, the wavelength should decrease , which is the same as saying the frequency should increase .

22.3

(a)

1 eV ⎛ E = hf = ( 6.63 × 10 −34 J ⋅s ) ( 5.00 × 1017 Hz ) ⎜ ⎝ 1.60 × 10 −19

(b)

E = hf =

⎞ 3 ⎟ = 2.07 × 10 eV = 2.07 keV J⎠

−34 8 hc ( 6.63 × 10 J ⋅ s ) ( 3.00 × 10 m s ) ⎛ 1 nm ⎞ −19 = ⎜⎝ −9 ⎟⎠ = 6.63 × 10 J l 3.00 × 10 2 nm 10 m

1 eV ⎛ ⎞ E = 6.63 × 10 −19 J ⎜ = 4.14 eV ⎝ 1.60 × 10 −19 J ⎟⎠ 22.4

c 3.00 × 108 m s = = 5.50 × 10 −7 m = 550 nm f 5.45 × 1014 Hz

(a)

l0 =

(b)

From Table 22.1, the index of refraction for benzene is n = 1.501. Thus, the wavelength in benzene is ln =

(c) (d)

22.5

22.6

l0 550 nm = = 366 nm 1.501 n

1 eV ⎛ ⎞ E = hf = ( 6.63 × 10 −34 J ⋅s ) ( 5.45 × 1014 Hz ) ⎜ = 2.26 eV ⎝ 1.60 × 10 −19 J ⎟⎠ The energy of the photon is proportional to the frequency, which does not change as the light goes from one medium to another. Thus, when the photon enters benzene, the energy does not change .

The speed of light in a medium with index of refraction n is v = c n , where c is its speed in vacuum. 3.00 × 108 m s = 2.25 × 108 m s . 1.333

(a)

For water, n = 1.333, and v =

(b)

For crown glass, n = 1.52, and v =

(c)

For diamond, n = 2.419, and v =

(a)

The energy of a photon is

3.00 × 108 m s = 1.97 × 108 m s . 1.52

3.00 × 108 m s = 1.24 × 108 m s . 2.419

⎛ c ⎞ hc E = hf = h ⎜ ⎟ = ⎝l⎠ l Thus, (b)

68719_22_ch22_p198-222.indd 203

l = hc E

Higher energy photons have shorter wavelengths.

1/7/11 2:48:57 PM

204

22.7

Chapter 22

From Snell’s law, n2 sinq 2 = n1 sinq1 . Thus, when q1 = 45.0° and the first medium is air (n1 = 1.00), we have sinq 2 = (1.00 ) sin 45.0° n 2 . (a)

⎛ (1.00 ) sin 45.0° ⎞ For quartz, n2 = 1.458, and q 2 = sin −1 ⎜ ⎟⎠ = 29.0° . ⎝ 1.458

(b)

⎛ (1.00 ) sin 45.0° ⎞ For carbon disulfide, n2 = 1.628, and q 2 = sin −1 ⎜ ⎟⎠ = 25.7° . ⎝ 1.628

(c)

⎛ (1.00 ) sin 45.0° ⎞ For water, n2 = 1.333, and q 2 = sin −1 ⎜ ⎟⎠ = 32.0° . ⎝ 1.333

(a)

From geometry, 1.25 m = d sin 40.0°, so d = 1.94 m .

(b)

50.0° above horizontal , or parallel to the incident ray

22.8

22.9

n1 sin q1 = n2 sin q 2 sin q1 = 1.333 sin 45.0° q1 = sin −1 (1.333 sin 45.0° ) = 70.5° Thus, the sun appears to be 19.5° above the horizontal .

22.10

In the sketch at the right, observe that the law of reflection is obeyed as the ray reflects from each of the mirrors. Also, note that the normal lines to the two mirrors intersect at a right angle since the two mirrors are perpendicular to each other. Considering the right triangle formed by the two normal lines and the ray, and recalling that the sum of the interior angles of any triangle is 180°, we find that q1 +q 2 + 90.0° = 180°

or

q 2 = 90.0° −q1

Mirror 2

q2 q2

q1

q1 a Mirror 1

Looking at the point where the incident ray strikes mirror 1, we see that a = 90° −q1 . Thus, both the incident ray and the final outgoing reflected ray are at angle 90° −q1 above the horizontal and hence, parallel to each other .

68719_22_ch22_p198-222.indd 204

1/7/11 2:49:00 PM

Reflection and Refraction of Light

22.11

22.12

(a)

From Snell’s law, n2 =

(b)

l2 =

(c)

f=

(d)

v2 =

(a)

205

n1 sinq1 (1.00 ) sin 30.0° = = 1.52 . sinq 2 sin19.24°

l0 632.8 nm = = 416 nm 1.52 n2 c 3.00 × 108 m s = = 4.74 × 1014 Hz in the air and also in the solution l0 632.8 × 10 −9 m c 3.00 × 108 m s = = 1.97 × 108 m s n2 1.52

When light refracts from air ( n1 = 1.00 ) into the crown glass, Snell’s law gives the angle of refraction as q 2 = sin −1 (sin 25.0° ncrown glass ) For first quadrant angles, the sine of the angle increases as the angle increases. Thus, from the above equation, note that q 2 will increase when the index of refraction of the crown glass decreases. From Figure 22.13, we see that the angle of refraction will increase with wavelength, so the longer wavelengths deviate the least from the original path.

(b)

From Figure 22.13, observe that the index of refraction of crown glass for the given wavelengths is l = 400 nm, ncrown glass = 1.53;

l = 500 nm, ncrown glass = 1.52;

l

Figure 22.13

and l = 650 nm, ncrown glass = 1.51 Thus, Snell’s law gives

22.13

l = 400 nm:

q 2 = sin −1 ( sin 25.0° 1.53) = 16.0°

l = 500 nm:

q 2 = sin −1 ( sin 25.0° 1.52 ) = 16.1°

l = 650 nm:

q 2 = sin −1 ( sin 25.0° 1.51) = 16.3°

From Snell’s law, ⎡ n sinq1 ⎤ −1 ⎡ (1.00)sin 40.0° ⎤ q 2 = sin −1 ⎢ 1 ⎥ = sin ⎢ ⎥⎦ = 29.4° 1.309 ⎣ ⎣ n2 ⎦ and from the law of reflection, f = q1 = 40.0°. Hence, the angle between the reflected and refracted rays is a = 180° −q 2 − f = 180.0° − 29.4° − 40.0° = 110.6°

68719_22_ch22_p198-222.indd 205

1/7/11 2:49:02 PM

206

22.14

Chapter 22

Consider the sketch at the right and note that the incident horizontal ray is parallel to the surface of mirror 2. Thus, the angle the incident ray makes with mirror 1 must be a = q1 = 50.0°. Since the ray must obey the law of reflection at mirror 1, the angle b must be b = a = 50.0°. Recalling that the sum of the interior angles of a triangle is always 180.0°, we find that

Mirror 1

a

b q1

g

q2

g = 180.0° −q1 − b = 180.0° − 50.0° − 50.0° = 80.0°

Mirror 2

Hence, in order to obey the law of reflection at mirror 2, the angle the outgoing reflected ray makes with the surface of mirror 2 must be q 2 = g = 80.0° . 22.15

22.16

The index of refraction of zircon is n = 1.923. c 3.00 × 108 m s = = 1.56 × 108 m s n 1.923

(a)

v=

(b)

The wavelength in the zircon is ln =

(c)

The frequency is f =

l0 632.8 nm = = 329.1 nm . 1.923 n

v c 3.00 × 108 m s = = = 4.74 × 1014 Hz . ln l0 632.8 × 10 −9 m

The sketch at the right shows the path of the ray inside the glass slab. Considering the reflections at points A and B, the law of reflection tells us that a = q A and b = q B . Also, we observe that q B = 90° − a = 90° −q A and

q C = 90° − b = 90° −q B = 90° − ( 90° −q A ) = q A

C b

qA

qC

Thus, if q A = 55°, then q B = 35° and q C = 55°.

a b qB

In order for part of the ray to leave the glass slab and enter the surrounding medium at a reflection point, the angle of incidence at that point must be less than the critical angle, q c = sin −1 (n2 nglass ) = sin −1 (n2 1.52).

22.17

A

a B

(a)

If the surrounding medium is air, then n2 = 1.00 and q c = sin −1 (1.00 1.52) = 41.1°. Thus, we see that total internal reflection will occur at points A and C, but part of the ray can refract into the surrounding air at point B .

(b)

When the surrounding medium is carbon disulfide, n2 = 1.628 > nglass . Thus, the critical angle does not exist, and total internal reflection will not occur when the ray attempts to go from the glass into the carbon disulfide. This means that part of the ray will enter the carbon disulfide at points A, B, and C .

The incident light reaches the left-hand mirror at distance d 2 = (1.00 m ) tan 5.00° = 0.087 5 m above its bottom edge. The reflected light first reaches the right-hand mirror at height d = 2 ( 0.087 5 m ) = 0.175 m

continued on next page

68719_22_ch22_p198-222.indd 206

1/7/11 2:49:04 PM

Reflection and Refraction of Light

207

It bounces between the mirrors with distance d between points of contact with a given mirror. Since the full 1.00 length of the right-hand mirror is available for reflections, the number of reflections from this mirror will be N right =

1.00 m = 5.71 → 0.175 m

5 full reflections

Since the first reflection from the left-hand mirror occurs at a height of d 2 = 0.087 5 m, the total number of that can occur from this mirror is N left = 1+ 22.18

(a)

1.00 m − 0.087 5 m = 6.21 → 0.175 m

6 full reflections

From Snell’s law, the angle of refraction at the first surface is ⎡ n sinq1 ⎤ −1 ⎡ (1.00 ) sin 30.0° ⎤ q 2 = sin −1 ⎢ air ⎥ = sin ⎢ ⎥ = 19.5° n 1.50 ⎦ ⎣ ⎢⎣ glass ⎥⎦

(b)

Since the upper and lower surfaces are parallel, the normal lines where the ray strikes these surfaces are parallel. Hence, the angle of incidence at the lower surface will be q 2 = 19.5° . The angle of refraction at this surface is then ⎡ nglass sinq glass q 3 = sin −1 ⎢ nair ⎣

⎤ −1 ⎡ (1.50 ) sin19.5° ⎤ ⎥ = sin ⎢ ⎥ = 30.0° 1.00 ⎦ ⎣ ⎦

Thus, the light emerges traveling parallel to the incident beam. (c)

Consider the sketch at the right, and let h represent the distance from point a to c (that is, the hypotenuse of triangle abc). Then, h=

2.00 cm 2.00 cm = = 2.12 cm cosq 2 cos19.5°

Also, a = q1 −q 2 = 30.0° − 19.5° = 10.5°, so d = h sina = ( 2.12 cm ) sin10.5° = 0.386 cm (d)

The speed of the light in the glass is v=

(e)

68719_22_ch22_p198-222.indd 207

nglass

=

3.00 × 108 m s = 2.00 × 108 m s 1.50

The time required for the light to travel through the glass is t=

(f )

c

h 2.12 cm ⎛ 1m ⎞ −10 = ⎜ ⎟ = 1.06 × 10 s v 2.00 × 108 m s ⎝ 10 2 cm ⎠

Changing the angle of incidence will change the angle of refraction and therefore the distance h the light travels in the glass. Thus, the travel time will also change .

1/7/11 2:49:07 PM

208

22.19

Chapter 22

From Snell’s law, the angle of incidence at the air-oil interface is ⎡ n sinq oil ⎤ −1 ⎡ (1.48 ) sin 20.0° ⎤ q = sin −1 ⎢ oil ⎥ = sin ⎢ ⎥ = 30.4° n 1.00 ⎦ ⎣ air ⎣ ⎦ and the angle of refraction as the light enters the water is ⎡ n sinq oil q ′ = sin −1 ⎢ oil ⎣ nwater

22.20

⎤ −1 ⎡ (1.48 ) sin 20.0° ⎤ ⎥ = sin ⎢ ⎥ = 22.3° 1.333 ⎦ ⎣ ⎦

Since the light ray strikes the first surface at normal incidence, it passes into the prism without deviation. Thus, the angle of incidence at the second surface (hypotenuse of the triangular prism) is q1 = 45.0° as shown in the sketch at the right. The angle of refraction is q 2 = 45.0° + 15.0° = 60.0° and Snell’s law gives the index of refraction of the prism material as n1 =

22.21

n2 sinq 2 (1.00 ) sin ( 60.0° ) = = 1.22 sinq1 sin ( 45.0° )

Applying Snell’s law where the ray first enters the glass gives ⎛ n sinq1 ⎞ −1 ⎡ (1.333 ) sin 42.0° ⎤ f = sin −1 ⎜ water ⎟ = sin ⎢ ⎥ = 35.9° n 1.52 ⎦ ⎣ ⎝ ⎠ glass

q1

water, n = 1.333 d f

d y

Thus, d = 90.0° − f = 90.0° − 35.9° = 54.1°. (a)

The distance down to point P, where the ray emerges from the glass, is now seen to be

d Crown glass, n = 1.52

P

q2

y = d tand = ( 3.50 cm ) tan 54.1° = 4.84 cm (b)

The angle of refraction as the ray leaves the block is given by Snell’s law as ⎛ nglass sind ⎞ ⎡ (1.52 ) sin 54.1° ⎤ = sin −1 ⎢ q 2 = sin −1 ⎜ ⎥ = 67.5° 1.333 ⎝ nwater ⎟⎠ ⎦ ⎣

22.22

⎞ ⎛n From Snell’s law, sinq = ⎜ m edium ⎟ sin 50.0°. ⎝ nliver ⎠ But so

nm edium c vm edium v = = liver = 0.900 nliver c vliver vm edium

nmedium nliver

q = sin −1 [( 0.900 ) sin 50.0° ] = 43.6°

continued on next page

68719_22_ch22_p198-222.indd 208

1/7/11 2:49:09 PM

Reflection and Refraction of Light

209

From the law of reflection, d= 22.23

(a)

12.0 cm = 6.00 cm 2

h=

and

d 6.00 cm = = 6.30 cm tanq tan ( 43.6° )

Before the container is filled, the ray’s path is as shown in figure (a) at the right. From this figure, observe that sinq1 =

d = s1

d h2 + d 2

=

q1

1

(h d )

h 2

d 2 = s2

d 2 h + ( d 2) 2

2

s1

+1

d

After the container is filled, the ray’s path is shown in figure (b). From this figure, we find that sinq 2 =

q1

(a)

q1

1

=

h q2

4 (h d ) + 1 2

s2

From Snell’s law, nair sinq1 = n sinq 2, or 1.00

(h d )

2

+1

=

d/2

n 4 (h d ) + 1 2

and

Simplifying, this gives ( 4 − n 2 ) ( h d ) = n 2 − 1 2

(b)

If d = 8.00 cm h = (8.00 cm )

22.24

and

(b)

4 (h d ) + 1 = n2 (h d ) + n2 2

2

or

h = d

n2 − 1 4 − n2

n = nwater = 1.333, then

(1.333)2 − 1 = 4.73 cm 2 4 − (1.333)

(a)

A sketch illustrating the situation and the two triangles needed in the solution is given below:

(b)

The angle of incidence at the water surface is ⎛ 90.0 m ⎞ q1 = tan −1 ⎜ = 42.0° ⎝ 1.00 × 10 2 m ⎟⎠

continued on next page

68719_22_ch22_p198-222.indd 209

1/7/11 2:49:11 PM

210

Chapter 22

(c)

Snell’s law gives the angle of refraction as ⎛ n sinq1 ⎞ ⎛ (1.333) sin 42.0° ⎞ = sin −1 ⎜ q 2 = sin −1 ⎜ water ⎟⎠ = 63.1° ⎟ ⎝ n 1.00 ⎝ ⎠ air

(d)

The refracted beam makes angle f = 90.0° −q 2 = 26.9° with the horizontal.

(e)

Since tanf = h (2.10 × 10 2 m), the height of the target is h = ( 2.10 × 10 2 m ) tan ( 26.9° ) = 107 m

22.25

As shown at the right, q1 + b +q 2 = 180°. When b = 90°, this gives q 2 = 90° −q1. Then, from Snell’s law sinq1 =

ng sinq 2 nair

= ng sin ( 90° −q1 ) = ng cosq1 Thus, when b = 90° , 22.26

22.27

( )

sinq1 = tanq1 = ng or q1 = tan −1 ng . cosq1

The index of refraction of the atmosphere decreases with increasing altitude because of the decrease in density of the atmosphere with increasing altitude. As indicated in the ray diagram at the right, the Sun located at S below the horizon appears to be located at S′.

Horizon plane Atmosphere

S´

S Earth

When the Sun is 28.0° above the horizon, the angle of incidence for sunlight at the air-water boundary is q1 = 90.0° − 28.0° = 62.0° Thus, the angle of refraction is ⎡ n sinq1 ⎤ q 2 = sin −1 ⎢ air ⎥ ⎣ nwater ⎦ ⎡ (1.00 ) sin 62.0° ⎤ = sin −1 ⎢ ⎥ = 41.5° 1.333 ⎦ ⎣ The depth of the tank is then h =

68719_22_ch22_p198-222.indd 210

3.00 m 3.00 m = = 3.39 m . tanq 2 tan ( 41.5° )

1/7/11 2:49:13 PM

Reflection and Refraction of Light

22.28

211

The angles of refraction for the two wavelengths are ⎛ n sinq1 ⎞ ⎛ 1.000 sin 30.00° ⎞ = sin −1 ⎜ q red = sin −1 ⎜ air ⎟⎠ = 18.03° ⎟ ⎝ n 1.615 ⎝ ⎠ red and

⎛ n sinq1 ⎞ ⎛ 1.000 sin 30.00° ⎞ = sin −1 ⎜ q blue = sin −1 ⎜ air ⎟⎠ = 17.64° ⎟ ⎝ 1.650 ⎝ nblue ⎠

Thus, the angle between the two refracted rays is Δq = q red −q blue = 18.03° − 17.64° = 0.39° 22.29

22.30

Using Snell’s law gives (a)

⎛ n sinq i ⎞ ⎛ (1.000)sin83.00° ⎞ = sin −1 ⎜ q blue = sin −1 ⎜ air ⎟⎠ = 47.79° ⎟ ⎝ n 1.340 ⎝ ⎠ blue

(b)

⎛ n sinq i ⎞ ⎛ (1.000)sin83.00° ⎞ = sin −1 ⎜ q red = sin −1 ⎜ air ⎟⎠ = 48.22° ⎟ ⎝ 1.331 ⎝ nred ⎠

Using Snell’s law gives ⎛ n sinq i ⎞ ⎛ (1.00)sin 60.0° ⎞ = sin −1 ⎜ q red = sin −1 ⎜ air ⎟⎠ = 34.9° ⎝ 1.512 ⎝ nred ⎟⎠ and

22.31

⎛ n sinq i ⎞ ⎛ (1.00)sin 60.0° ⎞ = sin −1 ⎜ q violet = sin −1 ⎜ air ⎟⎠ = 34.5° ⎟ ⎝ 1.530 ⎝ nviolet ⎠

Using Snell’s law gives ⎛ n sinq i ⎞ ⎛ (1.000)sin 50.00° ⎞ = sin −1 ⎜ q red = sin −1 ⎜ air ⎟⎠ = 31.77° ⎝ 1.455 ⎝ nred ⎟⎠ and

⎛ n sinq i ⎞ ⎛ (1.000)sin 50.00° ⎞ = sin −1 ⎜ q violet = sin −1 ⎜ air ⎟⎠ = 31.45° ⎟ ⎝ 1.468 ⎝ nviolet ⎠

Thus, the dispersion is q red −q violet = 31.77° − 31.45° = 0.32° . 22.32

For the violet light, nglass = 1.66, and ⎛ n sinq1i ⎞ q1r = sin ⎜ air ⎟ ⎝ nglass ⎠

nair = 1.00 60.0°

−1

a

q1i = 50.0°

⎛ 1.00 sin 50.0° ⎞ = sin ⎜ ⎟⎠ = 27.5° ⎝ 1.66

b q1r q2i

−1

q2r

nglass

a = 90° −q1r = 62.5°, b = 180.0° − 60.0° − a = 57.5°, and q 2i = 90.0° − b = 32.5°. The final angle of refraction of the violet light is ⎛ nglass sinq 2i ⎞ ⎛ 1.66sin 32.5° ⎞ = sin −1 ⎜ q 2r = sin −1 ⎜ ⎟⎠ = 63.1° ⎟ ⎝ nair 1.00 ⎝ ⎠

continued on next page

68719_22_ch22_p198-222.indd 211

1/7/11 2:49:15 PM

212

Chapter 22

(

)

Following the same steps for the red light nglass = 1.62 gives q1r = 28.2°, a = 61.8°, b = 58.2°, q 2i = 31.8°, and q 2r = 58.6° Thus, the angular dispersion of the emerging light is Dispersion = q 2 r 22.33

(a)

violet

−q 2 r

red

= 63.1° − 58.6° = 4.5°

The angle of incidence at the first surface is q1i = 30° , and the angle of refraction is ⎛ n sinq1i ⎞ q1r = sin −1 ⎜ air ⎟ ⎝ nglass ⎠

nair = 1.0 60°

⎛ 1.0 sin 30° ⎞ = sin −1 ⎜ ⎟⎠ = 19° ⎝ 1.5

b

a

q1i = 30°

q1r q2i

q2r

nglass = 1.5

Also, a = 90° −q1r = 71° and b = 180° − 60° − a = 49°. Therefore, the angle of incidence at the second surface is q 2 i = 90° − b = 41° . The angle of refraction at this surface is ⎛ n sinq 2i ⎞ −1 ⎛ 1.5sin 41° ⎞ q 2r = sin −1 ⎜ glass ⎟ = sin ⎜ ⎟ = 80° ⎝ 1.0 ⎠ nair ⎠ ⎝ (b)

The angle of reflection at each surface equals the angle of incidence at that surface. Thus,

(q )

1 reflection

22.34

22.35

22.36

= q1i = 30° , and (q 2 )reflection = q 2i = 41°

As light goes from a medium having a refractive index n1 to a medium with refractive index n2 < n1, the critical angle is given by the relation sinq c = n2 n1 . Table 22.1 gives the refractive index for various substances at l0 = 589 nm. (a)

For fused quartz surrounded by air, n1 = 1.458 and n2 = 1.00, giving q c = sin −1 (1.00 1.458) = 43.3° .

(b)

In going from polystyrene (n1 = 1.49) to air, q c = sin −1 (1.00 1.49) = 42.2° .

(c)

From sodium chloride (n1 = 1.544) to air, q c = sin −1 (1.00 1.544 ) = 40.4° .

When light is coming from a medium of refractive index n1 into water (n2 = 1.333), the critical angle is given by q c = sin −1 (1.333 n1 ). (a)

For fused quartz, n1 = 1.458, giving q c = sin −1 (1.333 1.458 ) = 66.1° .

(b)

In going from polystyrene (n1 = 1.49) to water, q c = sin −1 (1.333 1.49 ) = 63.5° .

(c)

From sodium chloride (n1 = 1.544) to water, q c = sin −1 (1.333 1.544 ) = 59.7° .

Using Snell’s law, the index of refraction of the liquid is found to be nliquid = (nair sinq i ) sinq r

continued on next page

68719_22_ch22_p198-222.indd 212

1/7/11 2:49:17 PM

Reflection and Refraction of Light

213

Thus, the critical angle for light going from this liquid into air is ⎡ ⎤ ⎛ n ⎞ nair −1 ⎡ sin 22.0° ⎤ q c = sin −1 ⎜ air ⎟ = sin −1 ⎢ = 48.5° ⎥ = sin ⎢ n ( n sinq ) sinq ⎣ sin 30.0° ⎥⎦ ⎝ liquid ⎠ i r ⎥ ⎢⎣ air ⎦ 22.37

22.38

When light attempts to cross a boundary from one medium of refractive index n1 into a new medium of refractive index n2 < n1, total internal reflection will occur if the angle of incidence exceeds the critical angle given by q c = sin −1 ( n2 n1 ). (a)

⎛ 1.00 ⎞ If n1 = 1.53 and n2 = nair = 1.00, then q c = sin −1 ⎜ = 40.8° . ⎝ 1.53 ⎟⎠

(b)

⎛ 1.333 ⎞ If n1 = 1.53 and n2 = nwater = 1.333, then q c = sin −1 ⎜ = 60.6° . ⎝ 1.53 ⎟⎠

The critical angle for this material in air is ⎛ n ⎞ ⎛ 1.00 ⎞ q c = sin −1 ⎜ air ⎟ = sin −1 ⎜ = 47.3° ⎝ 1.36 ⎟⎠ n ⎝ pipe ⎠ Thus, q r = 90.0° −q c = 42.7° and from Snell’s law, ⎛ npipe sinq r ⎞ ⎛ (1.36 ) sin 42.7° ⎞ = sin −1 ⎜ q i = sin −1 ⎜ ⎟⎠ = 67.3° ⎟ ⎝ nair 1.00 ⎝ ⎠

22.39

The angle of incidence at each of the shorter faces of the prism is 45°, as shown in the figure at the right. For total internal reflection to occur at these faces, it is necessary that the critical angle be less than 45°. With the prism surrounded by air, the critical angle is given by sinq c = nair nprism = 1.00 nprism, so it is necessary that sinq c =

or 22.40

(a)

nprism >

1.00 < sin 45° nprism

1.00 1.00 = = sin 45° 2 2

2

The minimum angle of incidence for which total internal reflection occurs is the critical angle. At the critical angle, the angle of refraction is 90°, as shown in the figure at the right. From Snell’s law, ng sinq i = na sin 90°, the critical angle for the glass-air interface is found to be ⎛ n sin 90° ⎞ −1 ⎛ 1.00 ⎞ q i = q c = sin −1 ⎜ a ⎟ = sin ⎜⎝ 1.78 ⎟⎠ = 34.2° n ⎝ ⎠ g

(b)

When the slab of glass has a layer of water on top, we want the angle of incidence at the water-air interface to equal the critical angle for that combination of media. At this angle, Snell’s law gives

air na = 1.00 glass ng = 1.78

qi

air na = 1.00 qc

qi

qc

water nw = 1.333 glass ng = 1.78

continued on next page

68719_22_ch22_p198-222.indd 213

1/7/11 2:49:21 PM

214

Chapter 22

nw sinq c = na sin 90° = 1.00 sinq c = 1.00 nw

and

Now, considering the refraction at the glass-water interface, Snell’s law gives ng sinq i = ng sinq c. Combining this with the result for sinq c from above, we find the required angle of incidence in the glass to be ⎛ n sinq c ⎞ ⎛ n (1.00 nw ) ⎞ ⎛ 1.00 ⎞ ⎛ 1.00 ⎞ q i = sin −1 ⎜ w = sin −1 ⎜ w = sin −1 ⎜ = sin −1 ⎜ = 34.2° ⎟ ⎟ ⎟ ⎝ 1.78 ⎟⎠ ng ⎠ ⎝ ng ⎠ ⎝ ⎝ ng ⎠ (c) and (d)

22.41

(a)

Observe in the calculation of part (b) that all the physical properties of the intervening layer (water in this case) canceled, and the result of part (b) is identical to that of part (a). This will always be true when the upper and lower surfaces of the intervening layer are parallel to each other. Neither the thickness nor the index of refraction of the intervening layer affects the result.

Snell’s law can be written as sinq1 sinq 2 = v1 v2. At the critical angle of incidence (q1 = q c ), the angle of refraction is 90°, and Snell’s law becomes sinq c = v1 v2. At the concrete-air boundary, ⎛v ⎞ ⎛ 343 m s ⎞ = 10.7° q c = sin −1 ⎜ 1 ⎟ = sin −1 ⎜ ⎝ 1 850 m s ⎟⎠ ⎝ v2 ⎠

22.42

(b)

Sound can be totally reflected only if it is initially traveling in the slower medium. Hence, at the concrete-air boundary, the sound must be traveling in air .

(c)

Sound in air falling on the wall from most directions is 100% reflected , so the wall is a good mirror.

(a)

The index of refraction for diamond is ndiam ond = 2.419, and the critical angle at a diamondair boundary is ⎛ n ⎞ ⎛ 1.000 ⎞ q c = sin −1 ⎜ air ⎟ = sin −1 ⎜ = 24.42° ⎝ 2.419 ⎟⎠ ⎝ ndiam ond ⎠

(b)

With the face of the diamond tilted up 35.0° from the horizontal, the normal line to this face at point P is tipped over 35.0° from the vertical. Thus, the angle of incidence at point P is q i = 35.0° > q c = 24.42°. Since this angle of incidence exceeds the critical angle, total internal reflection occurs at point P.

(c)

If the diamond is immersed in water (nwater = 1.333), the critical angle is ⎛ n ⎞ ⎛ 1.333 ⎞ q c = sin −1 ⎜ water ⎟ = sin −1 ⎜ = 33.44° ⎝ 2.419 ⎟⎠ ⎝ ndiam ond ⎠

(d)

The light continues to enter the top surface of the diamond at normal incidence, so the angle of incidence at point P continues to be q i = 35.0° > q c = 33.44°. Since this angle of incidence (barely) exceeds the critical angle for the diamond-water boundary, total internal reflection still occurs at point P .

(e)

To have light exit the diamond at point P, we need to decrease the angle of incidence at this point. Thus, we should rotate the diamond clockwise , thereby bringing the normal line closer to the incident ray at point P.

continued on next page

68719_22_ch22_p198-222.indd 214

1/7/11 2:49:23 PM

Reflection and Refraction of Light

(f )

Rotating the diamond clockwise by an angle q changes the angle of refraction at point A where the ray enters the diamond. To find what this new angle of refraction will be, we extend the line of the top of the diamond and the line of the face containing point P until they intersect at point B as shown at the right. Summing the interior angles in triangle ABP gives

q A a qr

and

35.0°

B

b

f

a + b + 35.0° = 180° or

215

P

35.0°

(90.0° −q ) + (90.0° − f ) + 35.0° = 180° r

q r = 35.0° − f

If the ray is to exit the diamond and enter the water at point P, the angle of incidence at this point must be less than or equal to the critical angle found in part (c). Thus, we require f ≤ 33.44° and see that we must have q r ≥ 35.0° − 33.44° = 1.6° Applying Snell’s law at point A gives the minimum required rotation as ⎛n sinq r ⎞ ⎡ 2.419sin (1.6° ) ⎤ = sin −1 ⎢ q m in = sin −1 ⎜ diam ond ⎥ = 2.9° ⎟ n 1.333 ⎝ ⎠ ⎦ ⎣ water 22.43

If q c = 42.0° at the boundary between the prism glass and the surrounding medium, then sinq c = n2 n1 gives

q1

nm = sin 42.0° nglass

nglass

Surrounding medium, nm b 60.0°

qr qc = 42.0° a

From the geometry shown in the figure at the right,

Surface 2

a = 90.0° − 42.0° = 48.0°, b = 180° − 60.0° − a = 72.0° and q r = 90.0° − b = 18.0°. Thus, applying Snell’s law at the first surface gives ⎛ sinq r ⎞ ⎛ nglass sinq r ⎞ ⎛ sin18.0° ⎞ = sin −1 ⎜ q1 = sin −1 ⎜ = sin −1 ⎜ = 27.5° ⎟ ⎟ ⎝ sin 42.0° ⎟⎠ nm ⎝ ⎠ ⎝ nm nglass ⎠ 22.44

The circular raft must cover the area of the surface through which light from the diamond could emerge. Thus, it must form the base of a cone (with apex at the diamond) whose half angle is q, where q is greater than or equal to the critical angle.

q

q

h q

rmin q

The critical angle at the water-air boundary is ⎛ n ⎞ ⎛ 1.00 ⎞ q c = sin −1 ⎜ air ⎟ = sin −1 ⎜ = 48.6° ⎝ 1.333 ⎟⎠ ⎝ nwater ⎠ Thus, the minimum diameter of the raft is 2rm in = 2 h tanq m in = 2 h tanq c = 2 ( 2.00 m ) tan 48.6° = 4.54 m

68719_22_ch22_p198-222.indd 215

1/7/11 2:49:25 PM

216

22.45

Chapter 22

At the air-ice boundary, Snell’s law gives the angle of refraction in the ice as sinq1r =

nair sinq1i nice

Since the sides of the ice layer are parallel, the angle of incidence at the ice-water boundary is q 2 i = q1r . Then, from Snell’s law, the angle of refraction in the water is ⎡ nice ⎛ n sinq 2 i ⎞ ⎛ nice sinq1r ⎞ −1 −1 ⎢ = sin = sin q 2 r = sin −1 ⎜ ice ⎜⎝ n ⎟⎠ ⎝ nwater ⎟⎠ ⎢⎣ nwater water or

⎛ n sinq ⎞ ⎤ air 1i ⎜ ⎟⎥ ⎝ nice ⎠ ⎥⎦

⎛ n sinq1i ⎞ ⎡ (1.00 ) sin 30.0° ⎤ = sin −1 ⎢ q 2 r = sin −1 ⎜ air ⎥ = 22.0° ⎟ n 1.333 ⎝ ⎠ ⎦ ⎣ water

Note that all of the properties of the ice canceled out in the above calculation, and the result is the same as if the ice had not been present. This will always be true when the intermediate medium has parallel sides. 22.46

When light coming from the surrounding medium is incident on the surface of the glass slab, Snell’s law gives ng sinq r = ns sinq i , or

(

)

sinq r = ns ng sinq i (a)

If q i = 30.0° and the surrounding medium is water (ns = 1.333), the angle of refraction is ⎡ 1.333sin ( 30.0° ) ⎤ q r = sin −1 ⎢ ⎥ = 23.7° 1.66 ⎦ ⎣

(b)

From Snell’s law given above, we see that as ns → ng we have sinq r → sinq i , or the angle of refraction approaches the angle of incidence, q r → q i = 30.0° .

22.47

(c)

If ns > ng, then sinq r = (ns ng )sinq i > sinq i , or q r > q i .

(a)

Given that the angle q shown in the figure at the right is 30.0°, the maximum distance the observer can be from the pool and continue to see the lower edge on the opposite side of the pool is

d q

d = (1.85 m ) tan 30.0° = 1.07 m

(b)

If the pool is now completely filled with water, the light ray coming to the observer’s eye from the lower opposite edge of the pool will refract at the surface of the water as shown in the second figure. The angle of refraction is ⎛ n sin 30.0° ⎞ ⎡ (1.333) sin 30.0° ⎤ = sin −1 ⎢ f = sin −1 ⎜ water ⎥ = 41.8° ⎟ nair 1.00 ⎝ ⎠ ⎦ ⎣

1.85 m q

d´ air nair

f

1.85 m

30.0°

nwater

The maximum distance the observer can now be from the pool and still see the same boundary is d ′ = (1.85 m ) tanf = (1.85 m ) tan 41.8° = 1.65 m

68719_22_ch22_p198-222.indd 216

1/7/11 2:49:27 PM

Reflection and Refraction of Light

22.48

(a)

217

For polystyrene surrounded by air, total internal reflection at the left vertical face requires that ⎛n ⎞ ⎛ 1.00 ⎞ q 3 ≥ q c = sin −1 ⎜ air ⎟ = sin −1 ⎜ = 42.2° ⎝ 1.49 ⎟⎠ n ⎝ p ⎠ From the geometry shown in the figure at the right, q 2 = 90.0° −q 3 ≤ 90.0° − 42.2° = 47.8° Thus, use of Snell’s law at the upper surface gives sinq1 =

n p sinq 2 nair

≤

(1.49 ) sin 47.8° 1.00

= 1.10

so it is seen that any angle of incidence ≤ 90° at the upper surface will yield total internal reflection at the left vertical face.

22.49

(b)

Repeating the steps of part (a) with the index of refraction of air replaced by that of water yields q 3 ≥ 63.5°, q 2 ≤ 26.5°, sinq1 ≤ 0.499, and q1 ≤ 29.9° .

(c)

Total internal reflection is not possible since npolystyrene < ncarbon disulfide .

(a)

From the geometry of the figure at the right, observe that q1 = 60.0°. Therefore, a = 90.0° −q1 = 30.0°

(q

and

2

+ 90.0° ) + a + 30.0° = 180.0°

Thus, q 2 = 180.0° − 120.0° − a = 30.0°

nglass

Since the prism is immersed in water, n2 = 1.333 and Snell’s law gives ⎛ nglass sinq 2 ⎞ −1 ⎛ (1.66 ) sin 30.0° ⎞ q 3 = sin −1 ⎜ ⎟⎠ = 38.5° ⎟⎠ = sin ⎜⎝ n2 1.333 ⎝ (b)

For refraction to occur at point P, it is necessary that q c > q1 . Thus,

⎛ n ⎞ q c = sin −1 ⎜ 2 ⎟ > q1 ⎝ nglass ⎠

which gives n2 > nglass sinq1 = (1.66 ) sin 60.0° = 1.44 . 22.50

Applying Snell’s law to this refraction, recognizing that nair = 1.00, gives nglass sinq 2 = nair sinq1 = sinq1 If q1 = 2q 2 , this becomes nglass sinq 2 = sin ( 2q 2 ) = 2sinq 2 cosq 2

continued on next page

68719_22_ch22_p198-222.indd 217

1/7/11 2:49:30 PM

218

Chapter 22

or

cosq 2 =

nglass 2

and

⎛ nglass ⎞ q 2 = cos −1 ⎜ ⎝ 2 ⎟⎠

Then, the desired angle of incidence is ⎛ nglass ⎞ ⎛ 1.56 ⎞ = 2 cos −1 ⎜ q1 = 2q 2 = 2 cos −1 ⎜ = 77.5° ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠ 22.51

In the figure at the right, observe that b = 90° −q1 and a = 90° −q1 . Thus, b =a. Similarly, on the right side of the prism, d = 90° −q 2 and g = 90° −q 2 , giving d = g . Next, observe that the angle between the reflected rays is B = (a + b ) + (g + d ) , so B = 2 (a + g ) . Finally, observe that the left side of the prism is sloped at angle a from the vertical, and the right side is sloped at angle g . Thus, the angle between the two sides is A = a + g , and we obtain the result B = 2 (a + g ) = 2 A .

22.52

(a)

Observe in the sketch at the right that a ray originally traveling along the inner edge will have the smallest angle of incidence when it strikes the outer edge of the fiber in the curve. Thus, if this ray is totally internally reflected, all of the others are also totally reflected. For this ray to be totally internally reflected it is necessary that q ≥ qc

But

sinq =

or R−d , R

sinq ≥ sinq c =

nair 1 = npipe n

so we must have

R−d 1 ≥ n R

which simplifies to R ≥ nd ( n − 1) , or Rm in = nd ( n − 1) . (b)

As d → 0, Rm in → 0. This is reasonable behavior. ⎛ nd d ⎞ As n increases, the minimum acceptable radius of curvature ⎜ Rm in = = n − 1 1 − 1 n ⎟⎠ ⎝ decreases. This is reasonable behavior. As n → 1, Rm in increases. This is reasonable behavior.

(c)

68719_22_ch22_p198-222.indd 218

Rm in =

(1.40 )(100 mm ) nd = = 350 mm n −1 1.40 − 1

1/7/11 2:49:32 PM

Reflection and Refraction of Light

22.53

219

Consider light which leaves the lower end of the wire and travels parallel to the wire while in the benzene. If the wire appears straight to an observer looking along the dry portion of the wire, this ray from the lower end of the wire must enter the observer’s eye as he sights along the wire. Thus, the ray must refract and travel parallel to the wire in air. The angle of refraction is then q 2 = 90.0° − 30.0° = 60.0°. From Snell’s law, the angle of incidence was ⎛ n sinq 2 ⎞ ⎛ (1.00 ) sin 60.0° ⎞ = sin −1 ⎜ q1 = sin −1 ⎜ air ⎟⎠ = 35.3° ⎝ 1.50 ⎝ nbenzene ⎟⎠ and the wire is bent by angle q = q 2 −q1 = 60.0° −q1 = 60.0° − 35.3° = 24.7° .

22.54

In the sketch at the right, the angle of incidence at A is the same as the prism angle at point O. This is true because tipping the line OA up angle q from the horizontal necessarily tips its normal line over at angle q from the vertical. Given that q = 60.0°, application of Snell’s law at point A gives 1.50 sin b = (1.00 ) sin 60.0° or

b = 35.3°

From triangle AOB, we calculate the angle of incidence and reflection, g , at point B: q + ( 90.0° − b ) + ( 90.0° − g ) = 180°

or

g = q − b = 60.0° − 35.3° = 24.7°

Now, we find the angle of incidence at point C using triangle BCQ:

(90.0° − g ) + (90.0° − d ) + (90.0° −q ) = 180° or

d = 90.0° − (q + g ) = 90.0° − 84.7° = 5.3°

Finally, application of Snell’s law at point C gives (1.00 ) sinf = (1.50 ) sin ( 5.3° ) , or f = sin −1 (1.50 sin 5.3° ) = 8.0° . 22.55

The path of a light ray during a reflection and/or refraction process is always reversible. Thus, if the emerging ray is parallel to the incident ray, the path which the light follows through this cylinder must be symmetric about the center line as shown at the right.

ncylinder

⎛ 1.00 m ⎞ ⎛ d 2⎞ = 30.0° Thus, q1 = sin −1 ⎜ = sin −1 ⎜ ⎝ R ⎟⎠ ⎝ 2.00 m ⎟⎠ Triangle ABC is isosceles, so g = a and b = 180° − a − g = 180° − 2a . Also, b = 180° −q1 , which gives a = q1 2 = 15.0°. Then, from applying Snell’s law at point A, ncylinder =

68719_22_ch22_p198-222.indd 219

nair sinq1 (1.00 ) sin 30.0° = = 1.93 sina sin15.0°

1/7/11 2:49:35 PM

220

Chapter 22

22.56

The angle of refraction as the light enters the left end of the slab is ⎛ n sinq1 ⎞ ⎛ (1.00 ) sin 50.0° ⎞ = sin −1 ⎜ q 2 = sin −1 ⎜ air ⎟⎠ = 31.2° ⎟ ⎝ 1.48 ⎝ nslab ⎠ Observe from the figure that the first reflection occurs at x = d, the second reflection is at x = 3d, the third is at x = 5d, and so forth. In general, the Nth reflection occurs at x = ( 2N − 1) d, where d=

( 0.310 cm 2) = tanq 2

0.310 cm = 0.256 cm 2 tan 31.2°

Therefore, the number of reflections made before reaching the other end of the slab at x = L = 42 cm is found from L = ( 2N − 1) d to be N= 22.57

(a)

1 ⎛ L ⎞ 1 ⎛ 42 cm ⎞ + 1⎟ = 82.5 or 82 complete reflections ⎜ + 1⎟ = ⎜ 2 ⎝ d ⎠ 2 ⎝ 0.256 cm ⎠

If q1 = 45.0°, application of Snell’s law at the point where the beam enters the plastic block gives

(1.00 ) sin 45.0° = n sinf

[1]

Application of Snell’s law at the point where the beam emerges from the plastic, with q 2 = 76.0°, gives n sin ( 90.0° − f ) = (1.00 ) sin 76.0°

or

(1.00 ) sin 76.0° = n cosf

[2]

Dividing Equation [1] by Equation [2], we obtain tanf =

sin 45.0° = 0.729 sin 76.0°

Thus, from Equation [1], (b)

f = 36.1°

sin 45.0° sin 45.0° = = 1.20 sinf sin 36.1°

Observe from the figure above that sinf = L d. Thus, the distance the light travels inside the plastic is d = L sinf , and if L = 50.0 cm = 0.500 m, the time required is Δt =

68719_22_ch22_p198-222.indd 220

n=

and

d L sinf nL 1.20 ( 0.500 m ) = = = = 3.39 × 10 −9 s = 3.39 ns v cn c sinf ( 3.00 × 108 m s ) sin 36.1°

1/7/11 2:49:38 PM

Reflection and Refraction of Light

22.58

221

Snell’s law would predict that nair sinq i = nwater sinq r , or since nair = 1.00, sinq i = nwater sinq r Comparing this equation to the equation of a straight line, y = mx + b, shows that if Snell’s law is valid, a graph of sinq i versus sinq r should yield a straight line that would pass through the origin if extended and would have a slope equal to nwater . q i ( deg ) q r ( deg ) sinq i

sinq r

10.0

7.50

0.174 0.131

20.0

15.1

0.342 0.261

30.0

22.3

0.500 0.379

40.0

28.7

0.643 0.480

50.0

35.2

0.766 0.576

60.0

40.3

0.866 0.647

70.0

45.3

0.940 0.711

80.0

47.7

0.985 0.740

The straightness of the graph line and the fact that its extension passes through the origin demonstrates the validity of Snell’s law. Using the end points of the graph line to calculate its slope gives the value of the index of refraction of water as nwater = slope = 22.59

0.985 − 0.174 = 1.33 0.740 − 0.131

Applying Snell’s law at points A, B, and C gives

and

1.40 sina = 1.60 sinq1

[1]

1.20 sin b = 1.40 sina

[2]

1.00 sinq 2 = 1.20 sin b

[3]

Combining Equations [1], [2], and [3] yields sinq 2 = 1.60 sinq1

[4]

Note that Equation [4] is exactly what Snell’s law would yield if the second and third layers of this “sandwich” were ignored. This will always be true if the surfaces of all the layers are parallel to each other. (a)

If q1 = 30.0°, then Equation [4] gives q 2 = sin −1 (1.60 sin 30.0° ) = 53.1° .

(b)

At the critical angle of incidence on the lowest surface, q 2 = 90.0°. Then, Equation [4] gives ⎛ sin 90.0° ⎞ ⎛ sinq 2 ⎞ q1 = sin −1 ⎜ = 38.7° = sin −1 ⎜ ⎝ 1.60 ⎟⎠ ⎝ 1.60 ⎟⎠

68719_22_ch22_p198-222.indd 221

1/7/11 2:49:40 PM

222

Chapter 22

Given Conditions and Observed Results

22.60

Case 1

Case 2

For the first placement, Snell’s law gives

n2 =

Case 3

n1 sin 26.5° sin 31.7°

In the second placement, application of Snell’s law yields n sin 36.7° ⎛ n sin 26.5° ⎞ n3 sin 26.5° = n2 sin 36.7° = ⎜ 1 sin 36.7°, or n3 = 1 ⎝ sin 31.7° ⎟⎠ sin 31.7° Finally, using Snell’s law in the third placement gives sinq R = and 22.61

⎛ sin 31.7° ⎞ n1 sin 26.5° = ( n1 sin 26.5° ) ⎜ = 0.392 n3 ⎝ n1 sin 36.7° ⎟⎠

q R = 23.1°

From the right triangle PAC in the figure at the right, observe that ⎛ 6.00 cm ⎞ ⎛ h⎞ q1 = sin ⎜ ⎟ = sin −1 ⎜ = 30.0° ⎝ 12.0 cm ⎟⎠ ⎝ R⎠ −1

nair = 1.00 nLucite = 1.50

q1

This is also the angle of incidence at point P where the light ray enters the Lucite. Snell’s law then gives ⎛ n sinq1 ⎞ sin 30.0° ⎤ = sin −1 ⎡⎢ q 2 = sin ⎜ air = 19.5° ⎟ ⎣ 1.50 ⎥⎦ ⎝ n ⎠ −1

P q2 h

f f

B q

R q1

A

C

Lucite

Now, by observing the vertical angles at point P, we find that q 2 + f = q1 , so the angle of incidence at point B where the ray exits from the Lucite is f = q1 −q 2 = 10.5°. The angle of refraction at point B is then given by Snell’s law as ⎛n sinf ⎞ −1 q = sin −1 ⎜ Lucite ⎟⎠ = sin [(1.50 ) sin10.5° ] = 15.9° nair ⎝

68719_22_ch22_p198-222.indd 222

1/7/11 2:49:43 PM

23 Mirrors and Lenses QUICK QUIZZES 1.

At C.

2.

Choice (c). Since nwater > nair, the virtual image of the fish formed by refraction at the flat water surface is closer to the surface than is the fish. See Equation 23.9 in the textbook.

3.

(a)

False. A concave mirror forms an inverted image when the object distance is greater than the focal length.

(b)

False. The magnitude of the magnification produced by a concave mirror is greater than 1 if the object distance is less than the radius of curvature.

(c)

True.

4.

Choice (b). In this case, the index of refraction of the lens material is less than that of the surrounding medium. Under these conditions, a biconvex lens will be divergent.

5.

Although a ray diagram only uses 2 or 3 rays (those whose direction is easily determined using only a straight edge), an infinite number of rays leaving the object will always pass through the lens.

6.

(a)

False. A virtual image is formed on the left side of the lens if p < f .

(b)

True. An upright, virtual image is formed when p < f , while an inverted, real image is formed when p > f .

(c)

False. A magnified, real image is formed if 2 f > p > f , and a magnified, virtual image is formed if p < f .

223

68719_23_ch23_p223-255.indd 223

1/7/11 2:50:25 PM

224

Chapter 23

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

The image formed by a flat mirror of a real object is always an upright, virtual image that is the same size as the object and located as far behind the mirror as the object is in front of the mirror. Thus, statements (b), (c), and (e) are all true, while statements (a) and (d) are false.

2.

Since the size of the image is 1.50 times the size of the object, the magnitude of the magnification is M = 1.50. Because the image is upright, M > 0. Thus, M = − q p = +1.50, which gives q = −1.50 p. Then from the mirror equation, we have f=

R qp ( −1.50 p) p = +3.00 p = +3.00 ( +30.0 cm ) = +90.0 cm = = 2 p+q p − 1.50 p

and the correct choice is (d). 3.

For a convergent lens, f > 0, and because the image is real, q > 0. The thin-lens equation, 1 p + 1 q = 1 f , then gives p=

qf (12.0 cm )(8.00 cm ) = = +24.0 cm q− f 12.0 cm − 8.00 cm

Since p > 0 , the object is in front (in this case, to the left) of the lens, and the correct choice is (c). 4.

For a converging lens, the focal length is positive, or f > 0. Since the object is virtual, we know that the object distance is negative, or p < 0 and p = − | p |. Thus, the thin-lens equation gives the image distance as q=

⎛ | p| ⎞ pf −| p| f f = = +⎜ p− f −| p|− f ⎝ | p | + f ⎟⎠

Since | p | and f are positive quantities, we see that q > 0 and the image is real. Also, since | p | (| p | + f ) < 1, we see that q < f . Thus, we have shown that choices (a) and (d) are false statements, while choices (b), (c), and (e) are all true. 5.

From the mirror equation, 1 p + 1 q = 2 R = 1 f , with f < 0 since the mirror is convex, the image distance is found to be q=

pf (16.0 cm )( −6.00 cm ) = = − 4.36 cm p − f 16.0 cm − ( −6.00 cm )

Since q < 0, the image is virtual and located 4.36 cm behind the mirror. Choice (d) is the correct answer. 6.

For a divergent lens, f < 0, and because the object is real, p > 0. The thin-lens equation, 1 p + 1 q = 1 f , then gives q=

pf (10.0 cm )( −16.0 cm ) = = − 6.15 cm p − f 10.0 cm − ( −16.0 cm )

Since q < 0, the image is in front (in this case, to the left) of the lens, and the correct choice is (b).

68719_23_ch23_p223-255.indd 224

1/7/11 2:50:26 PM

Mirrors and Lenses

225

7.

A concave mirror forms inverted, real images of real objects located outside the focal point ( p > f ), and upright, magnified, virtual images of real objects located inside the focal point ( p < f ) . Virtual images, located behind the mirror, have negative image distances by the sign convention of Table 23.1. Choices (d) and (e) are true statements, and all other choices are false.

8.

A convergent lens forms inverted, real images of real objects located outside the focal point ( p > f ). When p > 2 f , the real image is diminished in size, and the image is enlarged if 2 f > p > f . For real objects located inside the focal point ( p < f ) of the convergent lens, the image is upright, virtual, and enlarged. In the given case, p > 2 f , so the image is real, inverted, and diminished in size. Choice (c) is the correct answer.

9.

With a real object in front of a convex mirror, the image is always upright, virtual, diminished in size, and located between the mirror and the focal point. Thus, of the available choices, only choice (c) is a true statement.

10.

For a real object ( p > 0) and a diverging lens ( f < 0), the image distance given by the thin-lens equation is q=

p (− f ) p f pf = =− 0 p p

Thus, the image is always virtual and upright, meaning that choice (b) is a true statement while all other choices are false. 11.

Light rays reflecting from the fish, and passing through the flat refracting surface of the water (nwater > nair) on the way to the fisherman’s eye, form a virtual image located closer to the surface than the fish’s actual depth (see Example 23.6 in the textbook). Thus, the fisherman should aim below the apparent location of the fish, and (b) is the correct choice.

12.

The image that a plane reflecting surface forms of a real object is an upright, virtual image located as far behind the reflecting surface as the object is in front of it. Choices (a) and (d) must be eliminated because they portray inverted images. Also, the tip of the image arrow should be closer to the mirror than the base of the image arrow, just as is true of the ends of the object arrow. Thus, choice (c) must be eliminated, leaving (b) as best describing the image formed by the mirror.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The stream appears shallower than its true depth because refraction of light coming from pebbles on the bottom of the stream forms virtual images located closer to the surface than the actual objects.

4.

Chromatic aberration is produced when light passes through a material, as it does when passing through the glass of a lens. A mirror, silvered on its front surface, never has light passing through it, so this aberration cannot occur. This is only one of many reasons why large telescopes use mirrors rather than lenses for their primary optical elements.

68719_23_ch23_p223-255.indd 225

1/7/11 2:50:29 PM

226

Chapter 23

6.

Light rays diverge from the position of a virtual image just as they do from an actual object. Thus, a virtual image can be as easily photographed as any object can. Of course, the camera would have to be placed near the axis of the lens or mirror in order to intercept the light rays.

8.

Actually no physics is involved here. The design is chosen so your eyelashes will not brush against the glass as you blink. A reason involving a little physics is that with this design, when you direct your gaze near the outer circumference of the lens, you receive a ray that has passed through glass with more nearly parallel surfaces of entry and exit. Then the lens minimally distorts the direction to the object you are looking at.

10.

Both words are inverted. However, OXIDE looks the same right-side-up and upside-down. LEAD does not.

12.

(a)

No. The screen is needed to reflect the light toward your eye.

(b)

Yes. The light is traveling toward your eye and diverging away from the position of the image, the same as if the object were located at that position.

14.

The correct answer is choice (d). The entire image would appear because any portion of the lens can form the image. The image would be dimmer because the card reduces the light intensity on the screen by 50%.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a)

1.0 m behind nearest mirror

(b)

the palm

(c)

5.0 m behind nearest mirror

(d)

back of the hand

(e)

7.0 m behind nearest mirror

(f )

the palm

4.

4.58 m

6.

(a)

8.

R = − 0.790 cm

R = + 2.22 cm

(b)

(g)

all are virtual

M = +10.0

10.

The mirror is concave with R ≈ 60 cm and f ≈ 30 cm.

12.

(a)

p = +15.0 cm

14.

(a)

60.0 cm

(b)

q = 42.9 cm, M = −0.429, so the image is real, inverted, and 42.9 cm in front of mirror.

(c)

q = −15.0 cm, M = 1.50, so the image is virtual, upright, and 15.0 cm behind the mirror.

(a)

a real object located 16 cm in front of the mirror

(b)

upright and one-third the size of the object

(a)

p = 8.00 cm

16.

18.

68719_23_ch23_p223-255.indd 226

(b)

(b)

R = 60.0 cm

See Solution.

(c)

virtual

1/7/11 2:50:30 PM

Mirrors and Lenses

20.

227

(a)

As the ball moves from p = 3.00 m to p = 0.500 m, the image moves from q = + 0.600 m to q = + ∞. As the ball moves from p = 0.500 m to p = 0, the image moves from q = − ∞ to q = 0.

(b)

at t = 0.639 s when p = 1.00 m, and at t = 0.782 s when p = 0

22.

in the water, 9.00 cm inside the wall of the bowl

24.

(a)

26.

4.8 cm

28.

See Solution.

30.

(a)

at either 36.2 cm or 13.8 cm from the screen

(b)

When q = 36.2 cm, M = −2.62. When q = 13.8 cm, M = − 0.381.

(a)

See Solution.

(d)

The mirror equation yields q = − 40.0 cm and M = +2.00.

(a)

See Solution.

(c)

Graph (a) yields an upright, virtual image located 13.3 cm from the lens and one-third the size of the object. Graph (b) yields an upright, virtual image located 6.7 cm from the lens and two-thirds the size of the object. Both agree with the algebraic solutions.

32.

34.

1.50 m

(b)

(b)

(b)

1.75 m

40 cm behind the mirror

(c)

+2

See Solution.

36.

f = +5.68 cm

38.

(a)

12.3 cm to the left of the lens

(b)

M = +0.615

40.

(a)

q = p f (p − f )

(b)

q < 0 for all values of p > 0 and f < 0

(c)

q > 0 only if p > f when f > 0

(a)

f2 = −11.1 cm

(b)

M = +2.50

42.

(c)

See Solution.

(c)

virtual and upright

44.

(a)

13.3 cm

(b)

M = −5.91

46.

(a)

q1 = +30.0 cm

(b)

20.0 cm beyond the second lens

(c)

p2 = −20.0 cm

(d)

q2 = + 4.00 cm

(e)

M1 = −2.00

(f )

M 2 = + 0.200

(g)

M total = −0.400

(h)

real, inverted

48.

(a)

q = 5f 4

(b)

M = −1 4

(c)

real, inverted, opposite side

50.

(a)

f = −12.0 cm

(b)

q = f = −12.0 cm

(c)

q = −9.00 cm

(d)

q = −6.00 cm

(e)

q = − 4.00 cm

68719_23_ch23_p223-255.indd 227

(c)

inverted

1/7/11 2:50:32 PM

228

52.

Chapter 23

(a)

25.3 cm to the right of the mirror

(b)

virtual

(c)

54.

Δx = 21.3 cm

56.

See the Solution.

58.

See the Solution.

60.

(a)

p=4f 3

(c)

M a = −3, M b = + 4

62.

(b)

upright

(d)

M = + 8.05

(b)

See Solution.

(c)

f = −20.0 cm

p = 3f 4

fm irror = 11.7 cm

64.

(a)

0.708 cm in front or the spherical ornament

66.

(a)

convex

(b)

at the 30.0-cm mark

PROBLEM SOLUTIONS 23.1

(a)

Due to the finite value of the speed of light, the light arriving at your eye must have reflected from your face at a slightly earlier time. Thus, the image viewed in the mirror shows you younger than your actual age .

(b)

If you stand 40 cm in front of the mirror, the time required for light scattered from your face to travel to the mirror and back to your eye is Δt =

23.2

(a)

2d 2 ( 0.40 m ) = = 2.7 × 10 −9 s c 3.0 × 108 m s

or

∼ 10 −9 s

With the palm located 1.0 m in front of the nearest mirror, that mirror forms an image, I P 1 , located 1.0 m behind the nearest mirror .

68719_23_ch23_p223-255.indd 228

(b)

The image I P 1 is an image of the palm .

(c)

The woman’s left hand is located 2.0 m in front of the farthest mirror, which forms an image located 2.0 m behind this mirror (and hence, 5.0 m in front of the nearest mirror). This image, I B1 , serves as an object for the nearest mirror, which then forms an image, I B 2 , located 5.0 m behind the nearest mirror .

(d)

The images I B1 and I B 2 are both images of the back of the hand .

(e)

The image I P 1 [see part (a)] serves as an object located 4.0 m in front of the farthest mirror, which forms an image I P 2 , located 4.0 m behind that mirror and 7.0 m in front of the nearest mirror. This image then serves as an object for the nearest mirror, which forms an image I P 3 , located 7.0 m behind the nearest mirror .

(f)

The images I P 2 and I P 3 are both images of the palm .

(g)

Since all images are located behind the mirrors, all are virtual images .

1/7/11 2:50:35 PM

Mirrors and Lenses

23.3

23.4

229

(1)

The first image in the left-hand mirror is 5.00 ft behind the mirror or 10.0 ft from the person .

(2)

The first image in the right-hand mirror serves as an object for the left-hand mirror. It is located 10.0 ft behind the right-hand mirror, which is 25.0 ft from the left-hand mirror. Thus, the second image in the left-hand mirror is 25.0 ft behind the mirror or 30.0 ft from the person .

(3)

The first image in the left-hand mirror serves as an object for the right-hand mirror. It is located 20.0 ft in front of the right-hand mirror and forms an image 20.0 ft behind that mirror. This image then serves as an object for the left-hand mirror. The distance from this object to the left-hand mirror is 35.0 ft. Thus, the third image in the left-hand mirror is 35.0 ft behind the mirror or 40.0 ft from the person .

The virtual image is as far behind the mirror as the choir is in front of the mirror. Thus, the image is 5.30 m behind the mirror. The image of the choir is 0.800 m + 5.30 m = 6.10 m from the organist. Using similar triangles gives 6.10 m h′ = 0.600 m 0.800 m or

23.5

⎛ 6.10 m ⎞ h ′ = ( 0.600 m ) ⎜ = 4.58 m ⎝ 0.800 m ⎟⎠

When an object is in front of a plane mirror, that mirror forms an upright, virtual image that is the same size as the object and as far behind the mirror as the object is in front of the mirror. This statement is true even if the mirror is rotated, as shown in the ray diagrams given below. In figure (a), a real object O1 is distance p1 in front of the upper mirror in the periscope. This mirror forms the virtual image I1 at distance p1 behind the mirror. As shown in figure (b), this image serves as the object for the lower mirror in the periscope, and is distance p2 = p1 + h in front of the lower mirror. The lower mirror then forms the final image I2, an upright, virtual image, located distance p2 = p1 + h behind this mirror. O2 = I1

I1 Upper mirror

Lower mirror

| q1| = p1 O1

p2 = p1 + h I2

p1

(a)

| q2| = p2 = p1 + h

(b) continued on next page

68719_23_ch23_p223-255.indd 229

1/7/11 2:50:38 PM

230

Chapter 23

(a)

As shown in the above ray diagrams, the final image is distance q2 = p1 + h behind the lower mirror, where p1 is the distance from the original object to the upper mirror and h is the vertical distance between the two mirrors in the periscope.

(b)

The final image is behind the mirror and is virtual .

(c)

As seen from the ray diagrams, the final image I2 is oriented the same way as the original object, and is therefore upright .

(d)

The overall magnification is ⎛ q ⎞ ⎛ q ⎞ ⎛ −p ⎞ ⎛ −p ⎞ M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ( +1)( +1) = +1 ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ p1 ⎠ ⎝ p2 ⎠ The final image is therefore upright and the same size as the original object.

23.6

(e)

No . The images formed by plane mirrors are upright in both directions. Just as plane mirrors do not reverse up and down, neither do they reverse left and right.

(a)

Since the object is in front of the mirror, p > 0. With the image behind the mirror, q < 0. The mirror equation gives the radius of curvature as 1 1 10 − 1 2 1 1 = + = − = R p q 1.00 cm 10.0 cm 10.0 cm or

23.7

⎛ 10.0 cm ⎞ R = 2⎜ ⎟ = + 2.22 cm ⎝ 9 ⎠ q (−10.0 cm) =− = +10.0 . p 1.00 cm

(b)

The magnification is M = −

(a)

The center of curvature of a convex mirror is behind the mirror. Therefore, the radius of curvature, and hence the focal length f = R 2, is negative. With the image behind the mirror, the image is virtual and q = −10.0 cm. The mirror equation then gives p=

qf ( −10.0 cm )( −15.0 cm ) = = +30.0 cm q − f −10.0 cm − ( −15.0 cm )

The object should be placed 30.0 cm in front of the mirror . (b)

The magnification of the mirror is M=−

q ( −10.0 cm ) =− = + 0.333 p +30.0 cm

Therefore, the image is upright and one-third the size of the object. 23.8

The lateral magnification is given by M = − q p. Therefore, the image distance is q = −Mp = − ( 0.013 0 )( 30.0 cm ) = −0.390 cm The mirror equation,

2 pq 2 1 1 = + , or R = , gives p+q R p q

continued on next page

68719_23_ch23_p223-255.indd 230

1/7/11 2:50:39 PM

Mirrors and Lenses

R=

231

2 ( 30.0 cm )( −0.390 cm ) = − 0.790 cm 30.0 cm − 0.390 cm

The negative sign tells us that the surface is convex, as expected. 23.9

(a)

The center of curvature of a concave mirror is in front of the mirror. Therefore, both the radius of curvature and the focal length, f = R 2, are positive. Since the image is virtual, the image distance is negative and q = −20.0 cm. With R = + 40.0 cm and f = +20.0 cm, the mirror equation gives p=

( −20.0 cm )( + 20.0 cm ) qf = = +10.0 cm q − f −20.0 cm − ( + 20.0 cm )

Thus, the object should be placed 10.0 cm in front of the mirror . (b)

The magnification of the mirror is M=−

q ( −20.0 cm ) =− = + 2.00 p +10.0 cm

Therefore, the image is upright and twice the size of the object. 23.10

The image was initially upright but became inverted when Dina was more than 30 cm from the mirror. From this information, we know that the mirror must be concave , because a convex mirror will form only upright, virtual images of real objects. When the object is located at the focal point of a concave mirror, the rays leaving the mirror are parallel, and no image is formed. Since Dina observed that her image disappeared when she was about 30 cm from the mirror, we know that the focal length must be f ≈ 30 cm . Also, for spherical mirrors, R = 2 f . Thus, the radius of curvature of this concave mirror must be R ≈ 60 cm .

23.11

The enlarged, virtual images formed by a concave mirror are upright, so M > 0. Thus, M = −

q h ′ 5.00 cm = = = + 2.50, giving p h 2.00 cm

q = − 2.50 p = − 2.50 ( + 3.00 cm ) = −7.50 cm The mirror equation then gives f= 23.12

( 3.00 cm )( − 7.50 cm ) pq = = + 5.00 cm p+q 3.00 cm − 7.50 cm

Realize that the magnitude of the radius of curvature, R , is the same for both sides of the hubcap. For the convex side, R = − R ; and for the concave side, R = + R . The object distance p is positive (real object) and has the same value in both cases. Also, we write the virtual image distance as q = − q in each case. The mirror equation then gives: For the convex side,

2 1 1 = − −q −R p

or

q=

Rp R + 2p

[1]

For the concave side,

2 1 1 = − R p −q

or

q=

Rp R − 2p

[2] continued on next page

68719_23_ch23_p223-255.indd 231

1/7/11 2:50:41 PM

232

Chapter 23

Comparing Equations [1] and [2], we observe that the smaller magnitude image distance, q = 10.0 cm, occurs with the convex side of the mirror. Hence, we have 2 1 1 = − −10.0 cm − R p

[3]

and for the concave side, q = 30.0 cm gives 2 1 1 = − −30.0 cm R p

23.13

[4]

(a)

Adding Equations [3] and [4] yields

(b)

Subtracting [3] from [4] gives

3+1 2 = , or p = +15.0 cm . p 30.0 cm

3−1 4 = , or R = 60.0 cm . R 30.0 cm

The image is upright, so M > 0, and we have M=−

q = + 2.0, or q = − 2.0 p = − 2.0 ( 25 cm ) = − 50 cm p

The radius of curvature is then found to be 1 1 2 −1 2 1 1 = + = − = , or R = 2 ⎛ 0.50 m ⎞ = 1.0 m ⎜⎝ +1 ⎟⎠ R p q 25 cm 50 cm 50 cm 23.14

(a)

For spherical mirrors, both concave and convex, the radius of curvature is related to the focal length by R = 2 f . Hence, the radius of curvature of this concave mirror, having f = +30.0 cm, is R = + 60.0 cm .

(b)

When p = +100 cm (i.e., a real object located 100 cm in front of the mirror), q=

and

pf ( +100 cm )( +30.0 cm ) = = + 42.9 cm p− f 100 cm − 30.0 cm

M=−

q 42.9 cm =− = −0.429 p 100 cm

Thus, the image is real (q > 0), inverted (M < 0), diminished in size ( | M |< 1), and located 42.9 cm in front of the mirror . (c)

When p = +10.0 cm (i.e., a real object located 10.0 cm in front of the mirror), q=

and

pf ( +10.0 cm )( +30.0 cm ) = = −15.0 cm p− f 10.0 cm − 30.0 cm

M=−

q ( −15.0 cm ) =− = +1.50 p 10.0 cm

Thus, the image is virtual (q < 0), upright (M > 0), enlarged ( | M |> 1), and located 15.0 cm behind the mirror .

68719_23_ch23_p223-255.indd 232

1/7/11 2:50:43 PM

Mirrors and Lenses

23.15

233

The focal length of the mirror may be found from the given object and image distances as 1 f = 1 p + 1 q , or f=

pq (152 cm )(18.0 cm ) = = +16.1 cm p + q 152 cm + 18.0 cm

For an upright image twice the size of the object, the magnification is M = − q p = +2.00 , giving q = − 2.00 p Then, using the mirror equation again, 1 p + 1 q = 1 f becomes 1 2 −1 1 1 1 1 + = − = = p q p 2.00 p 2.00 p f

23.16

f 16.1 cm = = 8.05 cm 2.00 2.00

or

p=

(a)

The mirror is convex, so f < 0, and we have f = − f = −8.0 cm. The image is virtual, so q < 0, or q = − q . Since we also know that q = p 3, the mirror equation gives 1 1 1 3 1 + = − = p q p p f

or

−

2 1 = p − 8.0 cm

and

p = +16 cm

This means that we have a real object located 16 cm in front of the mirror .

23.17

(b)

The magnification is M = − q p = + q p = +1 3 . Thus, the image is upright and one-third the size of the object.

(a)

Since the mirror is convex, R < 0. Thus, R = − 0.550 m and f = R 2 = − 0.275 m. The mirror equation then yields q=

pf ( +10.0 m )( −0.275 m ) = = − 0.268 m = −26.8 cm p − f +10.0 m − ( −0.275 m )

The image is located 26.8 cm behind the mirror .

23.18

(b)

The magnification of the image is M = − q p. Since p > 0 and q < 0 in this case, we see that M > 0. Therefore, the image is upright .

(c)

M=−

(a)

Since the mirror is concave, R > 0, giving R = +24.0 cm and f = R 2 = +12.0 cm. Because the image is upright (M > 0) and three times the size of the object ( M = 3.00 ) , we have

q ( −26.8 cm ) ( −26.8 cm ) =− =− = +2.68 × 10 −2 = + 0.026 8 p 10.0 m 10.0 × 10 2 cm

M=−

q = +3.00 p

and

q = −3p

The mirror equation then gives 1 2.00 1 1 − = = p 3.00 p 3.00 p 12.0 cm

or

p = +8.00 cm

continued on next page

68719_23_ch23_p223-255.indd 233

1/7/11 2:50:46 PM

234

23.19

Chapter 23

(b)

The needed ray diagram, with the object 8.00 cm in front of the mirror, is shown below:

(c)

From a carefully drawn scale drawing, you should find that the image is upright, virtual, 24.0 cm behind the mirror, and three times the size of the object .

(a)

An image formed on a screen is a real image. Thus, the mirror must be concave since, of mirrors, only concave mirrors can form real images of real objects.

(b)

The magnified, real images formed by concave mirrors are inverted, so M < 0 and M =−

q 5.0 m q = 1.0 m = − 5 , or p = = 5 5 p

The object should be 1.0 m in front of the mirror . (a — revisited) Now that we have both the object distance and image distance, we can use the mirror equation to be more specific about the required mirror. The needed focal length of the concave mirror is given by f=

23.20

(a)

From

pq (1.0 m )( 5.0 m ) = = 0.83 m p + q 1.0 m + 5.0 m

Rp 1 1 2 (1.00 m ) p + = , we find q = = . 2 p − R 2 p − 1.00 m p q R

The table gives the image position at a few critical points in the motion. Between p = 3.00 m and p = 0.500 m, the real image moves from 0.600 m to positive infinity. From p = 0.500 cm to p = 0, the virtual image moves from negative infinity to 0.

Object Distance, p

Image Distance, q

3.00 m 0.500 m

0.600 m ±∞

0

0

Note the “jump” in the image position as the ball passes through the focal point of the mirror. (b)

The ball and its image coincide when p = 0 and when 1 p + 1 p = 2 p = 2 R , or p = R = 1.00 m. From Δy = v0 y t + 12 a y t 2 , with v0 y = 0, the times for the ball to fall from y = +3.00 m to these positions are found to be

68719_23_ch23_p223-255.indd 234

p = 1.00 m position:

t=

2 ( Δy ) = ay

2 ( −2.00 m ) = 0.639 s and −9.80 m s 2

p = 0 position:

t=

2 ( −3.00 m ) = 0.782 s −9.80 m s 2

1/7/11 2:50:50 PM

Mirrors and Lenses

23.21

From

235

n1 n2 n2 − n1 + = , with R → ∞, the image position is found to be p q R q =−

n2 p= n1

⎛ 1.00 ⎞ −⎜ ( 50.0 cm ) = − 38.2 cm ⎝ 1.309 ⎟⎠

or the virtual image is 38.2 cm below the upper surface of the ice . 23.22

The location of the image formed by refraction at this spherical surface is described by Equation 23.7 from the textbook, which states n1 p + n2 q = (n2 − n1 ) R and uses the sign convention of Table 23.2. As light crosses this surface, passing from the water into air, we have n1 = nwater = 1.333, n2 = nair = 1.00, p = +10.0 cm (object is in front of the surface), and R = −15.0 cm (center of curvature is in front of the nair = 1.00 surface). Thus, the image distance is found from

n2 n2 − n1 n1 = − q R p

or

p R nwater = 1.333

1.333 +0.666 − 3.999 −3.333 1.00 1.00 − 1.333 = − = = −15.0 cm +10.0 cm 30.0 cm 30.0 cm q

This yields q = 30.0 cm ( −3.333) = −9.00 cm. Thus, the goldfish appears to be in the water, 9.00 cm inside the wall of the bowl . 23.23

Since the center of curvature of the surface is on the side the light comes from, R < 0, giving n n n − n1 R = − 4.0 cm. Then, 1 + 2 = 2 becomes p q R 1.50 1.00 1.00 − 1.50 = − , or q = − 4.0 cm − 4.0 cm 4.0 cm q Thus, the magnification M =

⎛n ⎞q h′ = − ⎜ 1 ⎟ gives h ⎝ n2 ⎠ p

⎛ nq⎞ 1.50 ( −4.0 cm ) h′ = − ⎜ 1 ⎟ h = − ( 2.5 mm ) = 3.8 mm 1.00 ( 4.0 cm ) ⎝ n2 p ⎠ 23.24

For a plane refracting surface, R → ∞, and (a)

n1 n2 n2 − n1 n + = becomes q = − 2 p. p q R n1

Considering the bottom of the pool as the object, p = 2.00 m when the pool is full, and ⎛ 1.00 ⎞ q =−⎜ ( 2.00 m ) = −1.50 m ⎝ 1.333 ⎟⎠ or the pool appears to be 1.50 m deep.

(b)

68719_23_ch23_p223-255.indd 235

If the pool is half filled, then p = 1.00 m, and q = − 0.750 m. Thus, the bottom of the pool appears to be 0.75 m below the water surface or 1.75 m below ground level.

1/7/11 2:50:53 PM

236

23.25

Chapter 23

As parallel rays from the Sun ( object distance, p → ∞ ) enter the transparent sphere from air = 1.00 ) , the center of curvature of the surface is on the side the light is going toward (back 1 side). Thus, R > 0. It is observed that a real image is formed on the surface opposite the Sun, giving the image distance as q = +2R. Then,

(n

n1 n2 n2 − n1 + = p q R

becomes

0+

n n − 1.00 = 2R R

which reduces to n = 2n − 2.00, and gives n = 2.00 . 23.26

Light scattered from the bottom of the plate undergoes two refractions, once at the top of the plate and once at the top of the water. All surfaces are planes ( R → ∞ ) , so the image distance for each refraction is q = − ( n2 n1 ) p. At the top of the plate, ⎛n ⎞ ⎛ 1.333 ⎞ q1 B = − ⎜ water ⎟ p1 B = − ⎜ (8.00 cm ) = − 6.42 cm ⎝ 1.66 ⎟⎠ ⎝ nglass ⎠ or the first image is 6.42 cm below the top of the plate. This image serves as a real object for the refraction at the top of the water, so the final image of the bottom of the plate is formed at ⎛ n ⎞ ⎛ n ⎞ q2 B = − ⎜ air ⎟ p2B = − ⎜ air ⎟ 12.0 cm + q1B n ⎝ water ⎠ ⎝ nwater ⎠

(

)

⎛ 1.00 ⎞ = −⎜ (18.4 cm ) = −13.8 cm or 13.8 cm below the water surface. ⎝ 1.333 ⎟⎠ Now, consider light scattered from the top of the plate. It undergoes a single refraction, at the top of the water. This refraction forms an image of the top of the plate at ⎛ n ⎞ ⎛ 1.00 ⎞ qT = − ⎜ air ⎟ pT = − ⎜ (12.0 cm ) = − 9.00 cm ⎝ 1.333 ⎟⎠ ⎝ nwater ⎠ or 9.00 cm below the water surface. The apparent thickness of the plate is then Δy = q2B − qT = 13.8 cm − 9.00 cm = 4.8 cm . 23.27

In the drawing at the right, object O (the jellyfish) is located distance p in front of a plane water-glass interface. Refraction at that interface produces a virtual image I ′ at distance q ′ in front it. This image serves as the object for refraction at the glass-air interface. This object is located distance p′ = q ′ + t in front of the second interface, where t is the thickness of the layer of glass. Refraction at the glass-air interface produces a final virtual image, I, located distance q in front of this interface. From n1 p + n2 q = (n2 − n1 ) R with R → ∞ for a plane, the relation between the object and image distances for refraction at a flat surface is q = − ( n2 n1 ) p. Thus, the image distance for the refraction at the water-glass interface is q ′ = − ng nw p. This gives an object distance for the refraction at the glass-air interface of p′ = (ng nw ) p + t and a final image position (measured from the glass-air interface) of

(

)

continued on next page

68719_23_ch23_p223-255.indd 236

1/7/11 2:50:56 PM

Mirrors and Lenses

q=− (a)

na n p′ = − a ng ng

237

⎡⎛ n ⎞ ⎛ na ⎞ ⎤ ⎡ ⎛ ng ⎞ ⎤ a ⎢⎜ ⎟ p + t ⎥ = − ⎢⎜ ⎟ p + ⎜ ⎟ t ⎥ ⎝ ng ⎠ ⎥⎦ ⎢⎣ ⎝ nw ⎠ ⎥⎦ ⎢⎣⎝ nw ⎠

If the jellyfish is located 1.00 m (or 100 cm) in front of a 6.00-cm thick pane of glass, then p = +100 cm, t = 6.00 cm, and the position of the final image relative to the glass-air interface is ⎡⎛ 1.00 ⎞ ⎛ 1.00 ⎞ q = − ⎢⎜ 100 cm ) + ⎜ ( ( 6.00 cm )⎤⎥ = −79.0 cm ⎟ ⎝ 1.50 ⎟⎠ ⎣⎝ 1.333 ⎠ ⎦ It appears to be in the water, 79.0 cm back of the outer surface of the glass pane .

(b)

If the thickness of the glass is negligible (t → 0), the distance of the final image from the glass-air interface is q=−

na ng

⎡ ⎛ ng ⎞ ⎤ ⎛ na ⎞ ⎛ 1.00 ⎞ ⎢ ⎜ ⎟ p + 0 ⎥ = − ⎜ ⎟ p = − ⎜⎝ ⎟ (100 cm ) = −75.0 cm 1.333 ⎠ ⎝ nw ⎠ ⎢⎣ ⎝ nw ⎠ ⎥⎦

Now, it appears to be 75.0 cm back of the outer surface of the glass pane . (c)

23.28

Comparing the results of parts (a) and (b), we see that the 6.00-cm thickness of the glass in part (a) made a 4.00-cm difference in the apparent position of the jellyfish. We conclude that the thicker the glass, the greater the distance between the final image and the outer surface of the glass.

We assume the ray shown in the diagram at the right is a paraxial ray so q1 and q 2 are both sufficiently small to allow us to write Snell’s law as n1 tanq1 = n2 tanq 2

n1 | h | = +h

q1

P

n2 q2

O p

I | h′ | = –h′

q

Also, note that we are assuming transverse distances measured upward from the axis OPI are positive, while those measured downward from this axis are negative. Looking at the right triangle having distance OP as its base, we see that tanq1 = h p, and looking at the triangle having PI as a base, we have tanq 2 = h ′ q = − h ′ q. Thus, Snell’s law becomes n1 (h p) = −n2 (h ′ q), and the magnification becomes M= 23.29

nq h′ =− 1 n2 p h

With R1 = + 2.00 cm and R2 = + 2.50 cm, the lens maker’s equation gives the focal length as ⎛ 1 ⎞ ⎛ 1⎞ 1 1 1 = + 0.050 0 cm −1 = ( n − 1) ⎜ − ⎟ = (1.50 − 1) ⎜ − f ⎝ 2.00 cm 2.50 cm ⎟⎠ ⎝ R1 R2 ⎠ or

68719_23_ch23_p223-255.indd 237

f=

1 = + 20.0 cm 0.050 0 cm −1

1/7/11 2:50:59 PM

23.30

Chapter 23

Consider the figure at the right showing an object O located distance L in front of a screen. A convergent lens is positioned to focus an image I on the screen. Observe that the sum of the object and image distances is p + q = L, giving p = L − q. Also, from the thin-lens equation, we have p=

fq q− f

L−q =

giving

fq q− f

L screen

238

O

I

p

q

Simplifying the last result gives (L − q)(q − f ) = f q, which reduces to q 2 − Lq + Lf = 0. Solving by use of the quadratic formula, with L = 50.0 cm and f = 10.0 cm, gives q= (a)

(b)

23.31

L ± L2 − 4Lf 50.0 cm ± 2500 cm 2 − 2000 cm 2 = = 25.0 cm ± 11.2 cm 2 2

Thus, there are 2 locations, generally known as “conjugate positions,” where the lens could be positioned to form a clear image on the screen. These positions are: (i)

36.2 cm from the screen

→

( q = 36.2 cm and p = 13.8 cm )

(ii)

13.8 cm from the screen

→

( q = 13.8 cm and p = 36.2 cm )

If q = 36.2 cm, the magnification is M = −

q 36.2 cm =− = − 2.62 . p 13.8 cm

If q = 13.8 cm, the magnification is M = −

q 13.8 cm =− = − 0.381 . p 36.2 cm

The focal length of a converging lens is positive, so f = +10.0 cm. The thin-lens equation then p (10.0 cm ) pf . yields an image distance of q = = p − f p −10.0 cm (a)

When p = +20.0 cm, q =

( 20.0 cm )(10.0 cm ) 20.0 cm − 10.0 cm

= +20.0 cm, and M = −

q = −1.00, so the p

image is located 20.0 cm beyond the lens , is real (q > 0) , is inverted (M < 0) , and is the same size as the object ( M = 1.00 ) . (b)

When p = f = +10.0 cm, the object is at the focal point and no image is formed . Instead, parallel rays emerge from the lens .

(c)

When p = 5.00 cm, q =

( 5.00 cm )(10.0 cm )

q = +2.00, so the p 5.00 cm − 10.0 cm image is located 10.0 cm in front of the lens , is virtual (q < 0) , is upright (M > 0) , and is = −10.0 cm, and M = −

twice the size of the object ( M = 2.00 ) .

68719_23_ch23_p223-255.indd 238

1/7/11 2:51:01 PM

Mirrors and Lenses

23.32

(a)

To approximate paraxial rays, the rays should be drawn so they reflect at the vertical plane that passes through the vertex of the mirror, rather than at the mirror’s surface as is generally done in the textbook. For this reason, the concave surface of the mirror appears flat in the ray diagram given below.

C

F

I

h

O p

h′

q

(b)

In the diagram above, each square of the grid represents 5 cm. Thus, we see that the image I is located 40 cm behind the mirror, or q = − 40 cm .

(c)

From the above ray diagram, the height h ′ of the upright image I is seen to be twice the height h of the object O. Therefore, the magnification is M = h ′ h = + 2 .

(d)

From the mirror equation, the image distance is q = pf ( p − f ), or q=

( +20.0 cm )( + 40.0 cm ) 20.0 cm − 40.0 cm

M=−

and the magnification is 23.33

239

= − 40.0 cm q ( − 40.0 cm ) = +2.00 =− p +20.0 cm

For a divergent lens, the focal length is negative. Hence, f = −20.0 cm in this case. The thin-lens equation gives the image distance as q = pf ( p − f ) , and the magnification is given by M = − q p. (i)

(ii)

( 40.0 cm )( −20.0 cm ) = −13.3 cm 40.0 cm − ( −20.0 cm )

(a)

q=

(b)

q 0

(d)

M=−

(a)

q=

(b)

q 0

(d)

M=−

⇒

13.3 cm in front of the lens.

⇒

10.0 cm in front of the lens.

virtual image ⇒

( −13.3 cm ) + 40.0 cm

upright image

= + 0.333

( 20.0 cm )( −20.0 cm ) = −10.0 cm 20.0 cm − ( −20.0 cm ) ⇒

⇒

virtual image ⇒

( −10.0 cm ) + 20.0 cm

upright image

= + 0.500

continued on next page

68719_23_ch23_p223-255.indd 239

1/7/11 2:51:04 PM

240

Chapter 23

(iii) (a)

23.34

q=

(10.0 cm ) (−20.0 cm ) = −6.67 cm 10.0 cm − (−20.0 cm )

(b)

q 0

(d)

M=−

⇒

6.67 cm in front of the lens

virtual image ⇒

( − 6.67 cm ) = +10.0 cm

upright image + 0.667

(a) and (b) Your scale drawings should look similar to those given below:

Figure (a)

Figure (b) A carefully drawn-to-scale version of Figure (a) should yield an upright, virtual image located 13.3 cm in front of the lens and one-third the size of the object. Similarly, a carefully drawn-to-scale version of Figure (b) should yield an upright, virtual image located 6.7 cm in front of the lens and two-thirds the size of the object.

23.35

(c)

The results of the graphical solution are consistent with the algebraic answers found in Problem 23.33, allowing for small deviances due to uncertainties in measurement. Graphical answers may vary, depending on the size of the graph paper and accuracy of the drawing.

(a)

The real image case is shown in the ray diagram. Notice that p + q = 12.9 cm, or q = 12.9 cm − p. The thin-lens equation, with f = 2.44 cm, then gives 1 1 1 + = p 12.9 cm − p 2.44 cm or

p2 − (12.9 cm ) p + 31.5 cm 2 = 0

continued on next page

68719_23_ch23_p223-255.indd 240

1/7/11 2:51:06 PM

Mirrors and Lenses

241

Using the quadratic formula to solve gives p = 9.63 cm or p = 3.27 cm Both are valid solutions for the real image case. (b)

The virtual image case is shown in the second diagram. Note that in this case, q = − (12.9 cm + p ) , so the thin-lens equation gives 1 1 1 − = p 12.9 cm + p 2.44 cm or

p2 + (12.9 cm ) p − 31.5 cm 2 = 0

The quadratic formula then gives p = 2.10 cm or p = −15.0 cm . Since the object is real, the negative solution must be rejected, leaving p = 2.10 cm . 23.36

We must first realize that we are looking at an upright, enlarged, virtual image. Thus, we have a real object located between a converging lens and its front-side focal point, so q < 0, p > 0, and f > 0. q The magnification is M = − = + 2, giving q = − 2 p. Then, from the thin-lens equation, p 1 1 1 1 − =+ = , or f = 2 p = 2 ( 2.84 cm ) = 5.68 cm . p 2p 2p f

23.37

It is desired to form a magnified, real image on the screen using a single thin lens. To do this, a converging lens must be used, and the image will be inverted. The magnification then gives M=

−1.80 m q h′ = = − , or q = 75.0 p p h 24.0 × 10 −3 m

Also, we know that p + q = 3.00 m. Therefore, p + 75.0 p = 3.00 m , giving 3.00 m = 3.95 × 10 − 2 m = 39.5 mm 76.0

(b)

p=

(a)

The thin-lens equation then gives

1 1 76.0 1 + = = , or p 75.0 p 75.0 p f

⎛ 75.0 ⎞ ⎛ 75.0 ⎞ f =⎜ p=⎜ ( 39.5 mm ) = 39.0 mm ⎟ ⎝ 76.0 ⎠ ⎝ 76.0 ⎟⎠ 23.38

(a)

This is a real object, so the object distance is p = +20.0 cm. The thin-lens equation gives the image distance as q=

pf ( 20.0 cm )( −32.0 cm ) = = −12.3 cm p − f 20.0 cm − ( −32.0 cm )

F

O

I

so the image is 12.3 cm to the left of the lens .

68719_23_ch23_p223-255.indd 241

q ( −12.3 cm ) =− = + 0.615 . p +20.0 cm

(b)

The magnification is M = −

(c)

The ray diagram for this arrangement is shown above.

1/7/11 2:51:08 PM

242

23.39

Chapter 23

Since the light rays incident to the first lens are parallel, p1 = ∞, and the thin-lens equation gives q1 = f1 = −10.0 cm. The virtual image formed by the first lens serves as the object for the second lens, so p2 = 30.0 cm + q1 = 40.0 cm. If the light rays leaving the second lens are parallel, then q2 = ∞, and the thin-lens equation gives f2 = p2 = 40.0 cm .

23.40

(a)

Solving the thin-lens equation for the image distance q gives 1 1 1 p− f = − = pf q f p

(b)

or

q=

pf p− f

For a real object, p > 0 and p = p . Also, for a diverging lens, f < 0 and f = − f . The result of part (a) then becomes q=

p (− f

)

p − (− f

)

=−

p f p+ f

Thus, we see that q < 0 for all numeric values of p and f . Since negative image distances mean virtual images, we conclude that a diverging lens will always form virtual images of real objects. (c)

For a real object, p > 0 and p = p . Also, for a converging lens, f > 0 and f = f . The result of part (a) then becomes q=

p f >0 p− f

if

p − f >0

Since q must be positive for a real image, we see that a converging lens will form real images of real objects only when p > f (or p > f since both p and f are positive in this situation). 23.41

The thin-lens equation gives the image position for the first lens as q1 =

p1 f1 ( 30.0 cm )(15.0 cm ) = = + 30.0 cm p1 − f1 30.0 cm − 15.0 cm

and the magnification by this lens is M1 = −

q1 30.0 cm =− = −1.00. 30.0 cm p1

The real image formed by the first lens serves as the object for the second lens, so p2 = 40.0 cm − q1 = +10.0 cm. Then, the thin-lens equation gives q2 =

p2 f2 (10.0 cm )(15.0 cm ) = = − 30.0 cm p2 − f2 10.0 cm − 15.0 cm

and the magnification by the second lens is M2 = −

q2 ( − 30.0 cm ) = + 3.00 =− p2 10.0 cm

Thus, the final, virtual image is located 30.0 cm in front of the second lens , and the overall magnification is M = M1 M 2 = ( −1.00 ) ( + 3.00 ) = − 3.00 .

68719_23_ch23_p223-255.indd 242

1/7/11 2:51:10 PM

Mirrors and Lenses

23.42

(a)

243

With p1 = +15.0 cm, the thin-lens equation gives the position of the image formed by the first lens as q1 =

p1 f1 (15.0 cm )(10.0 cm ) = = + 30.0 cm p1 − f1 15.0 cm − 10.0 cm

This image serves as the object for the second lens, with an object distance of p2 = 10.0 cm − q1 = 10.0 cm − 30.0 cm = − 20.0 cm (a virtual object). If the image formed by this lens is at the position of O1 , the image distance is q2 = − (10.0 cm + p 1 ) = − (10.0 cm + 15.0 cm ) = − 25.0 cm The thin-lens equation then gives the focal length of the second lens as f2 = (b)

p2 q2 ( −20.0 cm )( −25.0 cm ) = = −11.1 cm p2 + q2 −20.0 cm − 25.0 cm

The overall magnification is ⎛ q ⎞ ⎛ q ⎞ ⎛ 30.0 cm ⎞ ⎡ ( −25.0 cm ) ⎤ M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ − ⎟ ⎢− ⎥ = + 2.50 ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ 15.0 cm ⎠ ⎣ ( −20.0 cm ) ⎦

(c) 23.43

Since q2 < 0, the final image is virtual ; and since M > 0, it is upright .

From the thin-lens equation, q1 =

p1 f1 ( 4.00 cm )(8.00 cm ) = = − 8.00 cm . p1 − f1 4.00 cm − 8.00 cm

The magnification by the first lens is M1 = −

q1 ( − 8.00 cm ) = + 2.00. =− p1 4.00 cm

The virtual image formed by the first lens is the object for the second lens, so p2 = 6.00 cm + q1 = +14.0 cm, and the thin-lens equation gives q2 =

(14.0 cm )( −16.0 cm ) p2 f2 = = − 7.47 cm p2 − f2 14.0 cm − ( −16.0 cm )

q2 ( − 7.47 cm ) = + 0.533, so the overall =− p2 14.0 cm magnification is M = M1 M 2 = ( + 2.00 ) ( + 0.533) = +1.07. The magnification by the second lens is M 2 = −

The position of the final image is 7.47 cm in front of the second lens , and its height is h ′ = M h = ( +1.07 )(1.00 cm ) = 1.07 cm . Since M > 0, the final image is upright ; and since q2 < 0, this image is virtual . 23.44

(a)

We start with the final image and work backward. From Figure P23.44, observe that q2 = − ( 50.0 cm − 31.0 cm ) = −19.0 cm. The thin-lens equation then gives p2 =

q2 f 2 ( −19.0 cm )( 20.0 cm ) = + 9.74 cm = q2 − f2 −19.0 cm − 20.0 cm

continued on next page

68719_23_ch23_p223-255.indd 243

1/7/11 2:51:13 PM

244

Chapter 23

The image formed by the first lens serves as the object for the second lens and is located 9.74 cm in front of the second lens. Thus, the image distance for the first lens is q1 = 50.0 cm − 9.74 cm = 40.3 cm, and the thin-lens equation gives p1 =

q1 f1 ( 40.3 cm )(10.0 cm ) = = +13.3 cm q1 − f1 40.3 cm − 10.0 cm

The original object should be located 13.3 cm in front of the first lens. (b)

The overall magnification is ⎛ q ⎞ ⎛ q ⎞ ⎛ 40.3 cm ⎞ ⎛ ( −19.0 cm ) ⎞ M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ − = − 5.91 ⎟ − 9.74 cm ⎟⎠ ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ 13.3 cm ⎠ ⎜⎝

(c) 23.45

Since M < 0, the final image is inverted .

Note: Final answers to this problem are highly sensitive to round-off error. To avoid this, we retain extra digits in intermediate answers and round only the final answers to the correct number of significant figures. Since the final image is to be real and in the film plane, q2 = + d . Then, the thin-lens equation gives p2 =

q2 f 2 d (13.0 cm ) = . q2 − f2 d − 13.0 cm

The object of the second lens ( L2 ) is the image formed by the first lens ( L1 ) , so 13.0 cm ⎞ d2 ⎛ q1 = (12.0 cm − d ) − p2 = 12.0 cm − d ⎜ 1+ = 12.0 cm − ⎟ ⎝ d − 13.0 cm ⎠ d − 13.0 cm If d = 5.00 cm, then q1 = +15.125 cm; and when d = 10.0 cm, q1 = + 45.3 cm . From the thin-lens equation, p1 =

q1 f1 q (15.0 cm ) = 1 . q1 − f1 q1 − 15.0 cm

When q1 = +15.125 cm ( d = 5.00 cm ) , then p1 = 1.82 × 10 3 cm = 18.2 m . When q1 = + 45.3 cm ( d = 10.0 cm ) , then p1 = 22.4 cm = 0.224 m . Thus, the range of focal distances for this camera is 0.224 m to 18.2 m . 23.46

(a)

From the thin-lens equation, the image distance for the first lens is q1 =

(b)

p1 f1 (15.0 cm )(10.0 cm ) = = +30.0 cm p1 − f1 15.0 cm − 10.0 cm

With q1 = +30.0 cm, the image of the first lens is located 30.0 cm in back of that lens. Since the second lens is only 10.0 cm beyond the first lens, this means that the first lens is trying to form its image at a location 20.0 cm beyond the second lens .

continued on next page

68719_23_ch23_p223-255.indd 244

1/7/11 2:51:16 PM

Mirrors and Lenses

(c)

The image the first lens forms (or would form if allowed to do so) serves as the object for the second lens. Considering the answer to part (b) above, we see that this will be a virtual object, with object distance p2 = −20.0 cm .

(d)

From the thin-lens equation, the image distance for the second lens is q2 =

23.47

245

p2 f2 ( −20.0 cm )( 5.00 cm ) = = + 4.00 cm p2 − f2 −20.0 cm − 5.00 cm

(e)

M1 = −

q1 30.0 cm =− = − 2.00 15.0 cm p1

(f )

M2 = −

q2 4.00 cm =− = + 0.200 p2 ( −20.0 cm )

(g)

M total = M1 M 2 = ( −2.00 )( +0.200 ) = − 0.400

(h)

Since q2 > 0, the final image is real , and since M total < 0, that image is inverted relative to the original object.

Since q = + 8.00 cm when p = +10.0 cm, we find that 1 1 18.0 1 1 1 = + = + = f p q 10.0 cm 8.00 cm 80.0 cm Then, when p = 20.0 cm, 18.0 1 18.0 − 4.00 14.0 1 1 1 = − = − = = 80.0 cm 80.0 cm q f p 80.0 cm 20.0 cm or

q=

80.0 cm = + 5.71 cm 14.0

Thus, a real image is formed 5.71 cm in front of the mirror . 23.48

(a)

We are given that p = 5 f , with both p and f being positive. The thin-lens equation then gives q=

68719_23_ch23_p223-255.indd 245

pf (5 f ) f = 5 f = p− f 5f − f 4

q (5 f 4 ) = − 1 =− p 5f 4

(b)

M=−

(c)

Since q > 0, the image is real . Because M < 0, the image is inverted . Since the object is real, it is located in front of the lens, and with q > 0, the image is located in back of the lens. Thus, the image is on the opposite side of the lens from the object.

1/7/11 2:51:19 PM

246

23.49

Chapter 23

incoming In the sketch at the right, the center of curvature of the left side of the biconcave lens is on the side of the light incoming light. Thus, by the convention of Table 23.2, the radius of curvature of this side is negative while the radius of curvature of the right side is positive. If R1 = −32.5 cm and R2 = 42.5 cm, the lens maker’s equation gives the focal length of the lens as

| R1|

R2

⎞ ⎛ 1 1 1 = (1− n )( 5.43 × 10 −2 cm −1 ) = ( n − 1) ⎜ − f ⎝ − 32.5 cm 42.5 cm ⎟⎠ (a)

For a very distant object ( p → ∞), the thin-lens equation gives the image distance as q = f . Thus, if the index of refraction of the lens material is n = 1.53 for violet light, q= f =

1 = −34.7 cm (1 − 1.53)( 5.43 × 10 −2 cm −1 )

and the image of violet light is formed 34.7 cm to the left of the lens . (b)

If n = 1.51 for red light, the image distance for very distance source emitting red light is q= f =

1 = −36.1 cm (1− 1.51)( 5.43 × 10 −2 cm −1 )

The image of the very distant red light source is formed 36.1 cm to the left of the lens . 23.50

(a)

Using the sign convention from Table 23.2, the radii of curvature of the surfaces are R1 = −15.0 cm and R2 = +10.0 cm. The lens maker’s equation then gives ⎛ 1 ⎛ 1⎞ 1 1 1 ⎞ or f = −12.0 cm = ( n − 1) ⎜ − ⎟ = (1.50 − 1) ⎜ − R f R −15.0 cm 10.0 cm ⎟⎠ ⎝ ⎝ 1 2 ⎠

(b)

If p → ∞, then q = f = −12.0 cm . The thin-lens equation gives q =

23.51

p ( −12.0 cm ) pf = and the following results: p− f p + 12.0 cm

(c)

If p = 3 f = + 36.0 cm, q = − 9.00 cm .

(d)

If p = f = +12.0 cm, q = − 6.00 cm .

(e)

If p = f 2 = + 6.00 cm, q = − 4.00 cm .

As light passes left to right through the lens, the image position is given by q1 =

p1 f1 (100 cm )(80.0 cm ) = = + 400 cm p1 − f1 100 cm − 80.0 cm

This image serves as an object for the mirror with an object distance of p2 = 100 cm − q1 = −300 cm (virtual object). From the mirror equation, the position of the image formed by the mirror is q2 =

p2 f2 ( − 300 cm ) ( − 50.0 cm ) = − 60.0 cm = p2 − f2 −300 cm − ( − 50.0 cm )

continued on next page

68719_23_ch23_p223-255.indd 246

1/7/11 2:51:21 PM

Mirrors and Lenses

247

This image is the object for the lens as light now passes through it going right to left. The object distance for the lens is p3 = 100 cm − q2 = 100 cm − ( − 60.0 cm ) , or p3 = 160 cm. From the thin- lens equation, q3 =

p3 f3 (160 cm )(80.0 cm ) = = +160 cm p3 − f3 160 cm − 80.0 cm

Thus, the final image is located 160 cm to the left of the lens . ⎛ q ⎞⎛ q ⎞⎛ q ⎞ The overall magnification is M = M1 M 2 M3 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ ⎜ − 3 ⎟ , or ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ p3 ⎠ ⎛ 400 cm ⎞ ⎡ ( − 60.0 cm ) ⎤ ⎛ 160 cm ⎞ M = ⎜− − = − 0.800 − ⎝ 100 cm ⎟⎠ ⎢⎣ ( −300 cm ) ⎥⎦ ⎜⎝ 160 cm ⎟⎠ Since M < 0, the final image is inverted . 23.52

(a)

Since the object is midway between the lens and mirror, the object distance for the mirror is p1 = +12.5 cm. The mirror equation gives the image position as 1 2 1 2 1 5−4 1 = − = − = = q1 R p1 +20.0 cm 12.5 cm 50.0 cm 50.0 cm

or

q1 = + 50.0 cm

This image serves as the object for the lens, so p2 = 25.0 cm − q1 = −25.0 cm. Note that since p2 < 0, this is a virtual object. The thin-lens equation gives the image position for the lens as q2 =

p2 f2 ( − 25.0 cm ) ( −16.7 cm ) = − 50.3 cm = p2 − f2 −25.0 cm − ( −16.7 cm )

Since q2 < 0, this image is located 50.3 cm in front of (to the right of) the lens or 25.3 cm behind (to the right of) the mirror . (b)

With q2 < 0, the final image is a virtual image.

(c) and (d) The overall magnification is ⎛ q ⎞ ⎛ q ⎞ ⎛ 50.0 cm ⎞ ⎡ ( − 50.3 cm ) ⎤ M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ − ⎢− ⎥ = +8.05 ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ 12.5 cm ⎟⎠ ⎣ ( −25.0 cm ) ⎦ Since M > 0, the final image is upright . 23.53

A hemisphere is too thick to be described as a thin lens. The light is undeviated on entry into the flat face. We next consider the light’s exit from the curved surface, for which R = − 6.00 cm. The incident rays are parallel, so p = ∞.

Then,

68719_23_ch23_p223-255.indd 247

n1 n2 n2 − n1 1.00 1.00 − 1.56 + = , from which q = 10.7 cm . = becomes 0 + q − 6.00 cm p q R

1/7/11 2:51:24 PM

248

23.54

Chapter 23

The diagram at the right shows a light ray traveling parallel to the principle axis of a plano-convex lens at distance h off the axis. This ray strikes the plane surface at normal incidence and passes into the glass undeviated. The angle of incidence at the spherical surface (of radius R) is q1 , where sinq1 = h R. If the index of refraction of the lens material is n, Snell’s law gives the angle of refraction at the spherical surface as

q2

q1 h

R

q1 f

h

f

C

h/tanf ¥ R2 − h2

⎛ n sinq1 ⎞ ⎛ nh ⎞ = sin −1 ⎜ ⎟ q 2 = sin −1 ⎜ ⎟ ⎝ R⎠ ⎝ nair ⎠

x

The distance x from the center of curvature of the spherical surface to the point where the refracted ray crosses the principle axis of the lens is x = R2 − h2 +

h tanf

f = q 2 −q1

where

If R = 20.0 cm, n = 1.60, and the first ray is distance h1 = 0.500 cm off the axis, we find ⎛ 0.500 cm ⎞ q1 = sin −1 ⎜ = 1.43° ⎝ 20.0 cm ⎟⎠ so

x1 =

( 20.0 cm )2 − ( 0.500 cm )2 +

and

⎡ 1.60 ( 0.500 cm ) ⎤ q 2 = sin −1 ⎢ ⎥ = 2.29° 20.0 cm ⎦ ⎣

0.500 cm = 53.3 cm tan ( 2.29° − 1.43° )

If the second ray is distance h2 = 12.0 cm from the axis, then ⎛ 12.0 cm ⎞ q1 = sin −1 ⎜ = 36.9° ⎝ 20.0 cm ⎟⎠ giving

x2 =

and

( 20.0 cm )2 − (12.0 cm )2 +

⎡ 1.60 (12.0 cm ) ⎤ q 2 = sin −1 ⎢ ⎥ = 73.7° ⎣ 20.0 cm ⎦

12.0 cm = 32.0 cm tan ( 73.7° − 36.9° )

The distance between the points where these two rays cross the principle axis is then Δx = x1 − x 2 = 53.3 cm − 32.0 cm = 21.3 cm 23.55

(a)

With light going through the piece of glass from left to right, the radius of the first surface is positive and that of the second surface is negative according to the sign convention of Table 23.2. Thus, R1 = + 2.00 cm, and R2 = − 4.00 cm. 1.50 1.50 − 1.00 n1 n2 n2 − n1 1.00 + + = , = to the first surface gives q1 + 2.00 cm p q R 1.00 cm which yields q1 = − 2.00 cm. The first surface forms a virtual image 2.00 cm to the left of that surface, and 16.0 cm to the left of the second surface. Applying

The image formed by the first surface is the object for the second surface, so p2 = +16.0 cm, 1.00 1.00 − 1.50 1.50 n n n − n1 + = , or q2 = + 32.0 cm. and 1 + 2 = 2 gives q2 − 4.00 cm 16.0 cm p q R The final image is located 32.0 cm to the right of the second surface . (b)

68719_23_ch23_p223-255.indd 248

Since q2 > 0, the final image formed by the piece of glass is a real image .

1/7/11 2:51:27 PM

Mirrors and Lenses

23.56

249

Consider an object O1 at distance p1 in front of the first lens. The thin-lens equation gives the image position for this lens as 1 1 1 = − . q1 f1 p1

The image, I1 , formed by the first lens serves as the object, O2 , for the second lens. With the lenses in contact, this will be a virtual object if I1 is real and will be a real object if I1 is virtual. In either case, if the thicknesses of the lenses may be ignored, p2 = −q1

and

1 1 1 1 =− =− + q1 f1 p1 p2

Applying the thin-lens equation to the second lens,

−

1 1 1 1 + + = f1 p1 q2 f2

or

1 1 1 + = becomes p2 q2 f2

1 1 1 1 + = + f1 f2 p1 q2

Observe that this result is a thin-lens-type equation relating the position of the original object, O1 , and the position of the final image, I 2 , formed by this two-lens combination. Thus, we see that we may treat two thin lenses in contact as a single lens having a focal length, f, given by 1 1 1 = + . f1 f2 f 23.57

From the thin-lens equation, the image distance for the first lens is q1 =

p1 f1 ( 40.0 cm )( 30.0 cm ) = = +120 cm p1 − f1 40.0 cm − 30.0 cm

and the magnification by this lens is M1 = −

q1 120 cm =− = − 3.00. 40.0 cm p1

The real image formed by the first lens serves as the object for the second lens, with object distance of p2 = 110 cm − q1 = −10.0 cm (a virtual object). The thin-lens equation gives the image distance for the second lens as q2 = (a)

p2 f2 ( −10.0 cm ) f2 = p2 − f2 −10.0 cm − f2

If f2 = − 20.0 cm, then q2 = + 20.0 cm, and the magnification by the second lens is M 2 = − q2 p2 = − ( 20.0 cm ) ( −10.0 cm ) = + 2.00. The final image is located 20.0 cm to the right of the second lens , and the overall magnification is M = M1 M 2 = ( −3.00 )( + 2.00 ) = − 6.00 .

(b)

Since M < 0, the final image is inverted .

continued on next page

68719_23_ch23_p223-255.indd 249

1/7/11 2:51:30 PM

250

Chapter 23

(c)

If f2 = + 20.0 cm, then q2 = + 6.67 cm, and M 2 = −

q2 6.67 cm =− = + 0.667. p2 ( −10.0 cm )

The final image is 6.67 cm to the right of the second lens , and the overall magnification is M = M1 M 2 = ( −3.00 )( + 0.667 ) = − 2.00 . Since M < 0, the final image is inverted . 23.58

The diagram at the right gives a ray diagram showing how this mirror system forms a final image I2 located just above the opening in the upper mirror. The focal point of each mirror is at the center of the opposite mirror, so their focal length f is the vertical distance between the centers of the mirrors. The original object O1 sits just above the surface at the center of the lower mirror. This places it on the principle axis and just inside the focal point of the upper mirror.

I2

O1

Therefore, the object distance for the upper mirror is p1 = f − e , where e is a very small number approaching zero. The mirror equation then gives q1 =

p1 f ( f − e ) f = − f 2 − e f = −d = f −e − f p1 − f e

and

d → ∞ when e → 0

Thus, the upper mirror forms a virtual image, on the principle axis, far above this mirror. This image serves as the object for the lower mirror and is located far in front of that mirror. With p2 ≈ q1 = d f , the image distance for the lower mirror is q2 =

p2 f df = p2 − f d − f

≈+

df =+f d

Since q2 > 0, the final image I2 is a real image located just above the focal point of the lower mirror (at the opening in the upper mirror). The overall magnification for this mirror system is ⎛ q ⎞⎛ q ⎞ M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ ⎝ p1 ⎠ ⎝ p2 ⎠

⎛ −d ⎞ ⎛ f ⎞ ⎜ − ⎟ = +1 ⎝ f ⎟⎠ ⎝ d ⎠

≈ ⎜−

This means that the final image is upright and the same size as the original object . When a flashlight beam is aimed at the final image, it passes through the opening in the upper mirror and reflects as shown in the ray diagram at the right. It emerges from the mirror as if it had reflected from that image, even obeying the law of reflection as it did so!

68719_23_ch23_p223-255.indd 250

1/7/11 2:51:33 PM

Mirrors and Lenses

23.59

(a)

251

⎛ 1 1⎞ 1 = ( n − 1) ⎜ − ⎟ , gives f ⎝ R1 R2 ⎠

The lens maker’s equation,

⎞ ⎛ 1 1 1 = ( n − 1) ⎜ − ⎟⎠ 5.00 cm 9.00 cm −11.0 cm ⎝ which simplifies to n = 1+ (b)

1 ⎛ 99.0 ⎞ ⎜⎝ ⎟ = 1.99 . 5.00 11.0 + 9.00 ⎠

As light passes from left to right through the lens, the thin-lens equation gives the image distance as q1 =

p1 f (8.00 cm )( 5.00 cm ) = = +13.3 cm p1 − f 8.00 cm − 5.00 cm

This image formed by the lens serves as an object for the mirror with object distance p2 = 20.0 cm − q1 = + 6.67 cm. The mirror equation then gives q2 =

p2 R ( 6.67 cm )(8.00 cm ) = = +10.0 cm 2 p2 − R 2 ( 6.67 cm ) − 8.00 cm

This real image, formed 10.0 cm to the left of the mirror, serves as an object for the lens as light passes through it from right to left. The object distance is p3 = 20.0 cm − q2 = +10.0 cm, and the thin-lens equation gives q3 =

p1 f (10.0 cm )( 5.00 cm ) = = +10.0 cm p3 − f 10.0 cm − 5.00 cm

The final image is located 10.0 cm to the left of the lens and its overall magnification is ⎛ q ⎞ ⎛ q ⎞ ⎛ q ⎞ ⎛ 13.3 ⎞ ⎛ 10.0 ⎞ ⎛ 10.0 ⎞ M = M1 M 2 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ ⎜ − 3 ⎟ = ⎜ − ⎟ ⎜− ⎟ ⎜− ⎟ = − 2.50 ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ p3 ⎠ ⎝ 8.00 ⎠ ⎝ 6.67 ⎠ ⎝ 10.0 ⎠ (c) 23.60

23.61

Since M < 0, the final image is inverted .

From the thin-lens equation, the object distance is p =

(4 f ) f

(a)

If q = + 4 f , then p =

(b)

When q = − 3 f , we find p =

(c)

In case (a), M = −

4f − f

=

qf . q− f

4f or 1.33 f . 3

( − 3 f ) f = 3 f 4 , or 0.750 f . −3f − f

q 4f q −3f =− = − 3 , and in case (b), M = − = − = +4 . p 4f 3 p 3f 4

If R1 = −3.00 m and R2 = −6.00 m, the focal length is given by ⎞⎛ 1 ⎛ n1 1 1 ⎞ ⎛ n1 − n2 ⎞ ⎛ −1 ⎞ = − 1⎟ ⎜ = + ⎜ ⎟ f ⎜⎝ n2 − 3.00 m 6.00 m ⎟⎠ ⎜⎝ n2 ⎟⎠ ⎝ 6.00 m ⎠ ⎝ ⎠

continued on next page

68719_23_ch23_p223-255.indd 251

1/7/11 2:51:35 PM

252

Chapter 23

( 6.00 m ) n2

or

f=

(a)

If n1 = 1.50 and n2 = 1.00, then f =

[1]

n2 − n1

The thin-lens equation gives q =

( 6.00 m )(1.00 )

= −12.0 m .

1.00 − 1.50

(10.0 m )( −12.0 m ) pf = = − 5.45 m . p− f 10.0 m + 12.0 m

A virtual image is formed 5.45 m to the left of the lens . (b)

If n1 = 1.50 and n2 = 1.33, the focal length is f = q=

( 6.00 m )(1.33) 1.33 − 1.50

= − 46.9 m, and

(10.0 m )( − 46.9 m ) pf = = − 8.24 m p− f 10.0 m + 46.9 m

The image is located 8.24 m to the left of the lens . (c)

When n1 = 1.50 and n2 = 2.00, f = q=

( 6.00 m )( 2.00 ) 2.00 − 1.50

= + 24.0 m, and

pf (10.0 m )( 24.0 m ) = = −17.1 m p− f 10.0 m − 24.0 m

The image is 17.1 m to the left of the lens . (d)

23.62

Observe from Equation [1] that f < 0 if n1 > n2 and f > 0 when n1 < n2. Thus, a diverging lens can be changed to converging by surrounding it with a medium whose index of refraction exceeds that of the lens material.

The inverted image is formed by light that leaves the object and goes directly through the lens, never having reflected from the mirror. For the formation of this inverted image, we have M=−

q1 = −1.50 p1

giving

q1 = +1.50 p1

The thin-lens equation then gives 1 1 1 + = p1 1.50 p1 10.0 cm

or

1 ⎞ ⎛ p1 = (10.0 cm ) ⎜ 1+ = 16.7 cm ⎝ 1.50 ⎟⎠

The upright image is formed by light that passes through the lens after reflecting from the mirror. The object for the lens in this upright image formation is the image formed by the mirror. In order for the lens to form the upright image at the same location as the inverted image, the image formed by the mirror must be located at the position of the original object (so the object distances, and hence image distances, are the same for both the inverted and upright images formed by the lens). Therefore, the object distance and the image distance for the mirror are equal, and their common value is qm irror = pm irror = 40.0 cm − p1 = 40.0 cm − 16.7 cm = +23.3 cm

continued on next page

68719_23_ch23_p223-255.indd 252

1/7/11 2:51:37 PM

Mirrors and Lenses

The mirror equation, 1 fm irror 23.63

(a)

=

253

1 1 2 1 + = = , then gives pm irror qm irror R fm irror

1 1 +2 + = 23.3 cm 23.3 cm 23.3 cm

fm irror = +

or

23.3 cm = + 11.7 cm 2

The lens-maker’s equation for a lens made of material with refractive index n1 = 1.55 and immersed in a medium having refractive index n2 is ⎞⎛ 1 1 ⎞ ⎛ 1.55 − n2 ⎞ ⎛ 1 1⎞ 1 ⎛ n1 = ⎜ − 1⎟ ⎜ − ⎟ = ⎜ − ⎟ ⎜ ⎟ f ⎝ n2 n2 ⎠ ⎝ R1 R2 ⎠ ⎠ ⎝ R1 R2 ⎠ ⎝ Thus, when the lens is in air, we have

1⎞ 1 ⎛ 1.55 − 1.00 ⎞ ⎛ 1 =⎜ ⎟⎠ ⎜ − ⎟ ⎝ 1.00 fair ⎝ R1 R2 ⎠

[1]

1⎞ ⎛ 1.55 − 1.33 ⎞ ⎛ 1 =⎜ − ⎟ ⎟ ⎜ ⎝ 1.33 ⎠ ⎝ R1 R2 ⎠

[2]

1

and when it is immersed in water,

fwater

Dividing Equation [1] by Equation [2] gives fwater ⎛ 1.33 ⎞ ⎛ 1.55 − 1.00 ⎞ =⎜ = 3.33 ⎝ 1.00 ⎟⎠ ⎜⎝ 1.55 − 1.33 ⎟⎠ fair If fair = 79.0 cm, the focal length when immersed in water is fwater = 3.33 ( 79.0 cm ) = 263 cm (b)

The focal length for a mirror is determined by the law of reflection, which is independent of the material of which the mirror is made and of the surrounding medium. Thus, the focal length depends only on the radius of curvature and not on the material making up the mirror or the surrounding medium. This means that, for the mirror, fwater = fair = 79.0 cm

23.64

(a)

The spherical ornament serves as a convex mirror, forming an image that is three-fourths the size of the object. Since convex mirrors only form upright, virtual images for real objects, the magnification is positive. Thus, M=

q 3 h′ =− =+ p 4 h

and

q=

− 3p 4

With R < 0 (convex mirror) and a diameter of 8.50 cm, we have R = − (8.50 cm) 2 = − 4.25 cm The mirror equation, 1 p + 1 q = 2 R , then gives 2 1 4 − = p 3p R

or

p=−

R ( − 4.25 cm ) = + 0.708 cm =− 6 6

Hence, the object is 0.708 cm in front of the spherical ornament .

continued on next page

68719_23_ch23_p223-255.indd 253

1/7/11 2:51:40 PM

254

Chapter 23

(b)

The ray diagram at the right shows the formation of the upright, virtual image by this convex mirror.

O

I F

23.65

The diagram at the right shows a bubble O, located 5.00 cm above the center of a glass sphere having radius R = 15.0 cm. This bubble is an actual distance of p = 10.0 cm below the refracting surface separating the glass and air. Refraction at this surface forms a virtual image I at distance q below the surface as shown. The object distance p and image distance q are related by n1 n2 n2 − n1 + = p q R

or

C

n2 = 1.00 p

| q| I O

5.00 cm

R n1 = 1.50

n2 n2 − n1 n1 = − q R p

Thus, with R < 0 for the concave surface the light crosses going from glass into the air, we have R = −15.0 cm, and 1.50 +1.00 − 4.50 −3.50 1.00 1.00 − 1.50 = − = = −15.0 cm 10.0 cm 30.0 cm 30.0 cm q

or

q=

30.0 cm = − 8.57 cm − 3.50

Therefore, the bubble has an apparent depth of 8.57 cm below the glass surface. 23.66

(a)

The only upright images that spherical mirrors can form of real objects are virtual images. Therefore, we know that the image described is virtual and diminished in size (4.00 cm high in comparison to the 10.0-cm height of the object). Since the virtual images formed by concave mirrors are always enlarged, we must conclude that this image is formed by a convex mirror .

(b)

The sketch at the right shows an upright, diminished, virtual image formed by a convex mirror. The distance between the object and this virtual image in this case is known to be p + q = 42.0 cm. The image distance for a virtual image is negative, so q < 0 and q = −q. Thus, we have

Convex surface I

O

p − q = 42.0 cm

p

| q| 42.0 cm

[1]

We also know that the magnification is positive (upright image) and M=

q 4.00 cm h′ =− = = 0.400 p 10.0 cm h

or

q = − 0.400 p

[2]

continued on next page

68719_23_ch23_p223-255.indd 254

1/7/11 2:51:42 PM

Mirrors and Lenses

255

Substituting Equation [2] into Equation [1] gives p − ( −0.400 ) p = 42.0 cm

or

p=

42.0 cm = 30.0 cm 1.40

With the object at the zero end of the meter stick, the mirror is at the 30.0-cm mark . (c)

The image distance is now seen to be q = −0.400 p = −0.400 ( 30.0 cm ) = −12.0 cm. The mirror equation (1 p + 1 q = 2 R = 1 f ) then gives the focal length of this mirror as f=

68719_23_ch23_p223-255.indd 255

pq ( 30.0 cm )( −12.0 cm ) = = −20.0 cm p+q 30.0 cm − 12.0 cm

1/7/11 2:51:44 PM

24 Wave Optics QUICK QUIZZES 1.

Choice (c). The fringes on the screen are equally spaced only at small angles, where tanq ≈ sinq is a valid approximation.

2.

Choice (c). The screen locations of the dark fringes of order m are given by (ydark )m = (lL d)(m + 12 ), with m = 0 corresponding to the first dark fringe on either side of the central maximum. The width of the central maximum is then 2(ydark )0 = 2(lL d)( 12 ) = lL d . Thus, doubling the distance d between the slits will cut the width of the central maximum in half.

3.

Choice (c). The screen locations of the bright fringes of order m are given by (ybright )m = (lL d)m , and the distance between successive bright fringes for a given wavelength is

(

Δybright = ybright

)

m +1

(

− ybright

)

m

= ( lL d )m +1 − ( lL d )m = lL d

Observe that this spacing is directly proportional to the wavelength. Thus, arranged from smallest to largest spacing between bright fringes, the order of the colors will be blue, green, red. 4.

Choice (b). The space between successive bright fringes is proportional to the wavelength of the light. Since the wavelength in water is less than that in air, the bright fringes are closer together in the second experiment.

5.

Choice (b). The outer edges of the central maximum occur where sinq = ± l a. Thus, as the width of the slit, a, becomes smaller, the width of the central maximum will increase.

6.

The compact disc. The tracks of information on a compact disc are much closer together than on a phonograph record. As a result, the diffraction maxima from the compact disc will be farther apart than those from the record.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

With phase changes occurring in the reflections at both the air-oil boundary and the oil-water boundary, the condition for constructive interference in the reflected light is 2noil t = ml, where m is any integer. Thus, the minimum nonzero thickness of the oil which will strongly reflect the 530-nm light is tm in = l 2noil = (530 nm) 2(1.25) = 212 nm, and (d) is the proper choice.

2.

The bright fringe of order m occurs where d = d sinq = ml. For small angles, the sine of the angle is approximately equal to the angle expressed in radians. Thus, the angular position of the second order bright fringe in the case described is ⎛ 5.0 × 10 −7 m ⎞ ⎛l⎞ = 0.050 radians q = 2⎜ ⎟ = 2⎜ ⎝d⎠ ⎝ 2.0 × 10 −5 m ⎟⎠ making (a) the correct choice. 256

68719_24_ch24_p256-281.indd 256

1/7/11 2:52:56 PM

Wave Optics

3.

257

In a single-slit diffraction pattern, formed on a screen at distance L from the slit, dark fringes occur where sinq dark = m(l a) ≈ tanq dark = ydark L and m is any nonzero integer. Thus, the width of the slit, a, in the described situation must be a=

(1) lL

(y ) dark

1

=

(5.00 × 10

−7

m ) (1.00 m )

5.00 × 10 −3 m

= 1.00 × 10 −4 m = 0.100 mm

and the correct answer is seen to be choice (b). 4.

From Malus’s law, the intensity of the light transmitted through a polarizer having its transmission axis oriented at angle q to the plane of polarization of the incident polarized light is I = I 0 cos 2 q . Therefore, the intensity transmitted through the first polarizer having q = 45° − 0 = 45° is I1 = I 0 cos 2 (45°) = 0.50I 0 , and the intensity passing through the second polarizer having q = 90° − 45° = 45° is I 2 = (0.50I 0 ) cos 2 (45°) = 0.25I 0 . The fraction of the original intensity making it through both polarizers is then I 2 I 0 = 0.25, which is choice (b).

5.

The spacing between successive bright fringes in a double-slit interference pattern is given by Δybright = ( lL d ) ( m + 1) − ( lL d ) m = lL d , where d is the slit separation. As d decreases, the spacing between the bright fringes will increase and choice (b) is the correct answer.

6.

As discussed in question 5 above, the fringe spacing in a double-slit interference pattern is Δybright = lL d . Therefore, as the distance L between the screen and the plane of the slits is increased, the spacing between the bright fringes will increase, and (a) is the correct choice.

7.

The fringe spacing in a double-slit interference pattern is Δybright = lL d . Note that this spacing is directly proportional to the wavelength of the light illuminating the slits. Thus, with l2 > l1 , the spacing is greater in the second experiment, and choice (c) gives the correct answer.

8.

The reflected light tends to be partially polarized, with the degree of polarization depending on the angle of incidence on the reflecting surface. Only if the angle of incidence equals the polarizing angle (or Brewster’s angle) will the reflected light be completely polarized. The better answer for this question is choice (d).

9.

The bright colored patterns are the result of interference between light reflected from the upper surface of the oil and light reflected from the lower surface of the oil film. Thus, the best answer is choice (e).

10.

In a single-slit diffraction pattern, dark fringes occur where ydark L ≈ sinq dark = m(l a). The width of the central maximum is the distance between the locations of the first dark fringes on either side of the center (i.e., between the m = ±1 dark fringes), giving width of central maximum = ( +1)

l l 2l − ( −1) = a a a

Thus, if the width of the slit, a, is made half as large, the width of the central maximum will double and (d) is the correct choice. 11.

68719_24_ch24_p256-281.indd 257

The colors observed on the oil film are the result of constructive interference of waves reflected from the upper and lower surfaces of the film. For this to produce bright colored fringes, the thickness of the film must be on the same order of magnitude as the wavelength of the light. If the thickness of the oil film were smaller than half of the wavelengths of visible light, no colors would appear. If the thickness of the oil film were much larger, the constructive fringes for different colors would overlap and mix to white or gray. The best choice for this question is (b).

1/7/11 2:52:58 PM

258

12.

Chapter 24

In principle, the ribs (or any set of parallel openings that waves pass through) can act as a diffraction grating. However, for any significant diffraction to occur as the waves pass through the openings, and also for the interference fringes to be separated sufficiently to be observed, the spacing between the slits must be on the same order of magnitude as the wavelength of the radiation. The spacing between the ribs is very large in comparison to the wavelength of x-rays, so the correct choice is (b).

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The wavelength of light is extremely small in comparison to the dimensions of your hand, so the diffraction of light around obstacles the size of your hand is totally negligible. However, sound waves have wavelengths that are comparable to the dimensions of the hand or even larger. Therefore, significant diffraction of sound waves occurs around hand-sized obstacles.

4.

The spacing between double-slit interference fringes on a screen is proportional to the wavelength of the light (Δybright = lL d ). Light having a wavelength l0 in vacuum will have wavelength l = l0 n in a medium with index of refraction n. Thus, the wavelength of the light is smaller in water than it is in air, meaning the interference fringes will be more closely spaced when the experiment is performed in water.

6.

Every color produces its own interference pattern, and we see them superimposed. The central maximum is white. The first maximum is a full spectrum with violet on the inside and red on the outside. The second maximum is also a full spectrum, with red in it overlapping with violet in the third maximum. At larger angles, the light soon starts mixing to white again.

8.

The skin on the tip of a finger has a series of closely spaced ridges and swirls on it. When the finger touches a smooth surface, the oils from the skin will be deposited on the surface in the pattern of the closely spaced ridges. The clear spaces between the lines of deposited oil can serve as the slits in a crude diffraction grating and produce a colored spectrum of the light passing through or reflecting from the glass surface.

10.

Suppose the index of refraction of the coating is intermediate between vacuum and the glass. When the coating is very thin, light reflected from its top and bottom surfaces will interfere constructively, so you see the surface white and brighter. Once the thickness reaches one-quarter of the wavelength of violet light in the coating, destructive interference for violet light will make the surface look red. Then other colors in spectral order (blue, green, yellow, orange, and red) will interfere destructively, making the surface look red, magenta, and then blue. As the coating gets thicker, constructive interference is observed for violet light and then for other colors in spectral order. Even thicker coatings give constructive and destructive interference for several visible wavelengths, so the reflected light starts looking white again.

12.

The reflected light is partially polarized, with the component parallel to the reflecting surface being the most intense. Therefore, the polarizing material should have its transmission axis oriented in the vertical direction in order to minimize the intensity of the reflected light from horizontal surfaces.

14.

Due to gravity, the soap film tends to sag in its holder, being quite thin at the top and becoming thicker as one moves toward the bottom of the holding ring. Because light reflecting from the front surface of the film experiences a phase change, and light reflecting from the back surface of the film does not (see Figure 24.7 in the textbook), the film must be a minimum of a half wavelength thick before it can produce constructive interference in the reflected light. Thus, the light must be striking the film at some distance from the top of the ring before the thickness is sufficient to produce constructive interference for any wavelength in the visible portion of the spectrum.

68719_24_ch24_p256-281.indd 258

1/7/11 2:53:00 PM

Wave Optics

259

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a) 1.77 mm

4.

2.40 mm

6.

(a)

8.

2.9 mm

1.52 cm

(b) 1.47 mm

(b)

2.13 cm

d = 3.00l

10.

(a)

1.93 mm

(b)

12.

(a)

2.3°, 4.6°, and 6.9°

(b) 1.1°, 3.4°, and 5.7°

(c)

At small angles, q ≈ sinq . The approximation breaks down at larger angles.

(a)

See Solution.

(c)

0.144°, 2.51× 10 −3 , sinq ≈ tanq only when q is small

(d)

603 nm

(a)

640 nm

(b)

make use of a higher order constructive interference (i.e., a larger value of m)

(c)

360 nm (m = 1), 600 nm (m = 2)

18.

(a)

512 nm

20.

233 nm

22.

(a)

541 nm

(b)

406 nm; The most strongly transmitted wavelengths are those which suffer destructive interference in reflection.

14.

16.

(b)

(e)

(b)

(c)

a maximum

tanq = 2.51× 10 −3

0.720°

(f )

2.26 cm

2.5m1 = 2m2 + 1

24.

8 dark fringes, including the one at the line of contact

26.

6.5 × 10 2 nm

28.

99.6 nm

30.

(a)

97.8 nm

(b) Yes, some of which are 293 nm, 489 nm, and 685 nm.

32.

(a)

2.3 mm

(b)

34.

91.1 cm

36.

0.227 mm

38.

659 nm

68719_24_ch24_p256-281.indd 259

4.5 mm

1/7/11 2:53:01 PM

260

Chapter 24

40.

(a)

42.

(a) Three maxima (for m = −1, m = 0, and m = +1) will be observed. (b)

(b)

2 complete orders

10.9°

for m = −1, q = − 45.2°; for m = 0, q = 0°; and for m = +1, q = + 45.2°.

44.

7.35°

46.

See Solution.

48.

(a)

2.381 × 10 −6 m

(b)

29.65°

(d)

26.69°, 1.140 m

(e)

2.000 × 10 −3 m

(c)

1.138 m

(f ) 1.537 × 10 −3 m; The two results agree to only 1 significant figure, showing that the calculation is sensitive to rounding of intermediate answers. 50.

78.1 nm and 469 nm

52.

(a)

54.

36.9°

56.

60.5°

58.

0.336

(b)

0.164

(a)

45.0°

(b)

60.0°

60.

(a)

I = I0 2

(b)

54.7°

62.

632.8 nm

64.

(a)

(b)

m1 = 5, m2 = 6; 2.52 cm from the central maximum

66.

maxima at 0°, ± 29.1°, and ± 76.4°, minima at ±14.1° and ± 46.8°

68.

113 dark fringes (counting the m = 0 order at the line of contact)

70.

(a)

72.

See Solution.

74.

a = 20.0 × 10 −6 ( °C )

6m1 = 5m2

I f Ii = 0

(b)

(c)

65.9°

I f I i = 0.25

−1

PROBLEM SOLUTIONS 24.1

The location of the bright fringe of order m (measured from the position of the central maximum) is (ybright )m = (lL d)m, m = 0, ± 1, ± 2, …. Thus, the spacing between successive bright fringes is

(

Δybright = ybright

)

m +1

(

− ybright

)

m

= ( lL d ) ( m + 1) − ( lL d ) m = lL d

The wavelength of the laser light must be l=

68719_24_ch24_p256-281.indd 260

( Δy ) d = (1.58 × 10 bright

L

−2

m ) ( 0.200 × 10 −3 m ) 5.00 m

= 6.32 × 10 −7 m = 632 nm

1/7/11 2:53:03 PM

Wave Optics

24.2

(a)

261

For a bright fringe of order m, the path difference is d = ml, where m = 0, 1, 2,…. At the location of the third order bright fringe, m = 3 and d = 3l = 3 ( 589 nm ) = 1.77 × 10 3 nm = 1.77 mm

(b)

1⎞ ⎛ For a dark fringe, the path difference is d = ⎜ m + ⎟ l, where m = 0, 1, 2,…. ⎝ 2⎠ At the third dark fringe, m = 2 and ⎛ d = ⎜2+ ⎝

24.3

(a)

1⎞ 5 3 ⎟⎠ l = ( 589 nm ) = 1.47 × 10 nm = 1.47 mm 2 2

The distance between the central maximum and the first order bright fringe is Δy = ybright Δy =

(b)

m=0

=

lL , d

or

The distance between the first and second dark bands is

m =1

− ydark

m=0

=

lL = 2.62 mm as in (a) above. d

In the double-slit interference pattern, the bright fringe of order m is found where d sinq = ml with m = 0, ± 1, ± 2,…. Here, q is the angle between the line to the central maximum and the line to the location of the bright fringe of interest. Thus, if q = 15.0° for the m = 1 bright fringe of light having wavelength l = 620 nm, the spacing between the slits must be d=

24.5

− ybright

−9 lL ( 546.1 × 10 m ) (1.20 m ) = = 2.62 × 10 −3 m = 2.62 mm d 0.250 × 10 −3 m

Δy = ydark 24.4

m =1

(a)

−9 ml (1)( 620 × 10 m ) = = 2.40 × 10 − 6 m = 2.40 mm sinq sin15.0°

From d sinq = ml, the angle for the m = 1 maximum for the sound waves is ⎡ ⎡m ⎛ v ⎞⎤ ⎛ 354 m s ⎞ ⎤ 1 ⎛m ⎞ q = sin −1 ⎜ l ⎟ = sin −1 ⎢ ⎜ sound ⎟ ⎥ = sin −1 ⎢ ⎟ ⎥ = 36.2° ⎜ ⎝d ⎠ ⎣ 0.300 m ⎝ 2 000 Hz ⎠ ⎦ ⎣ d ⎝ f ⎠⎦

(b)

For 3.00-cm microwaves, the required slit spacing is d=

(c)

The wavelength of the light is l = d sinq m , so the frequency is f=

24.6

ml (1)( 3.00 cm ) = = 5.08 cm sinq sin ( 36.2° )

(1)( 3.00 × 108 m s ) c mc = = = 5.08 × 1014 Hz l d sinq (1.00 × 10 −6 m ) sin 36.2°

In a double-slit interference pattern, the screen position of the mth order maximum for wavelength l is ym = (lL d)m. The separation between the maxima of orders m1 and m2 is then Δy = (lL d)(m2 − m1 ). continued on next page

68719_24_ch24_p256-281.indd 261

1/7/11 2:53:05 PM

262

Chapter 24

(a)

If l = 588 nm while m2 = 1 and m1 = 0 , the separation is −9 lL ( 588 × 10 m )(1.50 m ) lL = = 1.52 × 10 −2 m = 1.52 cm (1− 0 ) = d d 0.058 0 × 10 −3 m

Δy = (b)

If l = 412 nm, m2 = 4, and m1 = 2, the spacing between these maxima is Δy =

24.7

( 412 × 10

−9

m )(1.50 m )

0.058 0 × 10 −3 m

( 4 − 2 ) = 2.13 × 10 −2 m = 2.13 cm

As indicated in the sketch at the right, the path difference between waves from the two antennas that travel toward the car is given by d = d sinq . When d = ml, where m = 0, 1, 2,… , constructive interference produces the maximum of order m. Destructive interference produces the minimum of order m when d = ( m + 12 ) l. (a)

d

q

q

⎞ ( 300 m ) ⎡ d sinq d ⎛ y ⎢ = ⎜ 2 ⎟= ⎢ m 2 ⎝ x + y2 ⎠ 2 ⎣

x

O

d =dsinq

At the m = 2 maximum, d = d sinq = 2 l, or l=

(b)

y

400 m

(1000m ) + ( 400 m ) 2

2

⎤ ⎥ = 55.7 m ⎥ ⎦

The next minimum encountered is the m = 2 minimum, and it occurs where d = 5l 2, or ⎡ 5 ( 55.7 m ) ⎤ ⎛ 5l ⎞ q = sin −1 ⎜ ⎟ = sin −1 ⎢ ⎥ = 27.7° ⎝ 2d ⎠ ⎣ 2 ( 300 m ) ⎦ At this point, y = x tanq = (1 000 m ) tan 27.7° = 525 m, so the car must travel an additional 125 m .

24.8

The angular position of the bright fringe of order m is given by d sinq = ml . Thus, if the m = 1 bright fringe is located at q = 12° when l = 6.0 × 10 2 nm = 6.0 × 10 −7 m, the slit spacing is d=

24.9

The screen position of the mth order bright fringe of wavelength l in a double-slit interference pattern is ym = (lL d)m. Then, Δy = (lL d)(m2 − m1 )is the separation between the bright fringes of orders m1 and m2 . If Δy = 0.552 mm for the first and second order bright fringes when L = 1.75 m, and the slit separation is d = 0.552 mm, the wavelength incident upon the set of slits is l=

24.10

(1) ( 6.0 × 10 −7 m ) ml = = 2.9 × 10 −6 m = 2.9 mm sinq sin12°

( Δy ) d

L ( m2 − m1 )

=

( 0.552 × 10

−3

m ) ( 2.10 × 10 −3 m )

(1.75 m )( 2 − 1)

= 6.62 × 10 −7 m = 662 nm

The angular deviation from the line of the central maximum is given by ⎛ 1.80 cm ⎞ ⎛ y⎞ q = tan −1 ⎜ ⎟ = tan −1 ⎜ = 0.737° ⎝ 140 cm ⎟⎠ ⎝ L⎠ (a)

The path difference is then d = d sinq = ( 0.150 mm ) sin ( 0.737° ) = 1.93 × 10 −3 mm = 1.93 mm

continued on next page

68719_24_ch24_p256-281.indd 262

1/7/11 2:53:08 PM

Wave Optics

(b) (c) 24.11

l ⎛ ⎞ = 3.00 l d = (1.93 × 10 −6 m ) ⎜ ⎝ 643 × 10 −9 m ⎟⎠ Since the path difference for this position is a whole number of wavelengths (to three significant figures), the waves interfere constructively and produce a maximum at this spot.

The distance between the central maximum (position of A) and the first minimum is y=

lL ⎛ ⎜m + d ⎝

Thus, d = 24.12

(a)

1⎞ lL = ⎟⎠ 2 m=0 2 d

lL ( 3.00 m ) (150 m ) = = 11.3 m . 2y 2 ( 20.0 m )

The angular position of the bright fringe of order m is given by d sinq = ml, or q m = sin −1 (ml d). If d = 25l, the first three bright fringes are found at ⎛ 1⎞ q1 = sin −1 ⎜ ⎟ = 2.3° , ⎝ 25 ⎠

(b)

(c)

⎛ 2⎞ q 2 = sin −1 ⎜ ⎟ = 4.6° , ⎝ 25 ⎠

and

⎛ 3⎞ q 3 = sin −1 ⎜ ⎟ = 6.9° ⎝ 25 ⎠

The angular position of the dark fringe of order m is given by d sinq = (m + 12 )l, or q m = sin −1 [ (m + 12 )l d ] , m = 0, ± 1, ± 2,…. If d = 25l , the first three dark fringes are found at ⎛ 12 ⎞ q 0 = sin −1 ⎜ ⎟ = 1.1° , ⎝ 25 ⎠

24.13

263

⎛ 32 ⎞ q1 = sin −1 ⎜ ⎟ = 3.4° , ⎝ 25 ⎠

and

⎛ 52 ⎞ q 2 = sin −1 ⎜ ⎟ = 5.7° ⎝ 25 ⎠

The answers are evenly spaced because the angles are small and q ≈ sinq . At larger angles, the approximation breaks down and the spacing isn’t so regular.

As shown in the figure at the right, the path difference in the waves reaching the telescope is d = d 2 − d1 = d 2 (1 − sina ) . If the first minimum (d = l 2) occurs when q = 25.0°, then a = 180° − (q + 90.0° + q ) = 40.0° , and d2 =

d l 2 ( 250 m 2) = 350 m = = 1− sina 1− sina 1− sin 40.0°

Thus, h = d 2 sin 25.0° = 148 m . 24.14

(a)

(b)

tanq1 =

y1 4.52 × 10 −3 m = = 2.51 × 10 −3 L 1.80 m continued on next page

68719_24_ch24_p256-281.indd 263

1/7/11 2:53:13 PM

264

Chapter 24

(c)

q1 = tan −1 ( 2.51 × 10 −3 ) = 0.144° and sinq1 = sin ( 0.144° ) = 2.51 × 10 −3 . The sine and the tangent are very nearly the same but only because the angle is small.

(d)

From d = d sinq = ml for the order m bright fringe, l=

24.15

−4 d sinq1 ( 2.40 × 10 m ) sin ( 0.144° ) = = 6.03 × 10 −7 m = 603 nm 1 1

(e)

⎡ 5 ( 6.03 × 10 −7 m ) ⎤ ⎛ 5l ⎞ q 5 = sin −1 ⎜ ⎟ = sin −1 ⎢ ⎥ = 0.720° −4 ⎝ d ⎠ ⎣⎢ 2.40 × 10 m ⎥⎦

(f )

y5 = L tanq 5 = (1.80 m ) tan ( 0.720° ) = 2.26 × 10 −2 m = 2.26 cm

The path difference in the two waves received at the home is d = 2 d, where d is the distance from the home to the mountain. Neglecting any phase change upon reflection, the condition for destructive interference is 1⎞ ⎛ d = ⎜ m + ⎟ l with m = 0, 1, 2,… ⎝ 2⎠ so

24.16

(a)

d m in =

d m in ⎛ 1 ⎞ l l 300 m = ⎜0+ ⎟ = = = 75.0 m ⎝ 2⎠ 2 4 4 2

With a phase change in the reflection at the outer surface of the soap film and no change on reflection from the inner surface, the condition for constructive interference in the light reflected from the soap bubble is 2nfilm t = (m + 12 )l, where m = 0,1,2,…. For the lowest order reflection, m = 0, so the wavelength is l=

2nwater t = 4 (1.333)(120 nm ) = 6.4 × 10 2 nm = 640 nm ( 0 + 1 2)

(b)

To strongly reflect the same wavelength light, a thicker film will need to make use of a higher order constructive interference, i.e., use a larger value of m .

(c)

The next greater thickness of soap film that can strongly reflect 640 nm light corresponds to m = 1, giving t1 =

(1+ 1 2) l 2nfilm

=

3 ⎡ 640 nm ⎤ 2 ⎥ = 3.6 × 10 nm = 360 nm ⎢ 2 ⎣ 2 (1.333) ⎦

and the third such thickness (corresponding to m = 2) is t2 = 24.17

2nfilm

=

5 ⎡ 640 nm ⎤ 2 ⎥ = 6.0 × 10 nm = 600 nm ⎢ 2 ⎣ 2 (1.333) ⎦

Light reflecting from the first (glass-iodine) interface suffers a phase change, but light reflecting at the second (iodine-glass) interface does not have a phase change. Thus, the condition for constructive interference in the reflected light is 2nfilm t = (m + 12 )l, with m = 0,1,2,…. The smallest film thickness capable of strongly reflecting the incident light is tm in =

68719_24_ch24_p256-281.indd 264

( 2 + 1 2) l

(m

+ 1 2 ) l ( 0 + 1 2 ) λ 6.00 × 10 2 nm = = = 85.4 nm 2nfilm 4 (1.756 ) 2nfilm

m in

1/7/11 2:53:16 PM

Wave Optics

24.18

(a)

265

Phase changes are experienced by light reflecting at either surface of the oil film, a upper air-oil interface and a lower oil-water interface. Under these conditions, the requirement for constructive interference is 2nfilm t = m1 l , with m1 = 0,1,2,… , and the requirement for destructive interference is 2nfilm t = (m2 + 12 )l , with m2 = 0,1,2,…. To have the thinnest film that produces simultaneous constructive interference of l1 = 640 nm and destructive interference of l2 = 512 nm, it is necessary that 1⎞ ⎛ 2nfilm t = m1 ( 640 nm ) = ⎜ m2 + ⎟ ( 512 nm ) ⎝ 2⎠ where both m1 and m2 are the smallest integers for which this is true. It is found that m1 = m2 = 2 are the smallest integer values that will satisfy this condition, giving the minimum acceptable film thickness as tm in =

(b)

m1 l1 ( m2 + 12 ) l2 2 ( 640 nm ) ( 2.5)( 512 nm ) = = = = 512 nm 2nfilm 2nfilm 2 (1.25) 2 (1.25)

From the discussion above, it is seen that in order to have a film thickness that produces simultaneous constructive interference of 640-nm light and destructive interference of 512-nm light, it is necessary that 1⎞ ⎛ m1 ( 640 nm ) = ⎜ m2 + ⎟ ( 512 nm ) ⎝ 2⎠

or

1 ⎛ 640 nm ⎞ m1 ⎜ = m2 + ⎝ 512 nm ⎟⎠ 2

1 This gives (1.25) m1 = m2 + , or 2.5 m1 = 2m2 + 1 . 2 24.19

With nfilm = 1.52, > nair , light reflecting from the upper surface (air to glass transition) experiences a phase change, while light reflecting from the lower surface (a glass to air transition) does not experience such a change. Under these conditions, the requirement for constructive interference of normally incident waves reflecting from these two surfaces is 2nfilm t = (m + 12 )l, where m is any integer value. Thus, if the thickness of the film is t = 0.420 mm = 420 nm, the wavelengths that will strongly reflect from the film are l= yielding

2nfilm t 4 (1.52 )( 420 nm ) = m +1 2 2m + 1

m = 0, 1, 2,…

m = 0 ⇒ l = 2.55 × 10 3 nm (infrared);

m = 1 ⇒ l = 851 nm (infrared);

m = 2 ⇒ l = 511 nm (visible);

m = 3 ⇒ l = 365 nm (ultraviolet),

with the wavelengths for all higher values of m being even shorter. Thus, the only visible light wavelength that can strongly reflect from this film is 511 nm . 24.20

Since nair < noil < nwater , light reflected from both top and bottom surfaces of the oil film experiences a phase change, resulting in zero net phase difference due to reflections. Therefore, the condition for constructive interference in reflected light is 2 t = mln = m

l nfilm

or

⎛ l ⎞ t = m⎜ where m = 0, 1, 2,… ⎝ 2 nfilm ⎟⎠

Assuming that m = 1, the thickness of the oil slick is t = (1)

68719_24_ch24_p256-281.indd 265

600 nm l = = 233 nm . 2 nfilm 2 (1.29 )

1/7/11 2:53:19 PM

266

24.21

Chapter 24

There will be a phase change of the radar waves reflecting from both surfaces of the polymer, giving zero net phase change due to reflections. The requirement for destructive interference in the reflected waves is then 1⎞ ⎛ 2 t = ⎜ m + ⎟ ln ⎝ 2⎠

or

t = ( 2m + 1)

l where m = 0, 1, 2,… 4 nfilm

If the film is as thin as possible, then m = 0, and the needed thickness is t=

l 3.00 cm = = 0.500 cm 4 nfilm 4 (1.50 )

This anti-reflectance coating could be easily countered by changing the wavelength of the radar—to 1.50 cm—now creating maximum reflection! 24.22

(a)

With nair < nwater < noil , reflections at the air-oil interface experience a phase change, while reflections at the oil-water interface experience no phase change. With one phase change at the surfaces, the condition for constructive interference in the light reflected by the film is 2nfilm t = (m + 12 )l, where m is any positive integer. Thus, lm =

2nfilm t 2 (1.45)( 280 nm ) 812 nm = = m + 12 m + 12 m + 12

m = 0, 1, 2, 3,…

The possible wavelengths are: l0 = 1.62 × 10 3 nm, l1 = 541 nm, l2 = 325nm,…. Of these, only l1 = 541 nm (green) is in the visible portion of the spectrum. (b)

The wavelengths that will be most strongly transmitted are those that suffer destructive interference in the reflected light. With one phase change at the surfaces, the condition for destructive interference in the light reflected by the film is 2nfilm t = ml, where m is any positive, nonzero integer. The possible wavelengths are lm = or

2nfilm t 2 (1.45)( 280 nm ) 812 nm = = m m m

m = 1, 2, 3,…

l1 = 812 nm, l2 = 406 nm, l3 = 271 nm,…

of which only l2 = 406 nm (violet) is in the visible spectrum. 24.23

(a)

For maximum transmission, we want destructive interference in the light reflected from the front and back surfaces of the film. If the surrounding glass has refractive index greater than 1.378, light reflected from the front surface suffers no phase change, and light reflected from the back does undergo a phase change. Under these conditions, the condition for destructive interference in the reflected light is 2d = mln = m l nfilm , where m is a positive integer. Thus, for minimum nonzero film thickness, we require that m = 1 and find d=

(b)

l 656.3 nm = = 238 nm 2 nfilm 2 (1.378 )

The filter will expand. As d increases in 2nfilm d = l , so does l increase .

continued on next page

68719_24_ch24_p256-281.indd 266

1/7/11 2:53:22 PM

Wave Optics

(c)

267

Destructive interference of order 2 will occur for reflected light when 2 d = 2 l nfilm, or for the wavelength l = nfilm d = (1.378 )( 238 nm ) = 328 nm (near ultraviolet)

24.24

Light reflecting from the lower surface of the air layer experiences a phase change, but light reflecting from the upper surface of the layer does not. The requirement for a dark fringe (destructive interference) is then 2nfilm t = ml ,

where m = 0, 1, 2,…

At the thickest part of the film ( t = 2.00 mm ) , the order number is m=

−6 2nfilm t 2 (1.00 )( 2.00 × 10 m ) = = 7.32 l 546.1× 10 −9 m

Since m must be an integer, m = 7 is the order of the last dark fringe seen. Counting the m = 0 order along the edge of contact, a total of 8 dark fringes will be seen. 24.25

When the hair is inserted between one B end of the glass plates, which have A Δt length L = 14.0 cm, a wedge-shaped d tm+1 air film is created as shown at the tm Δx right. The maximum thickness of this air wedge is equal to the diameter d L = 14.0 cm of the hair. If the bright interference fringe of order m occurs at point A (where the air wedge has thickness tm ) and the adjacent bright fringe of order m + 1 occurs at B (where the thickness is tm +1 ), we may use the properties of similar triangles to write d Δt tm +1 − tm = = L Δx Δx

or

⎛ Δt ⎞ d =⎜ L ⎝ Δ x ⎟⎠

[1]

Since nair < nglass , light waves reflecting at the upper surface of the air film do not undergo a phase change, but those reflecting from the lower surface do experience such a change. Thus, the condition for constructive interference to produce a bright fringe is 2nair t = (m + 12 )l, or the thickness of the film at the location of the bright fringe of order m is tm = (m + 12 )l 2nair . The change in thickness between the locations of adjacent bright fringes is then Δt = tm +1 − tm =

[( m + 1) + 1 2] l − [ m + 1 2] l = 2nair

2nair

l l = 2nair 2

If the observed spacing between adjacent bright fringes is Δx = 0.580 mm when using light of wavelength l = 650 nm, Equation [1] gives the diameter of the hair as

(650 × 10−9 m ) (14.0 × 10 −2 m ) = 7.84 × 10 −5 m = 78.4 mm lL ⎛ l 2⎞ d =⎜ = L = ⎝ Δx ⎟⎠ 2 ( Δx ) 2 ( 0.580 × 10 −3 m )

68719_24_ch24_p256-281.indd 267

1/7/11 2:53:24 PM

268

Chapter 24

24.26

From the geometry shown in the figure, R 2 = h 2 + r 2 = ( R − t ) + r 2 , or 2

t = R − R 2 − r 2 = 3.0 m −

( 3.0 m )2 − ( 9.8 × 10 −3 m ) = 1.6 × 10 −5 m 2

With a phase change upon reflection at the lower surface of the air layer but no change with reflection from the upper surface, the condition for a bright fringe is 1⎞ 1⎞ l ⎛ 1⎞ ⎛ ⎛ 2 t = ⎜ m + ⎟ ln = ⎜ m + ⎟ = ⎜ m + ⎟ l , where m = 0, 1, 2,… ⎝ ⎠ ⎝ ⎠ ⎝ 2 2 nair 2⎠ At the 50th bright fringe, m = 49, and the wavelength is found to be l= 24.27

2 (1.6 × 10 −5 m ) 2t = = 6.5 × 10 −7 m = 6.5 × 10 2 nm m +1 2 49.5

There is a phase change due to reflection at the bottom of the air film but not at the top of the film. The requirement for a dark fringe is then 2nair t = ml

where

m = 0, 1, 2,…

At the 19th dark ring (in addition to the dark center spot), the order number is m = 19, and the thickness of the film is t= 24.28

−9 ml 19 ( 500 × 10 m ) = = 4.75 × 10 −6 m = 4.75 mm 2nair 2 (1.00 )

With a phase change due to reflection at each surface of the magnesium fluoride layer, there is zero net phase difference caused by reflections. The condition for destructive interference is then 1⎞ 1⎞ l ⎛ ⎛ 2 t = ⎜ m + ⎟ ln = ⎜ m + ⎟ , where m = 0, 1, 2,… ⎝ ⎝ 2⎠ 2 ⎠ nfilm For minimum thickness, m = 0, and the thickness is t = ( 2m + 1)

24.29

(550 × 10 m ) = 9.96 × 10 −8 m = 99.6 nm l = (1) 4 nfilm 4 (1.38 ) −9

Since the thin film has air on both sides of it, there is a phase change for light reflecting from the first surface but no change for light reflecting from the second surface. Under these conditions, the requirement to be met if waves reflecting from the two sides are to produce constructive interference and a strong reflection is 1⎞ ⎛ 2nfilm t = ⎜ m + ⎟ l ⎝ 2⎠

or

l=

4nfilm t 2m + 1

m = 0, 1, 2,…

continued on next page

68719_24_ch24_p256-281.indd 268

1/7/11 2:53:27 PM

Wave Optics

269

With nfilm = 1.473 and t = 542 nm, the wavelengths that produce strong reflections are given by l = 4 (1.473)( 524 nm ) (2m + 1), which yields m = 0 : l = 3.09 × 10 3 nm

m = 1: l = 1.03 × 10 3 nm

m = 2 : l = 617 nm

m = 3 : l = 441 nm

m = 4 : l = 343 nm

m = 5 : l = 281 nm

and many other shorter wavelengths. Of these, the only ones in the range 300 nm to 700 nm are 617 nm, 441 nm, and 343 nm . 24.30

With nair < nfilm < nglass , light undergoes a phase change when it reflects from either of the two surfaces of the film. Therefore, the condition for destructive interference of normally incident light reflecting from these two surfaces is ⎛ 2nfilm t = ⎜ m + ⎝ (a)

24.31

⎛ l ⎞ t = ( 2m + 1) ⎜ ⎝ 4nfilm ⎟⎠

or

where m = 0, 1, 2,…

[1]

The minimum thickness film that will minimize the light reflected from the lens is given by m = 0 in Equation [1], and has the value tm in =

(b)

1⎞ ⎟l 2⎠

l 540 nm = = 97.8 nm 4nfilm 4 (1.38 )

Yes. From Equation [1], we see than any film thickness given by t = (2m + 1)tm in , where tm in = (l 4nfilm ) = 97.8 nm, will produce destructive interference and hence minimum reflected light for wavelength l = 540 nm. For m = 1, 2, 3,… , we obtain thicknesses of 293 nm, 489 nm, 685 nm, … .

In a single-slit diffraction pattern, with the slit having width a, the dark fringe of order m occurs at angle q m , where sinq m = m(l a) and m = ±1, ± 2, ± 3,…. The location, on a screen located distance L from the slit, of the dark fringe of order m (measured from y = 0 at the center of the central maximum) is (ydark )m = L tanq m ≈ L sinq m = ml ( L a ) . (a)

The central maximum extends from the m = −1 dark fringe to the m = +1 dark fringe, so the width of this central maximum is ⎛ lL ⎞ ⎛ lL ⎞ 2lL − ( −1) ⎜ Central max. width = (ydark )1 − (ydark )−1 = 1⎜ = ⎟ ⎝ a ⎠ ⎝ a ⎟⎠ a =

(b)

2 ( 5.40 × 10 −7 m )(1.50 m ) 0.200 × 10 −3 m

= 8.10 × 10 −3 m = 8.10 mm

The first order bright fringe extends from the m = 1 dark fringe to the m = 2 dark fringe, or

( Δy ) = ( y ) − ( y ) bright

dark

1

=

2

(5.40 × 10

dark

−7

1

⎛ lL ⎞ ⎛ lL ⎞ lL − 1⎜ = = 2⎜ ⎝ a ⎟⎠ ⎝ a ⎟⎠ a

m ) (1.50 m )

0.200 × 10 −3 m

= 4.05 × 10 −3 m = 4.05 mm

Note that the width of the first order bright fringe is exactly one-half the width of the central maximum.

68719_24_ch24_p256-281.indd 269

1/7/11 2:53:29 PM

270

24.32

Chapter 24

(a)

Dark bands occur where sinq = m ( l a ) . At the first dark band, m = 1, and the distance from the center of the central maximum is ⎛ 600 × 10 −9 m ⎞ ⎛l⎞ = 2.3 mm y1 = L tanq ≈ L sinq = L ⎜ ⎟ = (1.5 m ) ⎜ ⎝ a⎠ ⎝ 0.40 × 10 −3 m ⎟⎠

(b)

The central maximum extends from the m = 1 dark band to the m = −1 dark band. Its width is (ydark )1 − (ydark )−1 = 1( lL a ) − ( −1) ( lL a ) = 2lL a , or width =

24.33

(a)

2 ( 600 × 10 −9 m )(1.5 m ) 0.40 × 10 −3 m

= 4.5 × 10 −3 m = 4.5 mm

Dark bands (minima) occur where sinq = m(l a). For the first minimum, m = 1, and the distance from the center of the central maximum is y1 = L tanq ≈ L sinq = L ( l a ) . Thus, the needed distance to the screen is ⎛ 0.75 × 10 −3 m ⎞ ⎛ a⎞ = 1.1 m L = y1 ⎜ ⎟ = ( 0.85 × 10 −3 m ) ⎜ ⎝l⎠ ⎝ 587.5 × 10 −9 m ⎟⎠

(b) 24.34

The width of the central maximum is 2 y1 = 2 ( 0.85 mm ) = 1.7 mm .

Note: The small angle approximation does not work well in this situation. Rather, you should proceed as follows. At the first order minimum, sinq1 = (1) l a , or ⎛l⎞ ⎛ 5.00 cm ⎞ q1 = sin −1 ⎜ ⎟ = sin −1 ⎜ = 7.98° ⎝ 36.0 cm ⎟⎠ ⎝ a⎠ Then, y1 = L tanq1 = ( 6.50 m ) tan 7.98° = 0.911 m = 91.1 cm

24.35

With the screen locations of the dark fringe of order m at (ydark )m = L tanq m ≈ L sinq m = m(lL a) the width of the central maximum is Δycentral

for

m = ±1, ± 2, ± 3,…

= (ydark )m = +1 − (ydark )m = −1 = 2(lL a), so

m axim um

⎛ ⎞ a ⎜ Δycentral ⎟ −3 −3 ⎝ m axim um ⎠ ( 0.600 × 10 m ) ( 2.00 × 10 m ) l= = = 4.62 × 10 −7 m = 462 nm 2L 2 (1.30 m ) 24.36

At the positions of the minima, sinq m = m ( l a ) and ym = L tanq m ≈ L sinq m = m ⎡⎣ L ( l a ) ⎤⎦ . Thus, y3 − y1 = ( 3 − 1) ⎡⎣ L ( l a ) ⎤⎦ = 2 ⎡⎣ L ( l a ) ⎤⎦ , and a=

68719_24_ch24_p256-281.indd 270

2 ( 0.500 m )( 680 × 10 −9 m ) 2Ll = = 2.27 × 10 −4 m = 0.227 mm y3 − y1 3.00 × 10 −3 m

1/7/11 2:53:33 PM

Wave Optics

24.37

271

The locations of the dark fringes (minima) mark the edges of the maxima, and the widths of the maxima equals the spacing between successive minima. At the locations of the minima, sinq m = m(l a), and ⎡ ⎛ 500 × 10 −9 m ⎞ ⎤ ym = L tanq m ≈ L sinq m = m ⎡⎣ L ( l a ) ⎤⎦ = m ⎢(1.20 m ) ⎜ ⎥ = m (1.20 mm ) ⎝ 0.500 × 10 −3 m ⎟⎠ ⎦ ⎣ Then, Δy = Δm(1.20 mm), and for successive minima, Δm = 1. Therefore, the width of each maximum, other than the central maximum, in this interference pattern is width = Δy = (1)(1.20 mm ) = 1.20 mm

24.38

In a single-slit diffraction pattern, minima are found where ⎛l⎞ sinq dark = m ⎜ ⎟ ⎝ a⎠

q L

where m = ± 1, ± 2,…

y

or, using the small angle approximation y = L tanq ≈ L sinq , at screen positions ym = m(lL a). Thus, if the m = 2 minimum is seen at y = 1.40 mm on a screen located distance L = 85.0 cm from a single slit of width a = 0.800 mm, the wavelength of the light passing through the slit must be l= 24.39

−3 −3 y a (1.40 × 10 m ) ( 0.800 × 10 m ) = = 6.59 × 10 −7 m = 659 nm mL 2 (85.0 × 10 −2 m )

The grating spacing is d = (1 3 660 )cm = (1 3.66 × 10 5 ) m, and bright lines are found where d sinq = ml . (a)

The wavelength observed in the first-order spectrum is l = d sinq1 , or 9 ⎛ 10 4 nm ⎞ ⎛ 1 m ⎞ ⎛ 10 nm ⎞ sinq1 = ⎜ sinq1 l =⎜ 5 ⎟ ⎜ ⎟ ⎝ 3.66 × 10 ⎠ ⎝ 1 m ⎠ ⎝ 3.66 ⎟⎠

This yields: at q1 = 10.1°, l = 698 nm . (b)

l = 479 nm ; at q1 = 13.7°, l = 647 nm ; and at q1 = 14.8°,

In the second order, m = 2. The second order images for the above wavelengths will be found at angles q 2 = sin −1 ( 2 l d ) = sin −1 [ 2sinq1 ] . This yields: for l = 479 nm, q 2 = 20.5° ; for l = 647 nm, q 2 = 28.3° ; and for l = 698 nm, q 2 = 30.7° .

24.40

(a)

The longest wavelength in the visible spectrum is 700 nm, and the grating spacing is d = 1 mm 600 = 1.67 × 10 −3 mm = 1.67 × 10 −6 m. The maximum viewing angle is q m ax = 90.0°, and maxima are found where ml = d sinq . Thus, mm ax =

−6 d sin 90.0° (1.67 × 10 m ) sin 90.0° = = 2.38 lred 700 × 10 −9 m

so 2 complete orders will be observed.

continued on next page

68719_24_ch24_p256-281.indd 271

1/7/11 2:53:36 PM

272

Chapter 24

(b)

From l = d sinq , the angular separation of the red and violet edges in the first order will be ⎡l Δq = sin −1 ⎢ red ⎣ d or

24.41

⎤ −1 ⎡ lviolet ⎥ − sin ⎢ d ⎣ ⎦

−9 −9 m⎤ m⎤ ⎤ −1 ⎡ 700 × 10 −1 ⎡ 400 × 10 = sin − sin ⎢ ⎥ ⎢ −6 −6 ⎥ m m ⎥⎦ 1.67 × 10 1.67 × 10 ⎦ ⎣ ⎦ ⎣

Δq = 10.9°

The grating spacing is d =

1.00 cm ⎛ 1.00 m ⎞ 1.00 m . ⎜⎝ 2 ⎟⎠ = 4 500 10 cm 4.50 × 10 5

From d sinq = ml, the angular separation between the given spectral lines will be Δq = sin −1 ⎡⎣ m lred d ⎤⎦ − sin −1 ⎡⎣ m lviolet d ⎤⎦ , or ⎡ m ( 434 × 10 −9 m ) ( 4.50 × 10 5 ) ⎤ ⎡ m ( 656 × 10 −9 m ) ( 4.50 × 10 5 ) ⎤ Δq = sin −1 ⎢ ⎥ ⎥ − sin −1 ⎢ 1.00 m 1.00 m ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ The results obtained are: for m = 1, Δq = 5.91° ; for m = 2, Δq = 13.2° ; and for m = 3, Δq = 26.5° . Complete orders for m ≥ 4 are not visible. 24.42

The array of wires will act as a diffraction grating for the ultrasound waves, with maxima located at those angles given by the grating equation d sinq = ml, where m = 0, ± 1, ± 2,…. The spacing between adjacent slits in this grating is d = 1.30 cm, and the wavelength of these ultrasound waves is l= (a)

vsound 343 m s = = 9.22 × 10 −3 m = 9.22 mm 37.2 × 10 3 Hz f

The order number of the maximum found at angle q is m = (d l)sinq , so the maximum order that can be found will be the largest integer satisfying the relation 1.30 × 10 −2 m ⎛d⎞ ⎛d⎞ m ≤ ⎜ ⎟ sin q m ax = ⎜ ⎟ sin 90.0° = = 1.41 ⎝l⎠ ⎝l⎠ 9.22 × 10 −3 m Thus, three maxima corresponding to m = −1, m = 0, and m = +1 can be found .

(b)

We use q = sin −1 (ml d) to find the direction for each maximum to be: for m = −1, q = − 45.2° ; for m = 0, q = 0° ; and for m = +1, q = + 45.2° .

24.43

For diffraction by a grating, the angle at which the maximum of order m occurs is given by d sinq m = ml, where d is the spacing between adjacent slits on the grating. Thus,

(1)( 632.8 × 10 ml d= = sinq m sin 20.5° 24.44

−9

m)

= 1.81× 10 −6 m = 1.81 mm

With 2 000 lines per centimeter, the grating spacing is d=

1 cm = 5.00 × 10 −4 cm = 5.00 × 10 −6 m 2 000

Then, from d sinq = ml , the location of the first order for the red light is ⎛ 1( 640 × 10 −9 m ) ⎞ ⎛ ml ⎞ −1 = sin q = sin −1 ⎜ ⎜ ⎟ = 7.35° −6 ⎝ d ⎟⎠ ⎝ 5.00 × 10 m ⎠

68719_24_ch24_p256-281.indd 272

1/7/11 2:53:40 PM

273

Wave Optics

24.45

The spacing between adjacent slits on the grating is d=

1 cm 10 −2 m = 5 310 slits 5 310

The maximum of order m is located where d sinq m = ml, so l=

y d d⎛ sinq m = ⎜ 2 m 2 m m ⎜⎝ L + ym

⎞ 10 −2 m ⎛ ⎜ ⎟= ⎟⎠ (1)( 5 310 ) ⎜⎝

0.488 m

(1.72 m )2 + ( 0.488 m )2

⎞ ⎟ ⎟⎠

= 5.14 × 10 −7 m = 514 nm 24.46

From d sinq m = ml, or sinq m = ml d , we see that sinq v 2 =

2lviolet 800 nm = d d

sinq v 3 =

3lviolet 1 200 nm = d d

sinq r 2 =

2lred 1 400 nm = d d

sinq r 3 =

3lred 2 100 nm = d d

Since, for 0° ≤ q ≤ 90°, q increases as sinq increases, we have that q v 2 < q v3 < qr 2 < qr 3 so the second and third order spectra overlap in the range q v 3 ≤ q ≤ q r 2 . 24.47

1 cm 10 −2 m = = 3.64 × 10 −6 m. From d sinq = ml , or 2 750 2.75 × 10 3 q = sin −1 ( ml d ) , the angular positions of the red and violet edges of the second-order spectrum are found to be The grating spacing is d =

⎛ 2 ( 700 × 10 −9 m ) ⎞ ⎛ 2l ⎞ q r = sin −1 ⎜ red ⎟ = sin −1 ⎜ ⎟ = 22.6° −6 ⎝ d ⎠ ⎝ 3.64 × 10 m ⎠ and

⎛ 2 ( 400 × 10 −9 m ) ⎞ ⎞ ⎛ 2l q v = sin −1 ⎜ violet ⎟ = sin −1 ⎜ ⎟ = 12.7° −6 ⎝ d ⎠ ⎝ 3.64 × 10 m ⎠

Note from the sketch at the right that yr = L tanq r and yv = L tanq v , so the width of the spectrum on the screen is Δy = L ( tanq r − tanq v ) .

Screen Grating

Since it is given that d = ( m + 1 2 ) l, the distance from the grating to the screen must be Δy 1.75 cm L= = tanq r − tanq v tan ( 22.6° ) − tan (12.7° )

Δy

yr qv

qr

yv

L

= 9.17 cm 24.48

(a)

d=

1 cm = 2.381 × 10 −4 cm = 2.381 × 10 −6 m 4.200 × 10 3 slits

continued on next page

68719_24_ch24_p256-281.indd 273

1/7/11 2:53:43 PM

274

Chapter 24

⇒

⎡ 2 ( 589.0 × 10 −9 m ) ⎤ ⎛ 2l ⎞ −1 = sin q 2 = sin −1 ⎜ ⎢ ⎥ = 29.65° −6 ⎝ d ⎟⎠ ⎣⎢ 2.381 × 10 m ⎥⎦

(b)

d sinq m = ml

(c)

y2 = L tanq 2 = ( 2.000 m ) tan ( 29.65° ) = 1.138 m

(d)

⎡ 2 ( 589.6 × 10 −9 m ) ⎤ ⎛ 2l′ ⎞ −1 = sin q ′2 = sin −1 ⎜ ⎢ ⎥ = 29.69° −6 ⎝ d ⎟⎠ ⎣⎢ 2.381× 10 m ⎥⎦ y2′ = L tanq 2′ = ( 2.000 m ) tan ( 29.69° ) = 1.140 m

(e)

Δy = y2′ − y2 = 1.140 m − 1.138 m = 2.000 × 10 −3 m

(f )

⎡ ⎛ ⎛ −1 ⎛ 2l ⎞ ⎞ ⎤ ⎛ 2l ′ ⎞ ⎞ Δy = y2′ − y2 = L ⎢ tan ⎜ sin −1 ⎜ ⎟⎠ ⎟ − tan ⎜ sin ⎜⎝ ⎟ ⎥ ⎝ d d ⎠ ⎟⎠ ⎦ ⎠ ⎝ ⎣ ⎝ ⎡ ⎛ ⎛ ⎛ 2 ( 589.6 × 10 −9 m ) ⎞ ⎞ ⎛ 2 ( 589.0 × 10 −9 m ) ⎞ ⎞ ⎤ −1 − tan sin = ( 2.000 m ) ⎢ tan ⎜ sin −1 ⎜ −2 ⎟ ⎜ ⎟ ⎜ −2 ⎟⎟ ⎥ ⎜⎝ 10 m 4.200 × 10 3 ⎠ ⎟⎠ 10 m 4.200 × 10 3 ⎠ ⎟⎠ ⎥ ⎢ ⎜⎝ ⎝ ⎝ ⎣ ⎦ = 1.537 × 10 −3 m The two answers agree to only one significant figure. The calculation is sensitive to rounding at intermediate steps.

24.49

(a)

For a diffraction grating having distance d between adjacent slits, primary maxima for wavelength l are produced at angles satisfying the grating equation, d sinq = ml. If the m = 3 maximum for l = 500 nm is observed at q = 32.0°, the grating spacing is d=

3 ( 500 × 10 −9 m ) ml = = 2.83 × 10 −6 m = 2.83 × 10 −4 cm sinq sin 32.0°

The number of rulings per centimeter on the grating is then n= (b)

1 cm 1 cm = = 3.53 × 10 3 grooves cm d 2.83 × 10 −4 cm

The order number of a maximum appearing at angle q is m = d sinq l . The magnitude of the largest order numbers visible is the largest integer satisfying the condition d 2.83 × 10 −6 m ⎛d⎞ ⎛d⎞ m ≤ ⎜ ⎟ sinq m ax = ⎜ ⎟ sin 90° = = = 5.66 ⎝l⎠ ⎝l⎠ l 500 × 10 −9 m Thus, 11 maxima for wavelength l = 500 nm will be observable with this grating. These are the maxima corresponding to m = 0, ± 1, ± 2, ± 3, ± 4, and ± 5.

24.50

The grating spacing is d = 1 cm 1 200 = 8.33 × 10 −4 cm = 8.33 × 10 −6 m. Using sinq = ml d and the small angle approximation, the distance from the central maximum to the maximum of order m for wavelength l is ym = L tanq ≈ L sinq = (lL d)m. Therefore, the spacing between successive maxima is Δy = ym +1 − ym = lL d. The longer wavelength in the light is found to be l long =

( Δy ) d = (8.44 × 10 L

−3

m ) (8.33 × 10 −6 m )

0.150 m

= 4.69 × 10 −7 m = 469 nm

continued on next page

68719_24_ch24_p256-281.indd 274

1/7/11 2:53:47 PM

Wave Optics

275

Since the third order maximum of the shorter wavelength falls halfway between the central maximum and the first order maximum of the longer wavelength, we have 3lshort L ⎛ 0 + 1 ⎞ llong L =⎜ ⎝ 2 ⎟⎠ d d 24.51

(a)

or

⎛ 1⎞ lshort = ⎜ ⎟ ( 469 nm ) = 78.1 nm ⎝ 6⎠

From Brewster’s law, the index of refraction is n2 = tanq p = tan ( 48.0° ) = 1.11

(b)

From Snell’s law, n2 sinq 2 = n1 sinq1 , the angle of refraction when q1 = q p is ⎛ n1 sinq p ⎞ ⎛ (1.00 ) sin 48.0° ⎞ q 2 = sin −1 ⎜ = sin −1 ⎜ ⎟⎠ = 42.0° ⎝ 1.11 ⎝ n2 ⎟⎠ Note that when q1 = q p , q 2 = 90.0° −q p as it should.

24.52

(a)

From Malus’s law, the fraction of the incident intensity of the unpolarized light that is transmitted by the polarizer is I ′ = I 0 ( cos 2 q )av = I 0 ( 0.500 ) The fraction of this intensity incident on the analyzer that will be transmitted is I = I ′ cos 2 ( 35.0° ) = ⎡⎣ 0.500I 0 ⎤⎦ cos 2 ( 35.0° ) = 0.336I 0 Thus, the fraction of the incident unpolarized light transmitted is I I 0 = 0.336 .

(b)

The fraction of the original incident light absorbed by the analyzer is I ′ − I 0.500I 0 − 0.336I 0 = = 0.164 I0 I0

24.53

The more general expression for Brewster’s angle is given in problem P24.57 as tanq p = n2 n1

24.54

(a)

⎛n ⎞ ⎛ 1.52 ⎞ When n1 = 1.00 and n2 = 1.52, q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜ = 56.7° . ⎝ 1.00 ⎟⎠ ⎝ n1 ⎠

(b)

⎛n ⎞ ⎛ 1.52 ⎞ When n1 = 1.333 and n2 = 1.52, q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜ = 48.8° . ⎝ 1.333 ⎟⎠ ⎝ n1 ⎠

The polarizing angle for light in air striking a water surface is ⎛n ⎞ ⎛ 1.333 ⎞ q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜ = 53.1° ⎝ 1.00 ⎟⎠ ⎝ n1 ⎠ This is the angle of incidence for the incoming sunlight (that is, the angle between the incident light and the normal to the surface). The altitude of the Sun is the angle between the incident light and the water surface. Thus, the altitude of the Sun is a = 90.0° −q p = 90.0° − 53.1° = 36.9°

68719_24_ch24_p256-281.indd 275

1/7/11 2:53:50 PM

276

24.55

Chapter 24

(a)

Brewster’s angle (or the polarizing angle) is ⎛ nquartz ⎞ ⎛n ⎞ ⎛ 1.458 ⎞ = tan −1 ⎜ q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜ = 55.6° ⎝ 1.000 ⎟⎠ ⎝ n1 ⎠ ⎝ nair ⎟⎠

(b) 24.56

When the angle of incidence is the polarizing angle, q p , the angle of refraction of the transmitted light is q 2 = 90.0° −q p . Hence, q 2 = 90.0° − 55.6° = 34.4° .

The critical angle for total internal reflection is q c = sin −1 ( n2 n1 ). Thus, if q c = 34.4° as light attempts to go from sapphire into air, the index of refraction of sapphire is nsapphire = n1 =

n2 1.00 = = 1.77 sinq c sin 34.4°

Then, when light is incident on sapphire from air, the Brewster’s angle is ⎛n ⎞ ⎛ 1.77 ⎞ q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜ = 60.5° ⎝ 1.00 ⎟⎠ n ⎝ 1⎠ 24.57

From Snell’s law, the angles of incidence and refraction are related by n1 sinq1 = n2 sinq 2 . If the angle of incidence is the polarizing angle (that is, q1 = q p), the refracted ray is perpendicular to the reflected ray (see Figure 24.28 in the textbook), and the angles of incidence and refraction are also related by q p + 90° +q 2 = 180°, or q 2 = 90° − q p . Substitution into Snell’s law then gives

(

)

n1 sinq p = n2 sin 90° − q p = n2 cosq p 24.58

sinq p cosq p = tanq p = n2 n1

From Malus’s law, the intensity of the light transmitted by the polarizer is I = I 0 cos 2 q , where I 0 is the intensity of the incident light, and q is the angle between the direction of the plane of polarization of the incident light and the transmission axis of the polarizing disk. Thus, q = cos −1 I I 0 .

(

24.59

or

)

(a)

1 I = I 0 2.00

⇒

⎛ 1 ⎞ = 45.0° q = cos −1 ⎜ ⎝ 2.00 ⎟⎠

(b)

1 I = I 0 4.00

⇒

⎛ 1 ⎞ q = cos −1 ⎜ ⎟⎠ = 60.0° 4.00 ⎝

(c)

1 I = I 0 6.00

⇒

⎛ 1 ⎞ = 65.9° q = cos −1 ⎜ ⎝ 6.00 ⎟⎠

From Malus’s law, the intensity of the light transmitted by the first polarizer is I1 = I i cos 2 q1 . The plane of polarization of this light is parallel to the axis of the first plate and is incident on the second plate. Malus’s law gives the intensity transmitted by the second plate as I 2 = I1 cos 2 (q 2 −q1 ) = I i cos 2 q1 cos 2 (q 2 −q1 ) . This light is polarized parallel to the axis of the second plate and is incident upon the third plate. A final application of Malus’s law gives the transmitted intensity as I f = I 2 cos 2 (q 3 − q 2 ) = I i cos 2 q1 cos 2 (q 2 − q1 ) cos 2 (q 3 − q 2 )

continued on next page

68719_24_ch24_p256-281.indd 276

1/7/11 2:53:52 PM

Wave Optics

277

With q1 = 20.0°, q 2 = 40.0°, and q 3 = 60.0°, this result yields I f = (10.0 units ) cos 2 ( 20.0° ) cos 2 ( 20.0° ) cos 2 ( 20.0° ) = 6.89 units 24.60

(a)

Using Malus’s law, the intensity of the transmitted light is found to be

(

I = I 0 cos 2 ( 45° ) = I 0 1 (b)

(

(a)

)

3 = 54.7°

If light has wavelength l in vacuum, its wavelength in a medium of refractive index n is ln = l n. Thus, the wavelengths of the two components in the specimen are ln =

l 546.1 nm = = 413.7 nm n1 1.320

ln =

l 546.1 nm = = 409.7 nm n2 1.333

1

and (b)

2

From Malus’s law, I I 0 = cos 2 q . Thus, if I I 0 = 1 3, we obtain cos 2 q = 1 3, or q = cos −1 1

24.61

)

2 , or I = I 0 2 .

2

The numbers of cycles of vibration each component completes while passing through the specimen of thickness t = 1.000 mm are N1 =

t 1.000 × 10 −6 m = = 2.417 ln 413.7 × 10 −9 m 1

N2 =

and

t 1.000 × 10 −6 m = = 2.441 ln 409.7 × 10 −9 m 2

Thus, when they emerge, the two components are out of phase by N 2 − N1 = 0.024 cycles. Since each cycle represents a phase angle of 360°, they emerge with a phase difference of Δf = ( 0.024 cycles ) ( 360° cycle ) = 8.6° 24.62

In a single-slit diffraction pattern, the first dark fringe occurs where sin(q dark )1 = (1) l a. If no diffraction minima are to be observed, the maximum width the slit can have is that which would place the first dark fringe at the maximum viewable angle of 90.0°. That is, when a = am ax , we will have sin(q dark )1 = (1) l am ax = sin 90.0° = 1.00, yielding am ax = l = 632.8 nm

24.63

The light has passed through a single slit since the central maximum is twice the width of other maxima (the space between the centers of successive dark fringes). In a double-slit pattern, the central maximum has the same width as all other maxima (compare Active Figures 24.1(b) and 24.16(b) in the textbook).

Figure P24.63

In single-slit diffraction the width of the central maximum on the screen is given by Δycentral m axim um

⎛ (1) l ⎞ ⎛ lL ⎞ = 2⎜ = 2L tan (q dark )m =1 ≈ 2L sin (q dark )m =1 = 2L ⎜ ⎝ a ⎟⎠ ⎝ a ⎟⎠

continued on next page

68719_24_ch24_p256-281.indd 277

1/7/11 2:53:55 PM

278

Chapter 24

The width of the slit is then a=

2lL Δycentral

=

2 ( 632.8 × 10 −9 m ) ( 2.60 m ) ⎛ 10 2 cm ⎞ −4 ⎜⎝ 1 m ⎟⎠ = 1.2 × 10 m = 0.12 mm (10.3 − 7.6 ) cm

m axim um

24.64

(a)

In a double-slit interference pattern, bright fringes on the screen occur where (ybright )m = m(lL d). Thus, if bright fringes of the wavelengths l1 = 540 nm and l2 = 450 nm are to coincide, it is necessary that m1

( 540 nm ) L d

= m2

( 450 nm ) L d

or, dividing both sides by 90 nm, 6m1 = 5m2 , where both m1 and m2 are integers. (b)

The condition found above may be written as m2 = (6 5)m1 . Trial and error reveals that the smallest nonzero integer value of m1 that will yield an integer value for m2 is m1 = 5 , yielding m2 = 6 . Thus, the first overlap of bright fringes for the two given wavelengths occurs at the screen position (measured from the central maximum) ybright =

5 ( 540 × 10 −9 m )(1.40 m ) 0.150 × 10 −3 m

=

6 ( 450 × 10 −9 m )(1.40 m ) 0.150 × 10 −3 m

= 2.52 × 10 −2 m = 2.52 cm 24.65

Dark fringes (destructive interference) occur where d sinq = ( m + 1 2 ) l for m = 0, 1, 2,…. Thus, if the second dark fringe ( m = 1) occurs at q = (18.0 min ) (1.00° 60.0 min ) = 0.300°, the slit spacing is ⎛ d = ⎜m + ⎝

24.66

−9 1⎞ l ⎛ 3 ⎞ ( 546 × 10 m ) = 1.56 × 10 −4 m = 0.156 mm = ⎟ ⎜ ⎟ 2 ⎠ sinq ⎝ 2 ⎠ sin ( 0.300° )

The wavelength is l = vsound f = ( 340 m s ) ( 2 000 Hz ) = 0.170 m, and maxima occur where d sinq = ml , or q = sin −1 ⎡⎣ m ( l d ) ⎤⎦ for m = 0, ± 1, ± 2,…. Since d = 0.350 m, l d = 0.486, which gives q = sin −1 [ m ( 0.486 )]. For m = 0, ± 1, and ± 2, this yields maxima at 0°, ± 29.1°, and ± 76.4° . No solutions exist for m ≥ 3, since that would imply sinq > 1. Minima occur where d sinq = ( m + 1 2 ) l, or q = sin −1 [( 2m + 1) l 2d ] for m = 0, ± 1, ± 2,…. With l d = 0.486 , this becomes q = sin −1 [( 2m + 1)( 0.243)]. For m = 0 and ± 1, we find minima at ± 14.1° and ± 46.8° . No solutions exist for m ≥ 2, since that would imply sinq > 1.

24.67

The source and its image, located 1.00 cm below the mirror, act as a pair of coherent sources. This situation may be treated as double-slit interference, with the slits separated by 2.00 cm, if it is remembered that the light undergoes a phase change upon reflection from the mirror. The existence of this phase change causes the conditions for constructive and destructive interference to be reversed. Therefore, dark bands (destructive interference) occur where y = m ( lL d ) for m = 0, 1, 2,…. The m = 0 dark band occurs at y = 0 (that is, at mirror level). The first dark band above the mirror corresponds to m = 1, and is located at −9 ⎛ lL ⎞ ( 500 × 10 m ) (100 m ) = y = (1) ⎜ = 2.50 × 10 −3 m = 2.50 mm ⎝ d ⎟⎠ 2.00 × 10 −2 m

68719_24_ch24_p256-281.indd 278

1/7/11 2:53:58 PM

Wave Optics

24.68

279

Assuming the glass plates have refractive indices greater than that of both air and water, there will be a phase change at the reflection from the lower surface of the film but no change from reflection at the top of the film. Therefore, the condition for a dark fringe is 2 t = mln = m ( l nfilm ) for m = 0, 1, 2,… If the highest order dark band observed is m = 84 (a total of 85 dark bands counting the m = 0 order at the edge of contact), the maximum thickness of the air film is tm ax =

mm ax ⎛ l ⎞ 84 ⎛ λ ⎞ = ⎜ ⎟ = 42 l 2 ⎜⎝ nfilm ⎟⎠ 2 ⎝ 1.00 ⎠

When the film consists of water, the highest order dark fringe appearing will be ⎛ 1.333 ⎞ ⎛n ⎞ mm ax = 2 tm ax ⎜ film ⎟ = 2 ( 42 l ) ⎜ = 112 ⎝ l ⎟⎠ ⎝ l ⎠ Counting the m = 0 order, a total of 113 dark fringes are now observed. 24.69

In the figure at the right, observe that the path difference between the direct and the indirect paths is d = 2x − d = 2 h 2 + ( d 2 ) − d 2

With a phase change (equivalent to a half-wavelength shift) occurring upon reflection at the ground, the condition for constructive interference is d = ( m + 1 2 ) l, and the condition for destructive interference is d = m l . In both cases, the possible values of the order number are m = 0, 1, 2,…. (a)

d The wavelengths that will interfere constructively are l = . The longest of these is m +1 2 for the m = 0 case and has a value of l = 2d = 4 h 2 + ( d 2 ) − 2d = 4 2

(b)

The wavelengths that will interfere destructively are l = d m , and the largest finite one of these is for the m = 1 case. That wavelength is l = d = 2 h2 + ( d 2) − d = 2 2

24.70

( 50.0 m )2 + ( 300 m )2 − 2 ( 600 m ) = 16.6 m

( 50.0 m )2 + ( 300 m )2 − 600 m = 8.28 m

From Malus’s law, the intensity of the light transmitted by the first polarizer is I1 = I i cos 2 q1 . The plane of polarization of this light is parallel to the axis of the first plate and is incident on the second plate. Malus’s law gives the intensity transmitted by the second plate as I 2 = I1 cos 2 (q 2 −q1 ) = I i cos 2 q1 cos 2 (q 2 −q1 ) . This light is polarized parallel to the axis of the second plate and is incident upon the third plate. A final application of Malus’s law gives the transmitted intensity as I f = I 2 cos 2 (q 3 − q 2 ) = I i cos 2 q1 cos 2 (q 2 − q1 ) cos 2 (q 3 − q 2 ) (a)

If q1 = 45°, q 2 = 90°, and q 3 = 0°, then I f I i = cos 2 45° cos 2 ( 90° − 45° ) cos 2 ( 0° − 90° ) = 0

continued on next page

68719_24_ch24_p256-281.indd 279

1/7/11 2:54:03 PM

280

Chapter 24

(b)

If q1 = 0°, q 2 = 45°, and q 3 = 90°, then I f I i = cos 2 0° cos 2 ( 45° − 0° ) cos 2 ( 90° − 45° ) = 0.25

24.71

If the signal from the antenna to the receiver station is to be completely polarized by reflection from the water, the angle of incidence where it strikes the water must equal the polarizing angle from Brewster’s law. This is given by ⎛n ⎞ q p = tan −1 ⎜ water ⎟ = tan −1 (1.333) = 53.1° ⎝ nair ⎠ From the triangle RST in the sketch, the horizontal distance from the point of refection, T, to shore is given by x = ( 90.0 m ) tanq p = ( 90.0 m )(1.333) = 120 m and from triangle ABT, the horizontal distance from the antenna to point T is y = ( 5.00 m ) tanq p = ( 5.00 m )(1.333) = 6.67 m The total horizontal distance from ship to shore is then x + y = 120 m + 6.67 m = 127 m .

24.72

There will be a phase change associated with the reflection at one surface of the film but no change at the other surface of the film. Therefore, the condition for a dark fringe (destructive interference) is ⎛ l ⎞ 2 t = mln = m ⎜ where m = 0, 1, 2,… ⎝ nfilm ⎟⎠ From the figure, note that 2 R 2 = r 2 + ( R − t ) = r 2 + R 2 − 2Rt + t 2 , which 2 reduces to r = 2Rt − t 2 . Since t will be very small in comparison to either r or R, we may neglect the term t 2 , leaving r ≈ 2Rt . For a dark fringe, t =

ml , so the radii of the dark rings will be 2nfilm

⎛ ml ⎞ r ≈ 2R ⎜ = ⎝ 2nfilm ⎟⎠ 24.73

mlR nfilm

for m = 0, 1, 2,…

In the single-slit diffraction pattern, destructive interference (or minima) occur where sinq = m ( l a ) for m = 0, ± 1, ± 2,…. The screen locations, measured from the center of the central maximum, of these minima are at ym = L tanq m ≈ L sinq m = m ( lL a )

continued on next page

68719_24_ch24_p256-281.indd 280

1/7/11 2:54:06 PM

Wave Optics

281

If we assume the first order maximum is halfway between the first and second order minima, then its location is y=

y1 + y2 (1+ 2 ) ( lL a ) 3lL = = 2 2 2a

and the slit width is −9 3lL 3 ( 500 × 10 m )(1.40 m ) a= = = 3.50 × 10 −4 m = 0.350 mm −3 2y 2 ( 3.00 × 10 m )

24.74

As light emerging from the glass reflects from the top of the air layer, there is no phase change produced. However, the light reflecting from the end of the metal rod at the bottom of the air layer does experience a phase change. Thus, the condition for constructive interference in the reflected light is 2t = ( m + 12 ) l nair = ( m + 12 ) l. As the metal rod expands, the thickness of the air layer decreases. The increase in the length of the rod is given by ΔL = Δt = ( mi + 12 )

l l l − m f + 12 = Δm 2 2 2

(

)

The order number changes by one each time the film changes from bright to dark and back to bright. Thus, during the expansion, Δm = 200, and the measured change in the length of the rod is ΔL = ( 200 )

(500 × 10 l = ( 200 ) 2 2

−9

m)

= 5.00 × 10 −5 m

From ΔL = L0a ( ΔT ) , the coefficient of linear expansion of the rod is a=

68719_24_ch24_p256-281.indd 281

ΔL 5.00 × 10 −5 m −1 = = 20.0 × 10 −6 ( °C ) L0 ( ΔT ) ( 0.100 m )( 25.0 °C )

1/7/11 2:54:08 PM

25 Optical Instruments QUICK QUIZZES 1.

Choice (c). The corrective lens for a farsighted eye is a converging lens, while that for a nearsighted eye is a diverging lens. Since a converging lens is required to form a real image of the Sun on the paper to start a fire, the campers should use the glasses of the farsighted person.

2.

Choice (a). We would like to reduce the minimum angular separation for two objects below the angle subtended by the two stars in the binary system. We can do that by reducing the wavelength of the light—this in essence makes the aperture larger, relative to the light wavelength, increasing the resolving power. Thus, we would choose a blue filter.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

The amount of light focused on the film by a camera is proportional to the area of the aperture through which the light enters the camera. Since the area of a circular opening varies as the square of the diameter of the opening, the light reaching the film is proportional to the square of the diameter of the aperture. Thus, increasing this diameter by a factor of 3 increases the amount of light by a factor of 9, and (c) is the correct choice.

2.

The power of a lens in diopters equals the reciprocal of the focal length when that focal length is expressed in meters. Hence, the power of a lens having a focal length of 25 cm is P=

1 1 = = 4.0 diopters f 0.25 m

and choice (b) is the correct answer. 3.

Diffraction of light as it passes through, or reflects from, the objective element of a telescope can cause the images of two sources having a small angular separation to overlap and fail to be seen as separate images. The minimum angular separation two sources must have in order to be seen as separate sources is inversely proportional to the diameter of the objective element. Thus, using a large diameter objective element in a telescope increases its resolution, making (c) the correct choice.

4.

When the eye is longer than normal, the lens-cornea system tends to form images of distant objects in front of the retina. Rays from near objects are more divergent and the lens-cornea system brings them into focus farther from the lens, on the retina. This means that the eye can see near objects clearly but is unable to focus on distant objects. Such an eye is nearsighted (myopia) and needs a diverging corrective lens to make the rays from distant objects more divergent before they enter the eye. Choice (b) is the correct answer.

5.

When the eye is shorter than normal, the lens-cornea system fails to bring light from near objects into focus by the time it reaches the retina, resulting in a blurry image. Light rays entering the pupil from distant objects are less divergent than those from near objects, and the lens-cornea system can focus them on the retina. Such an eye is farsighted, or has hyperopia, and needs a converging corrective lens to help bring rays from near objects to focus sooner. The correct choice is (c). 282

68719_25_ch25_p282-304.indd 282

1/7/11 2:55:10 PM

Optical Instruments

6.

283

The corrective lens must form an upright, virtual image located 55 cm in front of the lens (q = −55 cm) when the object is 25 cm in front of the lens (p = +25 cm). The thin-lens equation then gives the required focal length as f =

pq ( 25 cm )( −55 cm ) = = + 46 cm p+q 25 cm − 55 cm

and the correct choice is (c). 7.

Stars are very distant, and the reciprocal of the object distance p in the thin-lens or mirror equation 1 p +1 q = 1 f is essentially zero. This means that q = f , or the images are formed at a distance equal to the focal length from the objective element. Thus, the angular separation of the images, and hence the stars, is a =

s 20.0 × 10 −3 m ⎛ 360° ⎞ = 1.00 × 10 −2 rad ⎜ = = 0.573° ⎝ 2p rad ⎟⎠ 2.00 m f

and we see that choice (d) is the correct answer. 8.

When a compound microscope is adjusted for most relaxed viewing (i.e., the final image formed by the eyepiece is at infinity), the approximate overall magnification produced by the microscope is given by the expression m=−

L ⎛ 25 cm ⎞ fo ⎜⎝ fe ⎟⎠

where L is the length of the microscope, fo is the focal length of the objective lens, and fe is the focal length of the eyepiece. With the microscope described, the approximate magnification is m=−

15 cm ⎛ 25 cm ⎞ 2 ⎜ ⎟ = −1.2 × 10 0.80 cm ⎝ 4.0 cm ⎠

or

m ≈ 120

and the correct answer is choice (d). 9.

When using light of wavelength l, the resolving power needed to distinguish two closely spaced spectral lines having a difference in wavelength of Δl is R = l Δl . Thus, if two lines in the visible spectrum differ in wavelength by Δl = 0.1 nm, the minimum resolving power of a diffraction grating that might be used to separate them is Rmin =

lmin 400 nm = = 4 × 10 3 = 4 000 Δl 0.1 nm

and the correct choice is (b). 10.

The angular separation of the two stars is q = s r = (10 −5 ly) (200 ly) = 5 × 10 −8 radians . The limiting angle of resolution for a circular aperture is q min = 1.22(l D). Requiring that q min = q , and assuming a wavelength at the center of the visible spectrum (550 nm), the required diameter of the aperture is found to be D=

(

)

−9 1.22l 1.22 550 × 10 m = = 1 × 101 m = 10 m q 5 × 10 −8 rad

so (c) is the best choice of the listed possible answers.

68719_25_ch25_p282-304.indd 283

1/7/11 2:55:10 PM

284

11.

Chapter 25

Diffraction of the light as it passes through the opening of the detector limits the ability of that detector to resolve two closely spaced light sources. If the diameter of the opening of the detector does not change, the intensity of the light is not a factor in the resolution of the sources. Thus, if we neglect any contraction of the pupils of the eyes, your ability to resolve the two headlights will not change when they are switched to high beam, and choice (c) is the best response to the question. In reality, the natural response of the eyes to the increased light intensity is to contract the pupils to some degree, thereby somewhat decreasing their ability to resolve the two sources.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

Light nearly parallel to the horizontal ruler will scatter from rule marks, distance d apart, to produce a diffraction pattern on a vertical wall a y distance L away. At height y on the wall, where the scattering angle (in radians) is q = y L , light d q scattered from adjacent rule marks interferes constructively to produce a bright spot if the path d L difference is d = ml, where m = 1, 2, 3,…. The path difference is given by d = d cosq ≈ d(1 − q 2 2), where we have made use of a series expansion for cosq , valid for very small angles. Combining these equations gives ml = d(1 − ym2 2 L2 ). Thus, the wavelength of the light may be calculated by measurement of the heights ym of bright spots.

4.

There will be an effect on the interference pattern—it will be distorted. The high temperature of the flame will change the index of refraction of air for the arm of the interferometer in which the match is held. As the index of refraction varies randomly, the wavelength of the light in that region will also vary randomly. As a result, the effective difference in length between the two arms will fluctuate, resulting in a wildly varying interference pattern.

6.

The aperture of a camera is a close approximation to the iris of the eye. The retina of the eye corresponds to the film of the camera, and a close approximation to the cornea of the eye is the lens of the camera.

8.

Large lenses are difficult to manufacture and machine with accuracy. Also, their large weight leads to sagging, which produces a distorted image. In reflecting telescopes, light does not pass through glass; hence, problems associated with chromatic aberrations are eliminated. Large-diameter reflecting telescopes are also technically easier to construct. Some designs use a rotating pool of mercury as the reflecting surface.

10.

Under low ambient light conditions, a photoflash unit is used to ensure that light entering the camera lens will deliver sufficient energy for a proper exposure to each area of the film. Thus, the most important criterion is the additional energy per unit area (product of intensity and the duration of the flash, assuming this duration is less than the shutter speed) provided by the flash unit.

12.

The angular magnification produced by a simple magnifier is m = ( 25 cm ) f . Note that this is proportional to the optical power of a lens, P = 1 f , where the focal length f is expressed in meters. Thus, if the power of the lens is doubled, the angular magnification will also double.

14.

In a nearsighted person the image of a distant object focuses in front of the retina. The cornea needs to be flattened so that its focal length is increased.

68719_25_ch25_p282-304.indd 284

1/7/11 2:55:13 PM

Optical Instruments

285

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

2.5 mm to 46 mm

4.

The image is 19 mm across and easily fits on the 35-mm slide.

6.

(a) See Solution.

8.

1.05 m to 6.30 m

10.

12.

(b) 1 100 s

(a) farsighted

(b)

18.0 cm

(c)

38.0 cm

(d) virtual image, negative

(e)

34.2 cm

(f )

+2.92 diopters

(c)

−3.70 diopters

(c)

+4.2 cm

(g)

p = 20.0 cm, q = − 40.0 cm, f = + 40.0 cm, P = +2.50 diopters

(a)

+50.8 diopters ≤ P ≤ + 60.0 diopters

(b) –0.800 diopters, diverging 14.

(a) Yes, by using a bifocal or progressive lens. (b) +1.78 diopters

(c)

−1.18 diopters

16.

(a)

− 0.67 diopters

(b)

+0.67 diopters

18.

(a)

−25.0 cm

(b)

nearsighted

20.

(a) 4.17 cm in front of the lens

(b)

m = 6.00

22.

(a) +5.0

(b)

+6.0

24.

(a) mmax = 2.39

(b)

m = 1.39

26.

m = − 588

28.

0.809 mm

30.

(a)

(b)

2.00 cm toward the objective lens

32.

1.6 × 10 2 mi

34.

(a)

(b)

0.944 m

36.

(a) virtual image

L = fo [( m + 1) m ]

m = 7.50

(b) The final image is an infinite distance in front of the telescope. fe = −5.00 cm, fo = 15.0 cm

(c) 38.

1.7 m

40.

(a)

68719_25_ch25_p282-304.indd 285

2.29 × 10 − 4 rad

(b)

43.7 m

1/7/11 2:55:14 PM

286

Chapter 25

42.

2.2 × 1011 m

44.

38 cm

46.

(a)

48.

(a) 449 nm

50.

39.6 mm

52.

1.000 5

54.

q min ≈ 2.0 × 10 −3 radians

56.

(a)

58.

(a) 1.96 cm

60.

5.07 mm

62.

m = 10.7

3.6 × 10 3 lines

− 4.3 diopters

(b) 1.8 × 10 3 lines (b)

smaller, because the wavelength is longer

(b)

− 4.0 diopters, 44 cm

(b)

3.27

(c)

9.80

PROBLEM SOLUTIONS 25.1

The f-number (or focal ratio) of a lens is defined to be the ratio of focal length of the lens to its diameter. Therefore, the f-number of the given lens is f -number =

25.2

f 28 cm = = 7.0 D 4.0 cm

If a camera has a lens with focal length of 55 mm and can operate at f -numbers that range from f 1.2 to f 22 , the aperture diameters for the camera must range from Dmin =

f 55 mm = = 2.5 mm 22 ( f -number )max

Dmax =

f 55 mm = = 46 mm 1.2 ( f -number )min

to

25.3

The thin-lens equation, 1 + 1 = 1 , gives the image distance as p q f q=

pf (100 m )( 52.0 mm ) = = 52.0 mm p − f 100 m − 52.0 × 10 −3 m

From the magnitude of the lateral magnification, M = h ′ h = −q p , where the height of the image is h ′ = 0.092 0 m = 92.0 mm, the height of the object (the building) must be h = h′ −

68719_25_ch25_p282-304.indd 286

p 100 m = ( 92.0 mm ) − = 177 m q 52.0 mm

1/7/11 2:55:16 PM

Optical Instruments

25.4

287

Consider rays coming from opposite edges of the object and passing undeviated through the center of the lens as shown at the right. For a very distant object, the image distance equals the focal length of the lens. If the angular width of the object is q, the full image width on the film is ⎛ 20° ⎞ h = 2 ⎡⎣ f tan (q 2 ) ⎤⎦ = 2 ( 55.0 mm ) tan ⎜ = 19 mm ⎝ 2 ⎟⎠ so the image easily fits within a 23.5 mm by 35.0 mm area.

25.5

The exposure time is being reduced by a factor of t2 1 125 s 15 3 = = = 1 15 s 125 25 t1 Thus, to maintain correct exposure, the intensity of the light reaching the film should be increased by a factor of 25 3. This is done by increasing the area of the aperture by a factor of 25 3, so in terms of the diameter, p D22 4 = ( 25 3) p D12 4 , or D2 = D1 25 3.

(

)

The new f-number will be

( f -number )2 = 25.6

(a)

f f 3 4.0 3 = = ( f -number )1 = = 1.4 D2 D1 25 3 25 5

or

f 1.4

The intensity is a measure of the rate at which energy is received by the film per unit area of the image, or I ∝ 1 Aimage . Consider an object with horizontal and vertical dimensions hx and hy as shown at the right. If the vertical dimension intercepts angle q , the vertical dimension of the image is hy′ = qq , or hy′ ∝ q. Similarly for the horizontal dimension, hx′ ∝ q, and the area of the image is Aimage = hx′ hy′ Assuming a very distant object, q ≈ f , so Aimage ∝ f 2, and we conclude that I ∝ 1 f 2 .

∝ q 2.

The intensity of the light reaching the film is also proportional to the cross-sectional area of the lens and hence to the square of the diameter of that lens, or I ∝ D 2. Combining this with our earlier conclusion gives I (b)

∝

D2 1 = f 2 ( f D )2

or

I

∝

1 f -number ( )2

The total light energy delivered to the film is proportional to the product of intensity and exposure time, It. Thus, to maintain correct exposure, this product must be kept constant, or I 2 t2 = I1t1 , giving 2 ⎡ ( f -number )2 ⎤ ⎛I ⎞ ⎛ 4.0 ⎞ ⎛ 1 ⎞ t2 = ⎜ 1 ⎟ t1 = ⎢ 2 t = s ≈ 1 100 s ⎥ ⎜ ⎟ 1 2 ⎝ 1.8 ⎠ ⎜⎝ 500 ⎟⎠ ⎝ I2 ⎠ ⎢⎣ ( f1 -number ) ⎥⎦

68719_25_ch25_p282-304.indd 287

1/7/11 2:55:18 PM

288

25.7

Chapter 25

Since the exposure time is unchanged, the intensity of the light reaching the film must be doubled if the energy delivered is to be doubled. Using the result of Problem 25.6 (part a), we obtain ⎛

⎞

( f2 -number )2 = ⎜ II1 ⎟ ( f1 -number )2 = ⎛⎜⎝ 12 ⎞⎟⎠ (11)2 = 61, ⎝

2

⎠

or

f2 -number = 61 = 7.8

Thus, you should use the f 8.0 setting on the camera. 25.8

The image must always be focused on the film, so the image distance is the distance between the lens and the film. From the thin-lens equation, 1 p + 1 q = 1 f , the object distance is p = qf (q − f ), and the range of object distances this camera can work with is from pmin =

qmax f ( 210 mm )(175 mm ) = = 1.05 × 10 3 mm = 1.05 m qmax − f 210 mm − 175 mm

pmax =

qmin f (180 mm )(175 mm ) = 6.30 × 10 3 mm = 6.30 m = qmin − f 180 mm − 175 mm

to

25.9

The corrective lens must form an upright, virtual image at the near point of the eye (i.e., q = − 60.0 cm in this case) for objects located 25.0 cm in front of the eye (p = +25.0 cm). From the thin-lens equation, 1 p + 1 q = 1 f , the required focal length of the corrective lens is f =

pq ( 25.0 cm )( −60.0 cm ) = = + 42.9 cm p+q 25.0 cm − 60.0 cm

and the power (in diopters) of this lens will be P= 25.10

1 1 = = + 2.33 diopters fin meters + 0.429 m

(a)

The person is farsighted , able to see distant objects but unable to focus on objects at the normal near point for a human eye.

(b)

With the corrective lens 2.00 cm in front of the eye, the object distance for an object 20.0 cm in front of the eye is p = 20.0 cm − 2.00 cm = 18.0 cm .

(c)

The upright, virtual image formed by the corrective lens will serve as the object for the eye, and this object must be 40.0 cm in front of the eye. With the lens 2.00 cm in front of the eye, the magnitude of the image distance for the lens will be q = 40.0 cm − 2.00 cm = 38.0 cm .

(d)

The image must be located in front of the corrective lens, so it is a virtual image , and the image distance is negative . Thus, q = −38.0 cm.

(e)

From the thin-lens equation, 1 p + 1 q = 1 f , the required focal length of the corrective lens is f =

(f )

pq (18.0 cm )( −38.0 cm ) = + 34.2 cm = 18.0 cm − 38.0 cm p+q

The power of the corrective lens is then P=

1 fin meters

=

1 = + 2.92 diopters + 0.342 m

continued on next page

68719_25_ch25_p282-304.indd 288

1/7/11 2:55:21 PM

Optical Instruments

(g)

289

With a contact lens, the lens-to-eye distance would be zero, so we would have p = 20.0 cm and q = − 40.0 cm , giving a required focal length of f =

( 20.0 cm ) ( − 40.0 cm ) pq = + 40.0 cm = 20.0 cm − 40.0 cm p+q

and a power in diopters of P= 25.11

1 fin meters

=

1 = + 2.50 diopters + 0.400 m

His lens must form an upright, virtual image of a very distant object (p ≈ ∞) at his far point, 80.0 cm in front of the eye. Therefore, the focal length is f = q = −80.0 cm. If this lens is to form a virtual image at his near point (q = −18.0 cm), the object distance must be p=

25.12

(a)

qf ( −18.0 cm ) ( − 80.0 cm ) = 23.2 cm = q − f −18.0 cm − ( − 80.0 cm )

When the child clearly sees objects at her far point ( pmax = 125 cm ) , the lens-cornea combination has assumed a focal length suitable for forming the image on the retina ( q = 2.00 cm ). The thin-lens equation gives the optical power under these conditions as Pfar =

1 1 1 1 1 = + = + = + 50.8 diopters fin meters p q 1.25 m 0.020 0 m

When the eye is focused ( q = 2.00 cm ) on objects at her near point ( pmin = 10.0 cm ) , the optical power of the lens-cornea combination is Pnear =

1 1 1 1 1 = + = + = + 60.0 diopters fin meters p q 0.100 m 0.020 0 m

Therefore, the range of the power of the lens-cornea combination is +50.8 diopters ≤ P ≤ +60.0 diopters (b)

If the child is to see very distant objects ( p → ∞ ) clearly, her eyeglass lens must form an erect virtual image at the far point of her eye ( q = −125 cm ) . The optical power of the required lens is P=

1 1 1 1 = + = 0+ = − 0.800 diopters fin meters p q −1.25 m

Since the power, and hence the focal length, of this lens is negative, it is diverging . 25.13

(a)

The lens should form an upright, virtual image at the far point ( q = − 50.0 cm ) for very distant objects ( p ≈ ∞ ) . Therefore, f = q = − 50.0 cm, and the required power is P=

1 1 = = − 2.00 diopters f − 0.500 m

continued on next page

68719_25_ch25_p282-304.indd 289

1/7/11 2:55:23 PM

290

Chapter 25

(b)

If this lens is to form an upright, virtual image at the near point of the unaided eye

( q = −13.0 cm ) , the object distance should be p=

25.14

qf ( −13.0 cm ) ( − 50.0 cm ) = 17.6 cm = q − f −13.0 cm − ( − 50.0 cm )

(a)

Yes, a single lens can correct the patient's vision . The patient needs corrective action in both the near vision (to allow clear viewing of objects between 45.0 cm and the normal near point of 25 cm) and the distant vision (to allow clear viewing of objects more than 85.0 cm away). A single lens solution is for the patient to wear a bifocal or progressive lens. Alternately, the patient must purchase two pairs of glasses, one for reading, and one for distant vision.

(b)

To correct the near vision, the lens must form an upright, virtual image at the patient’s near point (q = − 45.0 cm) when a real object is at the normal near point ( p = +25.0 cm). The thin-lens equation gives the needed focal length as f =

pq ( 25.0 cm )( −45.0 cm ) = + 56.3 cm = 25.0 cm − 45.0 cm p+q

so the required power in diopters is P= (c)

fin meters

1 = +1.78 diopters 0.563 m

1 fin meters

=

1 = −1.18 diopters − 0.850 m

Considering the image formed by the cornea as a virtual object for the implanted lens, the object distance for this lens is p = − 5.33 cm, and the image distance is q = + 2.80 cm. The thin-lens equation then gives the focal length of the implanted lens as f =

pq ( − 5.33 cm ) ( 2.80 cm ) = + 5.90 cm = p+q − 5.33 cm + 2.80 cm

so the power is P = 25.16

=

To correct the distant vision, the lens must form an upright, virtual image at the patient’s far point (q = − 85.0 cm) for the most distant objects (p → ∞). The thin-lens equation gives the needed focal length as f = q = − 85.0 cm, so the needed power is P=

25.15

1

(a)

1 1 = = +17.0 diopters . f + 0.059 0 m

The upper portion of the lens should form an upright, virtual image of very distant objects ( p ≈ ∞ ) at the far point of the eye ( q = −1.5 m ). The thin-lens equation then gives f = q = −1.5 m, so the needed power is P=

1 1 = = − 0.67 diopters f −1.5 m

continued on next page

68719_25_ch25_p282-304.indd 290

1/7/11 2:55:26 PM

Optical Instruments

(b)

The lower part of the lens should form an upright, virtual image at the near point of the eye

( q = − 30 cm ) when the object distance is p = 25 cm. From the thin-lens equation, f =

( 25 cm ) ( − 30 cm ) pq = = +1.5 × 10 2 cm = +1.5 m p+q 25 cm − 30 cm

Therefore, the power is P = 25.17

(a)

f =

1 fin meters

=

1 = − 2.50 diopters − 0.400 m

1 1 = = − 0.250 m = − 25.0 cm P − 4.00 diopters

(b)

The corrective lens forms virtual images of very distant objects (p → ∞) at q = f = −25.0 cm. Thus, the person must be very nearsighted , unable to see objects clearly when they are more than (25.0 + 2.00) cm = 27.0 cm from the eye.

(c)

If contact lenses are to be worn, the far point of the eye will be 27.0 cm in front of the lens, so the needed focal length will be f = q = −27.0 cm, and the power is P=

25.19

1 1 = = + 0.67 diopters . f +1.5 m

The corrective lens should form an upright, virtual image at the woman’s far point (q = − 40.0 cm) for a very distant object (p → ∞). The thin-lens equation gives the required focal length as f = q = − 40.0 cm = − 0.400 m. Since f < 0, it is a diverging lens , and the required power is P=

25.18

291

(a)

1 fin meters

=

1 = − 3.70 diopters − 0.270 m

The simple magnifier (a converging lens) is to form an upright, virtual image located 25 cm in front of the lens ( q = −25 cm ) . The thin-lens equation then gives p=

qf ( −25 cm )( 7.5 cm ) = = +5.8 cm q− f −25 cm − 7.5 cm

so the stamp should be placed 5.8 cm in front of the lens . (b)

When the image is at the near point of the eye, the angular magnification produced by the simple magnifier is m = mmax = 1 +

25.20

(a)

25 cm 25 cm = 1+ = 4.3 f 7.5 cm

A simple magnifier produces maximum magnification when it forms an upright, virtual image at the near point of the eye (25.0 cm for a normal eye). If the focal length of the magnifier is f = +5.00 cm, the required object distance for maximum magnification with a normal eye is p=

qf ( −25.0 cm )( +5.00 cm ) = = + 4.17 cm q− f −25.0 cm − 5.00 cm

or the object should be placed 4.17 cm in front of the lens .

continued on next page

68719_25_ch25_p282-304.indd 291

1/7/11 2:55:28 PM

292

Chapter 25

(b)

Under the conditions described in part (a), the angular magnification produced is m = mmax = 1 +

25.21

(a)

From the thin-lens equation, f =

(b)

( 3.50 cm ) ( − 25.0 cm ) pq = + 4.07 cm = 3.50 cm − 25.0 cm p+q

With the image at the normal near point, the angular magnification is m = mmax = 1 +

25.22

(a)

25 cm 25 cm = = + 5.0 f 5.0 cm

When the object is positioned so the magnifier forms a virtual image at the near point of the eye (q = −25 cm), maximum magnification is produced and this is mmax = 1 +

(c)

(a)

25 cm 25 cm = 1+ = + 6.0 f 5.0 cm

From the thin-lens equation, the object distance needed to yield the maximum magnification computed in part (b) above is p=

25.23

25.0 cm 25.0 cm = 1+ = + 7.14 f 4.07 cm

When the object is at the focal point of the magnifying lens, a virtual image is formed at infinity and parallel rays emerge from the lens. Under these conditions, the eye is most relaxed and the magnification produced is m=

(b)

25.0 cm 25.0 cm = 1+ = 6.00 f 5.00 cm

qf ( −25 cm )( 5.0 cm ) = = + 4.2 cm q− f −25 cm − 5.0 cm

From the thin-lens equation, a real inverted image is formed at an image distance of q=

pf ( 71.0 cm )( 39.0 cm ) = = + 86.5 cm p− f 71.0 cm − 39.0 cm

so the lateral magnification produced by the lens is M=

h′ q 86.5 cm =− =− = −1.22 h p 71.0 cm

and the magnitude is M = 1.22 . (b)

If h is the actual length of the leaf, the small-angle approximation gives the angular width of the leaf when viewed by the unaided eye from a distance of d = 126 cm + 71.0 cm = 197 cm as q ≈

h h = d 197 cm

continued on next page

68719_25_ch25_p282-304.indd 292

1/7/11 2:55:31 PM

Optical Instruments

293

The length of the image formed by the lens is h ′ = M h = 1.22 h , and its angular width when viewed from a distance of d ′ = 126 cm − q = 39.5 cm is q′ ≈

h′ 1.22 h = d ′ 39.5 cm

The angular magnification achieved by viewing the image instead of viewing the leaf directly is q ′ 1.22 h 39.5 cm 1.22 (197 cm ) ≈ = = 6.08 q h 197 cm 39.5 cm 25.24

(a)

When a converging lens is used as a simple magnifier, maximum magnification is obtained when the upright, virtual image is formed at the near point of the eye ( q = 25.0 cm for a normal eye). For the given lens, this maximum magnification is mmax = 1 +

(b)

When this simple magnifier is positioned for relaxed viewing (virtual image formed at infinity), the magnification produced is mrelaxed =

25.25

M1 ( 25 cm ) ( −12 ) ( 25 cm ) = = 2.1 cm m −140

The approximate overall magnification of a compound microscope is given by m = −(L fo )(25.0 cm fe ), where L is the distance between the objective and eyepiece lenses, while fo and fe are the focal lengths of the objective and eyepiece lenses, respectively. Thus, the described microscope should have an approximate overall magnification of m=−

25.27

L fo

⎛ 25.0 cm ⎞ ⎛ 20.0 cm ⎞ ⎛ 25.0 cm ⎞ = −⎜ = −588 ⎜⎝ ⎝ 0.500 cm ⎟⎠ ⎜⎝ 1.70 cm ⎟⎠ fe ⎟⎠

The angular magnification produced by a telescope is m = fo fe , where fo is the focal length of the objective element, and fe is that of the eyepiece lens. Also, the optical power of a lens is P = 1 fmeters, where fmeters is the focal length of that lens, expressed in meters. Thus, fo = 1 Po and fe = 1 Pe, so the angular magnification of this telescope is m=

25.28

25.0 cm 25.0 cm = = 1.39 f 18.0 cm

The overall magnification is m = M1me = M1 ( 25 cm fe ) , where M1 is the lateral magnification produced by the objective lens. Therefore, the required focal length for the eyepiece is fe =

25.26

25.0 cm 25.0 cm = 1+ = 2.39 f 18.0 cm

fo 1 Po Pe 35.0 diopters = = = = 12.7 fe 1 Pe Po 2.75 diopters

It is specified that the final image the microscope forms of the blood cell is 29.0 cm in front of the eye and that the diameter of this image intercepts an angle of q = 1.43 mrad . The diameter of this final image must then be

(

)(

)

he = rq = 29.0 × 10 −2 m 1.43 × 10 −3 rad = 4.15 × 10 −4 m

continued on next page

68719_25_ch25_p282-304.indd 293

1/7/11 2:55:32 PM

294

Chapter 25

At this point, it is tempting to use Equation 25.7 from the textbook for the overall magnification of a compound microscope and compute h = he m as the size of the blood cell serving as the object for the microscope. However, the derivation of that equation is based on several assumptions, one of which is that the eye is relaxed and viewing a final image located an infinite distance in front of the eyepiece. This is clearly not true in this case, and the use of Equation 25.7 would introduce considerable error. Instead, we shall return to basics and use the thin-lens equation to find the size of the original object. The image formed by the objective lens is the object for the eyepiece, and we label the size of this image as h ′. The lateral magnification of the objective lens is M1 = h ′ h = − q1 p1 and that of the eyepiece is Me = he h ′ = − qe pe. The overall magnification produced by the microscope is M total =

he ⎛ h ′ ⎞ ⎛ he ⎞ =⎜ ⎟⎜ ⎟ h ⎝ h ⎠ ⎝ h′ ⎠

which gives the size of the original object as h = he M total . From the thin-lens equation, the required object distance for the eyepiece is pe =

qe fe ( − 29.0 cm ) ( 0.950 cm ) = 0.920 cm = qe − fe − 29.0 cm − 0.950 cm

and the magnification produced by the eyepiece is Me = −

qe ( − 29.0 cm ) = + 31.5 =− pe 0.920 cm

The image distance for the objective lens is then q1 = L − pe = 29.0 cm − 0.920 cm = 28.1 cm and the object distance for this lens is p1 =

q1 fo ( 28.1 cm )(1.622 cm ) = = 1.72 cm q1 − fo 28.1 cm − 1.622 cm

The magnification by the objective lens is M1 = −

q1 ( 28.1 cm ) =− = −16.3 p1 1.72 cm

and the overall lateral magnification is M total = M1 Me = ( −16.3) ( + 31.5) = − 513. The actual diameter of the red blood cell serving as the original object is found to be h= 25.29

he 4.15 × 10 −4 m = = 8.09 × 10 −7 m = 0.809 mm 513 M total

The sketch at the right shows an image of the nebula formed by an objective lens of diameter Do and focal length fo . The diameter of the image of the nebula formed on the film is given by h ′ = fo ⋅q

Objective, diameter = Do nebula h

image q

q

h′

fo

continued on next page

68719_25_ch25_p282-304.indd 294

1/7/11 2:55:35 PM

Optical Instruments

295

where q is the angular width of the nebula and its image as shown. If the intensity of the light arriving at the objective lens from the nebula is I, the energy entering the lens during an exposure time Δt is ⎛ p Do2 ⎞ E = IAobjective ( Δt ) = I ⎜ Δt ⎝ 4 ⎟⎠ The energy per unit area deposited on the film during the exposure time is then 2

4E E 4 ⎛ p IDo2 Δt ⎞ ⎛ I ⎞ ⎛ Do ⎞ = = = =⎜ ⎟ Δt p (h ′)2 4 p ( fo ⋅q )2 pq 2 ⎜⎝ 4 fo2 ⎟⎠ ⎝ q 2 ⎠ ⎜⎝ fo ⎟⎠

E Aimage

Telescope 1 has an objective diameter DO,1 = 200 mm, focal length fO,1 = 2 000 mm, and uses an exposure time (Δt)1 = 1.50 min. If telescope 2, with DO,2 = 60.0 mm, and fO,2 = 900 mm, is to deposit the same energy per unit area on the film as does telescope 1, it is necessary that E Aimage,2

⎛ I ⎞ ⎛ DO,2 ⎞ =⎜ 2⎟⎜ ⎝ q ⎠ ⎝ fO,2 ⎟⎠

2

( Δt )2

⎛ I ⎞ ⎛ DO,1 ⎞ =⎜ 2⎟⎜ ⎝ q ⎠ ⎝ fO,1 ⎟⎠

2

( Δt )1 =

E Aimage,1

The required exposure time for the second telescope is therefore

( Δt )2 =

(I q 2 ) (DO ,1 fO ,1 )2 (I q 2 ) (DO , 2 fO , 2 )

2

⎛ DO ,1 ⎞ ⎛ fO , 2 ⎞ ⎟ ⎜ ⎟ ⎝ fO ,1 ⎠ ⎝ DO , 2 ⎠

( Δt )1 = ⎜ 2

2

( Δt )1

2

⎛ 200 mm ⎞ ⎛ 900 mm ⎞ 2 =⎜ ⎜ ⎟ (1.50 min ) = 3.38 min ⎝ 2 000 mm ⎟⎠ ⎝ 60.0 mm ⎠ 25.30

(a)

For a refracting telescope, the overall length is L = fo + fe , and the magnification produced is m = fo fe , where fo and fe are the focal lengths of the objective element and the eyepiece, respectively. Thus, we may write fe = fo m to obtain L = fo +

(b)

fo 1⎞ ⎛ m + 1⎞ ⎛ = fo ⎜ 1 + ⎟ = fo ⎜ ⎝ m ⎟⎠ ⎝ m m⎠

Using the result of part (a), the required change in the length of the telescope will be ⎛ m′ + 1 m + 1⎞ ⎛ 101 51.0 ⎞ −2 ΔL = fo ⎜ − − ⎟ = −2.00 × 10 m = −2.00 cm ⎟⎠ = ( 2.00 m ) ⎜⎝ ⎝ m′ m 100 50.0 ⎠ or the telescope must be shortened by moving the eyepiece 2.00 cm forward toward the objective lens.

25.31

(a)

The magnification of a telescope is m = fo fe , where fo and fe are the focal lengths of the objective and eyepiece lenses, respectively. Thus, if m = 34.0, while fo = 86.0 cm, the focal length of the eyepiece must be fe =

(b)

fo 86.0 cm = = 2.53 cm m 34.0

When the telescope is adjusted for relaxed-eye viewing, the distance between the objective and eyepiece lenses is L = fo + fe = 86.0 cm + 2.53 cm = 88.5 cm

68719_25_ch25_p282-304.indd 295

1/7/11 2:55:37 PM

296

25.32

Chapter 25

The Moon may be considered an infinitely distant object ( p → ∞ ) when viewed with this lens, so the image distance will be q = fo = 1 500 cm. Considering the rays that pass undeviated through the center of this lens as shown in the sketch, observe that the angular widths of the image and the object are equal. Thus, if w is the linear width of an object forming a 1.00-cm-wide image, then q =

or 25.33

(a)

w 1.0 cm 1.0 cm = = 3.8 × 108 m fo 1 500 cm

⎛ 1.0 cm ⎞ ⎛ 1 mi ⎞ w = 3.8 × 108 m ⎜ = 1.6 × 10 2 mi ⎝ 1 500 cm ⎟⎠ ⎜⎝ 1 609 m ⎟⎠

(

)

From the thin-lens equation, q = p f ( p − f ), so the lateral magnification by the objective lens is M = h ′ h = − q p = − f ( p − f ). Therefore, the image size will be h′ = M h = −

fh . p

(b)

If p >> f , then f − p ≈ − p and h ′ ≈ −

(c)

Suppose the telescope observes the space station at the zenith. Then, h′ ≈ −

25.34

fh fh = p− f f−p

fh ( 4.00 m )(108.6 m ) =− = −1.07 × 10 −3 m = −1.07 mm p 407 × 10 3 m

Use the larger focal length (lowest power) lens as the objective element and the shorter focal length (largest power) lens for the eyepiece. The focal lengths are fo = (a)

1 1 = + 0.833 m, and fe = = + 0.111 m +1.20 diopters + 9.00 diopters

The angular magnification (or magnifying power) of the telescope is then m=

(b)

fo + 0.833 m = = 7.50 fe + 0.111 m

The length of the telescope is L = fo + fe = 0.833 m + 0.111 m = 0.944 m

25.35

The lens for the left eye forms an upright, virtual image at qL = − 50.0 cm when the object distance is pL = 25.0 cm, so the thin-lens equation gives its focal length as fL =

( 25.0 cm ) ( − 50.0 cm ) pL q L = 50.0 cm = pL + qL 25.0 cm − 50.0 cm

Similarly for the other lens, qR = −100 cm when pR = 25.0 cm, and f R = 33.3 cm. continued on next page

68719_25_ch25_p282-304.indd 296

1/7/11 2:55:39 PM

Optical Instruments

(a)

Using the lens for the left eye as the objective, fo f 50.0 cm = L = = 1.50 fe f R 33.3 cm

m= (b)

297

Using the lens for the right eye as the eyepiece and, for maximum magnification, requiring that the final image be formed at the normal near point ( qe = − 25.0 cm ) gives the object distance for the eyepiece as pe =

qe fe ( − 25.0 cm ) (33.3 cm ) = +14.3 cm = qe − fe − 25.0 cm − 33.3 cm

The maximum magnification by the eyepiece is then me = 1 +

25.0 cm 25.0 cm = 1+ = +1.75 fe 33.3 cm

and the image distance for the objective is q1 = L − pe = 10.0 cm − 14.3 cm = − 4.3 cm The thin-lens equation then gives the object distance for the objective as p1 =

q1 f1 ( − 4.3 cm ) (50.0 cm ) = + 4.0 cm = q1 − f1 − 4.3 cm − 50.0 cm

The magnification by the objective is then M1 = −

q1 ( − 4.3 cm ) = +1.1 =− p1 4.0 cm

and the overall magnification is m = M1me = ( +1.1) ( +1.75) = 1.9 . 25.36

Note: We solve part (b) before answering part (a) in this problem. (b)

The objective forms a real, inverted image, diminished in size, of a very distant object at q1 = fo . This image is a virtual object for the eyepiece at pe = − fe , giving 1 1 1 1 1 = − = + =0 qe pe fe − fe fe and

(a)

Fo

qo

qo

Fo

I L1

qe → ∞

Parallel rays emerge from the eyepiece, so the eye observes a virtual image .

Fe

q Fe

O L2

continued on next page

68719_25_ch25_p282-304.indd 297

1/7/11 2:55:42 PM

298

Chapter 25

(c)

The angular magnification is m =

fo = 3.00, giving fo = 3.00 fe . Also, the length of the fe

telescope is L = fo + fe = 3.00 fe − fe = 10.0 cm, giving fe = − fe = − 25.37

The angular separation of the lights is q = d h , where d = 1.00 m is their linear separation, and h is the altitude of the satellite. If the lights are just resolved according to the Rayleigh criterion, then q = q min = 1.22 (l D) , where l is the wavelength of the light, and D is the diameter of the lens. Thus, the altitude of the satellite must be h=

25.38

d d d⋅D (1.00 m )( 0.300 m ) = = = = 4.92 × 10 5 m = 492 km q 1.22(l D) 1.22l 1.22 500 × 10 −9 m

(

)

The angular separation of two objects seen on the ground is q = d h , where d is their linear separation and h is your altitude. If the objects are just resolved according to the Rayleigh criterion, then q = q min = 1.22 (l D) , where l is the wavelength of the light, and D is the diameter of the aperture (your pupil). Thus, the minimum separation of objects you can distinguish is d = h ⋅q min =

25.39

10.0 cm = − 5.00 cm and fo = 3.00 fe = 15.0 cm 2.00

(

)(

)

3 −9 1.22 ⋅ h ⋅ l 1.22 9.50 × 10 m 575 × 10 m = = 1.7 m D 4.0 × 10 −3 m

The limit of resolution in air is q m in

air

= 1.22

l = 0.60 mrad. D

In oil, the limiting angle of resolution will be q min oil = 1.22

or 25.40

(a)

q min oil =

( l noil ) = ⎛ 1.22 l ⎞ 1 loil = 1.22 ⎜⎝ ⎟ D D D ⎠ noil

q min air noil

(

)

500 × 10 −9 m ln l = 1.22 = 1.22 = 2.29 × 10 −4 rad D nD (1.33) 2.00 × 10 −3 m

(

)

From s = rq , the distance from the eye that two points separated by a distance s = 1.00 cm will intercept this minimum angle of resolution is r=

25.41

0.60 mrad = 0.40 mrad 1.5

The wavelength of the light within the eye is ln = l n. Thus, the limiting angle of resolution for light passing through the pupil (a circular aperture with diameter D = 2.00 mm) is q min = 1.22

(b)

=

s 1.00 cm = = 4.37 × 10 3 cm = 43.7 m q min 2.29 × 10 − 4 rad

The angular separation of the headlights when viewed from a distance of r = 10.0 km is q =

s 2.00 m = = 2.00 × 10 −4 rad r 10.0 × 10 3 m

continued on next page

68719_25_ch25_p282-304.indd 298

1/7/11 2:55:44 PM

Optical Instruments

299

If the headlights are to be just resolved, this separation must equal the limiting angle of resolution for the circular aperture, q min = 1.22l D, so the diameter of the aperture is D= 25.42

(

The angular separation of the two stars is q = d r , where d is their linear separation and r is their distance from Earth. If the stars are just resolved according to the Rayleigh criterion, then q = q min = 1.22 (l D) , where l is the wavelength of the light, and D is the diameter of the aperture (telescope objective). Thus, the linear separation of the stars must be d = r ⋅q min =

25.43

)

−9 1.22l 1.22l 1.22 885 × 10 m = = = 5.40 × 10 −3 m = 5.40 mm q min q 2.00 × 10 −4 rad

(

)

(

)

15 −9 r ⋅1.22l ⎡⎣( 23 ly ) 9.461 × 10 m ly ⎤⎦ 1.22 575 × 10 m = 2.2 × 1011 m = 0.68 m D

If just resolved, the angular separation of the objects is q = q min = 1.22

l D

⎡ ⎛ 500 × 10 −9 m ⎞ ⎤ and s = r q = 8.0 × 10 7 km ⎢1.22 ⎜ ⎥ = 9.8 km ⎝ 5.00 m ⎟⎠ ⎦ ⎣

(

25.44

)

If just resolved, the angular separation of the objects is q = q min = 1.22

λ D

⎡ ⎛ 550 × 10 −9 m ⎞ ⎤ and s = r q = 200 × 10 3 m ⎢1.22 ⎜ ⎥ = 0.38 m = 38 cm ⎝ 0.35 m ⎟⎠ ⎦ ⎣

(

25.45

)

The grating spacing is d = 1 cm 6 000 = 1.67 × 10 −4 cm = 1.67 × 10 −6 m, and the highest order of 600 nm light that can be observed is mmax

(

)

1.67 × 10 −6 m (1) d sin 90° = = = 2.78 → 2 orders l 600 × 10 −9 m

The total number of slits is N = (15.0 cm ) ( 6 000 slits cm ) = 9.00 × 10 4 , and the resolving power of the grating in the second order is

(

)

Ravailable = Nm = 9.00 × 10 4 2 = 1.80 × 10 5 The resolving power required to separate the given spectral lines is Rneeded =

l 600.000 nm = = 2.0 × 10 5 Δl 0.003 nm

These lines cannot be separated with this grating. 25.46

The resolving power of a diffraction grating is R = l Δl = Nm. (a)

The number of lines the grating must have to resolve the Ha line in the first order is N=

(b)

68719_25_ch25_p282-304.indd 299

R l Δl 656.2 nm = = = 3.6 × 10 3 lines m 0.18 nm (1)

In the second order ( m = 2 ) , N =

R 656.2 nm = = 1.8 × 10 3 lines . 2 2 ( 0.18 nm )

1/7/11 2:55:46 PM

300

25.47

Chapter 25

A fringe shift occurs when the mirror moves distance l 4 . Thus, if the mirror moves distance ΔL = 0.180 mm, the number of fringe shifts observed is N shifts

25.48

(a)

(

)

−3 ΔL 4 ( ΔL ) 4 0.180 × 10 m = = = = 1.31 × 10 3 l 4 l 550 × 10 −9 m

When the central spot in the interferometer pattern goes through a full cycle from bright to dark and back to bright, two fringe shifts have occurred, and the movable mirror has moved a distance of 2(l 4) = l 2. Thus, if N cycles = 1 700 such cycles are observed as the mirror moves distance d = 0.382 mm, it must be true that d = N cycles ( l 2 ) , or l = 2d N cycles. The wavelength of the light illuminating the interferometer is therefore l=

(

2 0.382 × 10 −3 m 1 700

) = 4.49 × 10

−7

m = 449 nm

which is in the blue region of the visible spectrum. (b)

25.49

Red light has a longer wavelength than blue light, so fewer wavelengths would cover the given displacement. Hence, N cycles would be smaller .

A fringe shift occurs when the mirror moves distance l 4 . Thus, the distance moved (length of the bacterium) as 310 shifts occur is ⎛ 650 × 10 −9 m ⎞ ⎛l⎞ −5 ΔL = N shifts ⎜ ⎟ = 310 ⎜ ⎟⎠ = 5.04 × 10 m = 50.4 mm ⎝ 4⎠ 4 ⎝

25.50

A fringe shift occurs when the mirror moves distance l 4 . Thus, the distance the mirror moves as 250 fringe shifts are counted is ⎛ 632.8 × 10 −9 m ⎞ ⎛l⎞ −5 ΔL = N shifts ⎜ ⎟ = 250 ⎜ ⎟⎠ = 3.96 × 10 m = 39.6 mm ⎝ 4⎠ 4 ⎝

25.51

When the optical path in one arm of a Michelson’s interferometer increases by one wavelength, four fringe shifts will occur (one shift for every quarter-wavelength change in path length). The number of wavelengths (in a vacuum) that fit in a distance equal to a thickness t is N vac = t l . The number of wavelengths that fit in this thickness while traveling through the transparent material is N n = t ln = t ( l n ) = nt l . Thus, the change in the number of wavelengths that fit in the path down this arm of the interferometer is ΔN = N n − N vac = ( n − 1)

t l

and the number of fringe shifts that will occur as the thin sheet is inserted will be # fringe shifts = 4 ( ΔN ) = 4 ( n − 1)

68719_25_ch25_p282-304.indd 300

⎛ 15.0 × 10 −6 m ⎞ t = 4 (1.40 − 1) ⎜ = 40 l ⎝ 600 × 10 −9 m ⎟⎠

1/7/11 2:55:49 PM

Optical Instruments

25.52

301

The wavelength of light within the tube decreases from l to ln = l n gas as the tube fills with gas. Thus, the number of wavelengths that will fit in the length L of the tube increases from L l to ngas L l . Four fringe shifts occur for each additional wavelength that fits within the tube, so the number of fringes shifts to be seen as the tube fills with gas is ⎡ ngas L L ⎤ 4L N shifts = 4 ⎢ − ⎥= ngas − 1 l⎦ l ⎣ l

(

⎡ 600 × 10 −9 m ⎛ l ⎞ ⎢ ngas = 1 + ⎜ = 1 + N −2 ⎝ 4L ⎟⎠ shifts ⎢⎣ 4 5.00 × 10 m

Hence, 25.53

(a)

(

25.54

)

⎤ ⎥ (160 ) = 1.000 5 ⎥⎦

For a refracting telescope, the magnification is m = fo fe , where fo and fe are the focal lengths of the objective lens and the eyepiece, respectively. Thus, when the Yerkes telescope uses an eyepiece with fe = 2.50 cm, the magnification is m=

(b)

)

fo 20.0 m = 8.00 × 10 2 = 800 = fe 2.50 × 10 −2 m

Standard astronomical telescopes form inverted images. Thus, the observer Martian polar caps are upside down .

When viewed from a distance of 50 meters, the angular length of a mouse (assumed to have an actual length of ≈ 10 cm) is q =

s 0.10 m = = 2.0 × 10 −3 radians r 50 m

Thus, the limiting angle of resolution of the eye of the hawk must be q min ≈ 2.0 × 10 −3 radians 25.55

With 485 lines equally spaced in a height , the distance separating adjacent lines is d = 485. When the screen is viewed from a distance L, the angular separation between adjacent lines is q = d L . If the individual lines are not to be seen (i.e., the lines are to be unresolved), this angular separation must be less than the minimum angle of resolution, q min = 1.22(l D) by the Rayleigh criterion. That is, we must have

or 25.56

(a)

485 1.22 ⋅ l < q min = L D

q =

d = L

L

D 5.00 × 10 −3 m = = 15.4 485 (1.22 ⋅ l ) 485 (1.22 ) 550 × 10 −9 m

>

(

)

Since this eye can already focus on objects located at the near point of a normal eye (25 cm), no correction is needed for near objects. To correct the distant vision, a corrective lens (located 2.0 cm from the eye) should form virtual images of very distant objects at 23 cm in front of the lens (or at the far point of the eye). Thus, we must require that q = −23 cm when p → ∞. This gives P=

1 1 1 1 = + = 0+ = − 4.3 diopters f p q − 0.23 m

continued on next page

68719_25_ch25_p282-304.indd 301

1/7/11 2:55:51 PM

302

Chapter 25

(b)

A corrective lens in contact with the cornea should form virtual images of very distant objects at the far point of the eye. Therefore, we require that q = −25 cm when p → ∞, giving P=

1 1 1 1 = + = 0+ = − 4.0 diopters f p q − 0.25 m

1 ⎛ ⎞ When the contact lens ⎜ f = = − 25 cm ⎟ is in place, the object distance which yields a ⎝ ⎠ P virtual image at the near point of the eye (that is, q = −16 cm) is given by p= 25.57

(a)

qf ( −16 cm )( −25 cm ) = = 44 cm q − f −16 cm − ( −25 cm )

The lens should form an upright, virtual image at the near point of the eye, q = −75.0 cm, when the object distance is p = 25.0 cm. The thin-lens equation then gives f =

pq ( 25.0 cm )( −75.0 cm ) = = 37.5 cm = 0.375 m p+q 25.0 cm − 75.0 cm

so the needed power is P = (b)

1 1 = = + 2.67 diopters . f 0.375 m

If the object distance must be p = 26.0 cm to position the image at q = −75.0 cm, the actual focal length is f =

and P =

pq ( 26.0 cm )( −75.0 cm ) = = 0.398 m p+q 26.0 cm − 75.0 cm 1 1 = = + 2.51 diopters f 0.398 m

The error in the power is ΔP = ( 2.67 − 2.51) diopters = 0.16 diopters too low 25.58

(a)

The image must be formed on the back of the eye (retina), so we must have q = 2.00 cm when p = 1.00 m = 100 cm. The thin-lens equation gives the required focal length as f =

(b)

pq (100 cm )( 2.00 cm ) = = 1.96 cm p + q 100 cm + 2.00 cm

The f-number of a lens aperture is the focal length of the lens divided by the diameter of the aperture. Thus, the smallest f-number occurs with the largest diameter of the aperture. For the typical eyeball focused on objects 1.00 m away, this is

( f -number )min = (c)

Dmax

=

1.96 cm = 3.27 0.600 cm

The largest f-number of the typical eyeball focused on a 1.00-m-distance object is

( f -number )max =

68719_25_ch25_p282-304.indd 302

f

f 1.96 cm = = 9.80 Dmin 0.200 cm

1/7/11 2:55:54 PM

Optical Instruments

25.59

(a)

The implanted lens should give an image distance of q = 22.4 mm for distant ( p → ∞ ) objects. The thin-lens equation then gives the focal length as f = q = 22.4 mm, so the power of the implanted lens should be Pimplant =

(b)

1 1 = = + 44.6 diopters f 22.4 × 10 −3 m

When the object distance is p = 33.0 cm, the corrective lens should produce parallel rays ( q → ∞ ) . Then the implanted lens will focus the final image on the retina. From the thinlens equation, the required focal length is f = p = 33.0 cm, and the power of this lens should be Pcorrective =

25.60

303

1 1 = = + 3.03 diopters f 0.330 m

We use n1 + n2 = n2 − n1 , with p → ∞ and q equal to the cornea to retina distance. p q R ⎛ n − n1 ⎞ ⎛ 1.34 − 1.00 ⎞ Then, R = q ⎜ 2 = ( 2.00 cm ) ⎜ = 0.507 cm = 5.07 mm . ⎟ ⎝ 1.34 ⎟⎠ ⎝ n2 ⎠

25.61

When a converging lens forms a real image of a very distant object, the image distance equals the focal length of the lens. Thus, if the scout started a fire by focusing sunlight on kindling 5.00 cm from the lens, f = q = 5.00 cm. (a)

When the lens is used as a simple magnifier, maximum magnification is produced when the upright, virtual image is formed at the near point of the eye (q = −15 cm in this case). The object distance required to form an image at this location is p=

qf ( −15 cm )( 5.0 cm ) 15 cm = = q− f −15 cm − 5.0 cm 4.0

q −15 cm =− = + 4.0 . Note p 15 cm 4.0 that adapting Equation 25.5 for use with this “abnormal” eye would give an angular magnification of mmax = 1 + q f = 1 + 15 cm 5.0 cm = + 4.0 . and the lateral magnification produced is M = −

(b)

When the object is viewed directly while positioned at the near point of the eye, its angular size is q 0 = h 15 cm . When the object is viewed by the relaxed eye while using the lens as a simple magnifier (with the object at the focal point so parallel rays enter the eye), the angular size of the upright, virtual image is q = h f . Thus, the angular magnification gained by using the lens in this manner is m=

25.62

q h f 15 cm 15 cm = = = = + 3.0 q 0 h 15 cm f 5.0 cm

The angular magnification is m = q q 0 , where q is the angle subtended by the final image, and q 0 is the angle subtended by the object as shown in the figure below. When the telescope is adjusted for minimum eyestrain, the rays entering the eye are parallel. Thus, the objective lens must form its image at the focal point of the eyepiece.

continued on next page

68719_25_ch25_p282-304.indd 303

1/7/11 2:55:56 PM

304

Chapter 25

From triangle ABC, q 0 ≈ tanq 0 = h ′ q1, and from triangle DEF, q ≈ tanq = h ′ fe . The angular h ′ fe q1 q magnification is then m = = = . q 0 h ′ q1 fe From the thin-lens equation, the image distance of the objective lens in this case is q1 =

p1 f1 ( 300 cm )( 20.0 cm ) = = 21.4 cm p1 − f1 300 cm − 20.0 cm

With an eyepiece of focal length fe = 2.00 cm, the angular magnification for this telescope is m=

68719_25_ch25_p282-304.indd 304

q1 21.4 cm = = 10.7 fe 2.00 cm

1/7/11 2:55:59 PM

26 Relativity QUICK QUIZZES 1.

False. One of Einstein’s two basic postulates in his special theory of relativity is the constancy of the speed of light. This postulate states that the speed of light in a vacuum has the same value in all inertial reference frames, regardless of the velocity of the observer or the velocity of the source emitting the light.

2.

Choice (a). Less time will have passed for you in your frame of reference than for your employer back on Earth. Thus, to maximize your paycheck, you should choose to have your pay calculated according to the elapsed time on a clock on Earth.

3.

False. Clocks, including biological clocks, in the observer’s own reference frame appear to run at normal speeds. It is only clocks in reference frames moving relative to the observer that the observer will judge to be running slower than normal.

4.

No. From your perspective you are at rest with respect to the cabin, so you will measure yourself as having your normal length and will require a normal-sized cabin.

5.

(i) Choices (a) and (e). The outgoing rocket will appear to have a shorter length and a slower clock. (ii) Choices (a) and (e). The answers for the incoming rocket are the same as for the outgoing rocket. Length contraction and time dilation depend only on the magnitude of the relative velocity, not on the direction.

6.

False. According to the mass-energy equivalence equation, the particle’s total energy is E = mc 2 1− v 2 c 2 , and it is seen that as v → c, the total energy becomes infinitely large. As the total energy becomes infinitely large, so will the kinetic energy given by KE = E − mc 2 and the relativistic momentum given by p = E 2 − (mc 2 )2 c.

7.

(a) False

(b) False

(c) True

(d) False

A reflected photon does exert a force on the surface. Although a photon has zero mass, a photon does carry momentum. When it reflects from a surface, there is a change in the momentum, just like the change in momentum of a ball bouncing off a wall. According to the momentum interpretation of Newton’s second law, a change in momentum results in a force on the surface. This concept is used in theoretical studies of space sailing. These studies propose building non-powered spacecraft with huge reflective sails oriented perpendicularly to the rays from the Sun. The large number of photons from the Sun reflecting from the surface of the sail will exert a force which, although small, will provide a continuous acceleration. This would allow the spacecraft to travel to other planets without fuel.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

Einstein based his special theory of relativity on the two postulates included in choices (d) and (e). The textbook refers to the postulate summarized in choice (d) as the principle of relativity and to the postulate in choice (e) as the constancy of the speed of light. 305

68719_26_ch26_p305-323.indd 305

1/7/11 2:56:39 PM

306

Chapter 26

2.

The length of the ship is parallel to the relative velocity between the ship and the observer. Thus, the observer at rest on Earth will measure the moving length of the ship to be shorter than its length when it was at rest. Choice (b) is the correct answer.

3.

According to the second postulate of special relativity (the constancy of the speed of light), both observers will measure the light speed to be c. Therefore, both choices (b) and (c) are true statements, and the other choices are false.

4.

The astronaut is moving with constant velocity and is therefore in an inertial reference frame. According to the principle of relativity, all the laws of physics are the same in her reference frame as in any other inertial reference frame. Thus, she should experience no effects due to her motion through space, and choice (e) is the correct answer.

5.

The astronaut in the spaceship is at rest relative to the pendulum and measures the proper time required for it to complete one oscillation (its period). The observer on Earth is in motion relative to the pendulum and measures the time required for a full oscillation to be longer than does the astronaut (i.e., T > P), and the correct choice is (c).

6.

If L p is the proper length of each edge of the cube, the volume measured by the observer at rest relative to it is V = L3p . The observer moving relative to the cube sees the two dimensions that are perpendicular to his velocity as unchanged, with each having value L p . The dimension parallel to his velocity is measured to have the contracted length L = L p g . Thus, the observer moving relative to the cube measures its volume to be V ′ = L p ⋅ L p ⋅ L = L2p ⋅ L p g = V 5, so the correct answer is (c).

7.

The relativistic momentum of a particle is p = E 2 − ER2 c, where E is the total energy of the particle and ER is its rest energy. In this case, each particle has the same total energy, E. Thus, the particle with the smallest rest energy has the greatest momentum, while the particle with the greatest rest energy has the smallest momentum. This means that the correct ordering of the particles, from smallest to greatest momentum, is proton, electron, photon, and (e) is the correct choice.

8.

The time interval between events appears to be dilated (or lengthened) to observers who are in motion relative to those events. Thus, the intervals between successive “ticks” of the clock in orbit will seem longer to observers on Earth than they do to an astronaut in orbit with the clock and at rest relative to it. This means the Earth-based observers detect the orbiting clock to run slower than the identical clock that was left at rest on Earth. The correct answer to this question is choice (c).

9.

When the ground observer is moving at speed v = 0.5c relative to the oscillating system, the time he will measure for the period is T ′ = g T , where g =

1 1− (v c)

2

=

1 1− (0.5)2

= 1.15

and the correct choice is (d). 10.

Choice (d) is the correct response. Each observer measures the intervals between “ticks” of the clock in motion relative to him to be lengthened by the same factor g = 1 1− v 2 c 2 , where v is the relative velocity between the two observers.

11.

According to the second postulate of special relativity, the speed of light is the same in all inertial reference frames, regardless of the velocity of the observer or the velocity of the source emitting the light. Thus, the speed of light from the rapidly moving quasar should be measured to be c, the same as the speed of light emitted by a stationary source. This means that (b) is the correct answer.

68719_26_ch26_p305-323.indd 306

1/7/11 2:56:40 PM

Relativity

307

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The two observers will agree on the speed of light and on the speed at which they move relative to one another.

4.

Special relativity describes inertial reference frames—that is, reference frames that are not accelerating. General relativity describes all reference frames.

6.

You would see the same thing that you see when looking at a mirror when at rest. The theory of relativity tells us that all experiments will give the same results in all inertial frames of reference.

8.

(a)

Attempts to apply special relativity to this situation would lead one to say that each observer (one on Earth and one in orbit) would find the other’s clock to run slow relative to their own clock. However, the observer in orbit is not in an inertial reference frame, and special relativity does not apply here. According to general relativity, the clock in orbit and undergoing an acceleration (the centripetal acceleration) will run slower.

(b)

When the moving clock returns to Earth, they are again in inertial reference frames and subject to the same laws of physics. Thus, the identical clocks should tick at the same rate once again. However, they will not be synchronized. The clock that underwent the accelerations (i.e., the orbiting clock) will have permanently lost time (or will have “aged less”) and will be behind the clock that remained on Earth.

10.

The 8 light-years represents the proper length of a rod from Earth to Sirius, measured by an observer seeing both the rod and Sirius nearly at rest. The astronaut sees Sirius coming toward her at 0.8c but also sees the distance contracted to d = (8 ly ) 1− ( v c ) = (8 ly ) 1− ( 0.8 ) = 5 ly 2

2

So the travel time measured on her clock is t =

d 5 ly ( 5 yr ) c = = = 6 yr. v 0.8c 0.8c

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a) 0.436 m

4.

(a)

6.

(a) 65.0 beats min

8.

5.0 s

70 beats min

(b)

less than 0.436 m by an undetectable amount

(b)

30 beats min

(b)

10.6 beats min

1.2 m, 2.0 m, 2.0 m

10.

26.0 min + 1.06 × 10 −11 s

12.

(a) rectangular box

(b)

14.

(a) 4.96 × 10 −19 kg ⋅ m s

(b) 3.52 × 10 −18 kg ⋅ m s

(c) No. Neglecting relativity in this case introduces an 86% error in the result. 16.

68719_26_ch26_p305-323.indd 307

0.94c toward the left

1/7/11 2:56:42 PM

308

Chapter 26

18.

+ 0.696c

20.

0.161 Hz

22.

(a) 939 MeV

24.

(a)

26.

(a) 0.582 MeV

28.

2.51× 10 −28 kg with speed 0.987c, and 8.84 × 10 −28 kg with speed 0.868c

30.

(a)

KE ≈ g ER = g m p c 2

KEi + mi c 2 = KE f + m f c 2

(b)

3.01 GeV

(b)

0.999 99c

(b)

2.45 MeV

(c)

2.07 GeV

(b)

236.052 588 u

(d) 0.190 976 u

(e)

177.893 MeV

32.

(a) 0.023 6c = 7.08 × 10 3 km s

(b)

(6.17 × 10 ) c = 185 km s

34.

(a) Yes. As the spring is compressed, positive work is done on it or energy is added to it. Since mass and energy are equivalent, mass has been added to the spring.

−4

(b)

Δm = kx 2 2c 2

(c) 2.5 × 10 −17 kg

36.

(a)

2.50 MeV c

(b)

4.60 GeV c

38.

1.2 × 1013 m

40.

(a) t =

(b)

t′ =

42.

(a) ∼ 108 km

(b)

∼ 10 2 s

44.

(a) See Solution.

(b)

0.943c

46.

Rg = 1.47 km for the Sun

48.

(a) 21.0 yr

(b)

14.7 ly

(c) 10.5 ly

(d)

35.7 yr

0.946c

(b)

0.160 ly

(d)

7.51× 10 22 J

50.

(a)

2d c+v

(c) 0.114 yr

(c) 235.861 612 u

2d c

c−v c+v

PROBLEM SOLUTIONS 26.1

(a)

Observers on Earth measure the time for the astronauts to reach Alpha Centauri as Δt E = 4.42 yr. But these observers are moving relative to the astronaut’s internal biological clock and hence experience a dilated version of the proper time interval Δt p measured on that clock. From Δt E = g Δt p , we find Δt p = Δt E g = Δt E 1− ( v c ) = ( 4.42 yr ) 1− ( 0.950 ) = 1.38 yr 2

2

continued on next page

68719_26_ch26_p305-323.indd 308

1/7/11 2:56:43 PM

Relativity

(b)

309

The astronauts are moving relative to the span of space separating Earth and Alpha Centauri. Hence, they measure a length-contracted version of the proper distance, L p = 4.20 ly. The distance measured by the astronauts is L = L p g = L p 1− ( v c ) = ( 4.20 ly ) 1− ( 0.950 ) = 1.31 ly 2

26.2

(a)

2

The length of the meterstick measured by the observer moving at speed v = 0.900 c relative to the meterstick is L = L p g = L p 1− ( v c ) = (1.00 m ) 1− ( 0.900 ) = 0.436 m 2

(b)

26.3

2

If the observer moves relative to Earth in the direction toward the meterstick, the velocity of the observer relative to the meterstick is greater than that in part (a). The measured length of the meterstick will be less than 0.436 m under these conditions, but at the speed the observer could run, the effect would be too small to detect.

The contracted length of the ship, L = L p − ΔL, as measured by the Earth-based observer is 2 2 L = L p − ΔL = L p 1− ( v c ) . This yields 1− ( v c ) = 1− ΔL L p , and solving for the speed v of

(

)

(

2

)

the ship, we find v = c 1− 1− ΔL L p . Thus, if the proper length of the ship is L p = 28.0 m, and the observed contraction is ΔL = 0.150 m, the speed of the ship must be 2

0.150 m ⎞ ⎛ v = c 1− ⎜ 1− ⎟ = 0.103c ⎝ 28.0 m ⎠ 26.4

(a)

The time for 70 beats, as measured by the astronaut and any observer at rest with respect to the astronaut, is Δt p = 1.0 min. The observer in the ship then measures a rate of 70 beats min .

(b)

The observer on Earth moves at v = 0.90 c relative to the astronaut and measures the time for 70 beats as Δt = g Δt p =

Δt p 1− ( v c )

2

=

1.0 min 1− ( 0.90 )

2

= 2.3 min

This observer then measures a beat rate of 70 beats 2.3 min = 30 beats min . 26.5

26.6

Δt p

Δt = g Δt p =

(b)

d = v ( Δt ) = ⎡⎣ 0.98 ( 3.0 × 108 m s ) ⎤⎦(1.3 × 10 −7 s ) = 38 m

(c)

d ′ = v Δt p = ⎡⎣ 0.98 ( 3.0 × 108 m s ) ⎤⎦( 2.6 × 10 −8 s ) = 7.6 m

(a)

As measured by observers in the ship (that is, at rest relative to the astronaut), the time required for 75.0 beats is Δt p = 1.00 min.

1− ( v c )

2

=

2.6 × 10 −8 s

(a)

1− ( 0.98 )

2

= 1.3 × 10 −7 s

( )

The time interval required for 75.0 beats as measured by the Earth observer is Δt = g Δt p = (1.00 min) 1 − (0.500)2 , so the Earth observer measures a pulse rate of 75.0 75.0 1− ( 0.500 ) = 65.0 min = 1.00 min Δt 2

rate =

continued on next page

68719_26_ch26_p305-323.indd 309

1/7/11 2:56:46 PM

310

Chapter 26

(b)

1.00 min

If v = 0.990 c, then Δt = g Δt p =

1− ( 0.990 )

2

and the pulse rate observed on Earth is 75.0 75.0 1− ( 0.990 ) = 10.6 min = 1.00 min Δt 2

rate =

That is, the life span of the astronaut (reckoned by the total number of his heartbeats) is much longer as measured by an Earth clock than by a clock aboard the space vehicle. 26.7

(a)

To the observer on Earth, the muon appears to have a lifetime of Δt =

d 4.60 × 10 3 m = = 1.55 × 10 −5 s v 0.990 ( 3.00 × 108 m s ) 1

1

(b)

g =

(c)

To an observer at rest with respect to the muon, its proper lifetime is

1− ( v c )

Δt p = (d)

2

=

= 7.09

1− ( 0.990 )

2

Δt 1.55 × 10 −5 s = = 2.19 × 10 −6 s g 7.09

The muon is at rest relative to the observer traveling with the muon. Thus, the muon travels zero distance as measured by this observer. However, during the observed lifetime of the muon, this observer sees Earth move toward the muon a distance of

( )

d ′ = v Δt p = ⎡⎣ 0.990 ( 3.00 × 108 m s ) ⎤⎦ ( 2.19 × 10 − 6 s ) = 6.50 × 10 2 m = 650 m (e)

As the third observer travels toward the incoming muon, his speed relative to the muon is greater than that of the observer at rest on Earth. Thus, his observed gamma factor (Δt = g Δt p) is higher, and he measures the muon’s lifetime as longer than that measured by the observer at rest with respect to Earth. Δt p

3.0 s

26.8

Δt = g Δt p =

26.9

The proper length of the faster ship is three times that of the slower ship L pf = 3L ps , yet they both appear to have the same contracted length, L. Thus,

1− ( v c )

=

2

= 5.0 s

1− ( 0.80 )

2

(

(

L = L ps 1− ( vs c ) = 3L ps 2

)

(

)

(

1− v f c , or 1− ( vs c ) = 9 − 9 v f c 2

2

)

)

2

This gives vf = 26.10

c 8 + ( vs c ) 3

2

8 + ( 0.350 ) c = 0.950 c 3 2

=

The driver is the observer at rest with respect to the clock measuring the 26.0-min time interval. Thus, this observer measures the proper time Δt p , and the Earth-based observer measures − the dilated time Δt = g Δt p = [1− (v c)2 ] ⋅ Δt p. In this case, v = 35.0 m s c, so we use the 1 2

continued on next page

68719_26_ch26_p305-323.indd 310

1/7/11 2:56:48 PM

Relativity

−1

binominal expansion [1− x] fixed on Earth as

2

≈ 1+ 12 x when x

2 Δt = ( 26.0 min ) ⎡⎣1− ( v c ) ⎤⎦

or Δt ≈ 26.0 min + 26.11

( 26.0

− 12

311

1. This gives the time measured by the observer

1 2⎞ ⎛ ≈ ( 26.0 min ) ⎜ 1+ ( v c ) ⎟ ⎝ ⎠ 2

)

2

min ⎛ 35.0 m s ⎞ ⎛ 60.0 s ⎞ −11 ⎜⎝ 3.00 × 108 m s ⎟⎠ ⎜⎝ 1 min ⎟⎠ = 26.0 min + 1.06 × 10 s . 2

The trackside observer sees the supertrain length-contracted as L = L p 1− ( v c ) = (100 m ) 1− ( 0.95) = 31 m 2

2

The supertrain appears to fit in the tunnel with 50 m − 31 m = 19 m to spare . 26.12

Length contraction occurs only in the dimension parallel to the motion. (a)

The sides labeled L2 and L3 in the figure at the right are unaffected, but the side labeled L1 will appear contracted, giving the box a rectangular shape, or more formally, the shape of a rectangular parallelepiped.

(b)

The dimensions of the box, as measured by the observer moving at v = 0.80 c relative to it, are L2 = L2 p = 2.0 m , L3 = L3 p = 2.0 m , and L1 = L1 p 1− ( v c ) = ( 2.0 m ) 1− ( 0.80 ) = 1.2 m 2

26.13

(a)

The classical expression for linear momentum is pclassical = mv, while the relativistic expression is p = g mv, where g = 1 1− (v c)2 . Thus, if p = 3pclassical, it is necessary that the gamma factor have a value g = 3. Solving for the speed of the particle gives v = c 1−

26.14

2

1 1 c 8 2c 2 = c 1− = = = 0.943c 2 g 9 3 3

(b)

Observe that the calculation above did not depend on the mass of the particle involved. Thus, the result is the same for a proton or any other particle.

(a)

Classically, p = mv = m ( 0.990c ) = (1.67 × 10 −27 kg )( 0.990 )( 3.00 × 108 m s ) = 4.96 × 10 −19 kg ⋅ m s

(b)

By relativistic calculations, p=

mv 1− (v c)2

=

m ( 0.990c ) 1− (0.990)2

=

(1.67 × 10

−27

kg )( 0.990 )( 3.00 × 108 m s ) 1− (0.990)2

= 3.52 × 10 −18 kg ⋅ m s (c)

68719_26_ch26_p305-323.indd 311

No , neglecting relativistic effects at such speeds would introduce an approximate 86% error in the result.

1/7/11 2:56:51 PM

312

26.15

Chapter 26

Momentum must be conserved, so the momenta of the two fragments must add to zero. Thus, their magnitudes must be equal, or g 2 m2 v2 = g 1 m1 v1 . This gives

(1.67 × 10

−27

kg ) v

1− ( v c )

2

=

( 2.50 × 10

−28

kg ) ( 0.893c )

1− ( 0.893)

2

⎡ (1.67 × 10 −27 kg ) 1− ( 0.893)2 and reduces to ⎢ ⎢ ( 2.50 × 10 −28 kg )( 0.893) ⎣

2 ⎤ v ⎥ ⎛⎜ ⎞⎟ = 1− ⎛⎜ v ⎞⎟ , or ⎝ c⎠ ⎥⎝ c ⎠ ⎦

12.3 ( v c ) = 1, and yields v = 0.285c . 2

26.16

We take to the right as the positive direction. Then, the velocities of the two ships relative to Earth are vR E = + 0.70c and vLE = − 0.70c. The velocity of ship L relative to ship R is given by the relativistic relative velocity relation (Equation 26.7 in the textbook) as vLR =

26.17

Taking to the right as the positive direction, the velocity of the electron relative to the laboratory is vEL = + 0.90c, and the velocity of the proton relative to the electron is vPE = − 0.70c. Thus, the relativistic addition of velocities (Equation 26.8 in the textbook) gives the velocity of the proton relative to the laboratory as vPL =

26.18

vLE − vR E −1.40c ( −0.70c ) − 0.70c = = = − 0.94c = 0.94c toward the left vLE vR E −0.70c )( 0.70c ) 1+ 0.49 ( 1− 1− c2 c2

vPE + vEL ( − 0.70c ) + 0.90c = + 0.20c = + 0.54c = 0.54c toward the right = vPE vEL ( −0.70c )( 0.90c ) 1− 0.63 1+ 1+ c2 c2

We choose the direction of the spaceship’s motion relative to Earth as the positive direction. Then, the spaceship’s velocity relative to Earth is vSE = + 0.750c. It is desired to have the velocity of the rocket relative to Earth be vR E = + 0.950c. The relativistic relative velocity relation (Equation 26.7 in the textbook) then gives the required velocity of the rocket relative to the ship as vR S =

26.19

Taking away from Earth as the positive direction, the velocity of ship A relative to Earth is vAE = + 0.800c, and the velocity of ship B relative to Earth is vB E = + 0.900c. The relativistic relative velocity relation (Equation 26.7 in the textbook) gives the velocity of ship B relative to ship A (and hence, the speed with which B is overtaking A) as vB A =

26.20

vR E − vSE 0.950c − 0.750c = = + 0.696c vR E vSE ( 0.950c )( 0.750c ) 1− 1 − c2 c2

vB E − vAE 0.900c − 0.800c = = + 0.357c vB E vAE ( 0.900c )( 0.800c ) 1− 1− 2 2 c c

We first determine the velocity of the pulsar relative to the rocket. Taking toward the Earth as the positive direction, the velocity of the pulsar relative to Earth is vPE = + 0.950c, and the velocity of the rocket relative to Earth is vR E = − 0.995c. The relativistic relative velocity relation (Equation 26.7 in the textbook) gives the velocity of the pulsar relative to the rocket as

continued on next page

68719_26_ch26_p305-323.indd 312

1/7/11 2:56:53 PM

Relativity

vPR =

313

vPE − vR E 0.950c − ( −0.995c ) = = + 0.999 87c vPE vR E ( 0.950c )( −0.995c ) 1− 1− 2 2 c c

The time interval between successive pulses in the pulsar’s own reference frame (i.e., on a clock at rest with respect to the event being timed) is Δt p = 1 10.0 Hz = 0.100 s. The duration of this interval in the rocket’s frame of reference is given by the time dilation relation as Δt = g Δt p =

Δt p 1− ( vPR c )

2

=

0.100 s 1− ( 0.999 87 )

2

= 6.20 s

and the frequency of the pulses as observed in the rocket’s reference frame is f= 26.21

1 1 = = 0.161 Hz Δt 6.20 s

Taking to the right as positive, it is given that the velocity of the rocket relative to observer A is vR A = + 0.92c. If observer B observes the rocket to have a velocity vR B = − 0.95c, the velocity of observer B relative to the rocket is vB R = + 0.95c. The relativistic velocity addition relation then gives the velocity of B relative to the stationary observer A as vB A =

26.22

0.998c toward the right

(a)

2 ⎛ 1 MeV ER = mc 2 = (1.67 × 10 −27 kg ) ( 3.00 × 108 m s ) ⎜ ⎝ 1.60 × 10 −13

(b)

E = g mc 2 = g ER =

=

26.23

v B R + vR A + 0.95c + 0.92c = = + 0.998c or v B R vR A ( 0.95c )( 0.92c ) 1+ 1+ c2 c2

939 MeV 1− ( 0.950 )

2

⎞ ⎟ = 939 MeV J⎠

ER 1− ( v c )

2

= 3.01× 10 3 MeV = 3.01 GeV

(c)

KE = E − ER = 3.01× 10 3 MeV − 939 MeV = 2.07 × 10 3 MeV = 2.07 GeV

(a)

The total energy is 400 times the rest energy, or E = g ER = 400ER , so it is necessary that g = 400. But g = 1 1− (v c)2 , and solving for the speed gives v = c 1−

(b)

1 1 = c 1− = 0.999 997c g2 ( 400 )2

The kinetic energy is KE = E − ER = 400ER − ER = 399ER . For a proton, ER = 938.3 MeV. Thus, KE = 399 ( 938.3 MeV ) = 3.74 × 10 5 MeV

26.24

(a)

The rest energy of a proton is ER = 938.3 MeV. If the kinetic energy is KE = 175 GeV we have KE = (g − 1)ER ER , or g − 1 1, and g − 1 ≈ g . Thus, we may write the approximate relation KE = (g − 1) ER ≈ g EE = g m p c 2 .

ER ,

continued on next page

68719_26_ch26_p305-323.indd 313

1/7/11 2:56:56 PM

314

Chapter 26

The gamma factor is g = 1 1− (v c)2 , so solving for the speed gives v = c 1− g Therefore, when KE ER and we may use the approximation KE ≈ g ER , we have g ≈ ER KE . Then, for the proton having KE = 175 GeV,

(b)

2

v = c 1−g 26.25

−2

−2

.

2

⎛E ⎞ ⎛ 938.3 MeV ⎞ ≈ c 1− ⎜ R ⎟ = c 1− ⎜ ⎟ = 0.999 99c ⎝ KE ⎠ ⎝ 175×10 3 MeV ⎠

The nonrelativistic expression for kinetic energy is KE = 12 mv 2, while the relativistic expression is KE = E − ER = (g − 1)ER = (g − 1)mc 2, where g = 1 1− ( v c ) . Thus, when the relativistic kinetic energy is twice the predicted nonrelativistic value, we have 2

⎛ ⎞ 1 1 ⎟ mc 2 = 2 ⎛⎜ mv 2 ⎞⎟ ⎜ − 1 2 ⎝2 ⎠ ⎜⎝ 1− ( v c ) ⎟⎠

2 2 1 = ⎡⎣1+ ( v c ) ⎤⎦ 1− ( v c )

or

Squaring both sides of the last result and simplifying gives

( v c ) ⎡⎣( v c ) + ( v c ) 2

4

2

− 1⎤⎦ = 0

Ignoring the trivial solution v c = 0, we must have ( v c ) + ( v c ) − 1 = 0. This is a quadratic 2 equation of the form x 2 + x − 1 = 0,with x = ( v c ) . Applying the quadratic formula gives 2 x = −1 ± 5 2. Since x = ( v c ) , we ignore the negative solution and find 4

(

2

)

2

−1+ 5 ⎛ v⎞ x=⎜ ⎟ = = 0.618 ⎝ c⎠ 2 which yields v = c 0.618 = 0.786 c . 26.26

The energy input to the electron will be W = E f − Ei = (g f − g i ) ER , or ⎛ 1 W =⎜ ⎜ ⎜⎝ 1− v f c

(

(a)

)

2

−

1 1− ( vi c )

2

⎞ ⎟E ⎟ R ⎟⎠

(a)

⎞ ⎟ ( 0.511 Mev ) = 0.582 MeV ⎟⎠

When v f = 0.990 c and vi = 0.900 c, we have ⎛ 1 1 W =⎜ − 2 ⎜⎝ 1− ( 0.990 )2 1− ( 0.900 )

26.27

ER = 0.511 MeV

If v f = 0.900 c and vi = 0.500 c, then ⎛ 1 1 W =⎜ − 2 ⎜⎝ 1− ( 0.900 )2 1− ( 0.500 )

(b)

where

⎞ ⎟ ( 0.511 Mev ) = 2.45 MeV ⎟⎠

From the work-energy theorem, Wnet = ΔKE. The ship starts from rest, so KEi = 0, and ΔKE = KE f = (g − 1)ER = (g − 1)mc 2. Thus, ⎛ ⎞ 2 1 Wnet = ⎜ − 1 ⎟ ( 2.40 × 10 6 kg ) ( 3.00 × 108 m s ) = 8.65 × 10 22 J 2 ⎜⎝ 1− ( 0.700 ) ⎟⎠

continued on next page

68719_26_ch26_p305-323.indd 314

1/7/11 2:56:59 PM

Relativity

(b)

If all the increase in the kinetic energy of the ship comes from the rest energy of the fuel (i.e., ΔKE = ER, fuel = mfuel c 2), the mass of the fuel required is ΔKE 8.65 × 10 22 J = = 9.61× 10 5 kg 2 2 8 c (3.00 × 10 m s)

mfuel = 26.28

315

Let m1 be the mass of the fragment moving at v1 = 0.987c, and m2 be the mass having a speed of v2 = 0.868c. From conservation of mass-energy, E1 + E2 = ER, i , or g 1 m1 c 2 + g 2 m2 c 2 = mi c 2 , giving m1 1− ( 0.987 )

2

m2

+

1− ( 0.868 )

2

= mi

and

6.22 m1 + 2.01m2 = 3.34 × 10 −27 kg

[1]

Since the original particle was at rest, the momenta of the two fragments after decay must add to zero. Thus, the magnitudes must be equal, giving p2 = p1 , or g 2 m2 v2 = g 1 m1 v1, and yielding 2.01m2 ( 0.868c ) = 6.22 m1 ( 0.987c )

or

m2 = 3.52 m1

[2]

Substituting Equation [2] into [1] gives

( 6.22 + 7.08 ) m1 = 3.34 × 10 −27 kg , or m1 = 2.51× 10 −28 kg Equation [2] then yields m2 = 3.52 ( 2.51× 10 −28 kg ) = 8.84 × 10 −28 kg . 26.29

The relativistic total energy is E = g ER = g mc 2, and the momentum is p = g mv, where g = 1 1 − v 2 c 2 . Thus, E 2 c 2 = g 2 m 2 c 2, and p2 = g 2 m 2 v 2, so subtracting yields ⎛ ⎞ E2 1 2 2 2 2 2 2 − p2 = g 2 m 2 ( c 2 − v 2 ) = g 2 m 2 c 2 (1− v 2 c 2 ) = ⎜ ⎟ m c 1− v c = m c 2 2 2 c ⎝ 1− v c ⎠

(

)

Rearranging, this becomes E2 = p2 + m 2 c 2 c2 26.30

(a)

E 2 = p2 c 2 + m 2 c 4

or

Since the total relativistic energy of a particle is E = KE + ER = KE + mc 2, requiring that total energy be conserved (i.e., total energy before reaction = total energy after reaction) gives KEi + mi c 2 = KE f + m f c 2 , where mi is the total mass of the particles present before the reaction, KEi is the total kinetic energy before the reaction, m f is the total mass of the particles present after the reaction, and KE f is the final total kinetic energy.

(b)

Using atomic masses from the tables in Appendix B of the textbook, the total mass of the initial particles is found to be mi = m

(c)

235 92 U

+m

1 0n

= 235.043 923 u + 1.008 665 u = 236.052 588 u

The total mass of the particles present after the reaction is mf = m

148 57 La

+m

87 35 B r

+m

1 0n

= 147.932 236 u + 86.920 71119 u + 1.008 665 u = 235.861 612 u

continued on next page

68719_26_ch26_p305-323.indd 315

1/7/11 2:57:03 PM

316

Chapter 26

(d)

The mass that must have been converted to energy in the reaction is Δm = mi − m f = ( 236.052 588 − 235.861 612 ) u = 0.190 976 u

(e)

Assuming that KEi = 0, the total kinetic energy of the product particles is then ⎛ 931.494 MeV c 2 ⎞ KE f = Δmc 2 + KEi = ( 0.190 976 u ) c 2 ⎜ ⎟⎠ + 0 = 177.893 MeV 1u ⎝

26.31

26.32

E 20.0 GeV ⎛ 10 3 MeV ⎞ = 3.91 × 10 4 = ER 0.511 MeV ⎜⎝ 1 GeV ⎟⎠

(a)

g =

(b)

L=

(a)

For an electron moving at ve = 0.750c, the gamma factor is

Lp g

=

3.00 × 10 3 m = 7.67 × 10 −2 m = 7.67 cm 3.91× 10 4

g e =1

1− (ve c)2 = 1

1− (0.750)2 = 1.51

The kinetic energy of a particle is KE = E − ER = (g − 1)ER, so if the kinetic energy of a proton (ER, p = 938.3 MeV) equals that of the electron (ER, e = 0.511 MeV), we must have (g p − 1)ER, p = (g e − 1)ER, e , or g p = 1+ (g e − 1) But g p = 1

1− (v p c)2 , so (v p c)2 = 1− 1 g p2 and the speed of the proton must be

v p = c 1− (b)

ER, e ⎛ 0.511 MeV ⎞ = 1+ (1.51− 1) ⎜ = 1 + 2.78 × 10 − 4 ⎝ 938.3 MeV ⎟⎠ ER, p

1 1 = c 1− = 0.023 6c = 7.08 × 10 3 km s −4 2 g p2 (1+ 2.78 × 10 )

As above, g e = 1.51 for an electron having speed ve = 0.750 c. The momentum of a particle is p = g mv =

g mc 2 (v c) g ER (v c) = c c

pc = g ER ( v c )

or

If the momentum of a proton equals that of the electron, then p p c = pe c, or

(

)

g p ER, p v p c = g e ER, e ( ve c )

( v c ) = g E ⎛ v ⎞ = (1.51)⎛ 0.511 MeV ⎞ ( 0.750 ) = 6.17 × 10 ⎜⎝ ⎟ ⎜ ⎟ E ⎝ c⎠ 938.3 MeV ⎠ 1− ( v c ) Thus, ( v c ) = 6.17 × 10 1− ( v c ) , and ( v c ) = (6.17 × 10 ) − (6.17 × 10 ) ( v c ) , which yields p

and

R, e

2

e

−4

e

R, p

p

2

−4

p

p

2

−4

2

−4

p

6.17 × 10 ( v c ) = 1+( 6.17 × 10 ) ( ) 2

p

2

2

p

−4

2

−4

2

= 3.81× 10 −7

and the speed of the proton must be v p = c 3.81× 10 −7 = ( 3.00 × 108 m s ) ( 6.17 × 10 − 4 ) = 1.85 × 10 5 m s = 185 km s

68719_26_ch26_p305-323.indd 316

1/7/11 2:57:07 PM

Relativity

26.33

KE = E − ER = (g − 1) ER so (a)

g = 1+

KE 1 = 2 ER 1− ( v c )

v = c 1−

giving

2

1

(1+ KE E )

2

= c 1−

1 = 0.979c (1+ 2.00 0.511)2

For a proton with KE = 2.00 MeV, the speed is v p = c 1−

1

(1+ KE E )

2

= c 1−

R,p

1 = 0.065 2c (1+ 2.00 938)2

(c)

ve − v p = 0.979c − 0.065 2c = 0.914 c

(a)

Yes. As the spring is compressed, positive work is done on it or energy is added to it. Since mass and energy are equivalent, mass has been added to the spring.

(b)

Δm =

ΔER ΔPEs 12 kx 2 kx 2 = 2 = 2 = 2 c c c 2c 2

(c)

Δm =

( 2.0 × 10 N m )( 0.15 m ) 2 ( 3.00 × 10 m s )

2

2

26.35

(1+ KE E ) R

R ,e

(b)

1

The speed of an electron having KE = 2.00 MeV will be ve = c 1−

26.34

317

2

8

= 2.5 × 10 −17 kg

An observer who moves at speed v relative to an object (or span of space) having proper length L p sees a contracted length given by L = L p g = L p 1− (v c)2 . Thus, if the proper distance to the star is L p = 5.00 ly, and this length is to have a contracted value of L = 2.00 ly in the reference frame of the spacecraft, the speed of the spacecraft relative to the star must be 2

2 ⎛ L⎞ ⎛ 2.00 ly ⎞ v = c 1− ⎜ ⎟ = c 1− ⎜ = 0.917c ⎝ 5.00 ly ⎟⎠ ⎝ Lp ⎠

26.36

26.37

From E 2 = ( pc)2 + ER2 , with E = 5 ER , we find that p = ER 24 c.

( 0.511 MeV ) 24

= 2.50 MeV c .

(a)

For an electron, p =

(b)

For a proton, p =

(a)

Observers on Earth measure the distance to Andromeda to be d = 2.00 × 10 6 ly = (2.00 × 10 6 yr)c. The time for the trip, in Earth’s frame of reference, is Δt = g Δt p = 30.0 yr 1 − (v c)2 . The required speed is then

c

( 938.3 MeV ) 24 c

= 4.60 × 10 3

MeV = 4.60 GeV c . c

( )

v=

( 2.00 × 106 yr ) c d = Δt 30.0 yr 1− ( v c )2

which gives (1.50 × 10 −5 ) ( v c ) = 1− ( v c ) . Squaring both sides of this equation and solv2

ing for v c yields v c = 1

1+ 2.25 × 10 −10 . Then, the approximation 1 1+ x = 1− x 2 gives continued on next page

68719_26_ch26_p305-323.indd 317

1/7/11 2:57:10 PM

318

Chapter 26

2.25 × 10 −10 v ≈ 1− = 1− 1.13 × 10 −10 c 2 (b)

KE = (g − 1) mc 2, and g =

1 1− ( v c )

2

1

=

1− (1− 1.13 × 10 −10 )

2

=

1 2.26 × 10 −10

Thus, ⎛ ⎞ 2 1 KE = ⎜ − 1⎟ (1.00 × 10 6 kg ) ( 3.00 × 108 m s ) = 5.99 × 10 27 J −10 ⎝ 2.26 × 10 ⎠ (c)

cost = KE × rate ⎡ ⎛ 1 kWh ⎞ ⎤ = ⎢( 5.99 × 10 27 J ) ⎜ ($0.13 kWh ) = $2.16 × 10 20 ⎝ 3.60 × 10 6 J ⎟⎠ ⎥⎦ ⎣

26.38

The clock, at rest in the ship’s frame of reference, will measure a proper time of Δt p = 10 h before sounding. Observers on Earth move at v = 0.75c relative to the clock and measure an elapsed time of

( )

Δt = g Δt p =

Δt p 1− ( v c )

2

=

10 h 1− ( 0.75)

2

= 15 h

The observers on Earth see the clock moving away at 0.75c and compute the distance traveled before the alarm sounds as ⎛ 3600 d = v ( Δt ) = ⎡⎣ 0.75 ( 3.0 × 108 m s ) ⎤⎦ (15 h ) ⎜ ⎝ 1h 26.39

s⎞ 13 ⎟⎠ = 1.2 × 10 m

(a)

Since Dina and Owen are at rest in the same frame of reference (S′), they both see the ball traveling in the negative x ′ direction with speed vball ′ = 0.800c . Note that the velocity of the ball relative to Dina is vB D = − 0.800c.

(b)

The distance between Dina and Owen, measured in their own rest frame, is L p = 1.80 × 1012 m. Therefore, the time required for the ball to reach Dina, measured on her own clock, is Δt p =

(c)

Lp vball ′

=

1.80 × 1012 m 2.25 × 1012 m = 7.50 × 10 3 s = 0.800c 3.00 × 108 m s

Ed sees a contracted length for the distance separating Dina and Owen. According to him, they are separated by a distance L = L p g = L p 1− (v c)2 , where v = 0.600c is the speed of the S′ frame relative to Ed’s reference frame, S. Thus, according to Ed, the ball must travel a distance L = (1.80 × 1012 m ) 1− ( 0.600 ) = 1.44 × 1012 m 2

(d)

The ball has a velocity of vB D = − 0.800c relative to Dina, and Dina moves at velocity vDE = + 0.600c relative to Ed. The relativistic velocity addition relation (Equation 26.8 from the textbook) then gives the velocity of the ball relative to Ed as

continued on next page

68719_26_ch26_p305-323.indd 318

1/7/11 2:57:13 PM

Relativity

vB E =

319

vB D + vDE − 0.800c + 0.600c −0.200c = = = − 0.385c vB D vDE 1+ ( − 0.800 )( 0.600 ) 0.520 1+ c2

Thus, the speed of the ball according to Ed is vball = vB E = 0.385c . 26.40

(a)

As seen by an observer at rest relative to the mirror in frame S, the light must travel distance d before it strikes the mirror and then a distance d − d1 back to the ship after reflection. Here, distance d1 = vt is the distance the ship moves toward the mirror in the time t between when the pulse was emitted from the ship and when the reflected pulse was received by the ship. Since all observers agree that light travels at speed c, the total travel time for the light is t=

d + ( d − d1 ) 2d − d1 2d ⎛ v ⎞ = = −⎜ ⎟t c c c ⎝ c⎠

or

⎛ ⎜⎝ 1+

v⎞ 2d ⎟⎠ t = c c

Solving for the travel time t of the light gives t = 2d ( c + v ) . (b)

From the viewpoint of observers in the spacecraft, the spacecraft is at rest, and the mirror moves toward it at speed v. At the time the pulse starts toward the mirror, these observers see a contracted initial distance d ′ = d 1− (v c)2 to the mirror. If the total time of travel for the light is t ′, during the time t ′ 2 while the light is traveling toward the mirror, the mirror moves a distance Δx = vt ′ 2 closer to the ship. Thus, the light must travel a distance d ′ − Δx before reflection and the same distance d ′ − Δx back to the stationary ship after reflection. The total distance traveled by the light is then D = 2 ( d ′ − Δx ) , and the time required (traveling at speed c) is D 2d ′ − 2v(t ′ 2) 2d ′ ⎛ v ⎞ t′ = = = − ⎜ ⎟ t′ c c c ⎝ c⎠

or

⎛ ⎜⎝ 1+

v⎞ 2d ′ 2d 1− ( v c ) = ⎟⎠ t ′ = c c c

2

Solving for the travel time t ′ measured by observers in the spacecraft then gives c2 − v2 c2

( c + v ) t ′ = 2d 26.41

and

t′ =

2d c

( c + v )( c − v ) 2d c − v = c c+v ( c + v )2

The length of the space ship, as measured by observers on Earth, is L = L p 1− ( v c ) . In Earth’s frame of reference, the time required for the ship to pass overhead is 2

L L p 1− ( v c ) = = Lp v v 2

Δt =

1 1 − v2 c2

2

Thus,

or

2 ⎛ 0.75 × 10 −6 s ⎞ 1 1 ⎛ Δt ⎞ 1 s2 −17 = + = + = 1.74 × 10 2 v 2 c 2 ⎜⎝ L p ⎟⎠ m2 (3.00 × 108 m s) ⎜⎝ 300 m ⎟⎠

v=

1 2

1.74 × 10 −17 26.42

s m2

⎞ m⎞ ⎛ c ⎛ = 0.80 c = ⎜ 2.4 × 108 ⎟ ⎝ s ⎠ ⎜⎝ 3.00 × 108 m s ⎟⎠

From KE = (g − 1) ER , we find g = 1+

KE 1013 MeV = 1.07 × 1010 = 1+ ER 938 MeV

continued on next page

68719_26_ch26_p305-323.indd 319

1/7/11 2:57:16 PM

320

Chapter 26

or g ~ 1010 . Thus, the speed of the proton (and hence, the speed of the galaxy as seen by the proton) is v = c 1− 1 g 2 ~ c 1− 10 −20 ≈ c. (a)

The diameter of the galaxy, as seen in the proton’s frame of reference, is L = L p 1− ( v c ) = 2

or (b)

26.44

~

g

10 5 ly = 10 −5 ly 1010

⎛ 9.461× 1015 m ⎞ ⎛ 1 km ⎞ 7 L ≈ 10 −5 ly ⎜ ⎟⎠ ⎜⎝ 10 3 m ⎟⎠ = 9.461× 10 km 1 ly ⎝

~ 108 km

The proton sees the galaxy rushing by at v ≈ c. The time, in the proton’s frame of reference, for the galaxy to pass is Δt =

26.43

Lp

1011 m L 108 km = = 333 s ~ c 3.00 × 108 m s v

~ 10 2 s

The difference between the relativistic momentum, p = g mv, and the classical momentum, mv, is Δp = g mv − mv = (g − 1)mv. (a)

The error is 1.00% when Δp p = 0.010 0, or (g − 1)mv = 0.010 0g mv. This gives 2 2 g = 1 0.990 , or 1− ( v c ) = ( 0.990 ) , and yields v = 0.141c .

(b)

When the error is 10.0%, we have g = 1 0.900 , and 1− ( v c ) = ( 0.900 ) . In this case, the speed of the particle is v = c 1− (0.900)2 = 0.436c . 2

2

The kinetic energy gained by the electron will equal the loss of potential energy, so KE = q ( ΔV ) = e (1.02 MV ) = 1.02 MeV (a)

If Newtonian mechanics remained valid, then KE = 12 mv 2, and the speed attained would be v=

(b)

2 (1.02 MeV )(1.60 × 10 −13 J MeV )

2 KE = m

9.11× 10 −31 kg

KE = (g − 1) ER, so g = 1+

= 5.99 × 108 m s ≈ 2c

KE 1.02 MeV = 1+ = 3.00 ER 0.511 MeV

The actual speed attained is v = c 1− 1 g 26.45

(a)

= c 1− 1 ( 3.00 ) = 0.943c 2

When at rest, muons have a mean lifetime of Δt p = 2.2 ms. In a frame of reference where they move at v = 0.95c, the dilated mean lifetime of the muons will be

( )

t = g Δt p = (b)

2

Δt p 1− ( v c )

2

=

2.2 ms 1− ( 0.95)

2

= 7.0 ms

In a frame of reference where the muons travel at v = 0.95c, the time required to travel 3.0 km is t=

d 3.0 × 10 3 m = = 1.05 × 10 −5 s = 10.5 ms v 0.95 ( 3.00 × 108 m s )

continued on next page

68719_26_ch26_p305-323.indd 320

1/7/11 2:57:19 PM

Relativity

321

If N 0 = 5.0 × 10 4 muons started the 3.0 km trip, the number remaining at the end is N = N 0 e − t t = ( 5.0 × 10 4 ) e − 10.5 m s 7.0 m s = 1.1× 10 4 26.46

The work required equals the increase in the gravitational potential energy, or W = GMSun m Rg . If this is to equal the rest energy of the mass removed, then mc 2 =

Rg = 26.47

GMSun m Rg

(6.67 × 10

−11

GMSun c2

Rg =

or

N ⋅ m 2 kg2 ) (1.99 × 10 30 kg )

(3.00 × 10

8

m s)

2

= 1.47 × 10 3 m = 1.47 km

According to Earth-based observers, the times required for the two trips are: For Speedo:

ΔtS =

For Goslo:

ΔtG =

Lp vS Lp vG

=

20.0 ly 20.0 yr = = 21.1 yr 0.950 c 0.950

=

20.0 ly 20.0 yr = = 26.7 yr 0.750 c 0.750

Thus, after Speedo lands, he must wait and age at the same rate as planet-based observers, for an additional ΔTage, S = ΔtG − ΔtS = (26.7 − 21.1) yr = 5.6 yr before Goslo arrives. The time required for the trip according to Speedo’s internal biological clock (which measures the proper time for his aging process during the trip) is Tage, S = Δt p, S =

ΔtS 2 2 = ΔtS 1− ( vS c ) = ( 21.1 yr ) 1− ( 0.950 ) = 6.59 yr gS

When Goslo arrives, Speedo has aged a total of TS = Tage, S + ΔTage, S = 6.59 yr + 5.6 = 12.2 yr The time required for the trip according to Goslo’s internal biological clock (and hence the amount he ages) is TG = Δt p, G =

ΔtG 2 2 = ΔtG 1− ( vG c ) = ( 26.7 yr ) 1− ( 0.750 ) = 17.7 yr gG

Thus, we see that when he arrives, Goslo is older than Speedo, having aged an additional TG − TS = 17.7 yr − 12.2 yr = 5.5 yr 26.48

(a)

The proper lifetime is measured in the ship’s reference frame, and Earth-based observers measure a dilated lifetime of Δt = g Δt p =

(b)

Δt p 1− ( v c )

2

=

15.0 yr 1− ( 0.700 )

2

= 21.0 yr

d = v ( Δt ) = [ 0.700 c ]( 21.0 yr ) = ⎡⎣( 0.700 )(1.00 ly yr ) ⎤⎦ ( 21.0 yr ) = 14.7 ly

continued on next page

68719_26_ch26_p305-323.indd 321

1/7/11 2:57:23 PM

322

Chapter 26

(c)

Looking out the rear window, the astronauts see Earth recede at a rate of v = 0.700 c. The distance it has receded, as measured by the astronauts, when the batteries fail is

( )

d ′ = v Δt p = ( 0.700 c ) (15.0 yr ) = ⎡⎣( 0.700 )(1.00 ly yr ) ⎤⎦ (15.0 yr ) = 10.5 ly (d)

26.49

Mission control gets signals for 21.0 yr while the battery is operating and then for 14.7 yr after the battery stops powering the transmitter, 14.7 ly away. The total time that signals are received is 21.0 yr + 14.7 yr = 35.7 yr .

Note: Excess digits are retained in some steps given below to more clearly illustrate the method of solution. We are given that L = 2.00 m and q = 30.0° (both measured in the observer’s rest frame). The components of the rod’s length as measured in the observer’s rest frame are Lx = L cosq = ( 2.00 m ) cos30.0° = 1.732 m and

L y = L sinq = ( 2.00 m ) sin 30.0° = 1.00 m

The component of length parallel to the motion has been contracted, but the component perpendicular to the motion is unaltered. Thus, L py = L y = 1.00 m and L px = (a)

Lx 1− ( v c )

2

=

1.732 m 1− ( 0.995)

2

The proper length of the rod is then

(17.34 m )2 + (1.00 m )2 = 17.4 m

L p = L2px + L2py = (b)

= 17.34 m

The orientation angle in the rod’s rest frame is ⎛ L py ⎞ −1 ⎛ 1.00 m ⎞ q p = tan −1 ⎜ ⎟ = tan ⎜⎝ 17.34 m ⎟⎠ = 3.30° L ⎝ px ⎠

26.50

(a)

Taking toward Earth as the positive direction, the velocity of the ship relative to Earth is vSE = + 0.600c, and the velocity of the lander relative to the ship is v = + 0.800c. The relativistic velocity addition relation (Equation 26.8 in the textbook) then gives the velocity of the lander relative to Earth as LS

vLE =

(b)

vLS + vSE 0.800c + 0.600c 1.40c = = = + 0.946c vLS vSE 1+ ( 0.800 )( 0.600 ) 1.48 1+ c2

Observers at rest on Earth measure the proper length between the ship and Earth as L p = 0.200 ly. The contracted distance measured by observers at rest relative to the spaceship is L = L p 1− ( vSE c ) = ( 0.200 ly ) 1− ( 0.600 ) = 0.160 ly 2

2

continued on next page

68719_26_ch26_p305-323.indd 322

1/7/11 2:57:25 PM

Relativity

(c)

Observers on the ship see the lander start toward Earth from an initial distance of L = 0.160 ly. They see this distance diminish as the lander nibbles into it from one end at vLS = 0.800c, and the Earth (as it appears to approach them) reducing it from the other end at vSE = 0.600c. The time they compute it will take the lander and Earth to meet is t=

(d)

323

L 0.160 ly 0.160 yr = = = 0.114 yr vLS + vSE 1.40c 1.40

The kinetic energy of the lander as observed in the Earth reference frame is 2 KE = E − ER = (g − 1) mc 2, where g = 1 1− ( vLE c ) . This gives ⎛ ⎞ 2 1 KE = ⎜ − 1 ⎟ ( 4.00 × 10 5 kg ) ( 3.00 × 108 m s ) = 7.51× 10 22 J 2 ⎜⎝ 1− ( 0.946 ) ⎟⎠

68719_26_ch26_p305-323.indd 323

1/7/11 2:57:27 PM

27 Quantum Physics QUICK QUIZZES 1.

True. When a photon scatters off an electron that was initially at rest, the electron must recoil to conserve momentum. The recoiling electron possesses kinetic energy that was gained from the photon in the scattering process. Thus, the scattered photon must have less energy than the incident photon.

2.

Choice (b). Some energy is transferred to the electron in the scattering process. Therefore, the scattered photon must have less energy (and hence, lower frequency) than the incident photon.

3.

Choice (c). Conservation of energy requires the kinetic energy given to the electron be equal to the difference between the energy of the incident photon and that of the scattered photon.

4.

False. The de Broglie wavelength of a particle of mass m and speed v is l = h p = h mv. Thus, the wavelength decreases when the momentum increases.

5.

Choice (c). Two particles with the same de Broglie wavelength will have the same momentum p = mv = h l. If the electron and proton have the same momentum, they cannot have the same speed because of the difference in their masses. For the same reason, remembering that KE = p2 2m, they cannot have the same kinetic energy. Because the kinetic energy is the only type of energy an isolated particle can have, and we have argued that the particles have different energies, the equation f = E h tells us that the particles do not have the same frequency.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

According to Einstein’s photoelectric effect equation, e ΔVs = KEm ax = Ephoton − f , and the work function of the metal is f = Ephoton − e ΔVs = 3.56 eV − e (1.10 V ) = 2.46 eV so the correct answer is choice (b).

2.

From Wien’s displacement law, T=

0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K = 6.10 × 10 3 K = 6 100 K = lm ax 475 × 10 −9 m

and choice (a) is correct. 3.

An electron accelerated from rest through a potential difference of 50.0 V will have a kinetic energy of KE = 50.0 eV ER, e = 0.511 MeV. Thus, it is classical, and its momentum may be expressed as p = 2me (KE). The de Broglie wavelength is then l=

h = p

h = 2me (KE)

h = 2me (e ΔV )

6.63 × 10 −34 J ⋅s

2 ( 9.11× 10 −31 kg ) (1.60 × 10 −19 C )( 50.0 V )

= 0.174 nm

and the correct choice is (b). 324

68719_27_ch27_p324-340.indd 324

1/7/11 2:58:04 PM

Quantum Physics

4.

325

The maximum energy, or minimum wavelength, photon is produced when the electron loses all of its kinetic energy in a single collision. Therefore, Em ax = hc lm in = KEi = e ΔV , or lm in =

(6.63 × 10−34 J ⋅s) (3.00 × 108 m s) = 4.14 × 10−7 m = 414 nm hc = e ΔV (1.60 × 10−19 C) (3.00 J C)

and we see that (c) is the correct choice. 5.

One form of Heisenberg’s uncertainty relation is ΔxΔpx ≥ h 4p , which says that one cannot determine both the position and momentum of a particle with arbitrary accuracy. Another form of this relation is ΔEΔt ≥ h 4p , which sets a limit on how accurately the energy can be determined in a finite time interval. Thus, both (a) and (c) are true statements, and the other listed choices are false.

6.

From the Compton shift formula, the change in the photon’s wavelength is Δl = l − l0 =

h 6.63 × 10 −34 J ⋅s (1− cosq ) = (1− cos180°) me c (9.11× 10−31 kg) (3.00 × 108 m s)

= 4.85 × 10 −12 m = 4.85 × 10 −3 nm and choice (d) is the correct answer. 7.

The comparative masses of the particles of interest are m p ≈ 1 840 me and mHe ≈ 4m p. Assuming the particles are all classical, their momenta are p p = m p v, pHe = mHe v = 4m p v, and pe = me ve = m p v 1 840. Their de Broglie wavelengths are l p = h p p = h m p v , lHe = h pHe = h 4m p v = l p 4 , and le = h pe = h (m p v 1 840) = 1 840l p . The ranking, from longest to shortest, of these wavelengths is le > l p > lHe , and choice (e) gives the correct order.

8.

From Bragg’s law, the angle q at which constructive interference of order m will be observed when x-rays reflect from crystalline planes separated by distance d is given by 2d sinq = ml. Thus, the separation distance of the planes in the crystal is d = ml 2sinq . With m and l the same for the two crystals, we see that d is inversely proportional to sinq . Since it is observed that q A > q B (and hence, sinq A > sinq B), it must be true that d A < d B , and the correct answer is (c).

9.

Diffraction, polarization, interference, and refraction are all processes associated with waves. However, to understand the photoelectric effect, we must think of the energy transmitted as light coming in discrete packets, or quanta, called photons. Thus, the photoelectric effect most clearly demonstrates the particle nature of light, and the correct choice is (b).

10.

Electron diffraction by crystals, first detected by the Davisson-Germer experiment in 1927, confirmed de Broglie’s hypothesis and, of the listed choices, most clearly demonstrates the wave nature of electrons. The correct answer is (e).

11.

During the scattering process, the photon will transfer some of its energy to the originally stationary electron, and the electron will recoil following this process. Since the energy of a photon is Ephoton = hf , where h is constant, the frequency f of the photon must decrease when the photon gives up energy. Thus, the correct choice is (a).

12.

The magnitude of the charge is q = e for both the electron and proton, and since the two particles are accelerated through the same potential difference, they are given identical kinetic energies KEe = KE p = eΔV. The momentum of a non-relativistic particle having kinetic energy KE = eΔV and mass m, is p = 2mKE = 2m(eΔV ). The de Broglie wavelength, l = h p, is then seen to be inversely proportional to m. Since the mass of an electron is smaller than that of a proton, we see that le > l p when the electron and proton have the same kinetic energy. Thus, the correct choice is (a).

68719_27_ch27_p324-340.indd 325

1/7/11 2:58:06 PM

326

Chapter 27

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

A microscope can see details no smaller than the wavelength of the waves it uses to produce images. Electrons with kinetic energies of several electron volts have wavelengths of less than a nanometer, which is much smaller than the wavelength of visible light (having wavelengths ranging from about 400 to 700 nm). Therefore, an electron microscope can resolve details of much smaller size as compared to an optical microscope.

4.

Measuring the position of a particle implies having photons reflect from it. However, collisions between photons and the particle will alter the velocity of the particle. Thus, the more accurately you try to measure one quantity, the more uncertainty you create in your knowledge of the other.

6.

Light has both wave and particle characteristics. In Young’s double-slit experiment, light behaves as a wave. In the photoelectric effect, it behaves like a particle. Light can be characterized as an electromagnetic wave with a particular wavelength or frequency, yet at the same time, light can be characterized as a stream of photons, each carrying a discrete energy, hf.

8.

Ultraviolet light has a shorter wavelength and higher photon energy, Ephoton = hf = hc l, than visible light.

10.

Increasing the temperature of the substance increases the average kinetic energy of the electrons inside the material. This makes it slightly easier for an electron to escape from the material when it absorbs a photon.

12.

Most stars radiate nearly as blackbodies. Of the two stars, Vega has the higher surface temperature and, in agreement with Wien’s displacement law, radiates more intensely at shorter wavelengths and has a bluish appearance, while Arcturus has a more reddish appearance.

14.

No, the crystal cannot produce diffracted beams of visible light. The angles where one could expect to observe diffraction maxima are given by Bragg’s law, 2d sinq = ml, where m is a nonzero integer. Because sinq ≤ 1, this equation gives m = (2d l)sinq ≤ 2d l . Since the wavelengths of visible light (l ∼ 10 2 nm) are much larger than the distances between atomic planes in crystals (d ∼ 1 nm), there are no nonzero integer values for m that can satisfy Bragg’s law for visible light.

16.

Any object of macroscopic size—including a grain of dust—has an undetectably small wavelength, so any diffraction effects it might exhibit are very small, effectively undetectable. Recall historically how the diffraction of sound waves was at one time well known, but the diffraction of light was not.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

(a)

∼ 100 nm, ultraviolet

(b)

∼ 10 −1 nm, gamma rays

4.

(a)

5.78 × 10 3 K

(b)

501 nm

6.

(a)

2.90 × 10 −19 J photon

(b)

4.23 × 10 3 K

(c)

1.65 × 10 26 W

(d)

5.69 × 10 44 photon s

(c)

1.19 eV

8.

5.7 × 10 3 photons s

10.

(a)

288 nm

(b)

1.04 × 1015 Hz

12.

(a)

only lithium

(b)

0.81 eV

68719_27_ch27_p324-340.indd 326

1/7/11 2:58:09 PM

Quantum Physics

14.

(a) 1.90 eV

16.

(a)

18.

0.455 nm

20.

m = 2, q = 25.8°; m = 3, q = 40.8°; m = 4, q = 60.8°

8.29 × 10 −11 m

(b)

0.216 V

(b)

1.24 × 10 −11 m

(c)

327

lm in decreases

Higher orders cannot be found since sinq cannot exceed a value of 1.00. 22.

1.03 × 10 −3 nm

24.

4.85 × 10 −3 nm

26.

(a) 0.101 nm

28.

(a)

30.

(a) 0.709 nm

32.

(a)

1.99 × 10 −11 m

p = 2mqΔV

(b)

80.9°

(b)

1.98 × 10 −14 m

(b)

414 nm

(b)

l=h

2mqΔV

(c) The proton, with the larger mass, will have the shorter wavelength. 34.

23 m s

36.

(a)

38.

(a) See Solution.

40.

v=c 2 2

42.

(a)

44.

(a)

46.

0.14 keV

48.

0.785 eV

50.

(a) 148 days (b)

0.250 m s

(b)

2.25 m

(b)

5.3 MeV

2.49 × 10 −11 m

(b)

0.29 nm

l = 2.82 × 10 −37 m

(b)

ΔE ≥ 1.06 × 10 −32 J

(c)

2.88 × 10 −35 %

This result is totally contrary to observations of the photoelectric effect.

PROBLEM SOLUTIONS 27.1

68719_27_ch27_p324-340.indd 327

From Wien’s displacement law, (a)

T=

0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K = 2.99 × 10 3 K, or ≈ 3 000 K = lm ax 970 × 10 −9 m

(b)

T=

0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K = = 2.00 × 10 4 K, or ≈ 20 000 K lm ax 145 × 10 −9 m

1/7/11 2:58:11 PM

328

27.2

27.3

Chapter 27

Using Wien’s displacement law, (a)

lm ax =

0.289 8 × 10 − 2 m ⋅K = 2.898 × 10 −7 m ~100 nm 10 4 K

ultraviolet

(b)

lm ax =

0.289 8 × 10 − 2 m ⋅K = 2.898 × 10 −10 m ~10 −1 nm 10 7 K

g -rays

From Wien’s displacement law, lm ax =

27.4

(a)

0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K = 9.47 × 10 −6 m = 9.47 mm (infrared) = T 306 K

The power radiated by an object with surface area A and absolute temperature T is given by Stefan’s law (see Chapter 11 of the textbook) as P = s AeT 4 , where s is the Stefan-Boltzmann constant equal to 5.669 6 × 10 −8 W m 2 ⋅K 4 . For an object that approximates a blackbody, the emissivity is e = 1. Thus, the temperature of our Sun’s surface should be 1

1

1

⎤4 ⎤4 ⎡ ⎡ P ⎤4 ⎡ P 3.85 × 10 26 W ⎢ ⎥ T=⎢ = = ⎢ ⎥ 2 ⎥ 2 ⎢ 4p ( 5.669 6 × 10 −8 W m 2 ⋅K 4 ) ( 6.96 × 108 m ) (1) ⎥ ⎣ s Ae ⎦ ⎢⎣ s ( 4p RSun ) e ⎥⎦ ⎣ ⎦ = 5.78 × 10 3 K (b)

Wien’s displacement law gives the peak wavelength of the radiation from the Sun as lm ax =

27.5

27.6

0.289 8 × 10 − 2 m ⋅K 0.289 8 × 10 − 2 m ⋅K = = 5.01× 10 −7 m = 501 nm T 5.78 × 10 3 K

The energy of a photon is given by E = hf = hc l , where h = 6.63 × 10 −34 J ⋅s is Planck’s constant. (a)

E=

− 34 8 hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1.00 eV = ⎜⎝ −2 l 5.00 × 10 m 1.60 × 10 −19

(b)

E=

− 34 8 hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1.00 eV ⎞ = ⎜⎝ ⎟ = 2.49 eV l 500 × 10 −9 m 1.60 × 10 −19 J ⎠

(c)

E=

− 34 8 hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1.00 eV ⎞ = ⎜⎝ ⎟ = 249 eV l 5.00 × 10 −9 m 1.60 × 10 −19 J ⎠

(a)

Epeak =

(b)

T=

(c)

The surface area of the spherical star is A = 4p r 2, and (considering it to be a blackbody) its emissivity is e = 1. Stefan’s law (see Chapter 11 of the textbook) gives the radiated power as

⎞ −5 ⎟ = 2.49 × 10 eV J⎠

(6.63 × 10−34 J ⋅s) (3.00 × 108 m s) = 2.90 × 10−19 J photon hc = lm ax 685 × 10 −9 m

0.289 8 × 10 −2 m ⋅K 0.289 8 × 10 −2 m ⋅K = = 4.23 × 10 3 K lm ax 685 × 10 −9 m

P = s AeT 4 = s ( 4p r 2 ) eT 4 = 4p ( 5.669 6 × 10 −8 W m 2 ⋅K 4 ) (8.50 × 108 m ) (1)( 4.23 × 10 3 K ) = 1.65 × 10 26 W 2

4

continued on next page

68719_27_ch27_p324-340.indd 328

1/7/11 2:58:13 PM

Quantum Physics

(d)

329

Assuming the average energy of an emitted photon equals the energy of photons at the peak of the radiation distribution, the number of photons emitted by the star each second is ΔN P 1.65 × 10 26 J s = = = 5.69 × 10 44 photon s Δt Epeak 2.90 × 10 −19 J photon

27.7

27.8

The energy of a photon having frequency f is given by Ephoton = hf, where Planck’s constant has a value of h = 6.63 × 10 −34 J ⋅s. This energy may be converted to units of electron volts by use of the conversion factor 1 eV = 1.60 × 10 −19 J. (a)

1 eV ⎛ Ephoton = hf = ( 6.63 × 10 −34 J ⋅s ) ( 620 × 1012 Hz ) ⎜ ⎝ 1.60 × 10 −19

(b)

1 eV ⎛ ⎞ Ephoton = hf = ( 6.63 × 10 −34 J ⋅s ) ( 3.10 × 10 9 Hz ) ⎜ = 1.28 × 10 −5 eV ⎝ 1.60 × 10 −19 J ⎟⎠

(c)

1 eV ⎛ Ephoton = hf = ( 6.63 × 10 −34 J ⋅s ) ( 46.0 × 10 6 Hz ) ⎜ ⎝ 1.60 × 10 −19

(a)

(b)

(c) (a)

2 J s ⋅ m 2 ) ⎡⎢p (8.5×10 −3 m ) 4⎤⎥ (500 ×10 −9 m ) ⎣ ⎦ = 5.7 ×10 3 photons s −34 8 (6.63×10 J ⋅ s) (3.00 ×10 m s)

(6.63 × 10

J ⋅s ) ( 3.00 × 108 m s ) ⎛ 1 eV ⎜⎝ −9 350 × 10 m 1.60 × 10 − 19

− 34

⎞ ⎟ − 1.31 eV = 2.24 eV J⎠

− 34 8 hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1 eV ⎞ −7 lc = = ⎜⎝ ⎟ = 5.55 × 10 m = 555 nm f 2.24 eV 1.60 × 10 − 19 J ⎠

fc =

c 3.00 × 108 m s = = 5.41× 1014 Hz lc 555 × 10 −9 m

At the cutoff wavelength, KEm ax = 0, so the photoelectric effect equation (KEm ax = hc l − f) gives the cutoff wavelength as lc =

(b)

−11

From the photoelectric effect equation, the work function is f = hc l − KEm ax , or f=

27.10

⎞ −7 ⎟ = 1.91× 10 eV J⎠

The energy entering the eye each second is P = I ⋅ A, where A is the area of the pupil. If the light has wavelength l, the energy of a single photon is Ephoton = hc l . Hence, the number of photons entering the eye in time Δt is given by N = ΔE Ephoton = P ⋅ ( Δt ) Ephoton , or N = IA(Δt) (hc l) = IA(Δt)l hc. With I = 4.0 × 10 −11 W m 2 , l = 500 nm, Δt = 1.0 s, and the opening of the pupil having an area of A = p d 2 4 , where d is the pupil diameter, we find N ( 4.0 ×10 = Δt

27.9

⎞ ⎟ = 2.57 eV J⎠

−34 8 hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1 ev ⎞ −7 = ⎜⎝ ⎟ = 2.88 × 10 m = 288 nm f 4.31 eV 1.60 × 10 −19 J ⎠

The lowest frequency of light that will free electrons from the material is fc = c lc , where lc is the cutoff wavelength (calculated above). Thus, fc =

c 3.00 × 108 m s = = 1.04 × 1015 Hz lc 2.88 × 10 −7 m

continued on next page

68719_27_ch27_p324-340.indd 329

1/7/11 2:58:16 PM

330

Chapter 27

(c)

The photoelectric effect equation may be written as KEm ax = Ephoton − f. Therefore, if the photons incident on this surface have energy Ephoton = 5.50 eV, the maximum kinetic energy of the ejected electrons is KEm ax = Ephoton − f = 5.50 eV − 4.31 eV = 1.19 eV

27.11

(a) (b)

⎛ 1.60 × 10 −19 J ⎞ −18 f = 6.35 eV ⎜ ⎟⎠ = 1.02 × 10 J 1 eV ⎝ The energy of a photon having the cutoff frequency or cutoff wavelength equals the work function of the surface, or Ephoton = hfc = hc lc = f . Thus, the cutoff frequency of a surface having a work function of f = 6.35 eV is fc =

(c)

The cutoff wavelength is lc =

27.12

⎛ 1.60 × 10 −19 J ⎞ f 6.35 eV 15 = −34 ⎟⎠ = 1.53 × 10 Hz h 6.63 × 10 J ⋅s ⎜⎝ 1 eV

hc c 3.00 × 108 m s = = = 1.96 × 10 −7 m = 196 nm f fc 1.53 × 1015 Hz

(d)

KEm ax = Ephoton − f = 8.50 eV − 6.35 eV = 2.15 eV

(e)

eVs = KEm ax , so the stopping potential is Vs = KEm ax e = 2.15 eV e = 2.15 V .

(a)

The energy of the incident photons is Ephoton =

− 34 8 hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1 eV ⎞ = ⎜⎝ ⎟ = 3.11 eV −9 l 400 × 10 m 1.60 × 10 − 19 J ⎠

For photoelectric emission to occur, it is necessary that Ephoton ≥ f. Thus, of the three metals given, only lithium will exhibit the photoelectric effect. (b) 27.13

For lithium, KEm ax = Ephoton − f = 3.11 eV − 2.30 eV = 0.81 eV .

The two frequencies of the light allowed to strike the surface are f1 =

and

c 3.00 × 108 m s = = 1.18 × 1015 Hz = 11.8 × 1014 Hz l1 254 × 10 −9 m

f2 =

c 3.00 × 108 m s = = 6.88 × 1014 Hz l2 436 × 10 −9 m

The graph you draw should look somewhat like that given at the right. The desired quantities, read from the axes intercepts of the graph line, should agree within their uncertainties with fc = 4.8 × 1014 Hz and f = 2.0 eV

68719_27_ch27_p324-340.indd 330

1/7/11 2:58:19 PM

Quantum Physics

27.14

(a)

The maximum kinetic energy of the ejected electrons is related to the stopping potential by the expression KEm ax = e ( ΔVs ) . Thus, if the stopping potential is Vs = 0.376 V when the incident light has wavelength l = 546.1 nm, the photoelectric effect equation gives the work function of this metal as f= =

(b)

331

hc hc − KEm ax = − e ( ΔVs ) l l

(6.63 × 10

J ⋅s ) ( 3.00 × 108 m s ) ⎛ 1 eV ⎞ ⎜⎝ ⎟ − 0.376 eV = 1.90 eV 546.1× 10 −9 m 1.60 × 10 −19 J ⎠ −34

If light of wavelength l = 587.5 nm is incident on this metal, the maximum kinetic energy of the ejected electrons is e ( ΔVs ) = KEm ax = =

hc −f l

(6.63 × 10

J ⋅s ) ( 3.00 × 108 m s ) ⎛ 1 eV ⎞ ⎜⎝ ⎟ − 1.90 eV = 0.216 eV 587.5 × 10 −9 m 1.60 × 10 −19 J ⎠ −34

Thus, the stopping potential will be ΔVs = 27.15

KEm ax 0.216 eV = = 0.216 V e e

Assuming the electron produces a single photon as it comes to rest, the energy of that photon is Ephoton = ( KE )i = e ( ΔV ). The accelerating voltage is then ΔV =

Ephoton e

=

−34 8 hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) 1.24 × 10 −6 V ⋅ m = = el l (1.60 × 10−19 C) l

For l = 1.0 × 10 −8 m, V =

1.24 × 10 −6 V ⋅ m = 1.2 × 10 2 V 1.0 × 10 −8 m

and for l = 1.0 × 10 −13 m, V = 27.16

1.24 × 10 −6 V ⋅ m = 1.2 × 10 7 V 1.0 × 10 −13 m

A photon of maximum energy and minimum wavelength is produced when the electron gives up all its kinetic energy in a single collision, or lm in = hc KE = hc e ΔV . (a)

If ΔV = 15.0 kV, lm in =

(6.63 × 10 (1.60 × 10

(b)

If ΔV = 100 kV, lm in =

(6.63 × 10 J ⋅s) (3.00 × 10 m s) = 1.24 × 10 (1.60 × 10 C) (100 × 10 V )

(c)

As seen above, lm in decreases when the potential difference, ΔV , increases.

−34

J ⋅s ) ( 3.00 × 108 m s )

−19

−34

27.17

= 8.29 × 10 −11 m .

8

−19

3

−11

m.

A photon of maximum energy or minimum wavelength is produced when the electron gives up all of its kinetic energy in a single collision within the target. Thus, Em ax = hc lm in = KEe = e ΔV. If lm in = 70.0 pm = 70.0 × 10 −12 m, the required accelerating voltage is ΔV =

68719_27_ch27_p324-340.indd 331

C ) (15.0 × 10 3 V )

(6.63 × 10−34 J ⋅s) (3.00 × 108 m s) = 1.78 × 104 V = 17.8 kV hc = elm in (1.60 × 10−19 C) ( 70.0 × 10−12 m )

1/7/11 2:58:21 PM

332

27.18

Chapter 27

From the Bragg equation, 2d sinq = ml, the interplanar spacing when 0.129-nm x-rays produce a first order diffraction maximum at q = 8.15° is d=

27.19

Using Bragg’s law, the wavelength is found to be l=

27.20

ml (1)( 0.129 nm ) = = 0.455 nm 2sinq 2sin8.15°

2 d sinq 2 ( 0.296 nm ) sin 7.6° = 0.078 nm = 1 m

From the Bragg equation, 2d sinq = ml, the angle at which the diffraction maximum of order m will be found when the first order maximum is at q1 = 12.6°, is given by sinq m = m ( l 2d ) = m sinq1 = m sin (12.6° ) Thus, additional orders will be found at m = 2:

sinq 2 = 2sin12.6° = 0.436

and

q 2 = sin −1 ( 0.436 ) = 25.8°

m =3:

sinq 3 = 3sin12.6° = 0.654

and

q 3 = sin −1 ( 0.654 ) = 40.8°

m=4:

sinq 4 = 4 sin12.6° = 0.873

and

q 4 = sin −1 ( 0.873) = 60.8°

m =5:

sinq 5 = 5sin12.6° = 1.09

Impossible, since sinq ≤ 1 for all q .

No orders higher than m = 4 will be seen because it is mathematically impossible for the sinq to be greater than one. 27.21

The interplanar spacing in the crystal is given by Bragg’s law as d=

27.22

ml (1)( 0.140 nm ) = = 0.281 nm 2sinq 2sin14.4°

From the Compton shift equation, the wavelength shift of the scattered x-rays is Δl =

h 6.63 × 10 −34 J ⋅s (1− cosq ) = (1− cos55.0°) me c (9.11× 10−31 kg) (3.00 × 108 m s)

⎛ 1 nm ⎞ = 1.03 × 10 −12 m ⎜ −9 ⎟ = 1.03 × 10 −3 nm ⎝ 10 m ⎠ 27.23

If the scattered photon has energy equal to the kinetic energy of the recoiling electron, the energy of the incident photon is divided equally between them. Thus, Ephoton =

(E ) 0

photon

2

⇒

hc hc = , so l = 2 l0 and Δl = 2 l0 − l0 = l0 = 0.001 60 nm l 2 l0

The Compton scattering formula, Δl = lC (1− cosq ) , then gives the scattering angle as ⎛ Δl ⎞ ⎛ 0.001 60 nm ⎞ −1 q = cos−1 ⎜1− ⎟ = cos ⎜1− ⎟ = 70.0° ⎝ 0.002 43 nm ⎠ ⎝ lC ⎠

68719_27_ch27_p324-340.indd 332

1/7/11 2:58:24 PM

Quantum Physics

27.24

At point A, the incident photon scatters at angle q from the first electron. The shift in wavelength that occurs in this scattering process is given by the Compton equation as

Incident photon l

Electron 1 A q

h l′ − l = (1− cosq ) me c

333

l′

l″

q

Reflected photon

B

Electron 2

q ′ = 180° − q

At point B, this intermediate photon scatters from the second electron at angle q ′ = 180° −q . Since cos(180° −q ) = cos180° cosq + sin180°sinq = − cosq , the Compton equation gives the wavelength shift that occurs in the scattering at B as l ′′ − l ′ =

h (1+ cosq ) me c

Hence, the total wavelength shift that occurs in the two scattering processes is given by l ′′ − l = ( l ′′ − l ′ ) + ( l ′ − l ) = or 27.25

l ′′ − l =

h h 2h (1+ cosq ) + (1− cosq ) = me c me c me c

2 ( 6.63 × 10 −34 J ⋅s )

(9.11× 10−31 kg) (3.00 × 108 m s)

= 4.85 × 10 −12 m = 4.85 × 10 −3 nm

The figure at the right shows the situation before and after the scattering process. Note that the scattering angle is q = 180°, so the Compton equation gives l′ − l =

Incident photon Before

+x

l, p, E l′, p′, E′

h 2h (1− cos180°) = me c me c

(a)

pe, KEe

After Scattered photon

l′ = l +

Electron at rest

Recoiling electron

+x

2 (6.63×10 −34 J ⋅ s) 2h = 0.110 ×10 −9 m + = 0.115×10 −9 m −31 8 me c 9.11×10 kg 3.00 ×10 m s ( )( )

The momentum of the incident photon is p = h l, while that of the scattered photon is p′ = − h l ′ (the negative sign is included since momentum is a vector quantity and the scattered photon travels in the negative x-direction). Thus, conservation of momentum gives pe − h l ′ = h l + 0, or the momentum of the recoiling electron is pe =

⎛ ⎞ h h 1 1 + = (6.63×10 −34 J ⋅ s) ⎜ + ⎟ −9 −9 ⎝ l l′ 0.110 ×10 m 0.115×10 m ⎠

= 1.18 ×10 −23 kg ⋅ m s (b)

Assuming the recoiling electron is nonrelativistic, its kinetic energy is given by

(1.18 ×10−23 kg ⋅ m s) = 7.64 ×10−17 J ⎛⎜ 1 eV ⎞⎟ = 478 eV p2 KEe = e = ⎝ 1.60 ×10 −19 J ⎠ 2me 2 (9.11×10 −31 kg) 2

continued on next page

68719_27_ch27_p324-340.indd 333

1/7/11 2:58:27 PM

334

Chapter 27

Note that this energy is very small in comparison to the rest energy of an electron. Thus, our assumption that the recoiling electron would be non-relativistic is seen to be valid. 27.26

First, observe that v = 2.18 × 10 6 m s 0, the decay may occur spontaneously. (a)

40 + For the decay 40 20 Ca → e + 19 K, the masses of the electrons do not automatically cancel. Thus, we add 20 electrons to each side of the decay to yield neutral atoms and obtain

(

40 20

Ca + 20e − ) → e + + ( 1940 K + 19e − ) + e −

(

Then, Q = m or (b)

or

(b)

40 19 K atom

40

Ca atom → e + + 40 K atom + e −

)

− 2me c 2 = ⎡⎣39.962 591 u − 39.963 999 u − 2 ( 0.000 549 u ) ⎤⎦ c 2 so the decay cannot occur spontaneously .

4 140 In the decay 144 60 Nd →2 He + 58 Ce, we may add 60 electrons to each side, forming all neutral atoms, and use masses from Appendix B to find

(

(a)

−m

Q = ( −0.002 506 u ) c 2 < 0

Q= m

29.28

40 20 C a atom

or

66 28

144 60

Nd

−m

−m

4 2 He

) c = (143.910 083 u − 4.002 603 u − 139.905 434 u) c 2

140 58 C e

2

Q = ( +0.002 046 u ) c 2 > 0 so the decay can occur spontaneously . Ni →

66 29

Cu + −10 e +n e

Because of the mass differences, neglect the kinetic energy of the recoiling daughter nucleus in comparison to that of the other decay products. Then, the maximum kinetic energy of the beta particle occurs when the antineutrino is given zero energy. That maximum is

(

KEm ax = m

66

Ni

−m

66

Cu

)c

2

= ( 65.929 1 u − 65.928 9 u ) ( 931.5 MeV u )

= 0.186 MeV = 186 keV 29.29

In the decay 13 H → 32 He + −10 e +n e , the antineutrino is massless. Adding 1 electron to each side of the decay gives ( 13 H + e − ) → ( 32 He +2 e − ) +n e , or 13 H atom → 32 Heatom +n e . Therefore, using neutral atomic masses from Appendix B, the energy released is

(

E = ( Δm ) c 2 = m

3

H

−m

3

He

)c

2

= ( 3.016 049 u − 3.016 029 u ) ( 931.5 MeV u )

= 0.018 6 MeV = 18.6 keV

68719_29_ch29_p360-376.indd 369

1/7/11 3:04:45 PM

370

29.30

Chapter 29

The initial activity of the 1.00-kg carbon sample would have been ⎛ 15.0 counts min ⎞ 4 −1 R0 = (1.00 × 10 3 g ) ⎜ ⎟⎠ = 1.50 × 10 min 1.00 g ⎝ From R = lN = lN 0 e − l t = R0 e − l t, and T1 2 = 5 730 yr for 14 C (Appendix B), the age of the sample is t=−

ln ( R R0 ) l

= − T1 2

= − ( 5 730 yr ) 29.31

ln 2

ln ⎡⎣( 5.00 × 10 2 min −1 ) (1.50 × 10 4 min −1 ) ⎤⎦ ln 2

= 2.81 × 10 4 yr

From R = lN = lN 0 e − l t = R0 e − l t, and T1 2 = 5 730 yr for 14 C (Appendix B), the age of the sample is t=−

29.32

ln ( R R0 )

ln ( R R0 ) λ

= − T1 2

ln ( R R0 ) ln 2

= − ( 5 730 yr )

ln ( 0.600 ) = 4.22 × 10 3 yr ln 2

The energy released in the reaction is given by

(

Q = ( Δm ) c 2 = m

1 1H

+m

27 13 Al

−m

27 14 Si

)

− mn c 2

= [1.007 825 u + 26.981 539 u − 26.986 721 u − 1.008 665 u ]( 931.5 MeV u ) = −5.61 MeV If we require total energy to be conserved and ignore the kinetic energy of the recoiling product nucleus, the kinetic energy of the emerging neutron must be KE f = KEi + Q = 6.61 MeV − 5.61 MeV = 1.00 MeV 29.33

We identify the missing particles by requiring that both the total mass number and the total charge number be the same on the two sides of the equation and by remembering that some form of neutrino always accompanies the emission of a beta particle or positron. (a)

29.34

21 10

Ne + 42 He → 1224 Mg + 10 n

U + 10 n → 90 Sr + 38

(b)

235 92

(c)

2 11 H → 12 H +

(a)

mi = m

(b)

mf = m

(c)

Q = mi − m f c 2 = [8.023 829 u − 8.025 594 u ]( 931.5 MeV u ) = −1.64 MeV

(d)

m p v = ( m B e + mn ) V

(e)

(

1 1H

+m

7 4 Be

0 +1

7 3 Li

144 54

Xe + 2 10 n

e + ne = 1.007 825 u + 7.016 004 u = 8.023 829 u

+ mn = 7.016 929 u + 1.008 665 u = 8.025 594 u

)

From conservation of energy, the total kinetic energy before the reaction, plus the energy released during the reaction, must equal the total kinetic energy after the reaction. That is, KE p + Q = KEB e + KEn , or

1 2

m p v2 =

1 2

(m

Be

+ mn ) V 2 − Q .

continued on next page

68719_29_ch29_p360-376.indd 370

1/7/11 3:04:48 PM

Nuclear Physics

(f )

371

⎛ mp ⎞ ⎛ 1.007 825 u ⎞ m⎞ ⎛ −1.64 MeV = 1.88 MeV KEm in = ⎜ 1+ ⎟ Q = ⎜ 1+ ⎟ Q = ⎜ 1+ ⎝ ⎠ ⎟ ⎜ 7.016 004 u ⎟⎠ m Li ⎠ M ⎝ ⎝ 7 3

29.35

29.36

We determine the product nucleus by requiring that both the total mass number and the total charge number be the same on the two sides of the reaction equation. The completed reaction equations are given below: B + 42 He → 11 H +

(a)

10 5

(a)

For the first reaction:

(

Q1 = mn + m

13 6

2 1H

C

(b)

−m

3 1H

)c

1 1H

+m

2 1H

−m

10 5

B

= [1.008 665 u + 2.014 102 u − 3.016 049 u ]( 931.5 MeV u ) exothermic

H + 21 H → 23 He

3 2 He

= 5.494 Mev > 0

2

⇒ 1 1

For the second reaction:

(

C + 11 H → 42 He +

n + 21 H → 31 H

= 6.258 Mev > 0

Q2 = m

13 6

)c

2

⇒

= [1.007 825 u + 2.014 102 u − 3.016 029 u ]( 931.5 MeV u ) exothermic

(b)

Since Q1 > Q1 , the first reaction released more energy . One reason for this is that the product nucleus in the second reaction contains 2 protons. Some energy had to be left stored as electrical potential energy of this system, leaving less energy to be released as kinetic energy of the product nucleus.

(c)

Assuming the electrical potential energy of the 2 protons in 23 He fully accounts for the difference in the Q-values of the two reactions, we have ΔQ = PEe = ke e 2 r, where r is the distance separating the 2 protons. Thus, 9 2 2 −19 ke e 2 (8.99 × 10 N ⋅ m C ) (1.60 × 10 C ) ⎛ 1 MeV ⎞ −15 r= = ⎜⎝ ⎟ = 1.88 × 10 m ΔQ 1.60 × 10 −13 J ⎠ ( 6.258 − 5.494 ) MeV 2

29.37

(a)

Requiring that both charge and the number of nucleons (atomic mass number) be conserved, 1 198 0 the reaction is found to be 197 79 Au + 0 n → 80 Hg + −1 e +n e . Note that the antineutrino has been included to conserve electron-lepton number, which will be discussed in the next chapter.

(b)

We add 79 electrons to both sides of the reaction equation given above to produce neutral atoms 1 198 so we may use mass values from Appendix B. This gives 197 79 Au atom + 0 n → 80 Hgatom +n e , and, remembering that the antineutrino is massless, the Q-value is found to be

(

Q = ( Δm ) c 2 = m

197 79 Au

+ mn − m

198 80 Hg

)c

2

= (196.966 552 u + 1.008 665 u − 197.966 750 u ) ( 931.5 MeV u ) = 7.89 MeV The kinetic energy carried away by the daughter nucleus is negligible. Thus, the energy released may be split in any manner between the electron and antineutrino, with the maximum kinetic energy of the electron being 7.89 MeV . 29.38

We complete the reaction equation in each case by requiring that both the total mass number and the total charge number be the same on the two sides of the equation. (a)

68719_29_ch29_p360-376.indd 371

4 2

He + 147 N → 11 H + 178 O

(b)

7 3

Li + 11 H → 42 He + 42 He

1/7/11 3:04:51 PM

372

29.39

Chapter 29

(a)

Determine the product of the reaction by requiring that both the total mass number and the total charge number be the same on the two sides of the equation. The completed reaction equation is 73 Li + 42 He → 105 B + 01 n.

(b)

Q = Δmc 2 = ⎡⎢ m ⎣

(

7 3 Li

+m

4 2 He

) − (m

10 5B

+m

1 0n

)⎤⎦⎥ c

2

= ⎡⎣( 7.016 004 u + 4.002 603 u ) − (10.012 937 u + 1.008 665 u ) ⎤⎦ ( 931.5 MeV u ) = −2.79 MeV 29.40

For equal amounts of biological damage, the two doses (in rem units) must be equal, or

( heavy ion dose in rad ) × RBE

heavy ions

= ( x-ray dose in rad ) × RBE x-rays

or ion dose in rad = 29.41

( x-ray dose in rad ) × RBE

x-rays

RBE heavy ions

=

(100 rad )(1.0 ) 20

= 5.0 rad

For each rad of radiation, 10 − 2 J of energy is delivered to each kilogram of absorbing material. Thus, the total energy delivered in this whole body dose to a 75.0-kg person is J kg ⎞ ⎛ E = ( 25.0 rad ) ⎜ 10 − 2 ⎟ ( 75.0 kg ) = 18.8 J ⎝ rad ⎠

29.42

(a)

Each rad of radiation delivers 10 − 2 J of energy to each kilogram of absorbing material. Thus, the energy delivered per unit mass with this dose is E J kg ⎞ ⎛ = ( 200 rad ) ⎜ 10 − 2 ⎟ = 2.00 J kg ⎝ m rad ⎠

(b)

From E = Q = mc ( ΔT ) , the expected temperature rise with this dosage is ΔT =

29.43

E m 2.00 J kg = = 4.78 × 10 − 4 °C c 4 186 J kg ⋅°C

The rate of delivering energy to each kilogram of absorbing material is J kg ⎛ E m⎞ ⎛ − 2 J kg ⎞ ⎜⎝ ⎟⎠ = (10 rad s ) ⎜⎝ 10 ⎟⎠ = 0.10 Δt rad s The total energy needed per unit mass is ⎛ J ⎞ E m = c ( ΔT ) = ⎜ 4 186 ( 50°C ) = 2.1× 105 J kg kg ⋅°C ⎟⎠ ⎝ so the required time will be Δt =

68719_29_ch29_p360-376.indd 372

energy needed 2.1× 10 5 J kg 1d ⎛ ⎞ = = 2.1× 10 6 s ⎜ = 24 d ⎝ 8.64 × 10 4 s ⎟⎠ delivery rate 0.10 J kg ⋅s

1/7/11 3:04:53 PM

Nuclear Physics

29.44

(a)

373

The number of x-rays taken per year is production = (8 x-ray d ) ( 5 d week ) ( 50 weeks yr ) = 2.0 × 10 3 x-ray yr so the exposure per x-ray taken is exposure rate =

(b)

exposure 5.0 rem yr = 2.5 × 10 −3 rem x-ray = production 2.0 × 10 3 x-ray yr

The exposure due to background radiation is 0.13 rem yr . Thus, the work-related exposure of 5.0 rem yr is 5.0 rem yr ≈ 38 times background levels 0.13 rem yr

29.45

(a)

(

)

From N = R l = R0 e − l t l = T1 2 R0 ln 2 e the 10-day period is ⎛ T1 2 R0 ⎞ − t ln 2 T 1− e ΔN = N 0 − N = ⎜ ⎟ ⎝ ln 2 ⎠

(

12

− t ln 2 T1 2

, the number of decays occurring during

)

⎡ (14.3 d )(1.31× 10 6 decay s ) ⎛ 8.64 × 10 4 s ⎞ − (10.0 d ) ln 2 14.3 d =⎢ ⎜⎝ ⎟⎠ 1− e ln 2 1d ⎢⎣

(

⎤

)⎥⎥ ⎦

= 8.97 × 1011 decays , and one electron is emitted per decay (b)

The total energy deposited is found to be ⎛ ⎛ 1.60 × 10 −16 J ⎞ keV ⎞ 8.97 × 1011 decays ) ⎜ E = ⎜ 700 ( ⎟⎠ = 0.100 J ⎟ decay ⎠ 1 keV ⎝ ⎝

(c)

The total absorbed dose (measured in rad) is given by dose =

29.46

(a)

energy deposited per unit mass ( 0.100 J 0.100 kg ) = = 100 rad energy deposition per rad ⎛ −2 J kg ⎞ 10 ⎜⎝ ⎟ rad ⎠

The dose (in rem) received in time Δt is given by rad ⎞ ⎤ rem ⎞ ⎡⎛ ⎛ dose = ( dose in rad ) × RBE = ⎢⎜ 100 × 10 − 3 ⎟⎠ Δt ⎥ × (1.00 ) = ⎜⎝ 0.100 ⎟ Δt ⎝ h h ⎠ ⎣ ⎦ If this dose is to be 1.0 rem, the required time is Δt =

(b)

1.0 rem = 10 h 0.100 rem h

Assuming the radiation is emitted uniformly in all directions, the intensity of the radiation is given by I = I 0 4p r 2 . Therefore, I 0 4p r 2 Ir (1.0 m ) = = 2 I1 I 0 4p (1.0 m ) r2

2

continued on next page

68719_29_ch29_p360-376.indd 373

1/7/11 3:04:54 PM

374

Chapter 29

r = (1.0 m )

and 29.47

(a)

I1 100 mrad h = 3.2 m = (1.0 m ) Ir 10 mrad h

The mass of a 116 C atom is matom = (11.011 u)(1.661× 10 −27 kg u) = 1.829 × 10 −26 kg = 1.829 × 10 −23 g and the mass of 1 mole of 116 C is M = matom NA = (1.829 × 10 −23 g atom ) ( 6.022 × 10 23 atoms mol ) = 11.01 g mol The number of moles in a 3.50 mg sample is then n=

(b)

msam ple M

=

3.50 × 10 −6 g = 3.18 × 10 −7 mol 11.01 g mol

The number of nuclei in the original sample is N0 = nNA = ( 3.18 × 10 −7 mol ) ( 6.022 × 10 23 atoms mol ) = 1.91× 1017 atoms or alternatively, N0 =

(c)

msam ple matom

=

3.50 × 10 −6 g = 1.91× 1017 atoms . 1.829 × 10 −23 g atom

The initial activity is ⎛ ln 2 ⎞ ln 2 R0 = lN 0 = ⎜ N0 = (1.91× 1017 ) = 1.08 × 1014 Bq ⎟ ( 20.4 min )( 60.0 s min ) ⎝ T1 2 ⎠

(d)

The activity after an elapsed time of t = 8.00 h = 480 min will be R = R0 e − l t = R0 e

29.48

− t ln 2 T1 2

= (1.08 × 1014 Bq ) e − ( 480 m in ) ln 2 ( 20.4 m in ) = 8.92 × 10 6 Bq

(

We must first calculate the Q value, given by Q = (Δm)c 2 = ⎡⎢ m n + m ⎣ 1 0

4 2 He

) − (m

2 1H

+m

)⎤⎥⎦ c . 2

3 1H

Q = ⎡⎣(1.008 665 u + 4.002 603 u ) − ( 2.014 102 u + 3.016 049 u ) ⎤⎦ ( 931.5 MeV u ) = −17.6 MeV The threshold kinetic energy for the incident neutron is then ⎛ mn ⎞ ⎛ 1.008 665 u ⎞ KEm in = ⎜ 1+ (17.6 MeV ) = 22.0 MeV ⎟ Q = ⎜ 1+ ⎜⎝ 4.002 603 u ⎟⎠ M He ⎟⎠ ⎝ 1 0

4 2

29.49

From R = R0 e − l t, the elapsed time is t=−

29.50

ln ( R R0 ) l

= − T1 2

ln ( R R0 ) ln 2

= − (14.0 d )

ln ( 20.0 mCi 200 mCi ) = 46.5 d ln 2

To compute the Q value for the reaction 146 C → 147 N + e − +n , we add 6 electrons to each side of the equation to obtain ( 146 C + 6e − ) → ( 147 N + 7e − ) +n , or 146 Catom → 147 N atom +n . Now, we use the atomic masses from Appendix B of the textbook. Since the neutrino is a massless particle, Q = (m C − m N )c 2, giving 14 6

atom

14 7

atom

Q = (14.003 242 u − 14.003 074 u ) ( 931.5 MeV u ) = 0.156 MeV = 156 keV

68719_29_ch29_p360-376.indd 374

1/7/11 3:04:56 PM

Nuclear Physics

29.51

(a)

If we assume all the 87 Sr nuclei in a gram of material came from the decay of 87 Rb nuclei, the original number of 87 Rb nuclei was N0 = 1.82 × 1010 + 1.07 × 10 9 = 1.93 × 1010. Then, from N = N 0 e − l t, the elapsed time is

t =−

29.52

375

T ln ( N N 0 ) ln ( N N 0 ) =− 12 =− l ln 2

(4.8 ×10

10

⎛ 1.82 ×1010 yr ) ln ⎜ 10 ⎝ 1.93×10 ln 2

⎞ ⎟ ⎠

= 4.1×10 9 yr

(b)

It could be no older. It could be younger if some 87 Sr were initially present.

(c)

We have assumed that all the

87

Sr present came from the decay of

87

Rb.

From R = lN = lN0 e − l t, if we have an activity of R after time t has passed, the number of unstable nuclei that must have been present initially is given by N0 = Re l t l . With R = 10 Ci = 10(3.7 × 1010 Bq) = 3.7 × 1011 decay s, l = ln 2 T1 2 = ln 2 5.2 yr , and t = 30 months = 2.5 yr, this yields ⎡ ( 5.2 yr ) ( 3.156 × 10 7 s 1 yr ) ⎤ ( ln 2 5.2 yr )( 2.5 yr ) N0 = ( 3.7 × 1011 s −1 ) ⎢ = 1.2 × 10 20 ⎥e ln 2 ⎢⎣ ⎥⎦ Thus, the initial mass of 60 C required is found to be (using Appendix B from the text) m = N 0 matom = (1.2 × 10 20 ) ( 59.933 822 u ) = 7.2 × 10 21 u or

29.53

⎛ 1.66 × 10 −27 kg ⎞ ⎛ 10 6 mg ⎞ m = 7.2 × 10 21 u ⎜ ⎟⎠ ⎜⎝ 1 kg ⎟⎠ = 12 mg 1u ⎝

The total activity of the working solution at t = 0 was

(R ) 0

total

= ( 2.5 mCi mL )(10 mL ) = 25 mCi

Therefore, the initial activity of the 5.0-mL sample, drawn from the 250-mL working solution, was

(R ) 0

sam ple

⎛ 5.0 mL ⎞ ⎛ 5.0 mL ⎞ = ( R0 )total ⎜ = ( 25 mCi ) ⎜ = 0.50 mCi = 5.0 × 10 − 4 Ci ⎝ 250 mL ⎟⎠ ⎝ 250 mL ⎟⎠

With a half-life of 14.96 h for 24 Na (Appendix B), the activity of the sample after 48 h is R = R0 e − l t = R0 e

− t ln 2 T1 2

= ( 5.0 × 10 − 4 Ci ) e − ( 48 h ) ln 2 (14.96 h )

= 5.4 × 10 − 5 Ci = 54 mCi 29.54

(a)

The mass of a single

40

K atom is

m = ( 39.964 u )(1.66 × 10 −27 kg u ) = 6.63 × 10 −26 kg = 6.63 × 10 −23 g Therefore, the number of 40 K nuclei in a liter of milk is N=

total mass of 40 K present ( 2.00 g L ) ( 0.011 7 100 ) = 3.53 × 1018 L = 6.63 × 10 −23 g mass per atom

continued on next page

68719_29_ch29_p360-376.indd 375

1/7/11 3:04:59 PM

376

Chapter 29

and the activity due to potassium is R = lN = (b)

Using R = R0 e − l t, the time required for the 131 I activity to decrease to the level of the potassium is given by t =−

29.55

(3.53 × 1018 L ) ln 2 N ln 2 = = 60.6 Bq L T1 2 (1.28 × 109 yr ) (3.156 × 107 s yr )

ln ( R R0 ) l

=−

T1 2 ln ( R R0 ) ln 2

=−

(8.04 d ) ln ( 60.6 2 000 ) ln 2

= 40.6 d

The decay constant for 235 U is l235 = ln 2 T1 2 = ln 2 0.70 × 10 9 yr = 9.9 × 10 −10 yr −1 , while that for 238 U is l238 = ln 2 4.47 × 10 9 yr = 1.55 × 10 −10 yr −1 . Assuming there were N 0 nuclei of each isotope present initially, the number of each type still present should be N 235 = N 0 e − l t and N 238 = N 0 e − l t . With the currently observed ratio of 235 U to 238 U being 0.007, we have − l − l )t N 235 N 238 = e ( = 0.007, or our estimate of the elapsed time since the release of the elements forming our Earth is 235

238

235

t=− 29.56

(a)

ln ( 0.007 ) ln ( 0.007 ) =− = 5.9 × 10 9 yr l235 − l236 9.9 × 10 −10 yr −1 − 1.55 × 10 −10 yr −1

Obtaining the mass of a single 239 94 Pu atom from the table of Appendix B in the text, we find N0 =

(b)

238

msam ple matom

=

1.0 kg

( 239.052 156 u ) (1.66 × 10

−27

kg u )

= 2.5 × 10 24

The initial activity of this sample is ⎡ ⎤ ⎛ ln 2 ⎞ ln 2 N0 = ⎢ R0 = lN 0 = ⎜ ⎥ ( 2.5 × 10 24 ) = 2.3 × 1012 Bq ⎟ 7 ⎝ T1 2 ⎠ ⎢⎣ ( 24 000 yr ) ( 3.156 × 10 s yr ) ⎥⎦

(c)

From R = R0 e − l t , the time that must elapse before this sample will have a “safe” activity level of 0.10 Bq is t=−

68719_29_ch29_p360-376.indd 376

ln ( R R0 ) l

=−

T1 2 ln ( R R0 ) ln 2

=−

( 24 000 yr ) ln ( 0.10 ln 2

2.3 × 1012 )

= 1.1× 10 6 yr

1/7/11 3:05:02 PM

30 Nuclear Physics and Elementary Particles QUICK QUIZZES 1.

Choice (a). This reaction fails to conserve charge and also fails to conserve baryon number. For each of these reasons, it cannot occur.

2.

Choice (b). This reaction fails to conserve charge and cannot occur.

ANSWERS TO MULTIPLE CHOICE QUESTIONS 1.

The total energy released was E = (17 × 10 3 ton) ( 4.0 × 10 9 J 1 ton ) = 6.8 × 1013 J, and according to the mass-energy equivalence (E = mc 2), the mass converted was m=

E 6.8 × 1013 J = = 7.6 × 10 −4 kg = 0.76 g 2 c 2 ( 3.00 × 108 m s )

or m ∼ 1 g, and the correct choice is seen to be (d). 2.

The energy released in the decay n → p + e − +n e is Q = (Δm)c 2 = (mn − m p − me )c 2, or combining the proton and electron to form a neutral hydrogen atom, Q = (mn − m H )c 2. We may then use the atomic masses from Appendix B in the textbook to obtain 2 1

atom

Q = (1.008 665 u − 1.007 825 u ) ( 931.5 MeV u ) = 0.782 MeV Alternately, we may use the particle masses (in energy units) from Table 30.2 in the textbook to obtain Q = (mn − m p − me )c 2 = (939.6 MeV c 2 − 938.3 MeV c 2 − 0.511 MeV c 2 )c 2 = 0.789 MeV From either approach, we see that the best choice is (a). 3.

Both the charge and mass of a particle are independent of its spin, so both choices (c) and (d) are false. A spin- 12 particle could be among the decay products, provided it is possible for the spins of all the decay products to couple to 32 and conserve angular momentum. Also, in a magnetic field, a spin- 32 particle could have spin states of ms = 3 2, 1 2, − 1 2, and − 3 2 so choices (a) and (e) are false, while choice (b) is true.

4.

The decay p → +10 e +n e would conserve charge (+1 → +1+ 0), electron lepton number (0 → −1+ 1), and strangeness (0 → 0 + 0), and can conserve energy if the total kinetic energy of the decay 377

68719_30_ch30_p377-408.indd 377

1/7/11 3:06:01 PM

378

Chapter 30

products equals the energy equivalent of the mass loss. However, it does not conserve baryon number (+1 → 0 + 0), and the decay cannot occur. The correct choice is then (b). 5.

Positively charged particles, such as protons and alpha particles, have difficulty approaching the target nuclei because of Coulomb repulsion. Fast-moving particles may not stay in close proximity with a uranium nucleus long enough to have a good probability of producing a reaction. The best particles to trigger a fission reaction of the uranium nuclei are slow-moving neutrons, so choice (d) is the correct answer.

6.

Slow neutrons have a much higher probability of causing fission in a collision with a nucleus in the fuel elements than do fast or high energy neutrons. The purpose of the moderator is to slow the neutrons down and without it the chain reaction would quickly die out. The correct choice is (c).

7.

The annihilation −10 e + +10 e → g can conserve energy [2(0.511 MeV) = 1.02 MeV], does conserve charge [−1+ 1 = 0], conserves baryon number [0 + 0 = 0], and conserves lepton number [+1− 1 = 0]. However, the total momentum is zero before annihilation, and the momentum of the single photon afterward is p = 1.02 MeV c ≠ 0. Thus, it cannot occur, and the correct choice is (b).

8.

1 137 96 1 In the fission reaction 235 92 U + 0 n → 53 I + 39 Y + n( 0 n), where n is some unknown number of neutrons, we see that charge is conserved (92 + 0 = 53 + 39 + 0) regardless of the value of n. The reaction must also conserve baryon number, so it is necessary that

235 + 1 = 137 + 96 + n

or

n=3

and (c) is seen to be the correct choice. 9.

The reaction of choice (c) fails to conserve charge [0 + 1 ≠ 0 + 0], while the reaction of choice (d) fails to conserve baryon number [+1 ≠ 2(+1) + 0 + 0], so neither of these reactions can occur. The reactions of choices (a), (b), and (e) satisfy all conservation laws and may occur. The correct answers for this question are choices (c) and (d).

10.

The reaction of choice (a) fails to conserve baryon number [1+ 1 ≠ 1+ 1− 1], while the reaction of choice (e) fails to conserve tau-lepton number [+1 ≠ 0 − 1+ 0], so neither of these reactions can occur. The reactions of choices (b), (c), and (d) satisfy all conservation laws and may occur. The correct answers for this question are choices (a) and (e).

11.

In a fusion reactor, the plasma must have a very high temperature so the nuclei have sufficient energy to overcome Coulomb repulsion and collide. Also, the plasma must be sufficiently dense to yield a good probability of collisions between nuclei and must be contained long enough to allow a large number of collisions to occur, so the reactions can be self-sustained. However, there are no requirements that the nuclei in the plasma be radioactive or that it consist only of hydrogen. Fusion of elements other than hydrogen does occur inside stars, but much higher temperatures are required for this to occur. The correct choices for this question are (a), (c), and (d).

12.

When a particle of mass m and charge q enters a magnetic field perpendicular to the direction of the field, it is deviated into a circular path of radius r = mv qB = 2m(KE) qB. In this case, the kinetic energy, KE, and the magnetic field strength, B, are the same for the electron and the alpha particle. However, the ratio m q is much smaller for the electron than for the alpha particle, so the path of the electron has a smaller radius. This means the electron is deflected more than the alpha particle, and the correct choice is (b).

68719_30_ch30_p377-408.indd 378

1/7/11 3:06:03 PM

Nuclear Physics and Elementary Particles

379

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.

The two factors presenting the most technical difficulties are the requirements of a high plasma density and a high plasma temperature. These two conditions must be met simultaneously and make containment of the plasma very difficult.

4.

Notice in the fusion reactions discussed in the text that the most commonly formed by-product of the reactions is helium, which is inert and not radioactive.

6.

They are hadrons. Such particles decay into other strongly interacting particles such as p, n, and p with very short lifetimes. In fact, they decay so quickly that they cannot be detected directly. Decays which occur via the weak force have lifetimes of 10 −13 s or longer; particles that decay via the electromagnetic force have lifetimes in the range of 10 −16 s to 10 −19 s.

8.

Each flavor of quark can have three colors, designated as red, green, and blue. Antiquarks are colored antired, antigreen, and antiblue. Baryons consist of three quarks, each having a different color. Mesons consist of a quark of one color and an antiquark with a corresponding anticolor. Thus, baryons and mesons are colorless or white.

10.

The decays of the neutral pion, eta, and neutral sigma occur by the electromagnetic interaction. These are the three shortest lifetimes in the table. All produce photons, which are the quanta of the electromagnetic force, and all conserve strangeness.

12.

A neutron inside a nucleus is stable because it is in a lower energy state than a free neutron and lower in energy than it would be if it decayed into a proton (plus electron and antineutrino). The nuclear force gives it this lower energy by binding it inside the nucleus and by favoring pairing between neutrons and protons.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.

192 MeV

4.

9.4 × 10 2 g

6.

2.63 kg

8.

(a) 3.1× 1010 g

(b)

1.3 × 108 mol; 7.8 × 10 31 atoms

(c)

2.6 × 10 21 J

(d)

5.5 yr

(e)

Fission alone cannot meet the world’s energy needs.

(a)

4 × 1015 g

10.

(b)

5 × 10 3 yr

(c) The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, but the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half-million years. 12.

5.49 MeV

14.

(a)

68719_30_ch30_p377-408.indd 379

13 7

N

(b)

13 6

C

(c)

14 7

N

1/7/11 3:06:05 PM

380

16.

Chapter 30

(d)

15 8

(a)

8.68 MeV

O

(e)

15 7

N

(f)

12 6

C

(b) The proton and the boron nucleus must each have enough kinetic energy to overcome the repulsive electrical force one exerts on the other. 18.

118 MeV

20.

(a) may occur via the weak interaction (b) violates conservation of baryon number – cannot occur by any interaction

22.

24.

(a)

cannot occur, violates baryon number

(b)

can occur, no laws violated

(c)

can occur, no laws violated

(d)

can occur, no laws violated

(e)

cannot occur, violates baryon number and muon-lepton number

(a)

See Solution.

(b)

The second reaction violates conservation of strangeness and cannot occur via the strong or electromagnetic interactions.

(c)

If a neutral kaon, K 0, was produced in addition to those shown, the second reaction could occur without violating any conservation laws. Because of the greater total mass of the product particles, the total kinetic energy of the incident particles in the this reaction would have to be greater than in the first reaction.

26.

See Solution.

28.

See Solution.

30.

(a) Q = −e, antiproton

32.

(a) conservation of charge

(b)

Q = 0, antineutron

(b) conservation of electron-lepton number and conservation of muon-lepton number (c) conservation of baryon number 34.

n m and ne

36.

a neutron

38.

(a)

7.85 × 1013 J

(b)

$1.74 × 10 6

(c)

$1.50 × 10 3

(d) At a 10% profit margin, the profit is $1.74 × 10 5 on each kilogram of deuterium.

68719_30_ch30_p377-408.indd 380

1/7/11 3:06:06 PM

Nuclear Physics and Elementary Particles

3.61× 10 30 J

381

1.63 × 108 yr

40.

(a)

(b)

42.

The first reaction has a net 1u and 2d quarks both before and after the reaction. The second reaction has a net 1u and 2d quarks before the reaction and 1u, 3d, and 1s quark present afterwards.

44.

0.827 9c

PROBLEM SOLUTIONS 30.1

The energy consumed by a 100-W lightbulb in a 1.0-h time period is ⎛ 3 600 s ⎞ = 3.6 × 10 5 J E = P ⋅ Δt = (100 J s )(1.0 h ) ⎜ ⎝ 1 h ⎟⎠ The number of fission events, yielding an average of 208 MeV each, required to produce this quantity of energy is n=

30.2

E 3.6 × 10 5 J ⎛ 1 MeV = ⎜ 208 MeV 208 MeV ⎝ 1.60 × 10 −13

The energy released in the reaction 10 n + Q = ( Δm ) c 2 = ⎡ m ⎣

235 92 U

− 2 mn − m

98 40 Zr

235 92

⎞ 16 ⎟ = 1.1× 10 J⎠

U →

−m

135 52 Te

98 40

1 Zr + 135 52 Te + 3 0 n is

⎤ c2 ⎦

= ⎡⎣ 235.043 923 u − 2 (1.008 665 u ) − 97.912 0 u − 134.908 7 u ⎤⎦ ( 931.5 MeV u ) = 192 MeV 30.3

The energy released in the reaction 10 n + Q = ( Δm ) c 2 = ⎡ m ⎣

235 92 U

− 11mn − m

88 38 Sr

235 92

U →

−m

136 54 Xe

88 38

1 Sr + 136 54 Xe + 12 0 n is

⎤ c2 ⎦

= ⎡⎣ 235.043 923 u − 11(1.008 665 u ) − 87.905 614 u − 135.907 220 u ⎤⎦ ( 931.5 MeV u ) = 126 MeV 30.4

The total energy released was ⎛ 4.0 × 10 9 J ⎞ ⎛ ⎞ 1 MeV = 5.0 × 10 26 MeV E = 20 × 10 3 ton TNT ⎜ −13 ⎜ ⎟ J ⎟⎠ ⎝ 1 ton TNT ⎠ ⎝ 1.60 × 10

(

)

The number of 235 U nuclei that must have undergone fission to yield this energy is n=

E 5.0 × 10 26 MeV = = 2.4 × 10 24 nuclei 208 MeV nucleus 208 MeV nucleus

The mass of 235 U which will contain this number of atoms is ⎛ n ⎞ ⎛ 2.4 × 10 24 atoms ⎞ ⎛ g ⎞ 2 m=⎜ M = ⎜⎝ 235 ⎟ = 9.4 × 10 g m ol 23 ⎜ ⎟ ⎟ mol ⎠ ⎝ 6.02 × 10 atoms mol ⎠ ⎝ NA ⎠

68719_30_ch30_p377-408.indd 381

1/7/11 3:06:08 PM

382

30.5

Chapter 30

(a)

With a specific gravity of 4.00, the density of soil is r = 4.00 × 10 3 kg m 3. Thus, the mass of the top 1.00 m of soil is 2 1m ⎞ ⎤ kg ⎞ ⎡ ⎛ 2 ⎛ m = rV = ⎜ 4.00 × 10 3 1.00 m 43 560 ft = 1.62 × 10 7 kg ( ) ( ) ⎜ ⎟ ⎟ ⎢ ⎝ 3.281 ft ⎠ ⎥⎦ ⎝ m3 ⎠ ⎣

At a rate of 1 part per million, the mass of uranium in this soil is then mU = (b)

235 Since 0.720% of naturally occurring uranium is 235 92 U, the mass of 92 U in the soil of part (a) is

m 30.6

m 1.62 × 10 7 kg = = 16.2 kg 10 6 10 6

235 92 U

= ( 7.20 × 10 −3 ) mU = ( 7.20 × 10 −3 ) (16.2 kg ) = 0.117 kg = 117 g

With a power output of Pout = 1.00 × 10 9 W and efficiency of e = 0.400, the total energy input required each day is Einput =

(

)(

)

1.00 × 10 9 J s 1.00 d ⎛ ⎞ ⎛ 8.64 × 10 4 s ⎞ Pout ⋅ Δt 1 MeV 27 = ⎜⎝ 1.60 × 10 −13 J ⎟⎠ ⎜⎝ ⎟⎠ = 1.35 × 10 MeV 0.400 1 d e

At 200 MeV per fission event, the number of 235 U atoms consumed each day is n=

Einput 200 MeV atom

=

1.35 × 10 27 MeV = 6.75 × 10 24 atoms 200 MeV atom

The mass of 235 U which will contain this number of atoms is ⎛ n ⎞ ⎛ 6.75 × 10 24 atoms ⎞ ⎛ g ⎞ 3 m=⎜ M = ⎜⎝ 235 ⎟ = 2.63 × 10 g = 2.63 kg m ol 23 ⎜ ⎟ ⎟ mol ⎠ ⎝ 6.02 × 10 atoms mol ⎠ ⎝ NA ⎠ 30.7

(a)

For a sphere of radius a, the surface area is A = 4p a 2, and the volume is V = 4p a 3 3. Thus, the surface-to-volume ratio is ratiosphere =

(b)

A 4p a 2 = = 3a V 4p a 3 3

For a cube of side s, A = 6 ⋅ Aface = 6s 2 and V = s ⋅ s ⋅ s = s 3. The surface-to-volume ratio for a cube is then ratiocube =

A 6s 2 6 = 3 = V s s

If the cube has the same volume as the sphere, then s 3 = 4p a 3 3, or s = ( 4p 3) a, and the surface-to-volume ratio for the cube becomes 1 3

ratiocube = (c)

68719_30_ch30_p377-408.indd 382

A 6s 2 6 = 3 = = 3.72 a s V ( 4p 3) a 1 3

The sphere has the better shape for minimum leakage since it has the smaller surface-tovolume ratio.

1/7/11 3:06:09 PM

Nuclear Physics and Elementary Particles

30.8

(a)

The mass of 235 U in the reserve is m

(b)

383

235

U

kg ⎞ ⎛ 3 g ⎞ ⎛ 0.70 ⎞ ⎛ 4.4 × 10 6 metric ton ) ⎜ 10 3 =⎜ = 3.1× 1010 g ⎟ 10 ⎝ 100 ⎟⎠ ( ⎝ ton ⎠ ⎜⎝ kg ⎟⎠

The number of moles is n = m M m ol = ( 3.1× 1010 g ) ( 235g mol ) = 1.3 × 108 mol , and the number of 235 U atoms in this reserve is N = nN A = (1.3 × 108 mol ) ( 6.02 × 10 23 atoms mol ) = 7.8 × 10 31 atoms

(c)

Assuming all atoms undergo fission and all released energy captured, the total energy available is MeV ⎞ MeV ⎞ ⎛ 1.60 × 10 −13 J ⎞ ⎛ ⎛ = ( 7.8 × 10 31 atoms ) ⎜ 208 E = N ⎜ 208 = 2.6 × 10 21 J ⎟ ⎟ ⎝ ⎝ atom ⎠ atom ⎠ ⎜⎝ 1 MeV ⎟⎠

(d)

At a consumption rate of 1.5 × 1013 J s, the maximum time this energy supply could last is t=

(e)

30.9

1 yr E 2.6 × 10 21 J ⎛ ⎞ = ⎜ ⎟ = 5.5 yr P 1.5 × 1013 J s ⎝ 3.156 × 10 7 s ⎠

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

The total energy required for one year is E = ( 2 000 kWh month ) ( 3.60 × 10 6 J kWh )(12.0 months ) = 8.64 × 1010 J The number of fission events needed will be N=

E Eevent

=

8.64 × 1010 J = 2.60 × 10 21 ( 208 MeV ) (1.60 × 10 −13 J MeV )

and the mass of this number of 235 U atoms is ⎛ N ⎞ ⎛ 2.60 × 10 21 atoms ⎞ m=⎜ = M m ol ⎜⎝ 6.02 × 10 23 atoms mol ⎟⎠ ( 235 g mol ) = 1.01 g ⎝ N A ⎟⎠ 30.10

(a)

At a concentration of c = 3 mg m 3 = 3 × 10 −3 g m 3, the mass of uranium dissolved in the oceans covering two-thirds of Earth’s surface to an average depth of hav = 4 km is mU = cV = c( 23 A) ⋅ hav = c[ 23 (4p RE2 )]⋅ hav , or 2 g ⎞ ⎛ 2⎞ ⎛ mU = ⎜ 3 × 10 −3 3 ⎟ ⎜ ⎟ 4p ( 6.38 × 10 6 m ) ( 4 × 10 3 m ) = 4 × 1015 g ⎝ ⎠ ⎝ ⎠ m 3

(b)

Fissionable 235 U makes up 0.7% of all uranium, so m U = 0.70mU 100. If we assume all of the 235 U is collected and caused to undergo fission, with the release of about 200 MeV per event, the potential energy supply is 235

E = ( number of

235

⎤⎛ MeV ⎞ MeV ⎞ ⎡⎛ 0.7mU 100 ⎞ ⎛ = ⎢⎜ N A ⎥ ⎜ 200 U atoms ) ⎜ 200 ⎟ ⎟ ⎟ ⎝ ⎝ M m ol atom ⎠ atom ⎠ ⎢⎣⎝ ⎠ ⎥⎦

continued on next page

68719_30_ch30_p377-408.indd 383

1/7/11 3:06:12 PM

384

Chapter 30

and at a consumption rate of P = 1.5 × 1013 J s, the time this could supply the world’s energy needs is t = E P, or ⎡ 0.7 ⎛ mU ⎞ ⎤ ( 200 MeV atom ) t=⎢ ⎜ ⎟⎥ P ⎢⎣ 100 ⎝ M m ol ⎠ ⎥⎦ ⎡ 0.7 ⎛ 4 × 1015 g ⎞ ⎛ ⎤ 200 MeV atom ⎛ 1.6 × 10 −13 J ⎞ ⎛ ⎞ 1 yr 23 atoms ⎞ =⎢ 6.02 × 10 ⎜ ⎟ ⎥ 13 7 ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ ⎠ mol ⎦ 1.5 × 10 J s ⎝ 1 MeV ⎠ ⎝ 3.156 × 10 s ⎟⎠ ⎣ 100 ⎝ 235 g mol ⎠ = 5 × 10 3 yr (c)

30.11

The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, but the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half-million years.

(a)

4 2

He + 42 He →

8 4

(b)

8 4

Be + 42 He →

12 6

(c)

Be + g C +g

The total energy released in this pair of fusion reactions is Q = ( Δm ) c 2 = ⎡⎣3m

4

He

−m

12

C

⎤⎦ c 2

= ⎡⎣3 ( 4.002 603 u ) − 12.000 000 u ⎤⎦ ( 931.5 MeV u ) = 7.27 MeV 30.12

The energy released in the reaction 11 H + 12 H → 32 He + g is Q = ( Δm ) c 2 = ⎡ m ⎣

1 1H

+m

2 1H

−m

3 2 He

⎤ c2 ⎦

= [1.007 825 u + 2.014 102 u − 3.016 029 u ]( 931.5 MeV u ) = 5.49 MeV 30.13

The energy released in the reaction 21 H + 21 H → 31 H + 11 H is Q = ⎡ 2m ⎣

2 1H

−m

3 1H

− m H ⎤ c 2, or ⎦ 1 1

Q = ⎡⎣ 2 ( 2.014 102 u ) − 3.016 049 u − 1.007 825 u ⎤⎦ ( 931.5 MeV u ) = 4.03 MeV 30.14

30.15

H + 126 C →

13 7

N +g

(b)

13 7

N→

13 6

C + 11 H →

14 7

N +g

(d)

14 7

N + 11 H →

15 8

O +g

15 8

O→

(f )

15 7

N + 11 H →

12 6

C + 42 He

(a)

1 1

(c) (e)

15 7

N +

0 +1

e +n

13 6

C+

0 +1

e +n

With the deuteron and triton at rest, the total momentum before the reaction is zero. To conserve momentum, the neutron and the alpha particle must move in opposite directions with equal magnitude momenta after the reaction, or pa = pn . Neglecting relativistic effects, we use the classical relationship between momentum and kinetic energy, KE = p2 2m, and write 2ma KEa = 2mn KEn , or KEa = (mn ma )KEn . To conserve energy, it is necessary that the kinetic energies of the reaction products satisfy the relation KEn + KEa = Q = 17.6 MeV. Then, using the result from above, we have KEn + (mn ma )KEn = 17.6 MeV, or the kinetic energy of the emerging neutron must be KEn =

68719_30_ch30_p377-408.indd 384

17.6 MeV = 14.1 MeV 1+ (1.008 665 u ) ( 4.002 603 u )

1/7/11 3:06:15 PM

Nuclear Physics and Elementary Particles

30.16

(a)

385

The energy released in the reaction 11 H + 115 B → 3( 42 He) is Q = ( Δm ) c 2 = ⎡ m ⎣

1 1H

+m

11 5B

− 3m

4 2 He

⎤ c2 ⎦

= ⎡⎣1.007 825 u + 11.009 306 u − 3 ( 4.002 603 u ) ⎤⎦ ( 931.5 MeV u ) = 8.68 MeV (b)

30.17

The proton and the boron nucleus both have positive charges but must come very close to one another in order for fusion to occur. Thus, they must have sufficient kinetic energy to overcome the repulsive Coulomb force one exerts on the other.

Note that pair production cannot occur in a vacuum. It must occur in the presence of a massive particle which can absorb at least some of the momentum of the photon and allow total linear momentum to be conserved. When a particle-antiparticle pair is produced by a photon having the minimum possible frequency, and hence minimum possible energy, the nearby massive particle absorbs all the momentum of the photon, allowing both components of the particle-antiparticle pair to be left at rest. In such an event, the total kinetic energy afterwards is essentially zero, and the photon need only supply the total rest energy of the pair produced. The minimum photon energy required to produce a proton-antiproton pair is Ephoton = 2 ( ER )proton. Thus, the minimum photon frequency is f= and l =

30.18

Ephoton h

=

2 ( ER )proton h

=

2 ( 938.3 MeV )(1.60 × 10 −13 J MeV ) 6.63 × 10 −34 J ⋅s

= 4.53 × 10 23 Hz

c 3.00 × 108 m s = = 6.62 × 10 −16 m = 0.662 fm f 4.53 × 10 23 Hz

The total kinetic energy after the pair production is KEtotal = Ephoton − 2 ( ER )proton = 2.09 × 10 3 MeV − 2 ( 938.3 MeV ) = 213 MeV The kinetic energy of the antiproton is then KE p = KEtotal − KE p = 213 MeV − 95.0 MeV = 118 MeV

30.19

The total rest energy of the p 0 is converted into energy of the photons. Since the total momentum was zero before the decay, the two photons must go in opposite directions with equal magnitude momenta (and hence equal energies). Thus, the rest energy of the p 0 is split equally between the two photons, giving for each photon Ephoton =

pphoton =

and

68719_30_ch30_p377-408.indd 385

f=

ER ,p

0

2 Ephoton c Ephoton h

=

135 MeV = 67.5 Mev 2

= 67.5 MeV c =

67.5 MeV ⎛ 1.60 × 10 −13 J ⎞ = 1.63 × 10 22 Hz 6.63 × 10 −34 J ⋅s ⎜⎝ 1 MeV ⎟⎠

1/7/11 3:06:18 PM

386

30.20

Chapter 30

Observe that the given reactions involve only mesons and baryons. With no leptons before or after the reactions, we do not have to consider the conservation laws concerning the various lepton numbers. All interactions always conserve both charge and baryon numbers. The strong interaction also conserves strangeness. Conservation of strangeness may be violated by the weak interaction but never by more than one unit. With these facts in mind consider the given interactions: K0 → p + + p − K 0 total before p + p − total after Charge

0

0

+1 −1

0

Baryon number

0

0

0

0

0

Strangeness

+1

+1

0

0

0

This reaction conserves both charge and baryon number but does violate strangeness by one unit. Thus, it can occur via the weak interaction but not other interactions. Λ0 → p + + p − + total before p

Λ0 Charge

p−

total after

0

0

+1

–1

0

Baryon number +1

+1

0

0

0

–1

0

0

0

Strangeness

–1

This reaction fails to conserve baryon number and cannot occur via any interaction. 30.21

(a)

Reaction p + p → m + + e−

(b)

p +p→ p+p

(c)

p+p→ p+p +

−

Conservation Law Violated Le : ( 0 + 0 → 0 + 1); and L m : ( 0 + 0 → −1+ 0 )

+

Charge, Q:

Baryon Number, B: Baryon Number, B:

(d) p + p → p + p + n (e)

( −1+ 1 → + 1+ 1)

g +p→ n+p 0

Charge, Q:

(1+ 1 → 1+ 0 ) (1+ 1 → 1+ 1 + 1)

(0 + 1 → 0 + 0)

30.22

(a)

Reaction p→p + +p 0

(b)

p+p→ p+p+p

(c)

p → m +n m

No violations − The reaction can occur.

(d)

n → p + e − +n e

No violations − The reaction can occur.

(e)

p + → m+ + n

+

+

Conservation Law Violated Baryon Number, B: 0

No violations − The reaction can occur.

Baryon Number, B:

68719_30_ch30_p377-408.indd 386

(1 → 0 + 0 )

( 0 → 0 + 1)

Muon-Lepton Number, L m :

( 0 → −1+ 0 )

1/7/11 3:06:19 PM

Nuclear Physics and Elementary Particles

387

30.23 Reaction

Conclusion

Baryon Number - violated: ( 0 + 1 → 0 )

Cannot Occur

p + p → 2h

(b)

K − + n → Λ0 + p −

All conservation laws observed.

May occur via the Strong Interaction

(c)

K →p +p

0

Strangeness violated by 1 unit ( −1 → 0 + 0 ). All other conservation laws observed.

Can occur via Weak Interaction, but not by Electromagnetic or Strong Interactions.

(d)

Ω → Ξ +p

0

Strangeness violated by 1 unit ( −3 → −2 + 0 ). All other conservation laws observed.

Can occur via Weak Interaction, but not by Electromagnetic or Strong Interactions.

All conservation laws observed.

Allowed via all interactions, but photons are the mediator of the electromagnetic interaction and the lifetime of the h 0 is consistent with decay by that interaction.

(e)

30.24

Conservation Law

(a)

(a)

−

−

−

−

−

0

h 0 → 2g

p + +p→K+ + Σ+ Baryon number, B:

0 +1→ 0 +1

ΔB = 0

Charge, Q:

1+ 1 → 1+ 1

ΔQ = 0

Baryon number, B:

0 +1→ 0 +1

ΔB = 0

Charge, Q:

1+ 1 → 1+ 1

ΔQ = 0

p + +p→p + + Σ+

(b)

Strangeness is conserved in the first reaction: Strangeness, S:

0 + 0 → 1− 1

ΔS = 0

The second reaction does not conserve strangeness: Strangeness, S:

0 + 0 → 0 −1

ΔS = −1

The second reaction cannot occur via the strong or electromagnetic interactions . (c)

If one of the neutral kaons were also produced in the second reaction, giving p + + p → p + + Σ + + K 0, then strangeness would no longer be violated: Strangeness, S:

0 + 0 → 0 − 1 +1

ΔS = 0

Because the total mass of the product particles in this reaction would be greater than that in the first reaction [see part (a)], the total incident energy of the reacting particles would have to be greater for this reaction than for the first reaction . 30.25

(a)

Λ0 → p + p − Strangeness, S:

−1 → 0 + 0

⇒ ΔS ≠ 0

Not Conserved continued on next page

68719_30_ch30_p377-408.indd 387

1/7/11 3:06:23 PM

388

Chapter 30

(b)

p − + p → Λ0 + K 0 Strangeness, S:

(c)

0 + 0 → +1− 1

⇒ ΔS = 0

Conserved

0 + 0 → 0 −1

⇒ ΔS ≠ 0

Not Conserved

−2 → −1+ 0

⇒ ΔS ≠ 0

Not Conserved

−2 → 0 + 0

⇒ ΔS ≠ 0

Not Conserved

Ξ− → Λ 0 + p − Strangeness, S:

(f)

Conserved

p − + p → p − + Σ+ Strangeness, S:

(e)

⇒ ΔS = 0

p + p → Λ0 + Λ0 Strangeness, S:

(d)

0 + 0 → −1+ 1

Ξ0 → p + p − Strangeness, S:

30.26

30.27

proton

u

u

d

total

strangeness

0

0

0

0

0

baryon number

1

1 3

1 3

1

charge

e

1 3 2e 3

2e 3

−e 3

e

neutron

u

d

d

total

strangeness

0

0

0

0

baryon number

1

1 3

0 1 3

1 3

1

charge

0

2e 3

−e 3

−e 3

0

The number of water molecules in one liter (mass = 1 000 g) of water is ⎛ 1 000 g ⎞ N =⎜ (6.02 × 1023 molecules mol ) = 3.34 × 1025 molecules ⎝ 18.0 g mol ⎟⎠ Each molecule contains 10 protons, 10 electrons, and 8 neutrons. Thus, there are N e = 10 N = 3.34 × 10 26 electrons , N p = 10 N = 3.344 ×10 26 protons, and

N n = 8 N = 2.68 × 10 26 neutrons

Each proton contains 2 up quarks and 1 down quark, while each neutron has 1 up quark and 2 down quarks. Therefore, there are N u = 2 N p + N n = 9.36 × 10 26 up quarks , and N d = N p + 2 N n = 8.70 × 10 26 down quarks .

68719_30_ch30_p377-408.indd 388

1/7/11 3:06:26 PM

Nuclear Physics and Elementary Particles

389

30.28 K 0 Particle K0

d

s

total

strangeness

1

0

1

1

baryon number

0

13

−1 3

0

charge

0

−e 3

e3

0

Λ0 Particle

30.29

30.30

30.31

L0

u

d

s

total

strangeness

−1

0

0

−1

−1

baryon number

1

13

13

13

1

charge

0

2e 3

−e 3

−e 3

0

Compare the given quark states to the entries in Table 30.4: (a)

suu = Σ +

(b) ud = p

(c)

sd = K 0

(d) ssd = Ξ −

(a)

⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ uud: charge = ⎜ − e ⎟ + ⎜ − e ⎟ + ⎜ + e ⎟ = − e . This is the antiproton . ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠

(b)

⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ udd: charge = ⎜ − e ⎟ + ⎜ + e ⎟ + ⎜ + e ⎟ = 0 . This is the antineutron . ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠

−

The reaction is Σ 0 + p → Σ + + g + X, or on the quark level, uds + uud → uus + 0 + x. The left side has a net 3 u, 2 d, and 1s. The right side has 2 u, 0 d, and 1s, plus the quark composition of the unknown particle. To conserve the total number of each type of quark, the composition of the unknown particle must be x = udd . Therefore, the unknown particle must be a neutron .

30.32

30.33

(a)

p − + p → Σ + + p 0 is forbidden by conservation of charge .

(b)

m − → p − +n e is forbidden by both conservation of electron-lepton number , and conservation of muon-lepton number .

(c)

p → p + + p + + p − is forbidden by conservation of baryon number .

To the reaction for nuclei, 11 H + 32 He → 42 He + +01 e + n e , we add three electrons to both sides to obtain 11 H atom + 32 Heatom → 42 Heatom + −01 e + +01 e + ne. Then we use the masses of the neutral atoms from Appendix B of the textbook to compute Q = ( Δm ) c 2 = ⎡ m ⎣

1 1H

+m

3 2

He

−m

4 2

He

− 2 me ⎤ c 2 ⎦

= ⎡⎣1.007 825 u + 3.016 029 u − 4.002 603 u − 2 ( 0.000 549 u ) ⎤⎦ ( 931.5 MeV u ) = 18.8 MeV

68719_30_ch30_p377-408.indd 389

1/7/11 3:06:30 PM

390

Chapter 30

30.34

For the particle reaction, m + + e − → 2n, the lepton numbers before the event are L m = −1 and Le = +1. These values must be conserved by the reaction so one of the emerging neutrinos must have L m = −1 while the other has Le = +1. The emerging particles are n m and n e .

30.35

(a)

K+ + p → X + p Since this occurs via the strong interaction, all conservation rules must be observed. Baryon Number:

0 +1→ B +1

ΔB = 0 ⇒ B = 0

so X is not a baryon.

Strangeness:

+1+ 0 → S + 0

ΔS = 0 ⇒ S = +1

or X has S = +1

Charge:

+e + e → Q + e

ΔQ = 0 ⇒ Q = +e

or X has Q = +e

Lepton Numbers:

0+0→ L+0

ΔL = 0 ⇒ L = 0

or X has Le = L m = Lt = 0

Of the particles in Table 30.2, the only non-baryon with S = +1 and Q = +e is the positive kaon, K + . Thus, this is an elastic scattering process. The weak interaction observes all conservation rules except strangeness, and ΔS = ±1. (b)

Ω− → X + p − Baryon Number:

1→ B + 0

ΔB = 0 ⇒ B = +1

or X has B = +1

Strangeness:

−3→ S + 0

ΔS = +1 ⇒ S = −2

or X has S = −2

Charge:

−e → Q − e

ΔQ = 0 ⇒ Q = 0

or X has Q = 0

Lepton Numbers:

0→ L+0

ΔL = 0 ⇒ L = 0

or X has Le = L m = Lt = 0

The particle must be a neutral baryon with strangeness S = −2. Thus, it is the Ξ0 . (c)

K + → X + m + +n m Baryon Number:

0→ B+0+0

ΔB = 0 ⇒ B = 0

so X is not a baryon.

Strangeness:

+1 → S + 0 + 0

ΔS = ±1 ⇒ S = 0, + 2

or X has S = 0 or + 2

Charge:

+e → Q + e + 0

ΔQ = 0 ⇒ Q = 0

or X has Q = 0

Lepton Numbers:

0 → Le + 0 + 0

ΔLe = 0 ⇒ Le = 0

or X has Le = 0

0 → L m − 1+ 1

ΔL m = 0 ⇒ L m = + 0

or X has L m = 0

0 → Lt + 0 + 0

ΔLt = 0 ⇒ Lt = + 0

or X has Lt = 0

The particle must be a neutral non-baryon with strangeness S = 0 or S = +2, and Le = L m = Lt = 0. This is the neutral pion, p 0 . 30.36

Assuming a head-on collision, the total momentum is zero both before and after the reaction p + p → p + p + + X. Therefore, since the proton and the pion are at rest after the reaction, particle X must also be left at rest.

continued on next page

68719_30_ch30_p377-408.indd 390

1/7/11 3:06:34 PM

Nuclear Physics and Elementary Particles

391

Particle X must be a neutral baryon in order to conserve charge and baryon number in the reaction. Conservation of energy requires that the rest energy this particle be ER, X = 2 ( ER, p + 70.4 MeV ) − ER, p − ER, π = ER, p − ER, π +140.8 MeV +

or

+

ER, X = 938.3 MeV −139.6 MeV +140.8 MeV = 939.5 MeV

Particle X is a neutron . 30.37

If a neutron starts with kinetic energy KEi = 2.0 MeV and loses one-half of its kinetic energy in each collision with a moderator atom, its kinetic energy after n collisions will be KE f = KEi 2n. The average kinetic energy associated with particles in a gas at temperature T = 20.0°C = 293 K (see Chapter 10 of the textbook) is KE f =

3 3 1 eV ⎛ ⎞ kB T = (1.38 × 10 −23 J K )( 293 K ) ⎜ = 0.0379 eV ⎝ 1.60 × 10 −19 J ⎟⎠ 2 2

Thus, the number of collisions the neutron must make before it reaches the energy associated with a room temperature gas is n ln 2 = ln(KEi KE f ), or 6 ⎛ 1 ⎞ ⎛ 2.0 × 10 eV ⎞ n=⎜ = 26 ln ⎝ ln 2 ⎟⎠ ⎜⎝ 0.0379 eV ⎟⎠

30.38

(a)

The number of deuterons in one kilogram of deuterium is N=

m 1 kg = = 3.00 × 10 26 matom ( 2.01 u )(1.66 × 10 −27 kg u )

Each occurrence of the reaction 21 D + 21 D → 23 He + 01 n consumes two deuterons and releases Q = 3.27 MeV of energy. The total energy available from the one kilogram of deuterium is then ⎛ 3.00 × 10 26 ⎞ ⎛ 1.60 × 10 −13 J ⎞ ⎛ N⎞ 13 E =⎜ ⎟Q =⎜ 3.27 Mev ( ) ⎟⎠ ⎜⎝ 1 MeV ⎟⎠ = 7.85 × 10 J ⎝ 2⎠ 2 ⎝ (b)

At a rate of eight cents per kilowatt-hour, the value of this energy is ⎛ $0.08 ⎞ ⎛ 1 kWh value = E × rate = ( 7.85 × 1013 J ) ⎜ ⎝ kWh ⎟⎠ ⎜⎝ 3.60 × 10 6

30.39

⎞ 6 ⎟ = $1.74 × 10 = $1 740 000 J⎠

(c)

Deuterium makes up four-twentieths or one-fifth of the mass of a heavy water molecule. Thus, five kilograms of heavy water are necessary to obtain one kilogram of deuterium. The cost for this water (5 kg)($300 kg) = $1 500 .

(d)

Whether it would be cost-effective depends on how much it cost to fuse the deuterium and how much net energy was produced. If the cost is nine-tenths of the value of the energy produced, each kilogram of deuterium would still yield a profit of $174 000.

(a)

The number of moles in 1.0 gal of water is ⎛ 3.786 L ⎞ ⎛ 10 3 cm 3 ⎞ g ⎞ ⎛ 1.0 gal ) ⎜ ⎜⎝ 1.0 3 ⎟( cm ⎠ ⎝ 1 gal ⎟⎠ ⎜⎝ 1 L ⎟⎠ m rV n= = = = 2.1× 10 2 mol M M 18 g mol continued on next page

68719_30_ch30_p377-408.indd 391

1/7/11 3:06:40 PM

392

Chapter 30

so the number of hydrogen atoms (2 per water molecule) will be N H = 2 ( nN A ) = 2 ( 2.1× 10 2 mol ) ( 6.02 × 10 23 mol −1 ) = 2.5 × 10 26 and one of every 6 500 of these contains a deuteron. Thus, the number of deuterons contained in 1.0 gal of water is N D = N H 6 500 = 2.5 × 10 26 6 500 = 3.8 × 10 22 deuterons and the available energy is E = ( 3.8 × 10 22 deuterons ) (1.64 MeV deuteron ) (1.6 × 10 −13 J MeV ) = 1.0 × 1010 J (b)

At a consumption rate of P = 1.0 × 10 4 J s, the time that this could supply a person’s energy needs is Δt =

30.40

(a)

E 1.0 × 1010 J ⎛ 1 d ⎞ = = 12 d P 1.0 × 10 4 J s ⎜⎝ 86 400 s ⎟⎠

The number of water molecules in the oceans is

NH

2O

⎛ mocean ⎞ ⎛ 1.32 × 10 21 kg ⎞ water ⎟ NA = ⎜ =⎜ 6.02 × 10 23 molecules mol ) ( −3 ⎟ M 18.0 × 10 kg mol ⎝ ⎠ ⎜⎝ m ol ⎟⎠ = 4.41× 10 46 molecules

Since there are 2 hydrogen atoms per water molecule, and the fraction of all hydrogen atoms which contain deuterons is 1.56 × 10 −4, the number of deuterons in the oceans is N D = ( 2N H

2O

) ×1.56 ×10

−4

= 2 ( 4.41×10 46 ) (1.56 ×10 −4 ) = 1.38 ×10 43

Each fusion event consumes 2 deuterons, so the number of fusion events possible is N events = 12 ND . The reaction 21 D+ 21 D → 23 He + 01 n releases 3.27 MeV per event, so the total energy that could be released is E = (3.27 MeV ) N events = (3.27 MeV ) ( 12 N D ) ⎛ 1.38 ×10 43 ⎞ ⎛ 1.60 ×10 −13 = (3.27 MeV ) ⎜ ⎟⎜ 2 ⎝ ⎠ ⎝ 1 MeV (b)

J⎞ 30 ⎟ = 3.61×10 J ⎠

One hundred times the current world electric power consumption is P = 100 ( 7.00 × 1012 W ) = 7.00 × 1014 J s At this rate, the time the energy computed in part (a) would last is t=

30.41

⎞ E 3.61× 10 30 J ⎛ 1 yr = = 1.63 × 108 yr 14 7 ⎜ P 7.00 × 10 J s ⎝ 3.156 × 10 s ⎟⎠

Conserving energy in the decay p − → m − +n m , with the p − initially at rest gives E m + En = E R ,p , or −

E m + En = 139.6 MeV

[1]

continued on next page

68719_30_ch30_p377-408.indd 392

1/7/11 3:06:43 PM

Nuclear Physics and Elementary Particles

393

Since the total momentum was zero before the decay, conservation of momentum requires the muon and antineutrino to recoil in opposite directions with equal magnitude momenta, or pm = pn . The relativistic relation between total energy and momentum of a particle gives for the antineutrino: En = pn c, or pn = En c. The same relation applied to the muon gives 2 2 E m2 = pm c + ER2 , m . Since we must have pm = pn , this may be rewritten as E m2 = ( pn c ) + E R2 , m , and using the fact that pn c = En , we have E m2 = En 2 + E R2 , m . Rearranging and factoring then gives

( )

(

E m2 − En2 = E m + En

)(E

m

)

− En = E R2 , m = (105.7 MeV )

2

and E m − En =

(105.7 MeV )2

[2]

E m + En

Substituting Equation [1] into [2] gives E m − En = (105.7 MeV)2 139.6 MeV, or E m − En = 80.0 MeV

[3]

Subtracting Equation [3] from Equation [1] yields 2En = 59.6 MeV, and En = 29.8 MeV 30.42

The reaction p − + p → K 0 + Λ 0 is ud + uud → d s + uds at the quark level. There is a net 1u and 2 d quarks both before and after the reaction . This reaction conserves the net number of each type quark. For the reaction p − + p → K 0 + n, or ud + uud → d s + udd, there is a net 1u and 2 d before the reaction and 1u, 3d, and 1 s quark afterwards . This reaction does not conserve the net number of each type quark.

30.43

To compute the energy released in each occurrence of the reaction 4p + 2e − → a + 2n e + 6g we add two electrons to each side to produce neutral atoms and obtain 4( 11 H atom ) → 42 Heatom + 2n e + 6g . Then, recalling that the neutrino and the photon both have zero rest mass, and using the neutral atomic masses from Appendix B in the textbook gives Q = ( Δm ) c 2 = ⎡ 4m ⎣

1 1 H atom

−m

4 2 He atom

⎤ c2 ⎦

= ⎡⎣ 4 (1.007 825 u ) − 4.002 603 u ⎤⎦ ( 931.5 MeV u ) = 26.7 Mev Each occurrence of this reaction consumes four protons. Thus, the energy released per proton consumed is E1 = 26.4 MeV 4 protons = 6.68 MeV proton. Therefore, the rate at which the Sun must be fusing protons to provide its power output is rate =

68719_30_ch30_p377-408.indd 393

P E1

=

3.85 × 10 26 J s ⎛ 1 MeV ⎞ = 3.60 × 10 38 proton s 6.68 MeV proton ⎜⎝ 1.60 × 10 −13 J ⎟⎠

1/7/11 3:06:46 PM

394

30.44

Chapter 30

With the kaon initially at rest, the total momentum was zero before the decay. Thus, to conserve momentum, the two pions must go in opposite directions and have equal magnitudes of momentum after the decay. Since the pions have equal mass, this means they must have equal speeds and hence, equal energies. The rest energy of the kaon is then split equally between the two pions, and the energy of each is Ep = ER ,K 2 = 497.7 Mev 2 = 248.9 MeV 0

The total energy of one of the pions is related to its rest energy by Ep = g ER ,p , where 2 g = 1 1− ( v c ) . Therefore, the speed of each of the pions after the decay will be 2

2 ⎛E ⎞ ⎛ 139.6 MeV ⎞ v = c 1− ⎜ R ,p ⎟ = c 1− ⎜ = 0.827 9c ⎝ 248.9 MeV ⎟⎠ ⎝ Ep ⎠

68719_30_ch30_p377-408.indd 394

1/7/11 3:06:50 PM