Physics Concepts and Connections, Book Two - Solutions Manual

B OOK TWO

Solutions Manual Brian Heimbecker Igor Nowikow Christopher T. Howes Jacques Mantha Brian P. Smith Henri M. van Bemmel

Physics: Concepts and Connections Book Two Solutions Manual Authors Brian Heimbecker Igor Nowikow Christopher T. Howes Jacques Mantha Brian P. Smith Henri M. van Bemmel

NELSON Director of Publishing David Steele

First Folio Resource Group Project Management Robert Templeton

Publisher Kevin Martindale

Composition Tom Dart

Project Editor Lina Mockus-O’Brien

Proofreading and Copy Editing Christine Szentgyorgi Patricia Trudell

Editor Kevin Linder

COPYRIGHT © 2003 by Nelson, a division of Thomson Canada Limited. Printed and bound in Canada. 1 2 3 4 05 04 03 02 For more information contact Nelson, 1120 Birchmount Road Toronto, Ontario, M1K 5G4. Or you can visit our Internet site at http://www.nelson.com

Illustrations Greg Duhaney Claire Milne

ALL RIGHTS RESERVED. No part of this work covered by the copyright hereon may be reproduced, transcribed, or used in any form or by any means—graphic, electronic, or mechanical, including photocopying, recording, taping, Web distribution, or information storage and retrieval systems—without the written permission of the publisher. For permission to use material from this text or product, contact us by Tel 1-800-730-2214 Fax 1-800-730-2215 www.thomsonrights.com Every effort has been made to trace ownership of all copyrighted material and to secure permission from copyright holders. In the event of any question arising as to the use of any material, we will be pleased to make the necessary corrections in future printings.

Table of Contents I

Solutions to Applying the Concepts Questions

Chapter 1 Section 1.3 1.4 1.6 1.7 1.8 1.11 1.12 1.13 1.14 1.15

1 1 1 3 3 4 5 6 6 7

Chapter 2 Section 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

8 9 10 11 12 13 14 15

Chapter 3 Section 3.3 3.4 3.5 3.6 3.7 3.8 3.9

17 18 18 20 21 22 22

Chapter 4 Section 4.2 4.3 4.4 4.5 4.6

24 24 25 25 26

Chapter 5 Section 5.2 5.3 5.4 5.5 5.6 5.7

28 28 29 29 30 31

Chapter 6 Section 6.1 6.2 6.3

33 33 34

Chapter 7 Section 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11

36 36 36 37 38 38 38 39 39 40

Chapter 8 Section 8.4 8.5 8.6 8.7 8.8 8.9

41 41 42 43 44 44

Chapter 9 Section 9.5

45

Chapter 10 Section 10.2 10.3 10.4 10.5

47 47 48 48

Chapter 11 Section 11.4 11.5 11.6 11.8 11.9 11.10

49 49 49 49 50 51

Chapter 12 Section 12.2 12.3 12.4 12.5 12.6 12.8

52 52 52 53 53 54

Chapter 13 Section 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8

55 55 55 56 56 57 57 58

Chapter 14 Section 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8

59 59 59 59 59 60 60 60

Table of Contents

II Answers to End-of-chapter Conceptual Questions Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14

61 63 65 66 67 68 69 71 75 77 79 80 81 83

III Solutions to Endof-chapter Problems Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14

87 95 107 120 126 134 140 151 160 165 170 178 183 191

iii

PART 1 Solutions to Applying the Concepts In this section, solutions have been provided only for problems requiring calculation.

Section 1.3

60 s 24 h 60 min 1. (30 days) 1h 1 day 1 min 2.6 106 s (units cancel to give answer in seconds) 1 mile 1 km 2. (7 furlongs) 1.4 km 8 furlong 0.63 mile (units cancel to give answer in kilometres) 27.5 mL 20 oz 3. (1 quart) 1 quart 1 oz 5.5 102 mL (units cancel to give answer in millilitres)

Section 1.4 1. Since the question is asking for velocity, the answer must include a direction. Since the direction in which the train travels is constant, d vavg t 2.5 104 m [N] vavg 1.8 103 s vavg 14 m/s [N] 2. a) Since the question is asking for average speed, direction is not required. d vavg t 8.0 km vavg 5.0 h vavg 1.6 km/h b) Since the question is asking for average velocity, direction is required. d vavg t 3.0 km [W] 5.0 km [E] vavg 5.0 h 3.0 km [E] 5.0 km [E] vavg 5.0 h 2.0 km [E] vavg 5.0 h vavg 0.40 km/h [E]

3. a) Since the question asks for the car’s velocity, direction is important. Since the direction is constant, d vavg t 9.0 m [E] 0 m [E] vavg 8.0 s vavg 1.1 m/s [E] b) The car’s instantaneous velocity at 5 s can be approximated by the difference between the distance travelled after 6 s and the distance travelled after 5 s, divided by the time during that interval: d vavg t 8.0 m [E] 8.0 m [E] vavg 6.0 s 5.0 s vavg 0 m/s

Section 1.6 1. v22 v12 2ad v22 v12 d 2a (600 m/s)2 (350 m/s)2 d 2(12.6 m/s2) d 9.4 103 m 2. 10 cm 1.0 101 m (v1 v2)t d 2 2d v 1 v2 t 2(1.0 101 m) v1 0.05 m/s 3.0 s v1 1.7 102 m/s 3. a) Igor: dI vIt 1 Brian: dB aBt2 2 If they meet, dI dB 8.0 m 1 vIt 8.0 m aBt2 2 1 0 (2.8 m/s2)t2 (7.0 m/s)t 8.0 m 2 0 (1.4 m/s2)t2 (7.0 m/s)t 8.0 m

Solutions to Applying the Concepts

1

2 b bac 4 t 2a

t

7.0 m/s (7.0 m/s)2 4(1 .4 m/s2 )(8.0 m) 2(1.4 m/s2)

7.0 m/s 2.05 m/s t 2.8 m/s2 t 3.2 s or t 1.8 s We will take the lower value: t 1.8 s. 1 b) dB (2.8 m/s2)(1.8 s)2 2 dB 4.4 m 4. 8.0 cm 8.0 102 m v22 v12 2ad v22 v12 a 2d (0 m/s)2 (350 m/s)2 a 2(8.0 102 m) a 7.7 105 m/s2 5. ttotal t1 t2 t3 ttotal 3.0 s 6.0 s 10 s ttotal 19.0 s In the first 3.0 s, the truck travels a distance of: 1 d1 (v1 v2)t1 2 1 d1 (0 m/s 8.0 m/s)(3.0 s) 2 d1 12 m Since the truck travels at a constant speed over the second interval, d2 v2t2 d2 (8.0 m/s)(6.0 s) d2 48 m For the final interval, 1 d3 v1t3 at32 2 1 d3 (8.0 m/s)(10 s) (2.5 m/s2)(10 s)2 2 2 d3 2.1 10 m dtotal d1 d2 d3 dtotal 12 m 48 m 2.1 102 m dtotal 2.7 102 m dtotal vavg ttotal 2.7 102 m vavg 19.0 s vavg 14 m/s 2

6. 100 km/h 27.8 m/s 1 d v1t at2 2 1 500 m (27.8 m/s)t (30 m/s2)t2 2 0 (15 m/s2)t2 (27.8 m/s)t 500 m t

27.8 m/s (27.8 m/s)2 4(1 5 m/s2 )(50 0 m) 2 2(15 m/s )

t 4.9 s 1 7. a) d v1t at2 2 1 80 m (17 m/s)t (9.8 m/s2)t2 2 2 2 0 (4.9 m/s )t (17 m/s)t 80 m t

17 m/s (17 m /s)2 4(4.9 m/s2) (80 m) 2 2(4.9 m/s )

t 2.7 s b) v22 v12 2ad 2 v2 vad 1 2 2 v2 (17 /s) m 2(9.8 (80 m/s2)) m v2 43 m/s v2 v1 8. a) a t v2 v1 t (eq.1) a v2 v1 d t (eq. 2) 2 Substituting equation 1 into equation 2, v2 v1 v2 v1 d 2 a 2 2 2ad v2 v1 v2v1 v1v2 v22 v12 2ad v2 v1 b) a t v1 v2 at (eq. 1) v2 v1 d t (eq. 2) 2 Substituting equation 1 into equation 2, v2 v2 at d t 2 1 d v2t at2 2


Section 1.7

Section 1.8

1. a) v2 v1 2ad Assuming up is positive, v22 v12 d 2a 0 (80.0 m/s)2 d 2(9.8 m/s2) d 330 m b) v2 v1 at v2 v1 t a 0 80.0 m/s t 9.8 m/s2 t 8.16 s c) 2(8.16 s) 16.3 s 1 2. a) d v1t at2 2 Assuming down is positive, 1 30.0 m (4.0 m/s)t (9.8 m/s2)t2 2 0 (4.9 m/s2)t2 (4.0 m/s)t 30.0 m 2 b bac 4 t 2a

vt vt 1. a) at7.0s t2 t1

2

t

2

4.0 m/s (4.0 m /s)2 4(4.9 m/s2) (30. 0 m) 2 2(4.9 m/s )

4.0 m/s 24.6 m/s t 9.8 m/s2 t 2.1 s 1 b) d v1t at2 2 Assuming down is positive, 1 30.0 m (4.0 m/s)t (9.8 m/s2)t2 2 0 (4.9 m/s2)t2 (4.0 m/s)t 30.0 m 2 b bac 4 t 2a t

2 4.0 m/s (4.0 m/s)2 4(4 .9 m/s )(30 .0 m) 2(4.9 m/s2)

4.0 m/s 24.6 m/s t 9.8 m/s2 t 2.9 s 1 3. d v1t at2 2 Assuming down is positive, 1 35 m v1(3.5 s) (9.8 m/s2)(3.5 s)2 2 v1 7.2 m/s or 7.2 m/s [up]

2

1

55.0 m/s 51.0 m/s at7.0s 8.0 s 6.0 s at7.0s 2.0 m/s2 60 m/s 60 m/s at12s 13 s 11 s at12s 0 m/s2 32.0 m/s 8.0 m/s at3.0s 4.0 s 2.0 s at3.0s 12 m/s2 b) The distance travelled by Puddles from t 5.0 s to t 13 s can be found by finding the area under the curve between those times. We must consider two separate intervals: between 5.0 s and 10 s, and between 10 s and 13 s. The area under the graph in the first interval can be expressed as the sum of the areas of a triangle and a rectangle: t1v1 t1v1 d1 2 (10 s 5.0 s)(60 m/s 50 m/s) d1 2 (10 s 5.0 s)(50 m/s) d1 275 m The area under the graph in the second interval can be expressed as a rectangle: d2 t2v2 d2 (13 s 10 s)(60 m/s 0 m/s) d2 180 m dT d 1 d 2 dT 275 m 180 m dT 455 m 2. a) For Super Dave, Sr., d vavg t d t vavg 50 m t 10 m/s t 5.0 s


3

We can find the acceleration of Super Dave, Jr. from the slope of his vt graph: v a t 6 m/s a 2s a 3 m/s2 at2 d v1t 2 But v1 0 m/s, so at2 d 2 t

2d a

2

3

1.0 m vavg 0.8 s vavg 1.25 m/s For segment 4, d4 2.2 m 1.0 m d4 1.2 m t4 2.6 s 1.8 s t4 0.8 s d vavg 4 t4 1.2 m vavg 0.8 s vavg 1.5 m/s dtotal b) vavg ttotal 3

3

2(50 m) t 3 m/s2 t 6 s b) Super Dave, Sr. wins the race by 1 s. c) Super Dave, Sr.: d vavg t d t vavg 100 m t 10 m/s t 10 s Super Dave, Jr.: at2 d v1t , where v1 0 m/s, so 2 2 at d 2 2d t a

t

For segment 2, d2 2.0 m 2.0 m d2 0 m vavg 0 m/s For segment 3, d3 1.0 m 2.0 m d3 1.0 m t3 1.8 s 1.0 s t3 0.8 s d vavg 3 t3

4

4

4

2.2 m 0.5 m vavg 2.2 s 0.0 s vavg 0.65 m/s

Section 1.11 1. a)

Fn

2(100 m) 3 m/s 2

t 8 s Super Dave, Jr. wins. 3. a) For segment 1, d1 2.0 m 0.5 m d1 1.5 m t1 0.6 s 0.0 s t1 0.6 s d vavg 1 t1 1.5 m vavg 0.6 s vavg 2.5 m/s

Fk

Fg

1

1

1

4

Ball


Forces are unbalanced as the force provided by the kicker, Fk, will cause the ball to accelerate.

b)

Fsupport

Fm

Gun

The forces are balanced. The force he provides on the gun, Fm, FB will balance the force of the bullet.

Fg

c)

Fbuoyant

The forces are not balanced, as the penny still accelerates downward, but at a slower rate.

Penny

Fg

d)

Fparachute

These forces are balanced, and the soldier falls downward at a constant speed.

Soldier

Fg

Section 1.12 1. a) F1 m1a1 F1 a1 m1 10 N a1 2.0 kg a1 5.0 m/s2 b) F1 2m1a2 F1 a2 2m1 F1 a1 m1 a a2 1 2 5.0 m/s2 a2 2 a2 2.5 m/s2

F c) 1 m1a3 2 F1 2m1a3 a a3 1 2 5.0 m/s2 a3 2 a3 2.5 m/s2 2. F ma Fg Ff ma Ff m( g a) Ff (90 kg)(9.8 m/s2 6.8 m/s2) Ff 270 N 2 3. v2 v12 2ad v22 v12 a 2d (0 m/s)2 (15 m/s)2 a 2(4.5 103 m) a 2.5 104 m/s2 F ma F (8.0 102 kg)(2.5 104 m/s2) F 2000 N 4. For the first kilometre, 1 d v1t a1t 2 2 1 d a1t2 2 2d a1 t2 2(1000 m) a1 (21.0 s)2 a1 4.54 m/s2 v22 v12 2ad v2 2ad v2 2(4.54 (1000 m/s2) m) v2 95.3 m/s For the last 1.4 km, the car’s acceleration is: v22 v12 2a2d v22 v12 a2 2d (0 m/s)2 (9.53 m/s)2 a2 2(1.40 103 m) a2 3.24 m/s2 Ff ma2 Ff (600 kg)(3.24 m/s2) Ff 1.94 103 N


5

During the first kilometre, the forces acting on the car are the force due to the engine and the frictional force: Fengine Ff ma1 Fengine ma1 Ff Fengine (600 kg)(4.54 m/s2) (1.94 103 N) Fengine 4.66 103 N 2 5. v2 v12 2ad v22 v12 a 2d (0 m/s)2 (28 m/s)2 a 2(0.35 m) a 1.12 103 m/s2 Fmitt ma Fmitt (0.25 kg)(1.12 103 m/s2) Fmitt 280 N

Section 1.13 1. a) Action: Foot striking the ball east Reaction: Ball pushing west on the foot b) Action: Paddle pushing backward on the water Reaction: Water pushing forward on the paddle c) Action: Balloon compressing and pushing air out Reaction: Air pushing back the other way on the balloon d) Action: Earth’s gravity pulling down on the apple Reaction: Apple’s gravity pulling up on Earth e) Action: Gravitational force downward of the laptop on the desk Reaction: Normal force upward of the desk on the laptop 3. a) FT mTa FT (6000 kg 5000 kg 4000 kg) (1.5 m/s2) FT 2.25 104 N b) The tension force in the rope between barges 1 and 2 is equal to the force required to accelerate barges 2 and 3 at a rate of 1.5 m/s2.

6

F1-2 (m2 m3)a F1-2 (5000 kg 4000 kg)(1.5 m/s2) F1-2 1.35 104 N The tension force in the rope between barges 2 and 3 can be found two ways: i) The difference between the force required to accelerate all the barges at a rate of 1.5 m/s2 minus the force required to accelerate the first two barges at the same rate: F2-3 FT F12 F2-3 2.25 104 N (6000 kg 5000 kg)(1.5 m/s2) F2-3 6.0 103 N ii) The force required to accelerate barge 3 at a rate of 1.5 m/s2: F2-3 m3a F2-3 (4000 kg)(1.5 m/s2) F2-3 6.0 103 N 4. a) FT mTa Fsled FT (m1 m2)a Fsled (Ff Ff ) a m1 m2 1

2

700 N 200 N a 600 kg a 0.83 m/s2 b) To find the tension force in the rope joining the two toboggans, we consider the forces acting on the second toboggan: FT m2a Frope Ff m2a Frope m2a Ff Frope (300 kg)(0.83 m/s2) 100 N Frope 350 N

Section 1.14 1. a) Friction is the only force acting on the truck, so Ff ma v2 v1 a t m(v2 v1) Ff t (4000 kg)(0 m/s 16.7 m/s) Ff 10 s 3 Ff 6.7 10 N


b) Ff k Fn Fn mg Ff kmg Ff k mg 6.7 103 N k (4000 kg)(9.8 m/s2) k 0.17 2. a) Since the toy duck is travelling at a constant velocity, it is not being acted upon by an unbalanced force. Therefore, the forces must have equal magnitudes and opposite directions. b) From a), we know that the applied force, Fapp, is equal in magnitude to the force due to friction, Ff . Fn mg Fapp Ff Fapp k Fn Fapp k mg Fapp a m a k g a (0.15)(9.8 m/s2) a 1.5 m/s2 3. Ff ma k Fn ma kmg ma a k g v22 v12 2ad v22 v12 d 2 k g (0 m/s)2 (22.2 m/s)2 d 2(0.60)(9.8 m/s2) d 42 m

Section 1.15 Gm1m2 1. Fg r2 (6.67 1011 N)(9.11 1031 kg)2 Fg (0.01 m)2 67 Fg 5.5 10 N

GmEarthmMoon 2. Fg r2 G(0.013)mEarth2 Fg r2 Fg

(6.67 1011 N m2kg2)(0.013)(5.97 1024 kg)2 (3.82 108 m)2

Fg 2.1 1020 N Gm1m2 3. a) Fg r2 1

1 Gm1m2 Fg 8 r2 1 Fg (Fg ) 8 G(2m1)m2 b) Fg (3r)2 2

2

1

2

2 Gm1m2 Fg 9 r2 2 Fg (Fg ) 9 G4m1m2 c) Fg (2r)2 Gm1m2 Fg r2 Fg Fg 1 (Fg ) Fg 4. 2 2

2

1

2

2

2

1

Earth

5.

2

1 GmyoumEarth GmyoumEarth 2 rEarth2 (rEarth r2)2 1 1 2 2 2(rEarth ) rEarth 2rEarthr2 r22 r22 2rEarthr2 rEarth2 0 2rEarth (2rEarth )2 4(1)( rEarth2) r2 2 r2 2.6 106 m GmyoumJupiter Fg rJupiter2 GmyoumJupiter myou gJupiter rJupiter2 GmJupiter gJupiter rJupiter2 gJupiter

(6.67 1011 N m2kg2)(1.9 1027 kg) (7.2 107 m)2

gJupiter 24 m/s2


7

Section 2.1 1. dA [E35°N] or [N55°E] dB [S12°E] or [E78°S] dC [S45°W] or [W45°S] dD [W80°N] or [N10°W] 2. a) In the N-S direction, dy d cos y (50 m) cos 14° [S] d y 49 m [S] d In the E-W direction, dx d sin x (50 m) sin 14° [E] d x 12 m [E] d b) In the N-S direction vy v sin vy (200 m/s) sin 30° [S] vy 100 m/s [S] In the E-W direction, vx v cos vx (200 m/s) cos 30° [W] vx 173 m/s [W] c) In the N-S direction, ay a sin y (15 m/s2) sin 56° [N] a y 12 m/s2 [N] a In the E-W direction, ax a cos ax (15 m/s2) cos 56° [E] ax 8.4 m/s2 [E] 3. Horizontally, vx v cos vx (5.0 m/s) cos 25° vx 4.5 m/s Vertically, vy v sin vy (5.0 m/s) sin 25° vy 2.1 m/s 4. vg vw vb vg 4.0 m/s [forward] 3.0 m/s [upward] Since vw and vb are perpendicular, 2 2 vg vv w b 2 vg (4.0 /s) m /s) (3.0 m 2 vg 5.0 m/s

8

vw tan vb

4.0 m/s tan1 3.0 m/s 53° vg 5.0 m/s [up 53° forward] 5. a) Component Method: vf v1 v2 For the x components, vfx v1x v2x vfx (50 m/s) cos 36° [W] (70 m/s) cos 20° [E] vfx (50 m/s) cos 36° (70 m/s) cos 20° vfx 25.3 m/s [E] For the y components, vfy v1y v2y vfy (50 m/s) sin 36° [N] (70 m/s) sin 20° [S] vfy (50 m/s) sin 36° (70 m/s) sin 20° vfy 5.45 m/s [N] 2 2 vf vv f f 2 vf (25.3 m/s) 5 (5.4 m/s)2 vf 26 m/s vf tan vf x

y

x

y

25.3 m/s tan1 5.45 m/s 78° vf 26 m/s [N78°E] Sine/Cosine Method: 54°

90° 54° 20°

16° vf2 v12 v22 2v1v2 cos vf2 (50 m/s)2 (70 m/s)2 2(50 m/s)(70 m/s) cos 16° vf 26 m/s To find direction, v1 vf sin sin 25.9 m/s 50 m/s sin sin 16° 32°


To find , 180° 54° 78° vf 26 m/s [N78°E] ϕ

θ

20°

For the y components, nety F 1y F 2y F 3y F Fnety (200 N) sin 30° [N]

v2 γ

v1 36°

β

vf

b) 37° (parallel line theorem)

180° 53° 37° (supplementary angles theorem)

90° Sine/cosine Method: df2 d12 d22 2d1d2 cos df2 (28 m)2 (40 m)2 2(28 m)(40 m) cos 90° df 49 m To find direction, df d1 sin sin 49 m 28 m sin 90° sin 28 m sin 49 m 35° To find , 180° 37° 18° f 49 m [W18°N] d γ

df d2 53°

β ϕ 37° θ

d1

c) Component Method: 1 F 2 F 3 Fnet F For the x components, 1x F 2x F 3x Fnetx F netx 140 N [W] (200 N) cos 30° [E] F (100 N) sin 35° [W] Fnetx 140 N (200 N) cos 30° (100 N) sin 35° Fnetx 24.15 N netx 24.15 N [W] F

(100 N) cos 35° [S] Fnety (200 N) sin 30° (100 N) cos 35° Fnety 18.08 N [N] 2 Fnet 2 Fnet F net Fnet (24.15 N)2 8 (18.0 N)2 Fnet 30.1 N Fnet tan Fnet x

y

x

y

24.15 N tan1 18.08 N 53° Fnet 30.1 N [N53°W] 6. vx v2 sin 40° v1 sin 15° vx 25.8 m/s [W] vy v2 cos 40° (v1 cos 15°) vy 1.17 m/s [N] (25.8 m/s)2 (1.17 m/s)2 v v 26 m/s 25.8 m/s tan1 1.17 m/s 87° v 26 m/s [N87°W]

Section 2.2

1. a) vog vmg vom vmg cos vom 5.0 km/h cos 20 km/h 76° The ship’s heading is [S76°E]. b) v2og v2om v2mg vog (20 km /h)2 (5.0 k m/h)2 vog 19 km/h [E] d c) t v 100 km t 19 km/h t 5.2 h


9

vmg 2. a) sin vom

0.50 m/s sin1 3.0 m/s 9.6° The girl’s heading is [N9.6°E]. b) The girl: (3.0 m /s)2 (0.50 m/s)2 vg vg 2.96 m/s [N] d t v 500 m t 2.96 m/s t 169 s The boy: d t v 500 m t 3.0 m/s t 167 s c) The boy travels an extra distance west of the girl’s landing point, caused by the horizontal component of his velocity (equal to the river’s current). d vt d (0.50 m/s)(167 s) d 83 m d) The time required for the boy to run the extra 83 m at 5.0 m/s is 17 s. The boy’s total time is 167 s 17 s 184 s. The girl’s time was 169 s. She wins the race. 3. vpw vsw vps v2pw v2sw v2ps v2pw (10 km /h)2 (6.0 k m/h)2 vpw 12 km/h 10 km/h tan 6.0 km/h 59° vpw 12 km/h [N59°E] 4. a) vog vom vmg vmg cos vog

0.50 m/s cos1 2.0 m/s 76° Terry must throw at [S76°E]. 10

(2.0 m /s)2 (0.50 m/s)2 b) vom vom 1.9 m/s [E] d t vom 5.0 m t 1.9 m/s t 2.6 s

Section 2.3 1. a)

b)

2. a)

b)

1 dy vi t ay t2 2 1 15 m (0 m/s)t (9.8 m/s2)t2 2 30 m t2 2 9.8 m/s t 1.7 s 1 dx vi t axt2 2 1 dx (25 m/s)(1.7 s) (0 m/s2)t2 2 dx 43 m v2 v1 ay t v2 v1 t a 0 m/s (35 m/s) sin 40° t 9.8 m/s2 t 2.3 s Since the curve Blasto travels is symmetrical (a parabola), the time he takes to reach maximum height is the same as the time he takes to reach the ground. ttotal 2(2.3 s) ttotal 4.6 s Solving for horizontal distance, 1 dx vi t axt2 2 dx (35 m/s) cos 40°(4.6 s) dx 120 m To find the time required for the bomb to reach the ground, 1 dy vi t ayt2 2 200 m (97.2 m/s) cos 25°t 1 (9.8 m/s2)t2 2 200 m (88.1 m/s)t (4.9 m/s2)t2 y

x

y

y

y

y

x

3. a)


y

0 (88.1 m/s)t (4.9 m/s2)t2 200 m t

88.1 m/s (88.1 m/s) 4(4 .9 m/s )(20 0 m) 2(4.9 m/s2) 2

2

t 2.0 s To calculate the horizontal distance, 1 dx vi t axt2 2 Since there is no horizontal acceleration, dx vi t dx (97.2 m/s) sin 25°(2.0 s) dx 82 m b) The y component of the final velocity, vfy, is vf 2 vi 2 2ad vf 2 [(97.2 m/s) cos 25°]2 2(9.8 m/s2)(200 m) vf 108 m/s vf (97.2 m/s) sin 25° vf 41.1 m/s 2 /s) m m/s) (41.1 2 vf (108 vf 115.6 m/s 41.1 m/s tan 108 m/s 21° vf 116 m/s inclined at 21° to the vertical 4. Since the time it takes for the ball to hit the green is not given, we can find two timerelated equations (one for the horizontal component and one for the vertical component), for the golf ball’s velocity, equate both equations, and solve for horizontal velocity. For the vertical component, 1 dy vi t ayt2 2 Since the change in height is 0 m, 1 0 (vi sin )t (9.8 m/s2)t2 2 (4.9 m/s2)t vi sin vi sin t (eq. 1) 4.9 m/s2 For the horizontal component, 1 dx vi t axt2 2 250 m (vi cos )t 250 m t (eq. 2) vi cos x

x

y

y

y

y

x

x

y

g

g

g

x

Equating equations 1 and 2, vi sin 250 m 2 4.9 m/s vi cos (250 m)(4.9 m/s2) vi 2 sin cos 1225 m2/s2 vi 2 sin 17°cos 17° vi 66 m/s vi 66 m/s, 17° above the horizontal g

g

g

g

g

g

Section 2.4

p 1. F F1 F2 Fp 200 N [N] 300 N [W] 2 2 F Fp F 1 2 2 Fp (200 ) N00 (3 N)2 Fp 361 N F F2 tan 2 Fp F1 θ 300 N tan 200 N F1 56°

Ff

p 361 N [N56°W] F For the frictional force, Ff kmg Ff 0.23(200 kg)(9.8 m/s2) Ff 451 N This is the maximum force of friction between the stove and the floor. However, friction only acts to oppose motion, so Ff 361 N [S56°E]. Fnet Fp Ff Fnet 361 N [N56°W] 361 N [S56°E] net 361 N[N56°W] 361 N [N56°W] F Fnet 0 N Fnet ma Fnet a m 0N a 200 kg a 0 m/s2 Since the frictional force is stronger than the force provided by the people’s pushing, the stove does not move.

g

g


11

1 F 2 F 3 2. a) Fnet F Fnet 25 N [S16°E] 35 N [N40°E] 45 N [W] Adding the x components, netx F 1x F 2x F 3x F Fnetx (25 N) sin 16° [E] (35 N) sin 40° [E] 45 N [W] Fnetx (25 N) sin 16° (35 N) sin 40° 45 N Fnetx 15.6 N netx 15.6 N [W] F Adding the y components, nety F 1y F 2y F 3y F Fnety (25 N) cos 16° [S] (35 N) cos 40° [N] Fnety (25 N) cos 16° (35 N) cos 40° Fnety 2.78 N [N] Fnet 2 Fnet 2 Fnet 2 Fnet (15.6 N) (2.78 N)2 Fnet 15.8 N Fnet tan Fnet 15.6 N tan 2.78 N 80° Fnet 15.8 N [N80°W] b) Fnet ma Fnet a m 16 N [N80°W] a 80 kg a 0.20 m/s2 [N80°W] net ma 3. F net (0.250 kg)(200 m/s2 [W15°S]) F net 50.0 N [W15°S] F net F 1 F 2 F net F 1 F2 F 2 50.0 N [W15°S] 100 N [N25°W] F 2 50.0 N [W15°S] 100 N [S25°E] F Adding the x components, 2x (50.0 N) cos 15° [W] F (100 N) sin 25° [E] F2x 6.03 N [W] x

x

y

12

y

Adding the y components, 2y (50.0 N) sin 15° [S] (100 N) cos 25° [S] F 2y 103.6 N [S] F 2 F2 2 F2 F 2 2 F2 (6.03 N) (103.6 N)2 F2 104 N F2 tan F2 103.6 N tan 6.03 N 86.7° 90° 86.7° 3.3° F2 104 N [S3.3°W] 4. The only two forces in the x direction are Fx and Ff. net F x F f F Fx F cos 45° Fx (250 N) cos 45° Fx 177 N Ff kFn g F y Fn F Fn mg F sin 45° Fn (20 kg)(9.8 m/s2) 177 N Fn 372 N Ff (0.40)(372 N) Ff 149 N Fnet 177 N 149 N Fnet 27.9 N Fnet ma Fnet a m 27.9 N a 20 kg a 1.38 m/s2 x

y

y

x

Section 2.5 1. The only two unbalanced forces are F|| and Ff. (eq. 1) Fnet F|| Ff (eq. 2) F|| Fg sin 25° Ff Fn Ff Fg cos 25° (eq. 3)


Substituting equations 2 and 3 into equation 1, Fnet Fg sin 25° Fg cos 25° Fnet Fg(sin 25° cos 25°) Fnet (2.0 kg)(9.8 m/s2)(sin 25° cos 25°) Fnet (19.6 N)(sin 25° cos 25°) Fnet 6.51 N Fnet ma 6.51 N (2.0 kg)a a 3.26 m/s2 1 d vit at2 2 1 4.0 m (3.26 m/s2)t2 2 8.0 m t 2 3.26 m/s t 1.6 s 2. Since there is no friction, the only force that prevents the CD case from going upward is the deceleration due to gravity, F||. Fnet F|| Fnet Fg sin 20° Since Fnet ma, ma mg sin 20° a g sin 20° a 3.35 m/s2 v2 v1 a t v2 v1 t a 4.0 m/s t 2 3.35 m/s t 1.2 s 3. To find the distance the skateboarder travels up the ramp, we need to find the velocity of the skateboarder entering the second ramp at v1. Since there is no change in velocity on the horizontal floor, v1 v2. For the acceleration on ramp 1, Fnet F|| ma mg sin 30° a g sin 30° a 4.9 m/s2 v22 v12 2ad v22 0 m/s 2(4.9 m/s2)(10 m) v2 9.9 m/s

For the deceleration on ramp 2, Fnet F|| Fn ma mg sin 25° (0.1)mg cos 25° a 5.02 m/s2 For d, v32 v22 2ad (0 m/s)2 (9.9 m/s)2 2(5.02 m/s2)d d 9.8 m 4. Fnet m(0.60g) Fnet also equals the sum of all forces in the ramp surface direction: net F || F f F engine F m(0.60g) mg sin 30° Fn Fengine m(0.60g) mg sin 30° (0.28)mg cos 30° Fengine Fengine (0.60)mg mg sin 30° (0.28)mg cos 30° Fengine mg(0.60 sin 30° (0.28) cos 30°) Fengine 3.36m N

Section 2.6 1. a) For m1, Fnet m1a (eq. 1) T m1 g m1a For m2, Fnet m2a (eq. 2) m2 g T m2a Adding equations 1 and 2, m2 g m1 g a(m1 m2) m2 g m1 g a m1 m2 a

(15 kg)(9.8 m/s2) 0.20(10 kg)(9.8 m/s2) 25 kg

a 5.1 m/s2 [right] Substitute a into equation 2: T m2 g m2a T 71 N b) For m1, Fnet m1a T m1 g sin 35° m1 g cos 35° m1a (eq. 1) For m2, Fnet m2a m2 g T m2a (eq. 2)


13

Adding equations 1 and 2, m2 g m1 g sin 35° m1 g cos 35° a(m1 m2) g(m2 m1 sin 35° m1 cos 35°) a m1 m2 a

(9.8 m/s2)[5.0 kg (3.0 kg) sin 35° 0.18(3.0 kg) cos 35°] 8.0 kg

a 3.5 m/s2 [right] Substitute a into equation 2: T m2 g m2a T (5.0 kg)(9.8 m/s2) (5.0 kg)(3.5 m/s2) T 32 N c) For m1, Fnet m1a T m1 g sin 40° 1m1 g cos 40° m1a (eq. 1) For m2, Fnet m2a m2 g sin 60° T 2m2 g cos 60° m2a (eq. 2) Adding equations 1 and 2, m2 g sin 60° 2m2 g cos 60° m1 g sin 40° 1m1 g cos 40° a(m1 m2) a

g(m2 sin 60° 2m2 cos 60° m1 sin 40° 1m1 cos 40°) m1 m2

a

(9.8 m/s2)[(30 kg) sin 60° 0.30(30 kg) cos 60° (20 kg) sin 40° 0.20(20 kg) cos 40°] 50 kg

a 1.1 m/s2 [right] Substitute a into equation 1: T m1a m1 g sin 40° 1m1 g cos 40° T (20 kg)(1.1 m/s2) (20 kg)(9.8 m/s2) sin 40° (0.20)(20 kg)(9.8 m/s2) cos 40° T 1.8 102 N d) For m1, Fnet m1a (eq. 1) m1 g sin 30° T1 m1a For m2, Fnet m2a (eq. 2) T1 T2 m2a For m3, Fnet m3a (eq. 3) T2 m3 g m3a

14

Adding equations 1, 2, and 3, m1 g sin 30° m3 g a(m1 m2 m3) m1 g sin 30° m3 g a m1 m2 m3 a

(30 kg)(9.8 m/s2)sin 30° (10 kg)(9.8 m/s2) 60 kg

a 0.82 m/s2 [left] Substitute a into equation 3: T2 m3a m3 g T2 (10 kg)(0.82 m/s2) (10 kg)(9.8 m/s2) T2 106 N Substitute a into equation 2: T1 m2a T2 T1 106 N (20 kg)(0.82 m/s2) T1 122 N

Section 2.7 v2 1. ac r (25 m/s)2 ac 30 m ac 21 m/s2 d 2. v t 2r v 25 t

v2 ac r 25002r ac t2 25002(1.3 m) ac (60 s)2 ac 8.9 m/s2 v2 3. ac r a) If v is doubled, ac increases by a factor of 4. b) If the radius is doubled, ac is halved. c) If the radius is halved, ac is doubled. 2r 4. a) v , where T r 3.8 105 km r 3.8 108 m T 27.3 days T 2.36 106 s


v2 ac r 42r ac T2 42(3.8 108 m) ac (2.36 106 s)2 ac 2.7 103 m/s2 b) The Moon is accelerating toward Earth. c) The centripetal acceleration is caused by the gravitational attraction between Earth and the Moon. 5. r 60 mm r 0.06 m ac 1.6 m/s2 v2 ac r v a cr v 0.31 m/s 6. Since d 500 m, r 250 m 2r v T 1 f T v 2rf ac g v2 ac r g 42rf 2 g f 42r 9.8 m/s2 f 42(250 m) f 0.0315 rotations/s f (0.0315 rotations/s) 60 s 60 min 24 h 1 min 1h 1 day f 2724 rotations/day

Section 2.8 d 1. a) v t 20(2r) v 180 s v 3.5 m/s

b) Fc mac

v2 Fc (10 kg) r Fc 24 N c) Friction holds the child to the merry-goround and causes the child to undergo circular motion. 2. Tension acts upward and the gravitational force (mg) acts downward. Fc Fnet and causes Tarzan to accelerate toward the point of rotation (at this instant, the acceleration is straight upward). Fc mac mv2 T mg r v2 Tm g r

(4 m/s)2 T (60 kg) 9.8 m/s2 2.5 m 2 T 9.7 10 N 3. Both tension and gravity act downward. Fc mac mv2 T mg r When T 0, mv2 mg r v gr 2 v (9.8 /s m 2 )(1. m) v 3.4 m/s 4. a) N cos 20°

N

N sin 20°

mg 20°

Fc mg tan 20° mv2 mg tan 20° r v rg tan 20° v (100 m )(9.8 m /s2) tan 20° v 19 m/s c) The horizontal component of the normal force provides the centre-seeking force. b)


15

d) If the velocity were greater (and the radius remained the same), the car would slide up the bank unless there was a frictional force to provide an extra centre-seeking force. The normal force would not be sufficient to hold the car along its path. e) Friction also provides a centre-seeking force. 5. G 6.67 1011 N·m2/kg2, mE 5.98 1024 kg Fc mMac mMv2 GmEmM 2 r r 2 GmE v r 42r3 2r GmE , where v T2 T T

10(2r) 4002r3 GmM , where v 2 T T r height of orbit rM r 1.9 105 m 1.74 106 m r 1.93 106 m 4002r3 T GmM

400 (1.93 10 m) T 11 2 2 22 2

(6.67 10

T 7.4 104 s

42r3 GmE

4 (3.4 10 m) T 11 2 2 24 2

(6.67 10

8

3

N m /kg )(5.98 10 kg)

T 1.97 10 s T 22.8 days 6. G 6.67 1011 N·m2/kg2, mE 5.98 1024 kg, rE 6.37 106 m Fc mHac GmEmH mHv2 2 r r 2 GmE v r GmE v r r height of orbit rE r 6.00 105 m 6.37 106 m r 6.97 106 m GmE v r 6

11

(6.67 10 N m /kg )(5.98 10 kg) v 6 2

2

24

6.97 10 m

v 7.57 10 m/s 7. G 6.67 1011 N·m2/kg2 mM (0.013)mE mM 7.77 1022 kg rM 1.74 106 m Fc mApolloac GmMmApollo mApollov2 r2 r 2 GmM v r 3

16


6

3

N m /kg )(7.77 10 kg)

c) Fnet 2Tv Fg Fnet ma Fnet 0 Fg 2Tv 2T sin m g 2(85 N) sin 1.5° m 9.8 N/kg

Section 3.3 1. Th = 1.0 × 104 N

Tv

60°

Th

m 0.45 kg

Horizontal: Th T cos 60° Th (1.0 104 N) cos 60° Th 5.0 103 N Vertical: Tv T sin 60° Tv (1.0 104 N) sin 60° Tv 8.7 103 N

4. a) FB

1.90 m

pail

0.65 m

Fg = mg

1.90 m tan 0.650 m 71.1° Fnet mg 2FBv Fnet ma Fnet 0 0 mg 2FB sin mg FB 2 sin (4.0 kg)(9.8 N/kg) FB 2 sin 71.1° FB 20.7 N b) Fh FB cos Fh (20.7 N) cos 71.1° Fh 6.71 N c) Fv FB sin Fv (20.7 N) sin 71.1° v 19.6 N [down] (not including the F weight of the beams)

Tv +

70°

70°

Ta = 100.0 N

Fnet Tv TA TA Fnet ma Fnet 0 Tv TA TA Tv 2TA Tv 2(100.0 N) cos 70° Tv 68.4 N 3. a) 5° 5° T = 85 N

T = 85 N bag

+

6.

Fn

Ff

Fg = mg

b) dv (1.5 m) sin 1.5° dv 0.039 m dv 3.9 cm

+ θ

2.

Ta = 100.0 N

FB θ

θ

T

boat F F||

F|| mg sin Ff mg cos Fnet T Ff F|| Fnet ma Solutions to Applying the Concepts

17

Fnet 0 T F|| Ff T mg sin mg cos T mg(sin cos ) T (400.0 kg)(9.8 N/kg) (sin 30° (0.25) cos 30°) T 1.11 103 N

w 1000 kg/m3 1000 cm3 Vw (10.0 L) 1L

1 m3 1.00 106 cm3 Vw 0.0100 m3 mw w·Vw mw (1000 kg/m3)(0.0100 m3) mw 10.0 kg Fg mg Fg (10.0 kg)(9.8 N/kg) Fg 98.0 N b) Position B provides the greatest torque because the weight is directed at 90° to the wheel’s rotation. c) A rF sin A (2.5 m)(10.0 kg)(9.8 N/kg) sin 45° A 1.7 102 N·m B rF sin B (2.5 m)(10.0 kg)(9.8 N/kg) sin 90° B 2.4 102 N·m C A C 1.7 102 N·m d) A larger-radius wheel or more and larger compartments would increase the torque.

Section 3.4 1. a)

1.50 m

Fg = mg 45.0 kg

50°

b) rF sin rmg sin (1.50 m)(45.0 kg)(9.8 N/kg) sin 40° 425 N·m 2. a) 2.0 103 N·m r 1.5 m 90° F? rF sin F r sin 2.0 103 N·m F (1.5 m) sin 90° F 1.3 103 N 3.

A

10.0 L

2.5 m B

C

a) Vw 10.0 L

18

Section 3.5 1. 20.0 kg

P

0.75 m 3.0 m

90° r1 ? m1 45.0 kg

0.75 m2 20.0 kg 3.0 m2 5.0 kg 0.75 m r2 2 r2 0.375 m m3 20.0 kg m2 m3 15.0 kg


3.0 m 0.75 m r3 2 r3 1.12 m 0 1 2 3 0 r1F1 sin 1 r2F2 sin 2 r3F3 sin 3 r3F3 r2F2 r1 F1 r1

At maximum height: H (1.75 m)(45 kg)(9.8 N/kg) sin 75.5° H 7.5 102 N·m (7.7 102 N·m 7.5 102 N·m) 100 % 2 7.7 10 N·m

% 2.6% 3.

P

(1.12 m)(15.0 kg)(9.8 N/kg) (0.375 m)(5.0 kg)(9.8 N/kg) (45.0 kg)(9.8 N/kg)

40°

r1 0.332 m 2. a) 1.7 m

Fg

1 2 0 r1F1 sin rcmFg sin rcmFg sin F1 r1 sin (0.375 m)(5.00 kg)(9.8 N/kg) F1 0.75 m F1 24.5 N b) Frv Fv2 0 Frv Fv2 Frv (5.00 kg)(9.8 N/kg) rv 49 N [up] F Frh Fh1 0 Frh Fh1 Frh 24.5 N rh 24.5 N [left] F The vertical reaction force is 49 N [up] and the horizontal reaction force is 24.5 N [left].

4.0 m

a)

t-t rF sin (1.0 m)(30.0 kg)(9.8 N/kg) t-t 2 t-t 147 N·m This torque applies to both sides of the teeter-totter, so the torques balance each other. b)

50°

40° F1

rh = 1.75 m

rl = ?

H L 0 L H rHmH g rL mL g (1.75 m)(45.0 kg) rL 30.0 kg rL 2.63 m c) 2.0 m 0.50 m

4.

F4

F3

P F1

θ

0.5 m cos 2.0 m 75.5° At the horizontal position: H (1.75 m)(45 kg)(9.8 N/kg) H 7.7 102 N·m

F2 1.6 m

0.4 m

1 2 3 0 r1F1 r2F2 r3FR3 0 0.75 m r1 2 r1 0.375 m


19

2.

2.0 m r2 2 r2 1.0 m r3 1.60 m 90° sin 1 r1F1 r2F2 F3 r3 F3

F2

θ2 F1 = Fn

+ θ1

3 306 N [up] F F4 F1 F2 F3 F4 (120.0 kg)(9.8 N/kg) (5.0 kg)(9.8 N/kg) 306 N FRP 919 N [up] Left saw horse: 919 N [up] Right saw horse: 306 N [up]

1 2 0 2 1 2 1 r2F2 sin 2 r1F1 sin 1 r1F1 sin 1 F2 r2 sin 2 F2

Section 3.6 τ

48 m

(8.0 102 m)(27 kg)(9.8 N/kg) sin (4.0 102 m) sin

F2 529.2 N F2 5.3 102 N The angle makes no difference — it cancels out.

45° +

3.

30 cm 45 cm

FL

Fm

11°

15°

P

Fw

45° rw 48.0 cm rw 0.480 m mw 10.0 kg 48.0 cm rL 2 rL 24.0 cm rL 0.240 m mL 5.00 kg w L 0 w L (rwFw sin 45°) (rLFL sin 45°) (0.480 m)(10.0 kg)(9.8 N/kg) sin 45° (0.240 m)(5.00 kg) (9.8 N/kg) sin 45° 41.6 N·m [clockwise]

20

4 cm

8 cm

(0.375 m)(120.0 kg)(9.8 N/kg) (1.0 m)(5.0 kg)(9.8 N/kg) 1.60 m

1.

P

a)

Fg–b Fg–s

m b s 0 m b s 0 m b s rmFm sin m rbFb sin b rsFs sin s rbFb sin b rsFs sin s Fm rm sin m


Fm

rbmb g sin b rsms g sin s rm sin m

rg sin (mb ms) Fm rm sin m Fm

(75 10–2 m)(9.8 N/kg) sin 75°[(0.57)85 kg19.0 kg] (45 10–2 m) sin 11°

Fm 5.57 103 N (tension)

Reaction forces: Fmy Fby Fsy 0 Fpy Fpy Fmy Fby Fsy Fpy (5.57 103 N)(sin 4°) (19.0 kg)(9.8 N/kg) (0.57)(85 kg)(9.8 N/kg) Fpy 1049.6 N Fpy 1.05 103 N [up] Fmx Fbx Fsx 0 Fpx Fpx Fmx Fbx Fsx Fpx (5.57 103 N)(cos 4°) 0 0 Fpx 5.55 103 N [right] Horizontal force: 1.49 103 N [right]; vertical force: 7.65 102 N [up]

Three-wheeled ATV:

θ

Top View

1.25 m

0.6 m τ

θ

0.55 m

Section 3.7

x

34.0 cm 1. a) sin 43° htipped 34.0 cm htipped sin 43° htipped 49.8 cm 34.0 cm b) tan 43° hstraight 34.0 cm hstraight tan 43° hstraight 36.5 cm 2. Four-wheeled ATV:

0.7 m

0.6 m

Back View

θ

T

m

1.0 m

1.0

θ θ

0.6 m

0.60 m tan T 1.0 m T 31.0°

x θT

0.60 m tan 1.25 m 25.64° x sin 0.55 m x (0.55 m)(sin 25.64°) x 0.237 m 0.237 m tan T 1.00 m tan T 13.3°


21

Section 3.8 1. k 16.0 N/m ∆x 30.0 cm ∆x 30.0 102 m a) F k∆x F (16.0 N/m)(30.0 102 m) F 4.80 N b) F ma F a m 4.80 N a 2.7 103 kg a 1.78 103 m/s2 2. Fg (67.5 kg)(9.8 N/m) Fg 661.5 N F kx F k x 661.5 N k 1.0 102 m

1. d 0.29 mm L 0.90 m ∆L 0.22 mm Esteel 200 109 N/m2

22

FL E AL AEL F L d 2 EL 2 F 4L

F

k 66150 N/m k 6.61 104 N/m Fg-truck mg Fg-truck (2.15 103 kg)(9.8 N/kg) Fg-truck 2.1 104 N This weight is distributed equally over four springs. 2.1 104 N Fs 4 springs Fs 5267.5 N/spring F x s k 5267.5 N/spring x 6.6150 104 N/m x 0.0796 m x 8.0 102 m 3. F kx F (120 N/m)(30.0 102 m) F 36 N

Section 3.9

d 2 2 A 4 F A E L L

0.29 103 m 2 (200 109 N/m2)(0.22 103 m) 2 4(0.90 m)

F 0.807 N For nylon, Enylon 5 109 N/m2 4FL d2 EL 4(0.807 N)(0.90 m) d 2 (5 109 N/m2)(0.22 103 m)

d 1.83 mm d 1.83 103 m 2. Emarble 50 109 N/m2 A 3.0 m2 m 3.0 104 kg F a) Stress A (3.0 104 kg)(9.8 N/kg) Stress 3.0 m2 Stress 9.8 104 N/m2 Stress b) E Strain Stress Strain E 9.8 104 N/m2 Strain 50 109 N/m2 Strain 2.0 106


c) L 15 m ∆L ? L Strain L L L(Strain) L (15 m)(2.0 106) L 3.0 105 m 3. a) Compressive strength of bone 17 107 N/m2 dbone 4.0 102 m Bone cross-sectional area is: A r2 A (2.0 102 m)2 A 1.26 103 m2

200 kg

Fg — 2

Fg — 2

F Fb g 2 mg Fb 2 F Breakage occurs if b Strength A mg 2 F b A A mg 2 Strength A 2(Strength)A m g 2(17 107 N/m2)(1.26 103 m2) m 9.8 N/kg 4 m 4.4 10 kg


23

Section 4.2

mv 1. p (8 kg)(16 m/s [W20°N]) p p 128 kg·m/s [W20°N] p 1.3 102 kg·m/s [W20°N] 9.0 104 kg·m/s [E] 2. p 1000 m 1h v (72 km/h [E]) 1 km 3600 s v 20 m/s [E] p m v 9.0 104 kg·m/s m 20 m/s m 4.5 103 kg mv 3. a) p (0.5 kg)(32 m/s [S]) p p 16 kg·m/s [S] Using a scale factor of 1 mm 1 kg·m/s,

p = 16 kg·m/s [S]

mv b) p (0.5 kg)(45 m/s [N]) p p 22.5 kg·m/s [N]

p = 22.5 kg·m/s [N]

p2 p1 c) p p 22.5 kg·m/s [N] 16 kg·m/s [S] 22.5 kg·m/s [N] 16 kg·m/s [N] p 38.5 kg·m/s [N] p

p = 38.5 kg·m/s [N]

24

Section 4.3

1. a) J Ft J (3257 N [forward])(1.3 s) J 4234.1 N·s [forward] J 4.2 103 N·s [forward] Ft b) J J ma t v2 v1 J m t t J m(v2 v1) J (0.030 kg)(200 m/s 0 m/s) J 6.0 N·s [out of gun] Ft c) J t J ma J (0.500 kg)(9.8 N/kg [down])(3.0 s) J 14.7 N·s [down] J 15 N·s [down] p1 p2 2. p p mv2 mv1 m(v2 v1) p (54 kg)(20 m/s [up] 25 m/s [down]) p (54 kg)(20 m/s [up] 25 m/s [up]) p (54 kg)(45 m/s [up]) p 2.4 103 N·s [up] p

J 3. a) F t 2.5 103 N·s F 0.2 s F 1.3 104 N b) v1 0 v2 120 km/h v2 33.3 m/s v2 v1 a t 33.3 m/s 0 m/s a 0.2 s a 166.7 m/s2 1 d v1t at2 2 1 d (0 m/s)(0.2 s) (166.7 m/s2)(0.2 s)2 2 d 3.3 m


1 4. a) J bh 2 1 J (5 s)(25 N [S]) 2 J 62.5 N·s [S] b) J Area under triangle rectangle J 1(500 250 N [W])(3 s) 2 (250 N [W])(6 s) J 1875 N·s [W] c) J Area above area below (counting the squares: approximately) J (13 squares above) (4 squares below) J 9 squares Multiplying 9 by the length and width of each square, J 9(0.05 s)(100 N [E]) J 45 N·s [E]

Section 4.4 1. m1 1.2 kg, v1o 6.4 m/s, v1f 1.2 m/s, m2 3.6 kg, v2o 0, v2f ? po pf m1v1o m2v2o m1v1f m2v2f (1.2 kg)(6.4 m/s) (1.2 kg)(1.2 m/s) (3.6 kg)v2f v2f 2.5 m/s [forward] 2. m1 30 g 0.03 kg, v1o 0, v1f 750 m/s, m2 1.9 kg, v2o 0, v2f ? po pf m1v1o m2v2o m1v1f m2v2f m1v1o m2v2o m1v1f v2f m2 v2f

(0.03 kg)(0) (1.9 kg)(0) (0.03 kg)(750 m/s) 1.9 kg

v2f 11.8 m/s [back] 3. m1 400 g 0.400 kg, v1o 3.0 m/s [forward], v1f 1.0 m/s [forward], m2 0.400 kg, v2o 0, v2f ? o pf p m1v1o m2v2o m1v1f m2v2f m1v1o m2v2o m1v1f v2f m2 v2f

(0.400 kg)(3.0 m/s [forward]) (0.400 kg)(0) (0.400 kg)(1.0 m/s [forward]) 0.400 kg

v2f 2.0 m/s [forward] 4. m1 m, m2 80m, m(12) 81m, v(12)o ?, v1f 1.5 106 m/s, v2f 4.5 103 m/s, po 7.9 1017 kg·m/s po pf po m1v1f m2v2f 7.9 1017 kg·m/s m(1.5 106 m/s) 80m(4.5 103 m/s) 7.9 1017 kg·m/s m[1.5 106 m/s 80(4.5 103 m/s)] 7.9 1017 kg·m/s m 1.5 106 m/s 80(4.5 103 m/s) m 6.9 1023 kg 5. m1 5m, v1o v, v(12)f ?, m2 4m, v2o 0 po pf m1v1 m2v2 (m1 m2)v(12)f (5m)(v) (4m)(0) (5m 4m)v(12)f 5mv 9mv(12)f 5 v(12)f v 9

Section 4.5

1. m1 m2 2.0 kg, v1o 5.0 m/s [W], v2o 0, v1f 3.0 m/s [N35°W], v2f ? 1o (2.0 kg)(5.0 m/s [W]) p 1o 10 kg·m/s [W] p 1f (2.0 kg)(3.0 m/s [N35°W]) p 1f 6.0 kg·m/s [N35°W] p o pf p 1o 2o 0 p p2o p1f p2f, where p p1o p1f p2f θ

p1f = 6.0 kg·m/s

p2f

35° θ

p1o = 10 kg·m/s

Using the cosine law, p2f2 (10 kg·m/s)2 (6.0 kg·m/s)2 2(10 kg·m/s)(6.0 kg·m/s) cos 55° p2f 8.2 kg·m/s p mv 8.2 kg·m/s v2f 2 kg v2f 4.1 m/s


25

Using the sine law to find direction, sin sin 55° 8.2 kg·m/s 6.0 kg·m/s 37° v2f 4.1 m/s [W37°S] 2. m1 85 kg, v1o 15 m/s [N], 1o 1275 kg·m/s [N], m2 70 kg, p 2o 350 kg·m/s [E] v2o 5 m/s [E], p o pf p p1o p2o pf p2o = 350 kg·m/s

p1o = 1275 kg·m/s θ

pf

Using Pythagoras’ theorem to solve for pf, pf2 (1275 kg·m/s)2 (350 kg·m/s)2 pf 1322 kg·m/s 350 kg·m/s tan 1275 kg·m/s 15.4° pf mfvf 1322 kg·m/s [N15°E] vf 85 kg 70 kg vf 8.5 m/s [N15°E] 3. m1 0.10 kg, v1f 10 m/s [N], 1f 1.0 kg·m/s [N], m2 0.20 kg, p v2f 5.0 m/s [S10°E], 2f 1.0 kg·m/s [S10°E], m3 0.20 kg, p v3f ? To 0 p pTf pTo 0 p1f p2f p3f Using the cosine law,

0.17 kg·m/s v3f 0.2 kg v3f 0.87 m/s Using the sine law to find direction, sin 10° sin 0.1743 kg·m/s 1.0 kg·m/s 85° v3f 0.87 m/s [S85°W] or 0.87 m/s [W5°S] 4. m1 0.5 kg, v1o 2.0 m/s [R], 1o 1.0 kg·m/s [R], m2 0.30 kg, v2o 0, p 2o 0, v1f 1.5 m/s [R30°U], p 1f 0.75 kg·m/s [R30°U], v2f ?, p 2f ? p To pTf p 2o 0 p1o p2o p1f p2f , where p p1f p2f p1o Using the cosine law, p1f = 0.75 kg·m/s

p1f = 1.0 kg·m/s p2f = 1.0 kg·m/s θ

p3f

p1o = 1.0 kg·m/s

p2f2 (1.0 kg·m/s)2 (0.75 kg·m/s)2 2(1.0 kg·m/s)(0.75 kg·m/s)cos 30° p2f 0.513 kg·m/s p mv 0.513 kg·m/s v2f 0.30 kg v2f 1.7 m/s Using the sine law to find direction, sin sin 30° 0.513 kg·m/s 0.75 kg·m/s 47° v2f 1.7 m/s [R47°D] or 1.7 m/s [D43°R]

Section 4.6 3.0 m 1. a) 1.5 m from both objects 2

2.0 kg b) (60 cm) 5.0 kg 2.0 kg 17.1 cm from the larger mass 200 c) (20 km) 600 6.67 km from the larger satellite

p3f2 (1.0 kg·m/s)2 (1.0 kg·m/s)2 2(1.0 kg·m/s)(1.0 kg·m/s)(cos 10°) p3f 0.1743 kg·m/s 26

θ

30°

10°

p2f


0.011 m 2. a) p1o (2.0 kg) 0.1 s p1o 0.22 kg·m/s [S20°E] 0.017 m p2o (1.0 kg) 0.1 s p2o 0.17 kg·m/s [S10°W] 0.013 m p1f (2.0 kg) 0.1 s p1f 0.26 kg·m/s [S5°W] 0.015 m p2f (1.0 kg) 0.1 s p2f 0.15 kg·m/s [S30°E] 0.013 m pcm (3.0 kg) 0.1 s pcm 0.39 kg·m/s [S8°E] b) i) 70°

p1o

10°

p To p2o

ii) 5°

p1f

p Tf

30°

p2f

c) The total momentum before and after collision is the same as the momentum of the centre of mass. The total momentum vectors have the same length and direction as the momentum of the centre of mass.


27

Section 5.2 1. a) W Fd W (40 N)(0.15 m) W 6.0 J b) W Fd W mgd W (50 kg)(9.8 N/kg)(1.95 m) W 9.6 102 J c) W Fd cos W (120 N)(4 m)(cos 25°) W 4.4 102 J 2. 45 km/h 12.5 m/s To find d, v22 v12 2ad (v22 v12) d 2a (12.5 m/s)2 0 d 2(2.5 m/s2) d 31.25 m W Fd W (5000 N)(31.25 m) W 1.6 105 J 3. W Fd cos W (78 N)(10 m)(cos 55°) W 4.5 102 J (v2 v1) 4. a t (14 m/s 25 m/s) a 5.0 s a 2.2 m/s2 F ma F (52 000 kg)(2.2 m/s2) F 114 400 N (v22 v12) d 2a [(14 m/s)2 (25 m/s)2] d 2(2.2 m/s2) d 97.5 m W Fd W (114 400 N)(97.5 m) W 1.1 107 J 5. a) W Fd W (175 N)(55 m) W 9625 J

28

b) The triangular areas above and below the axis are identical and cancel out, therefore, W (0.040 m)(20 N) W 0.80 J 6. F ma F (3 kg)(9.8 N/kg) F 29.4 N W d F 480 J d 29.4 N d 16 m

Section 5.3 1 1. a) Ek mv2 2 1 Ek (20 000 kg)(7500 m/s)2 2 Ek 5.6 1011 J b) 20 km/h 5.6 m/s 1 Ek mv2 2 1 Ek (1.0 kg)(5.6 m/s)2 2 Ek 15.4 J 1 c) Ek mv2 2 1 Ek (0.030 kg)(400 m/s)2 2 Ek 2.4 103 J 1 2. Ek mv2 2 1 3900 J (245 kg)v2 2 v

(3900 J)(2) 245 kg

v 5.6 m/s 1 3. Ek mv2 2 2Ek m v2 2(729 J) m 2 (15 m/s) m 6.5 kg 2mEk 4. p p 2(9.11 1031 kg)(6000 eV)(1.6 1019 J/eV) 23 p 4.2 10 N·s


5. Ek Ek2 Ek1 1 1 Ek (60 kg)(5.0 m/s)2 (60 kg)(14 m/s)2 2 2 3 Ek 5.1 10 J 1 6. a) Ek mv2 2 1 Ek (0.350 kg)(25.0 m/s)2 2 Ek 1.1 102 J (v22 v12) b) a 2d 0 (25.0 m/s)2 a 2(0.024 m) a 1.3 104 m/s2 F ma F (0.350 kg)(1.3 104 m/s2) F 4557 N W Fd W (4557 N)(0.024 m) W 1.1 102 J c) Favg ma Favg 4557 N Favg 4.6 103 N

Section 5.4 1. a) Eg mgh Eg (3.5 kg)(9.8 N/kg)(1.2 m) Eg 4.1 101 J b) Eg mgh Eg (2000 kg)(9.8 N/kg)(0) Eg 0 J c) Eg mgh Eg (2000 kg)(9.8 N/kg)(1.9 m) Eg 3.7 104 J 2. a) v22 v12 2ad (or use the conservation of energy) 2 v2 (0) 2(9.8 m/s2)(27 m) v2 23 m/s b) Efinal Einitial Ekf Ego Eko 1 (65 kg)vf2 (65 kg)(9.8 N/kg)(27 m) 2 1 (65 kg)(3.0 m/s)2 2 vf 23 m/s

3. a) Using the law of conservation of energy, Etotal 5460 J 1 mv2 mgh 5460 J 2 1 (3.0 kg)v2 (3.0 kg)(9.8 N/kg)(5.0 m) 5460 J 2 v 60 m/s b) Eg mgh 5460 J (3.0 kg)(9.8 N/kg)h h 185.7 m from the ground h 180.7 m from the pad c) v2 v1 at v2 (60 m/s) (9.8 N/kg)(2.0 s) v2 40.4 m/s 1 Ek mv2 2 1 Ek (3.0 kg)(40.4 m/s)2 2 Ek 2.4 103 J Ep Etotal Ek Ep 5460 J 2448.24 J Ep 3.0 103 J 4. F kx F k x mg k x (5000 kg)(9.8 N/kg) k 0.04 m k 1.2 106 N/m For only one spring: 1 225 000 N/m k 4 k 3.0 105 N/m

Section 5.5 rise 1. a) k run 20 N k 0.1 m k 200 N/m k 2.0 102 N/m


29

2.

3.

4.

5.

30

b) Maximum elastic potential energy occurs at x 0.1 m. 1 Ep kx2 2 1 Ep (200 N/m)(0.1 m)2 2 Ep 1.0 J c) Ee Ee2 Ee1 1 E (200 N/m)(0.04 m)2 2 1 (200 N/m)(0.03 m)2 2 Ee 7.0 102 J Fg Fe mg kx (0.500 kg)(9.8 N/kg) k(0.04 m) k 122.5 N/m a) W E W E2 E1 where E1 0 W E2 1 W kx2 2 1 W (55 N/m)(0.04 m)2 2 W 4.4 102 J b) W E W E2 E1 where E1 0 W E2 1 W kx2 2 1 W (85 N/m)(0.08 m)2 2 W 2.7 101 J Ee Ek 1 1 kx2 mv2 2 2 2 (200 N/m)(0.08 m) (0.02 kg)v2 v 8.0 m/s Ee Ek 1 2 1 2 kx mv 2 2 6 2 (5 10 N/m)x (2000 kg)(4.5 m/s)2 x 9 cm

6. The loss in elastic potential energy is equal to the gain in kinetic energy. Ee Ek Let the subscript 1 represent the initial compressed spring and subscript 2 represent the moment after the spring has been released when the cart has a velocity of 0.42 m/s. (Ee2 Ee1) Ek2 Ek1 1 1 1 kx12 kx22 mv22 0 2 2 2 x2

kx12 mv22 k

(65 N/m)(0.08 m) (1.2 kg)(0.42 m/s) x2 2

2

65 N/m

x2 0.056 m x2 5.6 cm

Section 5.6 1. The energy required to heat the water is Ew (4.2 103 J/°C/L)(65°C 10°C)(2.3 L) Ew 5.31 105 J The energy expended by the stove is E P t Es Pt Es (1000 W)(600 s) Es 6.0 105 J The energy lost to the environment is E Es Ew E 6.9 104 J 2. a) Ep mgh Ep (83.0 kg)(9.8 N/m)(13.0 m) Ep 1.057 104 J E P t 1.057 104 J P 18.0 s P 590 W b) Ep 1.057 104 J Ep 10 600 J 3. Once the radiation of the Sun reaches Earth, it has spread out into a sphere surrounding the Sun. This sphere has a surface area of: SA 4 r2 SA 4 (1.49 1011 m)2 SA 2.79 1023 m2


The ratio of this area to the area of Earth exposed to the radiation will be equal to the ratio of the power radiated by the Sun to the power absorbed by Earth. SASun Sun’s radiation absorbed radiation AEarth 23 2 3.9 1026 W 2.79 10 m dEarth 2 x () 2 2.79 1023 m2 3.9 1026 W (6.87 106 m)2 x 26 (3.9 10 W)(1.48 1014 m2) x 2.79 1023 m2 x 2 1017 W Therefore, Earth intercepts 2 1017 J of energy from the Sun each second. 4. The total time is 3(20 min)(60 s/min) 3600 s The time the player spends on ice is (3600 s)(0.25) 900 s E P t E Pt E (215 W)(900 s) E 1.935 105 J While sitting on the bench, the player expends 100 W of power. He spends 3600 s 900 s 2700 s on the bench. E (100 W)(2700 s) E 2.7 105 J ET (1.935 105 J) (2.7 105 J) ET 4.6 105 J

Section 5.7 3. a) m1 3000 kg v1o 20 m/s [W] v1f 10 m/s [W] m2 1000 kg v2o 0 v2f ? pTo pTf m1v1o m2v2o m1v1f m2v2f (3000 kg)(20 m/s) 0 (3000 kg)(10 m/s) (1000 kg)v2f v2f 30 m/s

b) Since Eko Ekf, the collision is elastic (EkTotal 6 105 J). c) W Ektruck 1 W (3000 kg)(10 m/s)2 2 1 (3000 kg)(20 m/s)2 2 W 4.5 105 J 4. mp 0.5 kg mg 75 kg dp 0.03 m vpo 33.0 m/s vgo 0 vgf 0.30 m/s a) pgo mv pgo (75 kg)(0) pgo 0 Ekgo 0 ppo mv ppo (0.5 kg)(33.0 m/s) ppo 16.5 kg·m/s 1 Ekpo (0.5 kg)(33.0 m/s)2 2 Ekpo 272.25 J b) po pf ppo pgo ppf pgf mpvpo 0 mpvpf mgvgf (0.500 kg)(33.0 m/s) (0.500 kg)vpf (75 kg)(0.30 m/s) vpf 12 m/s 1 c) Ekp mpvpf2 2 1 Ekp (0.500 kg)(12 m/s)2 2 Ekp 36 J 1 Ekg mgvgf2 2 1 Ekg (75 kg)(0.30 m/s)2 2 Ekg 3.4 J d) The collision is inelastic due to the loss of kinetic energy.


31

5. m1 10 g m2 50 g v1o 5 m/s v2o 0 m1 m2 v1f v1o m1 m2 10 g 50 g v1f (5 m/s) 10 g 50 g v1f 3.3 m/s 2m1 v2f v1o m1 m2 2(10 g) v2f (5 m/s) 10 g 50 g v2f 1.7 m/s 6. m1 0.2 kg m2 0.3 kg v1o 0.32 m/s v2o 0.52 m/s Changing the frame of reference, v1o 0.84 m/s v2o 0 m/s 0.2 kg 0.3 kg v1f (0.84 m/s) 0.2 kg 0.3 kg v1f 0.168 m/s 2(0.2 kg) v2f (0.84 m/s) 0.2 kg 0.3 kg v2f 0.672 m/s Returning to the original frame of reference, v1f 0.168 m/s 0.52 m/s v1f 0.69 m/s v2f 0.672 m/s 0.52 m/s v2f 0.15 m/s 1 8. a) Estored bh 2 1 Estored (0.06 m 0.02 m)(50 N) 2 Estored 1.0 J 1 b) Elost 1.0 J (0.005 m)(30 N) 2 (0.005 m)(20 N) 1 (0.035 m)(20 N) 2 Elost 1.0 J 0.075 J 0.1 J 0.35 J Elost 0.475 J

32

9. a) Counting the squares below the top curve, there are about 16.5 squares, each with an area of (0.01 m)(166.7 N) 1.6667 J. The amount of energy going into the shock absorber is (16.5)(1.6667 J) 27.5 J. b) There are roughly 6 squares below the lower curve. The energy returned to the shock absorber is (6)(1.6667 J) 10 J 27.5 J 10 J c) % energy lost 100 27.5 J % energy lost 64%


Section 6.1 1. mE 5.98 10 kg, mS 1.99 10 kg, r 1.50 1011 m GMm a) Ek 2r 24

30

(6.67 1011 N·m2/kg2)(1.99 1030 kg)(5.98 1024 kg) 2(1.50 1011 m)

Ek

Ek 2.65 1033 J GMm b) Ep r Ep

(6.67 1011 N·m2/kg2)(1.99 1030 kg)(5.98 1024 kg) (1.50 1011 m)

Ep 5.29 1033 J c) ET Ek Ep ET 2.65 1033 J (5.29 1033 J) ET 2.65 1033 J GM 2. ag r2 (6.67 1011 N·m2/kg2)(5.98 1024 kg) ag (6.38 106 m 1 106 m)2 ag 7.32 m/s2 3. v1000 km 6.0 km/s 6.0 103 m/s, h 1000 km 1 106 m a) vesc

2GM r 11

2(6.67 10 N·m /kg )(5.98 10 kg) vesc 6 6 2

2

24

(6.38 10 m 1 10 m)

vesc 10 397 m/s Since the rocket has only achieved 6000 m/s, it will not escape Earth. 1 b) Ek 1000 km mv2 2 1 Ek 1000 kmm(6000 m/s)2 2 Ek 1000 km1.8 107m J Since all kinetic energy is converted to gravitational potential energy at maximum height, Ek Ep Ek E2 E1 GMm GMm 1.8 107m J r2 r1

GM r2 GM 1.8 107 J r1 r2

(6.67 1011 N·m2/kg2)(5.98 1024 kg) (6.67 1011 N·m2/kg2)(5.98 1024 kg) 1.8 107 J 6.38 106 m 1 106 m

r2 1.1 107 m hmax r2 rE hmax 1.1 107 m 6.38 106 m hmax 4.7 106 m

Section 6.2 1. a) MSun 1.99 1030 kg, T 76.1 a 2.4 109 s T 2 ka3 (2.4 109 s)2 a 4 2 GM a

1 3

(2.4 109 s)2 4 2 (6.67 1011 N·m2/kg2)(1.99 1030 kg)

1 3

a 2.7 1012 m b) 0.97 d c) v t 2 (2.69 1012 m) v 2.4 109 s v 7031 m/s 2. raltitude 10 000 km 1 107 m, rJupiter 7.15 107 m, mJupiter 1.9 1027 kg vesc

2GM r 11

2(6.67 10 N·m /kg )(1.9 10 kg) vesc 7 7 2

2

27

7.15 10 m 1 10 m

vesc 56 000 m/s 3. mMoon 7.36 1022 kg, mEarth 5.98 1024 kg, r 3.82 108 m a) vesc

2GM r 11

2(6.67 10 N·m /kg )(5.98 10 kg) vesc 8 2

2

24

3.82 10 m

vesc 1445 m/s


33

To find the current speed of the Moon, 1 GMm mv2 2 2r v

GM r

2

(6.67 10

11

2

2

24

3.82 10 m

v 1022 m/s To find the additional speed required for escape, vadd esc 1445 m/s 1022 m/s vadd esc 423 m/s 1 1 b) Ek mvesc2 mv 2 2 2 1 Ek (7.36 1022 kg)[(1445 m/s)2 2 (1022 m/s)2] Ek 3.84 1028 J c) This value is comparable to a 900-MW nuclear power plant (e.g., Darlington) running for 2.35 1011 years! 4. Geostationary Earth satellites orbit constantly above the same point on Earth because their period is the same as that of Earth. 5. M 5.98 1024 kg, r 6.378 106 m, v 25 m/s To find the semimajor axis, ET Ep Ek GMm GMm 1 mv2 2 2a r GM 2GM v2 a r 1 2 v2 a r GM 1 2GM v2r a GMr GMr a 2GM v2r (6.67 1011 N·m2/kg2)(5.98 1024 kg)(6.378 106 m) 2(6.67 1011 N·m2/kg2)(5.98 1024 kg) (25 m/s)2(6.378 106 m)

a 3.19 106 m

34

4 2 T2 ka3, where k GM 4 (3.19 10 m) T 11 2 2 24

(6.67 10 N·m /kg )(5.98 10 kg) v 8

a

To find the period,

6

3

N·m /kg )(5.98 10 kg)

T 1792 s

Section 6.3 1. a) At the equilibrium point, the bob’s kinetic energy accounts for all the energy in the system. This total energy is the same as the maximum elastic potential energy. Ek equilET Ek equilEpmax 1 Ek equilkx2 2 1 Ek equil(33 N/m)(0.23 m)2 2 Ek equil0.87 J b) 0 1 c) Ek mv2 2 v

2E m

v

2(0.87 J) 0.485 kg

k

v 1.9 m/s 2. a) To find the period of an object in simple harmonic motion, k 0.485 kg T 2 33 N/m T 2

m

T 0.76 s b) At 0.16 m, the elastic potential energy of the bob is 1 Ep 0.16m kx2 2 1 Ep 0.16m (33 N/m)(0.16 m)2 2 Ep 0.16m 0.42 J ET Ek Ep Ek ET Ep Ek 0.87 J 0.42 J Ek 0.45 J


1 Ek mv2 2 v

2E m

v

2(0.45 J) 0.485 kg

k

v 1.36 m/s c) Ek 0.45 J, from part b 3. Displacement (m)

Position vs. Time 0.4 0.2 0 –0.2 –0.4 –0.6 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

Time (s)


35

Section 7.2 10° 1. a) 0.17 rad 57.3°/rad 60° b) 1.0 rad 57.3°/rad 90° c) 1.6 rad 57.3°/rad 176° d) 3.07 rad 57.3°/rad 256° e) 4.47 rad 57.3°/rad 2. a) ( rad)(57.3°/rad) 180° b) rad (57.3°/rad) 45° 4 c) (3.75 rad)(57.3°/rad) 675° d) (11.15 rad)(57.3°/rad) 639° e) (40 rad)(57.3°/rad) 2.3 103° 3. a) Earth rotates 2 radians every 24 h. 2 rad 6.0 h 1.57 rad 24 h b) Earth moves 2 rad every 365 days. 2 rad 265 d 4.56 rad 365 d c) The second hand moves 2 rad every 60 s. 2 rad 25 s 2.62 rad 60 s d) A runner moves 2 rad for every lap. 2 rad 25.6 laps 161 rad 1 lap

Section 7.3 v2 2. a) ac r acr v 2 v (9.8 m/s )(12 00 m) v 108 m/s v 1.1 102 m/s v b) r 108 m/s

1200 m

0.090 rad/s The angular acceleration is zero because the angular velocity is constant.

36

1 min 2 rad 1.2 rev 3. a) 1 min 60 s 1 rev

0.12566 rad/s

0.13 rad/s b) r 1500 m ac r 2 ac (1500 m)(0.12566 rad/s)2 ac 24 m/s2 c) The angular acceleration is zero because the angular velocity is constant. d) ac-space-station 24 m/s2 ac-Earth 9.8 m/s2 24 m/s2 2 2.4 9.8 m/s

Section 7.4 1. a) (3.35 rev/s)(2 rad/rev)

21.0 rad/s 60 s t 2 min (50 sec) 1 min t 170 s

t t (21.0 rad/s)(170 s) 3.58 103 rad

b) t (22.0 rad/s 0) 0.5 s 44 rad/s2

2. a) t (0 1.75 rad/s) t 0.21 rad/s2 t 8.3 s ( 1 2) b) t 2 (1.75 rad/s 0) (8.3 s) 2 7.3 rad c) There are 2 radians in one cycle. 7.3 rad number of cycles 2 rad/cycle number of cycles 1.16 number of cycles 1.2


1.16 cycles d) 0.58 cycles 2 (0.58 cycles)(2 rad/cycle) 3.6 rad 1 2t t2 2 1 3.6 rad 0 (0.21 rad/s2)t2 2 t 5.9 s 3. a) t ( 1 2) 2 92.2 rad t 2 (16.1 rad/s 14.5 rad/s) t 6.026 s t 6.03 s

b) t 14.5 rad/s 16.1 rad/s 6.026 s 0.266 rad/s2

Section 7.5 2. a) I (0.045 kg·m2)(1.90 rad/s2) 0.086 N·m b) For 78 rpm:

1 0 1 min 2 rad 78 rev

2 1 min 60 s 1 rev

2 8.2 rad/s

22 12 2 ( 22 12) 2 (8.2 rad/s)2 0 2(1.90 rad/s2) 17.69 rad 18 rad 17.69 rad number of turns 2 rad/turn number of turns 2.8 For 45 rpm:

1 0 1 min 2 rad 45 rev

2 1 min 60 s 1 rev

2 4.7 rad/s

22 12 2 ( 22 12) 2 (4.7 rad/s)2 0 2(1.90 rad/s2) 5.813 rad 5.8 rad 5.813 rad number of turns 2 rad/turn number of turns 0.93 1 For 33 rpm: 3

1 0 100 rev 1 min 2 rad 3

2 60 s 1 min 1 rev

2 3.5 rad/s

22 12 2 ( 22 12) 2 (3.5 rad/s)2 0 2(1.90 rad/s2) 3.223 rad 3.2 rad 3.223 rad number of turns 2 rad/turn number of turns 0.51 3. I 8.45 N·m I 2 12.2 rad/s I 0.693 kg·m2 1 4. a) I mr2 (moment of inertia for a disk) 2 1 I (5.55 kg)(1.22 m)2 2 I 4.13 kg·m2 b) rF (1.22 m)(15.1 N) 18.4 N·m c) I 18.4 N·m 4.13 kg·m2 4.46 rad/s2


37

b) v r

v (0.320 m)(33.3 rad/s) v 10.7 m/s 1 Ek mv2 2 1 Ek (1000 kg)(10.7 m/s)2 2 Ek 5.7 104 J

Section 7.6 1. a) rF (0.20 m)(23.1 N) 4.62 N·m 4.6 N·m WR WR (4.62 N·m)(2 rad) WR 29 J b) WR WR (4.62 N·m)(1.5 rad) WR 6.9 J c) 95° 1.66 rad WR WR (4.62 N·m)(1.66 rad) WR 7.7 J 2. a) 45° rad 4 WR WR rF WR (0.556 m)(12.2 N) rad 4 WR 5.3 J b) The work done does not change.

Section 7.8

Section 7.7 2 1. I mr2 5 2 I (0.0350 kg)(0.035 m)2 5 I 1.7 105 kg·m2 1 Erot I 2 2 1 Erot (1.7 105 kg·m2)(165 rad/s)2 2 Erot 0.23 J 2. a) (5.3 rev/s)(2 rad/rev)

33.3 rad/s 1 Erot 4 I 2 2 Erot

1 4(0.900 kg·m )(33.3 rad/s) 2 2

Erot 2.0 10 J 3

38

v 1. a) r 25 m/s

0.320 m

78 rad/s 1 Erot 4 I 2 2 Erot 2(0.900 kg·m2)(78 rad/s)2 Erot 1.1 104 J 1 b) Ek mv2 2 1 Ek (1300 kg)(25 m/s)2 2 Ek 4.1 105 J c) ET Ek Erot ET (4.1 105 J) (1.1 104 J) ET 4.2 105 J 2. v1 0

1 0 h1 12.0 m m 2.2 kg r 0.056 m I mr2 (moment of inertia for a hollow cylinder) a) ET mgh1 ET (2.2 kg)(9.8 m/s2)(12.0 m) ET 2.6 102 J b) To find the gravitational potential energy halfway down: Eg mgh2 h Eg mg 1 2

2

12.0 m Eg (2.2 kg)(9.8 m/s2) 2 2 Eg 1.29 10 J


To find the velocity halfway down: ET1 ET2 1 1 mgh1 mv22 I 2 mgh2 2 2 v 2 1 1 mgh1 mv22 mr2 2 mgh2 r 2 2 2 mgh1 mv2 mgh2 mv22 mgh1 mgh2 h v22 gh1 g 1 2 2 2v2 2gh1 gh1 2v22 gh1

v2

gh 2

v2

(9.8 m/s )(12.0 m) 2

1

2

v2 7.67 m/s v c) r 10.8 m/s

0.056 m

1.9 102 rad/s

Section 7.9 1 rev 1d 2 rad 1h 1. 365 d 24 h 3600 s 1 rev 7

1.99 10 rad/s 2 I mr2 (moment of inertia for a sphere) 5 L I

2 L mr2

5 2 L (5.98 1024 kg)(6.38 106 m)2 5 (1.99 107 rad/s) L 1.94 1031 kg·m2/s 2 rad 4.5 cycles 2. 1.1 s 1 cycle

25.7 rad/s 1.8 m r 2 r 0.9 m 2 I mr2 (moment of inertia for a sphere) 5

L I

2 L mr2

5 2 L (85 kg)(0.9 m)2(25.7 rad/s) 5 L 7.1 102 kg·m2/s 3. At perihelion, v 5472.3 m/s r 4.4630 1012 m m 1.027 1026 kg v

r 5472.3 m/s

4.4630 1012 m

1.2261 109 rad/s L I

2 L mr2

5 2 L (1.027 1026 kg)(4.4630 1012 m)2 5 (1.2261 109 rad/s) L 1.003 1042 kg·m2/s At aphelion: v 5383.3 m/s r 4.5368 1012 m m 1.027 1026 kg v

r 5383.3 m/s

4.5368 1012 m

1.1866 109 rad/s L I

2 L mr2

5 2 L (1.027 1026 kg)(4.5368 1012 m)2 5 (1.1866 109 rad/s) L 1.003 1042 kg·m2/s

Section 7.10 2. 1 t 2 rad

1 2.14 106 s

1 2.94 106 rad/s


39

I1 1 I2 2 2 2 mr12 1 mr22 2 5 5 2 r1 1 r22 2 r12 1

2 r22

2

(6.95 108 m)2(2.94 106 rad/s) (5500 m)2

2 4.69 104 rad/s 2 rad T2

2 2 rad T2 4.69 104 rad/s T2 1.34 104 s 3. ra 1.52 1011 m rp 1.47 1011 m vp 30 272 m/s Ia a Ip p v v mra2 a mrp2 p ra rp rava rpvp rpvp va ra (1.47 1011 m)(30 272 m/s) va 1.52 1011 m va 2.93 104 m/s va 29.3 km/s

Section 7.11 3. R 0.040 m r 0.0070 m g a I 2 1 mr g a 1 mR2 2 1 mr2 9.8 m/s2 a 1 (0.040 m)2 2 1 (0.0070 m)2 a 0.64 m/s2

40


Section 8.4 6

From the force vector diagram we see that, Fe tan mg Fe mg tan kq1q2 mg tan r2 r2mg tan q1 kq2

6

1. q1 3.7 10 C, q2 3.7 10 C, d 5.0 102 m, k 9.0 109 N·m2/C2 kq1q2 F d2 F

(9.0 109 N·m2/C2)(3.7 106 C)(3.7 106 C) (5.0 102 m)2

F 49 N F 49 N (attraction) 2. F 2(49 N) F 98 N r

q1

q1 1.1 1015 C The dust balls are 0.20 m apart, and the charge on the tethered dust ball is 1.1 1015 C.

kq q F 1 2

6

6

(9.0 10 N·m /C )(3.7 10 C)(3.7 10 C) r 9

2

2

98 N

r 3.5 102 m 3. a) T

(0.20 m)2(2.0 1010 kg)(9.8 N/kg)(tan 21°) (9.0 109 N·m2/C2)(3.0 106 C)

Section 8.5 1. a)

Fe

mg

b)

Fe

T

b) mg

c) How close do the dust balls get and what is the charge on the tethered dust ball? m 2.0 1010 kg , l 0.42 m, dwall-1 0.35 m, q 3.0 106 C, 21° dwall-2 0.35 m 0.42 m(sin 21°) dwall-2 0.35 m 0.15 m dwall-2 0.20 m

c)

+


41

Section 8.6 6

1. a) q 1.0 10 C,

1.7 106 N/C [right] Let right be the positive direction. Fe q Fe (1.0 106 C)(1.7 106 N/C) Fe 1.7 N Fe 1.7 N [left] b) q 1.0 106 C,

2(1.7 106 N/C) [right] If right is still the positive direction, Fe q Fe (1.0 106 C)[2(1.7 106 N/C)] Fe 3.4 N Fe 3.4 N [right] 2. T

Fe

–

mg

– Stationary charge creating a field

q 1.0 106 C, 1.7 106 N/C [right] Fe mg tan Fe 1.7 N [left] 3. a)

+

The field lines radiate outward, away from the charge. b) k 9.0 109 N·m2/C2, q 3.0 106 C At 2 cm away from the charge: kq

r2 (9.0 109 N·m2/C2)(3.0 106 C)

(2.0 102 m)2

At 4 cm away: kq

r2 (9.0 109 N·m2/C2)(3.0 106 C)

(4.0 102 m)2

1.7 107 N/C At 6 cm away: kq

r2 (9.0 109 N·m2/C2)(3.0 106 C)

(6.0 102 m)2

7.5 106 N/C c) Doubling the distance, kq

1 2 (2r) 1

1 4 Tripling the distance, kq

2 2 (3r) 1

2 9 1 1

1 decreases to and 2 decreases to of 4 9 the original strength. 1 d) 2 . The field strength varies as the r inverse square of the distance away from the charge. e) q1 1.0 106 C, q2 3.0 106 C, r 8.0 102 m kq1

r2 (9.0 109 N·m2/C2)(3.0 106 C)

(8.0 102 m)2

4.22 106 N/C e q F Fe (1.0 106 C)(4.22 106 N/C) Fe 4.22 N [right] 4. a) q1 q2 1.0 106 C, r 0.20 m Let the positive direction be left. At point A: r1 0.05 m, r2 0.25 m

6.8 107 N/C

42


TA

1

2 kq1 kq2

TA 2 r1 r 22

TA (9.0 109 N·m2/C2)(1.0 106 C) 1 1 2 2 (0.05 m) (0.25 m)

TA 3.7 106 N/C [left] At point B: r1 0.10 m, r2 0.10 m The addition of these two distances as was done in the previous question will yield a zero quantity.

TB 0 N/C At point C: r1 0.15 m, r2 0.05 m

1

2

TC kq2 kq1

TC r 22 r 21

TC (9.0 109 N·m2/C2)(1.0 106 C) 1 1 2 2 (0.05 m) (0.15 m)

TC 3.2 106 N/C [left] b) At the centre point, 1 is equal in magnitude but opposite in direction to 2, therefore there is no net field strength as the fields cancel out. c) For all field strengths to cancel out, the q magnitudes of the ratio of 2 must be r equal and pointing in opposite directions.

Section 8.7 kq1q2 1. a) Ee r Ee

(9.0 109 N·m2/C2)(5.0 106 C)(1.5 106 C) 10 102 m

Ee 6.8 101 J E b) V e q 6.8 101 J V 1.5 106 C V 4.5 105 V kq c) V r (9.0 109 N·m2/C2)(5.0 106 C) V 5.0 102 m V 9.0 105 V

∆V V2 V1 ∆V 9.0 105 V (4.5 105 V) ∆V 4.5 105 V 2. a) m1 m2 5.0 109 g 5.0 1012 kg, q1 4.0 1010 C, q2 1.0 1010 C On particle 1: W1 qV W1 (4.0 1010 C)(50 V) W1 2.0 108 J On particle 2: W2 qV W2 (1.0 1010 C)(50 V) W2 5.0 109 J b) W Ek 1 W mv2 2 v

2W m

v m The similar masses cancel. v 2W m v W v W 2W1

1

1

2

2

2

1

1

2

2

v 1 v2

2.0 108 J 5.0 109 J

v 1 2 v2 3. a) Extensive: electric force, potential energy Intensive: field strength, electric potential b) Electric force — Charge and the field strength Potential energy — Charge and the electric potential c) Extensive properties Product cost (per package) Mass Volume Length Force of gravity Etc. Intensive properties Unit product cost (per unit weight or measure) Density Heat capacity


43

Indices of refraction Gravitational field strength Etc.

Section 8.8 1. qA 2e, qB 79e, Ek 7.7 MeV (7.7 106 eV)(1.602 1019 J) Ee Ek kqAqB Ee r kqAqB r Ee r

(9.0 109 N·m2/C2)(1.602 1019 C)2(2)(79) (7.7 106 eV)(1.602 1019 J)

r 2.96 1014 m r 3.0 1014 m 3. q 1.5 105 C 1 mv2 q(V2 V1) 2 v v

1

2(1.5 105 C)(12 V) (1.0 105 kg)

v 6.0 m/s [left] 4. a) V 1.5 103 V, m 6.68 1027 kg, q 2e 3.204 1019 C Ek Ee 1 2 mv Vq 2 v

2Vq m 19

2(1.5 10 V)(3.204 10 C) v 27 3

6.68 10

kg

v 3.8 105 m/s 1 1 b) mv2 Vq 2 2 v v

v v

k

2(3.2 1015 J) 9.11 1031 kg

v 8.4 107 m/s

Section 8.9

V 3.7 102 V 2. d 7.5 103 m, V 350 V, V

d 350 V

7.5 103 m

4.7 104 N/C 3. m 2.166 1015 kg, V 530 V, d 1.2 102 m Fe Fg qV mg d mgd q V q

(2.166 1015 kg)(9.8 N/kg)(1.2 102 m) 530 V

q 4.8 1019 C

Vq m (1.5 103 V)(3.204 1019 C) 6.68 1027 kg

v 2.7 105 m/s

44

2E m

1. W 2.4 104 J, q 6.5 107 C W V q 2.4 104 J V 6.5 107 C

2q(V V ) m 2

5. a) V 20 kV 2.0 104 V, q 1.602 1019 C, m 9.11 1031 kg Ek Ee Ek Vq Ek (2.0 104 V)(1.602 1019 C) Ek 3.2 1015 J 1 b) Ek mv2 2


5. a) I 100 A L 50 m B 3.0 105 T 45° I r 2 B (4 107 T·m/A)(100 A) r 2 (3.0 105 T)

Section 9.5 1. L 0.30 m I 12 A B 0.25 T 90° F BIL sin F (0.25 T)(12 A)(0.30 m) sin 90° F 0.90 N 2. L 0.15 m F 9.2 102 N B 3.5 102 T 90° F I BL sin (9.2 102 N) I (3.5 102 T)(0.15 m) sin 90°

r 0.67 m b) Referring to the diagram in question 3, Earth’s field lies in a line that is crossing the wire at 45° below the horizontal. The magnetic field would form a circular ring in the clockwise direction (rising on the south side of the wire, descending on the north with a radius of 0.67 m). Therefore, the field will cancel that of Earth on the south side below the wire, as shown in the diagram.

I 18 A 3. a) L 50 m I 100 A F 0.25 N 45° F B IL sin (0.25 N) B (100 A)(50 m) sin 45° B 7.1 105 T b) Wire (cross-section) 45°

45°

0.67 m 45° x x N

Earth's Magnetic Field

Direction of Force

Tower

S

4. B 3.0 105 T L 0.20 m N 200 4 107 T·m/A BL I N (3.0 105 T)(0.20 m) I (4 107 T·m/A)(200) I 2.4 102 A

2x (0.67 m) x 0.47 m The fields will cancel 4.7 101 m south and 4.7 101 m below the wire. 6. a) r 2.4 103 m I 13.0 A L1m I2L F 2 r (4 107 T·m/A)(13.0 A)2(1 m) F 2 (2.4 103 m) 2

N

2

F 1.4 102 N/m 7. q 20 C B 4.5 105 T v 400 m/s 90° F qvB sin F (20 C)(400 m/s)(4.5 105 T) sin 90° F 0.36 N Solutions to Applying the Concepts

45

8. q 1.602 1019 C v 4.3 104 m/s B 1.5 T 90° F qvB sin F (1.602 1019 C)(4.3 104 m/s)(1.5 T) sin 90° 1.0 1014 N [south] F

46


Section 10.2 t 1. a) T cycles 375 min T 5 T 75 min 6.7 s b) T 10 T 0.67 s 60 s c) T 33.3 T 1.80 s 57 s d) T 68 T 0.838 s cycles 2. a) f t 120 f 2.0 s f 60 Hz 45 b) f 60 s f 0.75 Hz 40 c) f 1.2 60 60 s f 0.009 26 Hz 65 d) f 48 s f 1.35 Hz 1 3. a) i) f T 1 f 75 60 s f 2.22 104 Hz 1 ii) f 0.67 s f 1.49 Hz 1 iii) f 1.80 s f 0.556 Hz 1 iv) f 0.838 s f 1.19 Hz

b) i)

5. a)

b) c) d) e)

1 T f

1 T 60 Hz T 0.0167 s 1 ii) T 0.75 Hz T 1.33 s 1 iii) T 0.009 26 Hz T 108 s 1 iv) T 1.35 Hz T 0.74 s x (30 cm) cos x (30 cm) cos 30° x 26 cm x (30 cm) cos 180° x 30 cm x (30 cm) cos 270° x 0 cm (equilibrium) x (30 cm) cos 360° x 30 cm x (30 cm) cos 4 x 21 cm

Section 10.3 4. a) v f v f 3.0 108 m/s f 640 109 m f 4.7 1014 Hz 3.0 108 m/s b) f 1.2 m f 2.5 108 Hz 3.0 108 m/s c) f 2 109 m f 1.5 1017 Hz 5. a) v f v f 3.0 108 m/s 1.5 1013 Hz 2.0 105 m


47

3.0 108 m/s b) 2.0 109 Hz 0.15 m 3.0 108 m/s c) 3.0 1022 Hz 1.0 1014 m

Section 10.5 5. a)

Section 10.4 c 4. a) n v c v n 3.0 108 m/s v 1.33 v 2.26 108 m/s 3.0 108 m/s b) v 2.42 v 1.24 108 m/s 3.0 108 m/s c) v 1.51 v 1.99 108 m/s c 5. a) n v 3.0 108 m/s n 2.1 108 m/s n 1.43 3.0 108 m/s b) n 1.5 108 m/s n 2.0 3.0 108 m/s c) n 0.79(3.0 108 m/s) n 1.27 6. a) n1 sin 1 n2 sin 2 n1 sin 1 2 sin1 n2 2 sin1

48

c vo ray no ray 3.0 108 m/s vo ray 1.658 vo ray 1.81 108 m/s c ve ray ne ray 3.0 108 m/s ve ray 1.486 ve ray 2.02 108 m/s 2.02 108 m/s ve ray b) 100% 1.81 108 m/s vo ray ve ray 111.6% vo ray Therefore, the speed of the e ray is 11.6% greater than the speed of the o ray.

sin 25° 1.33

2 18.5° sin 25° b) 2 sin1 2.42 2 10.1° sin 25° c) 2 sin1 1.51 2 16.3°

c n v


Section 11.4

Section 11.6

2. d 5.6 m x2 28 cm L 1.1 m m2 dxm

mL (5.6 106 m)(0.28 m)

(2)(1.1 m) 7.13 107 m 713 nm 3. 510 nm d 5.6 m L 1.1 m L x

d (5.10 107 m)(1.1 m) x

5.6 106 m

2. m 22 625 nm m t

2 (22)(6.25 107 m) t

2 6 t 6.87 10 m t 6.9 m 3. t 1.75 105 m 625 nm 1 2t m 2 2t 1 m

2 1 2(1.75 105 m) m

7 2 (6.25 10 m)

x 0.10 m x 10 cm 4. m 3 d 5.6 m L 1.1 m 713 nm mL xm

d (3)(7.13 107 m)(1.1 m) x3

(5.6 106 m) x3 0.42 m x3 42 cm

Section 11.5 2. ∆PD 3 ng 1.52 624 nm PD t

2(ng 1) (6.24 107 m)(3) t

2(0.52) t 1.8 106 m t 1.8 m

m 55.5 m 55

Section 11.8 1. w 5.5 106 m 550 nm L 1.10 m m2 m a) sin m

w (2)(5.50 107 m) sin 2

5.5 106 m sin 2 0.2 2 11.5° b) xm L sin m xm (1.10 m)(0.2) xm 0.22 m xm 22 cm 2L 2. a) x

w 2(5.50 107 m)(1.10 m) x

(5.5 106 m) x 0.22 m x 22 cm


49

1 b) sin

w 2

5.50 10 m 1 sin

2 5.5 10 m 7

6

1 sin

0.1 2 11.5° L 3. x

w (5.50 107 m)(1.10 m) x

(5.5 106 m) x 0.11 m x 11 cm 6. R 1 107 rad d 2.4 m Rd a)

1.22 (1 107 rad)(2.4 m)

1.22 1.97 107 m 197 nm x b) sin =

L 1

(1.0 103 m) 2 L

sin(1 107 rad) L 5000 m L 5 km

Section 11.9 1. N 8500 w 2.2 cm 530 nm w d

N 2.2 102 m d

8500 d 2.59 106 m m sin m

d 5.30 107 m sin 1

2.59 106 m sin 1 0.205 1 12°

50

m sin m

d 2(5.30 107 m) sin 2

2.59 106 m sin 2 0.410 2 24° m sin m

d 3(5.30 107 m) sin 3

2.59 106 m sin 3 0.614 3 38° d 2. a) m

2.59 106 m m

6.50 107 m m4 d b) m

2.59 106 m m

5.50 107 m m 4.7 m4 d c) m

2.59 106 m m

4.50 107 m m 5.7 m5 3. m 2 2 8.41o 614 nm m a) d

sin m (2)(6.14 107 m) d

sin 8.41° d 8.396 106 m d 8.40 m b) w 1.96 cm w N

d 1.96 102 m N

8.396 106 m N 2334 slits


Section 11.10 1.

3000 lines 100 cm

300 000 lines/m 1 cm 1m 100 cm 20 000 lines

100 000 lines/m 20 cm 1m

Therefore, 3000 lines/cm produces the best resolution. m 3. sin Red

d

6.

2d sin

m m sin

2d (5.2 1011 m)(2) sin

2(2.5 1010 m) sin 0.208 168°, 192°

(1)(7.30 107 m) sin Red

1.89 106 m sin Red 0.386 Red 22.7° m sin Violet

d (1)(4.00 107 m) sin Violet

1.89 106 m sin Violet 0.211 Violet 12.2° m sin Green

d (1)(5.10 107 m) sin Green

1.89 106 m sin Green 0.269 Green 15.6° This can be similarly proven for the next 3 orders using the appropriate m. The sequence is violet, green, red. At the fourth order, green and red maxima are no longer visible. 5. d 2.5 1010 m 12o m2 2d sin

m 2(2.5 1010 m) sin 12°

2 5.198 1011 m 52 pm


51

Section 12.2

Vstop vs. f0

Section 12.3

h W V f0 0 e e eV hf0 W0 Choosing two pairs of values from the table and subtracting, (1.6 1019 C)(0.95 V) h(7.7 1014 Hz) W0 (1.6 1019 C)(0.7 V) h(7.2 1014 Hz) W0 (1.6 1019 C)(0.25 V) h(0.5 1014 Hz) h 8 1034 J·s W0 4.64 1019 J W0 2.9 eV

1.

52

3

Vstop (V)

1. T 12 000 K a) The maximum wavelength can be found using Wien’s law: 2.898 103 max

T 2.898 103 max

12 000 K max 2.4 107 m The peak wavelength of Rigel is 2.4 107 m. It is in the ultraviolet spectrum. b) It would appear violet. c) No: the living cells would be damaged by the highly energetic UV photons. 2. T 900 K a) The maximum wavelength can be found using Wien’s law: 2.898 103 max

T 2.898 103 max

900 K max 3.2 106 m The peak wavelength of the light is 3.2 106 m. b) It would appear in the infrared spectrum. c) Since the peak is in infrared, more energy is required to produce the light in the visual spectrum.

2

1

0

7

8

9

10 11 12 f0 (×1014 Hz)

13

2. a) Increasing the work function by 1.5 would cause a vertical shift of the line. Hence, potential would have to be greater, but the frequencies would not change. h b) The term

is constant and hence the e slope would not change. 3. 230 nm 2.3 107 m The energy can be found as follows: hc E W0 (8 1034 J·s)(3.0 108 m/s) E

2.3 107 m 4.64 1019 J E 5.79 1019 J

Section 12.4 2. E 85 eV, 214 nm 2.14 107 m a) Momentum of the original electron can be found using: E p

c (85 eV)(1.6 1019 C) p

3.0 108 m/s p 4.53 1026 N·s b) Momentum of the resultant electron can be found using: h p

6.626 1034 J·s p

2.14 107 m p 3.1 1027 N·s c) The energy imparted can be found by: hc E E


E (85 eV)(1.6 1019 C)

The wavelength of the spectral lines is: hc 81

E81

(6.626 1034 J·s)(3.0 108 m/s)

2.14 107 m E 1.27 1017 J The energy imparted to the electron was 1.27 1017 J. d) The energy imparted increased the speed of the electron. Hence, it can be found using: 2E v

m

v

2(1.27 1017 J)

9.11 1031 kg

(6.626 1034 J·s)(3.0 108 m/s) 81

2.15 1018 J 81 9.25 108 m Similarly, the energy change when the electron transfers from 7 to 2 is: E72 E7 E2 2.18 1018 J 2.18 1018 J E72

2 7 22 E72 5 1019 J hc 72

E72 (6.26 1034 J·s)(3.0 108 m/s) 72

5 1019 J

v 5.27 106 m/s The speed increase of the electron is 5.27 106 m/s.

Section 12.5 1. v 1 km/s 1000 m/s The wavelength can be found using de Broglie’s equation: h

mv 6.626 1034 J·s

(9.11 1031 kg)(1000 m/s) 7.27 107 m Hence, the wavelength of the electron is 7.27 107 m.

Section 12.6 2. We shall first compute the change in energies and the wavelength of spectral lines emitted in each case. From that, the wavelength separation can be computed. The energy change when the electron transfers from 8 to 1 is: E81 E8 E1 2.18 1018 J 2.18 1018 J

E81

82 12 E81 2.15 1018 J

72 3.98 107 m Hence the wavelength separation is ∆ 72 81 ∆ 3.98 107 m 9.25 108 m ∆ 3.05 107 m 3. The change in energy can be computed using: E Ef Ei 13.6 eV 13.6 eV E

nf2 ni2 For the Lyman series, the lower boundary is when the electron jumps from the second to the first orbital: 13.6 eV 13.6 eV

Emin

2 12 2 Emin 10.2 eV The higher boundary for the Lyman series is when the electron jumps from infinity to the first orbital: 13.6 eV 13.6 eV

Emax

12 2 Emax 13.6 eV For the Balmer series, the lower boundary is when the electron jumps from the third to the second orbital: 13.6 eV 13.6 eV

Emin

22 32 Emin 1.89 eV


53

The higher boundary for the Balmer series is when the electron jumps from infinity to the second orbital: 13.6 eV 13.6 eV

Emax

22 2 Emax 3.4 eV For the Paschen series, the lower boundary is when the electron jumps from the fourth to the third orbital: 13.6 eV 13.6 eV

Emin

32 42 Emin 0.66 eV The higher boundary for the Paschen series is when the electron jumps from infinity to the third orbital: 13.6 eV 13.6 eV

Emax

2 32 Emax 1.51 eV For the Brackett series, the lower boundary is when the electron jumps from the fifth to the fourth orbital: 13.6 eV 13.6 eV

Emin

2 42 5 Emin 0.31 eV The higher boundary for the Brackett series is when the electron jumps from infinity to the fourth orbital: 13.6 eV 13.6 eV

Emax

42 2 Emax 0.85 eV Thus, the boundaries for the four series are: Lyman: 10.2 eV to 13.6 eV Balmer: 1.89 eV to 3.4 eV Paschen: 0.66 eV to 1.51 eV Brackett: 0.31 eV to 0.85 eV

Hence, the uncertainty in position is 6.3 102 m. 2. In the equation ∆E∆t ≥ –h, the units are J·s. h This coincides with the units of h in –h , 2 where 2 is a constant. 6. Ek 1.2 keV 1.92 1016 J, mp 1.673 1027 kg First we shall find the velocity using: 2Ek v

mp

v

v 4.8 105 m/s The uncertainty in position can be found using: –h y

mv 1.0546 1034 J·s y

(1.673 1027 kg)(4.8 105 m/s) y 1.32 1013 m The uncertainty in the position is 1.32 1013 m. 7. The uncertainty does not affect the object at a macroscopic level.

Section 12.8 1. ∆v 1 m/s 1 106 m/s, mp 1.673 1027 kg The uncertainty in position can be found using: –h y

mv 1.0546 1034 J·s y

(1.673 1027 kg)(1 106 m/s) y 6.3 102 m

54

2(1.92 1016 J)

1.673 1027 kg


Section 13.1 1. For Nadia: mRLv0 mRvR2 mLvR2 (6 kg)(0 m/s) (3 kg)(2 m/s) (3 kg)(2 m/s) 0 kg·m/s 0 kg·m/s For Jerry: mRLv0 mRvR2 mLvR2 (6 kg)(2 m/s) (3 kg)(2 2 m/s) (3 kg)(2 2 m/s) 12 kg·m/s 12 kg·m/s

Section 13.2 1. v 0.5c or 1.5 108 m/s 2. The classical addition of velocities gives: kvp kvu uvp kvp 0.5c [R] 0.6c [R] kvp 1.1c This answer violates the second postulate of special relativity. 3. Ek-gained Ee-lost 1

mv2 Vq 2 2Vq v

m

v

2(1.00 106 V)(1.6 1019 C)

(9.11 1031 kg)

v 5.93 108 m/s This value is almost double the speed of light.

Section 13.3 1. The muon travels farther due to the time dilation from 2.2 s to 3.1 s that occurs at its speed of v 0.7c. The extra path length is: ∆d d2 d1 ∆d vt2 vt1 ∆d v(t2 t1) ∆d (0.7c)(3.1 s 2.2 s) ∆d 189 m

2. For Phillip, at rest relative to the experiment: d t

v 2h t0

c 2(3.0 m) t0

3.0 108 m/s t0 2.0 108 s For Barb, the stationary observer watching the experiment travel by at v 0.6c: t0 t

v2 1

2 c 2.0 108 s t

(0.6c)2 1

c2 t 2.5 108 s 3. For Marc, the time for one beat is: 60 s/min

1.1538 s 52 beats/min The dilated time for the earthly observers is: t0 t

v2 1

2 c

1.1538 s t

(0.28c)2 1

c2 t 1.2019 s The new rate is: 60 s/min

49.9 bpm 1.20 s/beat 4. The contracted distance L, measured by Katrina, is given by: L 0.5L0 v2 L L0 1

c2

v2 0.5L0 L0 1

c2 v2 0.25 1 2 c v

0.75 c v 0.866c v 2.60 108 m/s


55

3. L0 200 ca v 0.9986c

6. L0 5.75 1012 m The time you take: distance measured t0

velocity

v2 L0 1

2 c t0

v

v2 (vt0)2 L02 1 2 c

L v c v ct0

2

2

2

2

0

v

v

c2

ct 1

0 L0

(3.0 108 m/s)2

8 s) 1 (3.0 10 m/s)(3600 (5.75 1012 m)

v 2.95 108 m/s

Section 13.4 1. L0 7 ca L t

0 v 7 ca 7 a 3 a

v 7 ca v

10 a v 0.7c 2. The age or time difference for the twins is: 5 a tS tT d dT 5 a

S

v v v2 2L0 1

2 c 2L 5 a 0

v v v2 10 ca 1

2 c 10 ca 5 a

v v

v2 v 2c 2c 1

c2

v2 v 2c 2c 1

c2 v2 4vc 4c2 4c2 4v2 v(5v 4c) 0 v 0.8c since v ≠ 0 56

v2 L0 1

2 c t0

v 200 ca1 (0.9986) 2 t0

0.9986c t0 10.59 a 6. For Rashad: (∆s)2 c2(∆t)2 (∆x)2 (∆s)2 (3 108 m/s)2(1.5 s)2 02 (∆s)2 2.05 1017 m2 For Kareem: (s)2 c2(t)2 (x)2 x c2(t)2 (s )2 2 x (3 1 08 m/s )2(2 s) (2 .025 1017 m2) 8 x 3.97 10 m

Section 13.5 1. m0 5.98 1024 kg v 2.96 104 m/s m0 m

v2 1

2 c

5.98 1024 kg m

(2.96 104 m/s)2 1

(3.0 10 8 m/s)2 m 5.980 000 03 1024 kg m0 2. m

v2 1

2 c At 0.9c: m0 m

2 1 (0 .9) m 2.294m0 At 0.99c: m0 m

1 (0 .99)2 m 7.089m0 At 0.999c: m0 m

1 (0 .999)2 m 22.366m0 Therefore, there is a much greater increase in mass when accelerating from 0.99c to 0.999c.


3. Using the low-speed mass dilation approximation, 1 m0v2 m

2 c2 1 (60 kg)(10 m/s)2 m

2 (3.0 108 m/s)2 m 3.3 1014 kg 4. Since cost, C, is proportional to energy3, E3, C E 3

2 2 C1 E1

5000 MeV 3 C2 ($100 million)

500 MeV C2 $100 billion 5. The radius for charges moving at right angles mv to a magnetic field is r

. The ratio of the Bq r m vf radius of a fast to slow electron is

f f . rs msvs Assuming ms m0 (its rest mass), and m0 vf mf

, the ratio becomes

. 2 v v2 1

2 vs 1

2 c c 6. As in question 5, the ratio of radii is: r mpvp

p

and since vp ve: re meve

mpre rp

me (1.67 1027 kg)re rp

9.11 1031 kg

r 1.76 1010 ca

Section 13.7

rp 1833re 7. m0 1.67 1027 kg v 0.996c B 5.0 105 T m0v r

v2 qB 1

2 c

1. For momentum dilation, m0v p

v2 1

2 c At v 0.2c: m0(0.2c) p

(0.2c)2 1

c2 p 0.204m0c At v 0.5c: m0(0.5c) p

(0.5c)2 1

c2 p 0.577m0c At v 0.8c: m0(0.8c) p

(0.8c)2 1

c2 p 1.33m0c

r

3. The speed of the bullet relative to Earth is: bvc cvE bvE

bvccvE 1

c2 c c

3 2 bvE c c

3 2 1

c2 5c

6 bvE

1 1

6 5c bvE

7 bvE 0.714c Therefore, the bullet will never reach the bandits because its speed is less than 0.75c. 4. Putting the limiting velocity v c into Hubble’s law: v Hr gives the limiting case of: c r

H 3.0 108 m/s r

1.7 102 m/s/ca

(1.67 1027 kg)(0.996c)

19 (1.6 10 C)(5.0 105 T) 1 (0 .996)2

r 6.98 105 m

Section 13.6 2. Using the relativistic formula for velocity addition: vN NvL vL

vNNvL 1

c2 c 0.999c vL

(c)(0.999c) 1

c2 vL c


57

2. E m0c2 Ek Case A: 125 J m0c2 87 J m0 38c2 J Case B: 54 J m0c2 15 J m0 39c2 J Therefore, B has the greater rest mass. 3. The energy used by the bulb is: E mc2 E Pt Pt m

c2 (80 W)(365 24 60 60 s) m

(3.0 108 m/s)2 m 2.80 108 kg m 2.80 105 g 4. E mc2 E (65 kg)(3 108 m/s)2 E 5.85 1018 J

Section 13.8 1. E mc2 E (106 MeV/c 2)c 2 E 106 MeV 1.6 1019 J 1 106 eV E 106 MeV

1 MeV 1 eV 11 E 1.696 10 J The equivalent mass is: E m 2 c 1.696 1011 J m

(3.0 108 m/s)2 m 1.88 1028 kg 2. A mass, m, is equivalent to an energy: E mc2 E (1.67 1027 kg)(3 108 m/s)2 E 1.503 1010 J 1 eV 1.6 1019 J 1.503 1010 J m

1.6 1019 J/eV m 939.37 106 eV/c 2 m 939.4 MeV/c 2

58

E2 (pc)2 (m0c)2 (mvc)2 (mc2)2 (m0c)2 mc2 (m0c2 Ek) mc2 m0c2 5m0c2 mc2 6m0c2 (mvc)2 (6m0c2)2 (m0c2)2 (mvc)2 35m02c4 m2v2 35m02c2 m0 Since m

, v2 1

2 c 2 v v2 35 1 2 c2 c 2 2 v 35c 35v2 36v2 35c 2 35 v c2 36 v 0.986c v 2.96 108 m/s 4. Given the dilated mass of the proton, m 4 106m0 m0 m

v2 1

2 c 1 4 106

v2 1

2 c Since v2 ≈ c2, we can use the high-speed approximation: v v2 1

1 2 2 c c 1 4 106

v 2 1

c 1 v 1

(4 106) 2 c v 1 3.13 1014 c c v 3.13 1014c c v 9.38 106 m/s The protons are travelling 9.38 106 m/s slower than c. 3.


Section 14.1 3. a) Binding energy is: B [Zm(1H) Nmn m(2H)]c2 B 938.78 MeV 939.57 MeV 1876.12 MeV B 2.23 MeV 2.23 MeV B b)

1.12 MeV/nucleon A 2 nucleons 4. Average atomic mass of Cl is 0.758(35 u) 0.242(37 u) 35.48 u, compared to 35.453 u in the periodic table.

3. T235 7.04 108 a, T238 4.45 109 a, 235 235 N N 0.0044,

0.030 238 238 N N 0

1 ( N)

2 N

1 N ( N)

2 1 ( 0.0044 (0.030)

2 0.0044 log t(1.196 10 a 0.030 235

235 238

238

1. The amount eaten is: 1 1 1 1 1 1 1

2 4 8 16 32 128 256 255 255 1 . The amount left is 1

256 256 256

1 8 or . 2

1

7.04 108 a

9

t

Section 14.3

t

T238

0

t

Section 14.2 2. Since AZX → A4 Z2 Y: 234 219 240 60 a) 90Th b) 244 94Pu c) 84Po d) 92U e) 27 Co 3. Since AZ X → AZ1Ye: 23 45 35 64 a) 32 16S b) 11 Na c) 17Cl d) 21 Sc e) 30Zn 4. Since AZ X → AZ1Ye: 46 239 64 a) 199F b) 22 10 Ne c) 23V d) 92U e) 28 Ni

t

T235

0

1

4.45 10 a 9

)

1 ) log

2

1

0.8337

(0.3010)(1.196 109 a1)

t 2.3 109 a

Section 14.4 1. Bismuth or 209 83 Bi 360 mSv unobservable annual dosage 2.

dosage per dental x-ray 0.20 mSv 1800 doses 4. annual dose dose equivalent activity time 6 D (1.3 10 eV)(1.602 1019 J/eV)(1) (29 000 Bq/kg) (365 24 60 60 s) D 0.1905 J/kg D 191 mSv

Section 14.5

2. T 1.28 109 a, N0 5 mg, N 1 mg 1

2

1 N N0

2

1u3u v v

1u3u

t

T 1

2

t 1 N log log

2 N0 T 1

2

N log

N0 t T

1 log

2 1

2

2. In a head-on elastic collision with the target, 3 H at rest, the recoil velocity is: mn mx v v

mn mx

1 mg log

5 mg t (1.28 109 a)

1 log

2 t 2.97 109 a

v 0.5v Tritium is 50% effective in slowing down the fast neutrons. 4. power amount of energy/mole number of moles used/12 h (12 3600)1 h/s 400 g 600 g P (1699 GJ/mol) or

2 g/mol 3 g/mol 1

43200 s P 7.87 GW


59

5. Since 1 neutrino is created along with 1 deuterium atom, and 2 deuterium atoms are needed to create an 4He ion, 2 neutrinos are created.

Section 14.6 1. Using Einstein’s energy triangle: (mvc)2 (m0c2Ek)2 (m0c2)2 mvc (0.511 MeV 310 0 MeV )2 (0 .511 M eV)2 13 mvc (3100.5 MeV)(1.602 10 J/MeV) 4.9670 1010 J mv

3.0 108 m/s mv 1.6557 1018 N·s The de Broglie wavelength is: h

mv 6.63 1034 J·s

1.6557 1018 N·s 4.0 1016 m v 2. f

2r 3.0 108 m/s f

2(4300 m) f 11 kHz 3. a) In Einstein’s energy triangle, (mc2)2 (m0c2)2 (mvc)2 [see Chapter 13] m0c2 938.27 MeV mc2 m0c2 Ek mc2 938.27 MeV 10 MeV mc2 948.27 MeV In the triangle, m0c2 cos

mc2 938.27 MeV cos

948.27 MeV 8.328° mvc sin

mc2 v

sin 8.328° c v 0.1448c v 4.35 107 m/s

60

v b) r

2f 4.35 107 m/s r

2(32 106 Hz) r 0.216 m

Section 14.7 2 2 1 2. a) uud 1 3 3 3 2 2 1 b) u ud 1 3 3 3 2 1 c) ud 1 3 3 2 1 1 d) udd 0 3 3 3 2 1 e) su 1

3 3 3. a) proton (baryon) b) antiproton (baryon) c) pion (meson) d) neutron (baryon) e) kaon (meson) 2 1 1 4. udd 0 3 3 3 5. The mass “defect” of a 0 meson is: md mb m (8 4700 5279) MeV/c 2 571 MeV/c 2

Section 14.8 1. i) An electron and a positron annihilate each other, releasing two gamma rays. ii) A neutron undergoes decay to an antineutrino, a positron, and an electron. iii)A planet orbits the Sun via the exchange of a graviton.


PART 2 Answers to End-of-chapter Conceptual Questions Chapter 1

Position in Metres

1. It is possible for an object to be accelerating and at rest at the same time. For example, consider an object that is thrown straight up in the air. During its entire trajectory it is accelerating downward. At its maximum height it has a speed of zero. Therefore, at that point it is both accelerating and at rest. 2. A speedometer measures a car’s speed, not its velocity, since the speedometer gives no indication as to the direction of the car’s motion. 3. Position-Time m 5 4 3 2 1 1 2 3 4 5 6 7

t

Time in Seconds Time in Seconds Velocity in Metres per Second

1 2 3 4 5 6 7

t

–1 –2 –3 –4 –5 –6 –7 –8 –9 –10 m

4. Displacement, velocity, and acceleration are all vector quantities. Therefore, a negative displacement, velocity, or acceleration is a negative vector quantity, which indicates that the vector’s direction is opposite to the direction designated as positive. 5. The seconds are squared in the standard SI unit for acceleration, m/s2, because acceleration is the change in velocity per unit of time. Therefore, the standard SI unit for acceleration is (m/s)/s, which is more conveniently written as m/s2.

6. Assume for all cases that north is positive and south is negative. a) Positiontime graph: The object sits motionless south of the designated zero point. The object then moves northward with a constant velocity, crossing the zero point and ending up in a position north of the zero point. Velocitytime graph: The object moves southward with a constant velocity. The object then slows down while still moving southward, stops, changes direction, and speeds up northward with a constant acceleration. b) Positiontime graph: The object starts at the zero point and speeds up while moving northward, then continues to move northward with a constant velocity. Velocitytime graph: The object starts at rest and speeds up with an increasing acceleration while moving northward. The object then continues to speed up with a constant acceleration northward. c) Positiontime graph: The object starts north of the zero point and moves southward past the zero point with a constant velocity. The object then abruptly slows down and continues to move southward with a new constant velocity. Velocitytime graph: The object slows down while moving northward, stops, changes direction, and speeds up southward with a constant acceleration. The object then abruptly reduces the magnitude of its acceleration and continues to speed up southward with a new constant acceleration. d) Positiontime graph: The object starts at the zero point and moves northward and slows down to a stop, where it sits motionless for a period of time. The object then quickly speeds up southward and moves

Answers to End-of-chapter Conceptual Questions

61

southward with a constant velocity, going past the zero point. Velocitytime graph: The object starts at rest and speeds up while moving northward. The acceleration in this time period is decreasing. The object then continues to move northward with a constant velocity. The object then slows down while moving northward, stops, changes direction, and speeds up southward with a constant acceleration. dtot 7. a) vavg ttot 1000 m (5)(60 s) 3.3 ms dtot b) vavg ttot 1000 m (4)(60 s) 4.2 ms dtot c) vavg ttot 2000 m (9)(60 s) 3.7 ms d) The answer for c) is the average speed of the bus over the whole trip, whereas half the sum of its speed up the hill and its speed down the hill is an average of the average speeds up and down the hill. 8. In flying from planet A to planet B, you would need to burn your spacecraft’s engines while leaving planet A in order to escape its gravitational pull and then to make any necessary course corrections, and while arriving at planet B in order to slow down and stop. Assuming there were no forces acting on the craft in between, it would travel with constant velocity once the engines were turned off. 9. A free-body diagram shows the forces acting on an object, as these are the only forces that can cause the body to accelerate. Since, by Newton’s third law, for every action force there is a reaction force, equal in magnitude and opposite in direction, then each of the

62

forces acting on an object is half of an actionreaction pair. If both the action forces and the reaction forces were included in a free-body diagram, then all the forces would cancel. For example, a free-body diagram for a ball being kicked must not include the reaction force provided by the ball on the foot, or else the forces would cancel and the ball would not accelerate. 10.

Fn

Fm

Motorcycle

Ff

Fg

11. Dear Cousin, You asked me to explain Newton’s first law of motion to you. Newton’s first law of motion states that an object will keep moving at a constant speed in the same direction unless a force makes it slow down, speed up, or change direction. Here’s an example. Suppose you’re pushing a hockey puck across the carpet. When you let go, the puck quickly stops moving. This is because the carpet is not very slippery; we say that it has a lot of friction. The force of friction is making the puck slow down. What if you slide the puck across a surface with less friction, like ice? The puck will take longer to stop moving, because the force of friction is much less than on the carpet. Now suppose you slide the puck across an air hockey table. The force of friction is so small that the puck will slide for a much, much longer time. So, you can imagine sliding a puck on a surface with no friction at all. The puck never stops, because there is no force to slow it down! Perhaps you’re wondering about a motionless object that isn’t experiencing a force — why isn’t it moving at a constant speed in the same direction? But it is! Zero is a constant speed.


Fn

Fn

Puck on carpet

Ff

Fg

Fg

Ff

Fg

Fn

Puck on air table

Puck on ice

Fn

Ff

Puck on frictionless surface

Fg

12. The gravitational force applied by the Moon on Earth does not cancel with the gravitational force applied by Earth on the Moon because these forces act on different bodies. Only forces applied on the same body can possibly cancel one another. 13. When you fire a rifle, the forces applied to the bullet and the rifle make up an actionreaction pair. By Newton’s third law, the force applied to the bullet is equal and opposite to its reaction force, the force applied to the rifle. This reaction force causes the rifle and you to recoil in the opposite direction. 14. While in the air, the ball’s vertical acceleration is constant and equal to g 9.8 m/s2. The ball travels the same distance upward as downward, and therefore the ball’s speed is the same when it reaches the ground as when it leaves the ground, since its acceleration is constant. Suppose the lengths of time it takes the ball to travel upward and downward are t1 and t2, respectively. We can use the equations t2(v2i v2f) t1(v1i v1f) d1 and d2 2 2 for the distances travelled upward and downward, respectively, where v1i and v1f are the initial and final velocities during the upward

flight, respectively, and v2i and v2f are the initial and final velocities downward, respectively. Since d1 d2, we can write the following equation: t2(v2i v2f) t1(v1i v1f) 2 2 On the left side, the final velocity upward, v1f, is equal to zero. On the right side, the initial velocity downward, v2i, is equal to zero. The equation simplifies: t1(v1i) t2(v2f) 2 2 But v1i is equal to v2f and is not zero, and therefore t1 t2. 15. The ball is undergoing uniform circular motion, as it is travelling in a circle at a constant speed. Because its trajectory is curved, it cannot be undergoing uniform motion, which requires an object to be travelling at a constant speed in a straight line.

Chapter 2 1. Frictional forces are forces that oppose motion. A frictional force will only try to prevent an object from moving, it will not actually cause an object to move. 2. It is not possible to swing a mass in a horizontal circle above your head. Since gravity is always pulling down on the mass, an upward component of the tension force is required to balance gravity. As the speed of rotation increases, the angle relative to the horizontal may approach 0° but will never reach 0°. 3. If the gravitational force downward and the normal force upward are the only two vertical forces acting on an object, we can be certain that they are balanced if the object is not accelerating. If one of these forces were greater than the other, the object would accelerate in the direction of the greater force. 4. The most common way to describe directions in three dimensions is by the use of three unit vectors (and their opposites). Traditionally, the three unit vectors used are labelled as i, j, and k. One of these unit vectors will represent right, one will represent up, and one will


63

represent coming out of the plane of the page toward you. 5. The bullets reach the ground in the same amount of time. Recall that the horizontal and vertical motions of each bullet are independent of each other. Since both identical objects are accelerating downward at the acceleration due to gravity and they are both dropped from the same height, it takes the same time for them to reach the ground. 6. Dear Wolfgang, You asked whether the time it takes to paddle a canoe across a river depends on the strength of the current. When you are paddling a canoe across a river, the variables that determine how long it takes are the width of the river and the forward velocity of the canoe due to your paddling. The canoe’s forward velocity and the current velocity are perpendicular to each other, so they don’t affect each other. As a result, the current does not affect the length of time required to cross the river. The only effect of the current on the motion of the canoe is to cause it to move downstream from where it would otherwise have landed. 7. The student who wants to apply the force above the horizontal has the better idea. The horizontal component of the applied force in the direction of motion will be the same regardless of whether the force is applied above or below the horizontal. It is in the students’ best interest to minimize the amount of friction. Recall that the frictional force is directly proportional to the normal force. If they apply the force above the horizontal, this will reduce the magnitude of the normal force needed to be supplied by the floor on the sofa, which will therefore reduce the frictional force and make it easier to move the sofa. On the other hand, if they apply the force below the horizontal, this will increase the normal force required and thereby increase the frictional force, making it harder to move the sofa.

64

8. a) The baseball’s velocity will be upward with a magnitude less than its initial velocity. The acceleration will be downward at 9.8 m/s2. b) The baseball’s velocity will be zero. The acceleration will be downward at 9.8 m/s2. c) The baseball’s velocity will be downward with the same magnitude as in a). The acceleration will be downward at 9.8 m/s2. 9. You would still need a pitcher’s mound on the Moon because the ball would still accelerate downward due to gravity. Since the Moon has a smaller mass than Earth, the acceleration due to gravity on the Moon is less than that on Earth. As a result, the height of the mound would not have to be as great as that on Earth. 10. She could jump twice as far on a planet that has one-half the gravity of Earth. If we assume that her initial speed and the direction for launch are the same, and that her initial vertical displacement is zero, we can write the following. 1 dy v1 t ayt2 2 1 0 v1 ayt 2 2v1 t ay If the acceleration, ay, is halved, then the time in flight, t, will be doubled. Therefore, the horizontal distance travelled will also be doubled, assuming that her horizontal speed is constant. 11. As your bicycle’s rear tire spins, it takes water with it due to adhesion. Inertia causes the water to try to move in a straight line. As a result, the water leaves the wheel with a velocity tangential to the tire and may spray your back if your bicycle does not have a protective rear fender. 12. Inertia causes the water in your clothing to try to move in a straight line. If the drum in the washing machine were solid, it would apply a centripetal force on the water, which would keep it moving in a circle. Since the drum has holes in it, however, the water is able to leave the drum as it spins. y

y

y


13. The aircraft can be flown in one of two ways, or a combination of these, to provide “weightlessness.” If the aircraft accelerates downward at the acceleration due to gravity, the astronauts inside the aircraft will experience “weightlessness.” The other possibility is to travel in a vertical arc. If the aircraft flies in a vertical arc at such a speed that at the top of the arc the gravitational force provides all the centripetal force required to keep the aircraft and its occupants travelling in a circle, they will experience “weightlessness.”

Chapter 3 1. Hydro lines and telephone cables cannot be run completely horizontally because the force of gravity acts downward on the entire wire and there is very little means of counterbalancing this force using supports. 2. a) The ladder is pushing directly into the wall on which it is resting, normal to the surface of the wall. With no friction, there is no force to prevent the ladder from sliding down the wall. b) The force exerted by the ladder on the ground is exactly equal to the force of gravity (weight) of the ladder because there is no vertical force due to friction. The only force that acts vertically, upward or downward, is the force of gravity. 3. Standing with your feet together or wide apart makes no difference to the condition of static equilibrium, since in both cases all forces are balanced. In terms of stability, the wider stance is more stable. A wider stance means a lower centre of mass and a wider “footprint.” This means there is a greater tipping angle for this wider stance. 4. High-heeled shoes force the centre of mass of the person wearing to move forward from its normal position. To maintain balance, the person must move the centre of mass back again, usually by leaning the shoulders backward. This effort can cause fatigue in the back muscles.

5. Line installers allow a droop in their lines when installing them because the droop allows a moderate upward vertical application of force as the wire curves upward to the support standards. This allows an upward force to support the wire when loaded with freezing rain and ice buildup. This droop means that the tension to support the load can be much less because of the greater angle. 6. A wrench can be made to more easily open a rusty bolt by adapting the wrench so as to apply more torque. More torque can be applied by the same force by adding length to the wrench handle. 7. The higher up on a ladder a person is, the farther he is from the pivot point, which is the point where the ladder touches the ground. Therefore, the ladder will be more likely to slide down the wall if the person stands on a higher rung. 8. The torque varies as sin , where is the angle between the pedal arm and the applied force. The torque is at a minimum (zero) when the pedals are vertical (one on top of the other), because the force (weight) is applied at 0° to the pedal arm, and sin 0° 0. The maximum torque is applied when the pedals are horizontal, because the angle between the pedal arm and the applied force is 90°, and sin 90° 1. 9. There is no extra benefit for curls to be done to their highest position. As the forearm is raised, the angle of the force of gravity vector decreases at the same rate as the angle between the muscle of effort and the arm. As the forearm is raised, the effort required to lift the arm decreases, but so does the muscle’s ability to provide the effort. 10. Your textbook is sitting in stable equilibrium when flat on your desk. When the book is balanced on its corner, it is in unstable equilibrium. Motion in any direction will cause a lowering of the centre of mass and a release of gravitational potential energy, making the tipping motion continue and thus making the book fall.


65

11. In terms of stability, a walking cane provides a wider base (footprint) over which the person is balancing. It is harder to force the person’s centre of mass outside this wider support base. 12. When standing up from a sitting position, we first must lean forward to move our centre of mass over our feet to maintain stability. Unless we first lean forward, our centre of mass is already outside our support base and it is impossible to stand up. 13. A five-legged chair base is more stable because of the wider support base (footprint). The extra leg effectively increases the tipping angle, making the chair more stable. 14. Tall fluted champagne glasses must have a wide base to improve the stability of the glass. Recall that the tipping angle is given by the (0.5)(width of base) expression tan1 . height of centre of mass Therefore, the taller the glass, the greater the height of the centre of mass, and the smaller the tipping angle. A wider base increases the tipping angle by compensating for the taller glass. 15. The extra mass helps to mimic the mass of the cargo and lowers the centre of mass of the ship. Without this extra mass, the ship would be top-heavy and more prone to capsizing, especially in rough weather. 16. This figure is so stable because the design of the toy places the effective centre of mass below the balance point. A gentle push actually raises the centre of mass like a pendulum, which increases the gravitational potential energy, which tends to return the toy to its stable equilibrium position. 17. The bone that has the smaller length will fracture first if the same twisting stress is applied to two bones of equal radius but different lengths. This is due to the fact that the strain L on the longer bone will be much L smaller than that on the shorter bone, because the length term appears in the denominator of the expression for strain.

66

18. Lumber is used this way to support greater spans because of the greater dimensions of wood in the vertical direction. More wood provides a means of supporting a greater weight through a tension force throughout the wood. 19. Concrete would not be an ideal material for a cantilevered structure because of the difference in the way that this material deals with tension and compression forces. A cantilever would require a great tensile strength in the upper layer and a great compressive strength in the lower layer. Concrete has great compressive strength but poor tensile strength.

Chapter 4 1. Momentum is the product of mass and velocity; mv. Since velocity is a vector quantity, so p is momentum. 2. A system represents all the objects involved in a collision. In a closed system, the boundary is closed (that is, there are no interactions with the external environment) and therefore the net external force acting on the system’s objects as a group is zero. In an open isolated system, the boundary is not closed but the net external force acting on the system is zero. 3. The net force is used in the calculation of impulse; J F∆t. 4. Impulse is the change in momentum; J ∆p. 5. In an isolated system, the net external force, F, acting on the system is zero. Therefore, the impulse, J, is zero (J F∆t), and the change in momentum, ∆p, is zero (J ∆p). 6. The law of conservation of (linear) momentum states that the total momentum of an isolated system before a collision is equal to the total momentum of the system after the collision. This can be expressed algebraically as ptotalinitial ptotalfinal. Equivalently, in an isolated system the change in momentum is zero; ∆p 0. 7. Yes, a ball thrown upward loses momentum as it rises because there is a net external force downward (gravity) acting on the ball, slowing it down.


8. Assuming that the net external force acting on the grenade during the explosion is zero (ignoring gravity), the sum of the 45 momentum vectors after the explosion is equal to the momentum vector of the grenade before the explosion, since ptotalinitial ptotalfinal. 9. Assume that the astronaut’s initial momentum is zero as he floats in space. By throwing the monkey wrench in the opposite direction of the space station, he would be propelled toward the space station. This is an example of Newton’s third law: The total momentum of the astronaut–wrench system would still be zero after he threw the wrench. 10. A rocket can change its course in space by ejecting any object or matter such as a gas. Assuming that the total momentum of the rocket–gas system is conserved, the momentum of the rocket will change as the gas is ejected. This change in momentum will correspond to an impulse, which will change the course of the rocket. 11. Assume that the total momentum of the system is conserved: pTo pTf p1o p2o p1f p2f mv1o mv2o mv1f mv2f mv1o m(–v1o) mv1f mv2f (substituting v2o v1o) 0 m(v1f v2f) Therefore, the general equation for the total momentum before and after the collision is pTo 0 m(v1f v2f) pTf. 12. As rain falls into the open-top freight car, the car will slow down. Assuming that momentum is conserved as the rain falls into the car, the combined mass of the car and the water will move along the track at a slower speed. 13. Object A is moving faster before the collision. Assuming that the momentum of the A-B system is conserved, the final velocity of the objects, vf, is equal to the average of their initial velocities, vAo and vBo:

pTo pTf mvAo mvBo mvAf mvBf vAo vBo vf vf vAo vBo vf 2 Since the angle between vBo and vf is greater than the angle between vAo and vf, the magnitude of vAo is greater than the magnitude of vBo. 14. The component method would be preferred for solving momentum problems in which trigonometry could not be used readily — for instance, problems involving more than two objects colliding, or non-linear problems. 15. a) Grocery clerks lean back when carrying heavy boxes so that their centres of mass stay in line with their feet. b) The centre of mass of a system of masses is the point where the masses could be considered to be concentrated or “balanced” for analyzing their motion. This concept can simplify momentum problems since the momentum of the centre of mass is equal to the total momentum before, and after, a collision, and is conserved during the collision.

Chapter 5 1. When you are holding your physics book steady in your outstretched arm, there is no work done because there is no displacement (W F∆d). 2. The momentum, p, of an object with mass m is related to its kinetic energy, Ek, according to k. If a golf ball and a the equation p 2mE football have the same kinetic energy then the football has the greater momentum, since the mass of the football is greater than the mass of the golf ball. 3. A negative area under a force–displacement graph represents negative work, which means that the displacement is in the opposite direction of the force applied. For example, when friction is slowing down a car, there is a positive displacement but a negative force. 4. After work is done on an object, it has gained energy.


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5. When a spring diving board is compressed by a diver jumping on it, the diving board possesses elastic potential energy. As the diving board straightens out, it transfers its elastic potential energy to the diver, who gains kinetic and gravitational potential energy. As the diver rises in the air, her kinetic energy is transformed into potential energy until she only has gravitational potential energy as she reaches her highest point. As she descends toward the pool, her potential energy is transformed into kinetic energy and she increases her speed as she falls. As she enters the pool and slows down in the water, her kinetic energy is transferred to the water as kinetic energy, potential energy, and heat energy. 1 6. Ek mv2 2 ( J) (kg)[(ms)2] ( J) (kg·m2s2) ( J) (kg·ms2·m) ( J) (N·m) ( J) ( J) 7. The equation ∆Ee ∆Ek means that a loss of elastic potential energy becomes a gain in kinetic energy. 8. Yes, since gravitational potential energy is measured relative to a point which could change. That point could be the ground level, the basement level, or any other arbitrary point. 9. In an elastic collision the total kinetic energy is conserved, whereas in an inelastic collision the total kinetic energy is not conserved. An example of an (almost) elastic collision is a collision between two billiard balls. An example of an inelastic collision is a collision between two vehicles in which their kinetic energy is transferred to heat energy, sound energy, and energy used to permanently deform the vehicles.

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p2 10. No, the equation Ek shows that if an 2m object has momentum then it must have kinetic energy. The converse is also true, as the equation also shows.

Chapter 6 1. We do not require the more general form of Newton’s law of universal gravitation because for situations on or near the surface of Earth, the values of G, M, and r can be assumed to be specified constants. After these simplifications are made, the general form becomes equivalent to the simpler form. 2. Due to the direction in which Earth rotates, more energy would be required to reach the same orbit if a spacecraft was launched westward, since an eastward launch aids the spacecraft. 3. The near side of the Moon is more massive than the far side, possibly due to impacted meteors. Over time this side was more attracted to Earth, so that eventually the more massive side came to face Earth all the time. This is also true for the moons of Jupiter and Saturn relative to their planets. 4. The force of gravity is the derivative of gravitational potential energy, Ep. Equivalently, the force of gravity is the slope of the graph of Ep versus x. 5. Assuming that the spacecraft is initially in orbit and that jettisoning a large piece of itself does not significantly alter its momentum, it will continue in the same orbit. 6. The velocity of a spacecraft in orbit is constantly changing due to the centripetal force acting on it. Therefore, if one spacecraft points toward another and rockets in that direction, the two spacecraft will not meet because the “added velocity” vector of the first spacecraft does not change as is required for convergence. 7. a) The escape speed required to leave Earth is approximately 11 km/s. The necessary upward acceleration, a, of a spacecraft during firing from an 80-m cannon is given


(11 000 m/s)2 by a 756 250 m/s2. 2(80 m) This is more than 77 000 times the magnitude of the acceleration due to gravity, and would be experienced for about 11 000 m/s ∆t 2 0.015 s. The 756 250 m/s mission would not be survivable. b) The downward force of the gun’s recoil would be roughly equal to the upward force on the spacecraft. If the spacecraft had a mass of 5000 kg, the force of the recoil would be approximately (5000 kg)(756 250 m/s2) 3.781 109 N. 8. Given: hmax 2 m k 500 N/m xmax 0.45 m m 80 kg First, calculate the maximum energy that our knees can absorb without damage. Vg mgh Vg (80 kg)(9.8 m/s2)(2 m) Vg 1568 J Next, calculate the maximum energy that the springs can absorb. 1 Ve kx2 2 1 Ve (500 N/m)(0.45 m2) 2 Ve 51 J Finally, calculate the maximum height from which we could survive a fall without damage. Vg mgh Vg h mg 1568 J 51 J h (80 kg)(9.8 m/s2) h 2.07 m With the springs attached, we could survive a fall of at most 2.07 m without damage. 9. The force of gravity would be (9.8 m/s2)(80 kg) 1837 N downward, whereas the force of the springs would be only (500 N/m)(0.45 m) 225 N upward. The net force acting on us would act downward, so we would not bounce off the ground.

10. Three everyday examples of SHM are: an idling engine, as periodic power from combustion keeps piston movement in a state of SHM; someone rocking in a rocking chair, where periodic “foot pushes” or shifts in the centre of mass counteract dampening; the motion of a toy bird that “drinks” water, provided that there is a constant supply of water. 11. Three examples of damping in oscillatory systems are: engine braking (desired) — as the fuel supply to the cylinders is lessened, so is the power, which dampens piston movement; swinging on a swing (undesired) — the height of successive swings becomes smaller and smaller due to friction and air resistance; air bags (desired) — when deployed, they gradually dampen the effects of a collision on a person’s body, as opposed to a steering wheel or dashboard, which do so almost instantaneously.

Chapter 7 1. All objects on Earth that are stationary relative to Earth’s surface have the same angular velocity, since they all complete one rotation about Earth’s axis in the same amount of time. However, they do not all have the same tangential velocity, since they are not all the same distance from Earth’s axis of rotation. If

is the angular velocity of an object on Earth’s surface and r is the object’s distance from Earth’s axis of rotation then the object’s tangential velocity, v, is given by v r . 2. A differential mechanism is necessary to allow a car to turn smoothly. The wheels on the inside of a turn move through a smaller radius than the wheels on the outside, thus travelling a smaller arc distance in the same amount of time. Therefore, the inside wheels rotate at a smaller angular speed. In the absence of a differential, however, the inside and outside drive wheels (connected to the motor) must rotate at the same angular speed. To turn, you would have to lock up the inside drive wheel, causing an uncontrolled turn.


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3.

4.

5.

6.

7.

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The differential allows the drive wheels to turn at different angular speeds. The top of the CN Tower, with the tower located at the highest altitude on the equator, would have the greatest tangential speed, because in that case the top of the tower would be the greatest distance from the axis of rotation of Earth. However, since the height of the CN Tower is negligible compared to the radius of Earth, the variation in tangential speed among different parts of the tower is negligible. A larger car tire has a greater moment of inertia (greater radius and mass), thus in principle more energy would be needed to start turning the tire. Once the tire was moving, the law of inertia would apply and a greater force would be needed to slow and stop the tire, thus less energy would be needed to keep the tire moving. a) Yes, changing the tire size affects the odometer reading. For example, a tire with a larger radius than the calibrating tire covers a greater distance in the same number of turns. In that case, the car will travel a greater distance than what the odometer indicates. b) Yes, the speedometer reading is affected, for the same reason. For example, a car with larger tires will travel at a greater speed than what the speedometer indicates. The angular equivalents to force and displacement are torque and angular displacement. No linear work is done on an object if an applied force does not change the displacement of the object in the direction that the force is applied. No rotational work is done if an applied torque does not result in a change in angular displacement. No, angular momentum is conserved because the diver is in fact still rotating as she enters the water. There is no external torque applied to the diver after she leaves the diving board. Because the diver increases her moment of inertia by extending out straight from a tuck, her angular spin decreases. This is

8.

9.

10.

11.

not visually apparent as the diver then enters the water out of sight of the spectators and judges. No, the centripetal force acting on a rider varies depending on the radius of turn: the larger the radius, the larger the centripetal force. The riders on the outer part of the ride swing out farther than the inner riders because of the larger centripetal force. According to the law of conservation of angular momentum, the total angular momentum before the tape recorder was turned on high speed was equal to the total angular momentum after. When the tape recorder was turned on high speed, the angular momentum of the system had an added component in the angular direction of the turning tape. Voyager 2 rotated in the opposite direction to compensate, although not as fast, since its moment of inertia was much larger than the tape recorder’s. a) The hollow cylinder has a greater moment of inertia than the solid one because the hollow cylinder’s mass is concentrated farther from the axis of rotation. However, since there is no friction, there is no force available to create the torque necessary to turn the cylinders. Translational motion does not depend on the distribution of mass, so both objects accelerate at the same rate and reach the base of the incline at the same time. b) As in part a), in the absence of friction the cylinder does not roll. Therefore, both objects slide down the ramp, accelerating at the same rate (ignoring the effect of wind resistance on the different shapes). Translational motion does not depend on the distribution of mass, so both objects reach the ground at the same time. A spinning projectile behaves like a gyroscope. The spin means that the object possesses angular momentum about its axis of rotation. This allows the object to resist forces acting on it as it travels, which in turn allows the object to maintain its projectile motion.


12.

13.

14.

15.

16.

Without the spin, uneven airflow over the surface of the object would make it tumble, experience greater air resistance, and travel a shorter distance. If the wheel does not slip as it rolls then the translational distance, d, that the axle moves is equal to the arc length, s, along the outside of the wheel. This is not true in the case of “squealing your tires.” Rotation axes can be anywhere, but for simplicity’s sake consider only some symmetric ones. Ranked from least to greatest moment of inertia, the rotation axis can pass through the centre of the top and bottom (shown), through the centre of the spine, through the centre of the front and back cover, or run diagonally from one corner to another. The angular momentum of a Sun–planet system is conserved. The force acting on the planet is that of gravity due to the Sun. At any instant in time, this force acts through the axis about which the planet instantaneously rotates. This means that the moment arm is zero and no torque acts on the planet. Therefore, the angular momentum of the planet remains constant and the total momentum of the system does not change. It is easier to balance on a moving bike than on a stationary one because of a combination of the aspect called “trail” and gyroscopic action. The law of conservation of angular momentum applies when a motorcycle is in mid-air. In the absence of an external torque, the increased angular momentum of the fasterspinning rear wheel causes the entire motorcycle to rotate in the other direction in order to keep the total angular momentum the same as it was when the motorcycle left the ground.

Chapter 8 1. A neutral object is attracted to a charged object because the charged object induces a charge separation in the neutral object. The electrons in the neutral object are forced away from or toward the charged object, depending

on whether the charged object has a negative or positive charge, inducing an opposite charge which acts to attract the two objects by way of the law of electric forces. 2. The function of an electroscope is to detect an electric field. An electric field will cause the movement of electrons within an electroscope, inducing similar charges to cluster at each of the two pieces of dangling foil. The two pieces of foil will repel each other, indicating the presence of the electric field. 3. Rubbing the balloon against your dry hair charges the balloon electrostatically. When the balloon approaches the wall, the negative charge forces the electrons in the ceiling away, leaving the positive charges close to the surface. The result is that the negatively charged balloon attracts the now positively charged ceiling surface. –

–

–

Ceiling – – –

–

–

–

+

+

+

+

+

+

+

+

+

Force of Attraction –

– – Balloon + +– –+ – – –

4. The electrostatic series identifies silk as having a greater affinity for electrons than acetate does. When acetate and silk are rubbed together, electrons move from the acetate to the silk because of the different affinity the materials have for electrons. 5. Choose two materials listed at either end of the electrostatic series, such as acetate and silk, and rub them together to place the predictable negative charge on the silk. Neutralize the acetate and then rub it with the mystery substance. Place the mystery substance next to the silk and judge whether the mystery substance has a negative charge (repulsion) or a positive charge (attraction). A negative charge would place the mystery substance below acetate in the electrostatic series. Similarly, rubbing the mystery substance with silk would help to place the


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mystery substance in the series compared to silk. By selectively choosing different substances, you could narrow down the appropriate spot for the mystery substance in the electrostatic series. 6. Computer technicians touch the metallic part of a computer before repair, assuming it is still plugged into the wall outlet, so that they ground themselves from any excess charge. Otherwise, a static electric discharge could damage the computer’s micro-circuitry. Newton’s law Coulomb’s law 7. Criterion Equation Constant of proportionality Type of force(s) Conditions for use

of universal gravitation

of electrostatic forces

Gm1m2 F 2 r

kq q F 122 r

G 6.67 1011 N·m2/kg2

k 9.0 109 N·m2/C2

Attraction only

Attraction and repulsion

Acts between any two masses

Acts between any two electrostatic charges

field lines. Otherwise, the rod will tend to rotate 180° and point in the opposite direction (still parallel to the field lines). 11. Each point charge experiences an identical force of repulsion from all of the other point charges, so that they are all repelled symmetrically outward from the centre of the orientation. A test charge placed outside of the circle would experience a net force directed along radial lines inward to the centre of the circle, as shown in the diagram. A test charge placed inside of the circle would experience no net force, and therefore there would be no electric field inside the circle at all.

–

8. Field lines show the direction of the net force on a test charge in an electric field. Two crossed field lines would mean that there would be two net forces acting on a test charge in two different directions at the same time. This is impossible, since there is only one net force at any point, which is only one force in one direction by definition. 9. In an electric field, charges always move along the direction described by the field lines. The direction in which a charge moves along a field line depends on the sign of the charge. A positive charge will move in the direction described by the arrows in a field diagram, whereas a negative charge will move in the opposite direction. 10. a) When a polar charged rod is placed perpendicular to electric field lines, the rod will tend to rotate such that it will become parallel to the field lines. The positive end of the rod will point in the same direction in which the field lines are oriented. b) When a polar charged rod is placed parallel to electric field lines, the rod will tend to stay in the same orientation if its positive end is pointing in the same direction as the

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–

– –

–

– – –

This charge distribution models the electric field inside a coaxial cable because the outer braided conductor in a coaxial cable acts as the site modelled by the ring of charge described above. This ring acts to eliminate the field within the entire cable. 12. By definition, the electric potential is the same at any point along an equipotential line. Therefore, no force is required, and no work is done, to move a test charge along this line. In a situation like this, a constant force causes the constant acceleration of the test charge. 13. We use the term “point charge” to imply that the charge has no larger physical dimensions. Larger dimensions would mean that the charge would exist within a region of space instead of at a specific location. This implication reduces the number of variables and simplifies questions that deal with the distribution of charges within a three-dimensional space. Any other approach would require some way of accounting for the variability of distances between charges.


14. Statement: In each case the field gets stronger as you proceed from left to right. False Reasoning: The field lines remain the same distance apart as you move from left to right in the field in (b), so the field does not change in strength. Statement: The field strength in (a) increases from left to right but in (b) it remains the same everywhere. True Reasoning: The field lines become closer together as you move from left to right in the field in (a), so the field does increase in strength, whereas the field lines in (b) are parallel, so the field strength does not change. Statement: Both fields could be created by a series of positive charges on the left and negative ones on the right. False Reasoning: Although true for (b), (a) must be created by a single positive point charge at the base of the four arrows. Statement: Both fields could be created by a single positive point charge placed on the right. False Reasoning: As described above, a point charge could be responsible for (a), but (b) would require rows of parallel opposites such as those in oppositely charged parallel plates. 15. Electric fields are more complicated to work with because the forces that charges exert on each other are all significant. In contrast, the gravitational force between small masses is negligible compared with the gravitational force exerted on them by large masses like Earth. 16. The field shape around a single negative point charge is exactly like that around a single positive point charge with the exception that for a negative point charge, the arrows are all pointing inward instead of outward, as shown in the following diagram.

–

17. Doubling the value of the test charge will do nothing to the measurement of the strength of the electric field. The force on the test charge will double because of the change to the test charge, but the field strength is measured as F the force experienced per unit charge, . qt Therefore, the doubling of the test charge and the doubling of the force will cancel, leaving the measurement of the field strength unchanged. 18. The stronger an electric field is, the closer together the field lines are. Therefore, a weak electric field has field lines that are farther apart than the field lines of a strong electric field. 19. Both gravitational fields and electric fields are made up of lines of force that are directed in a way that a test “item” would be forced. Gravitational fields are created by and influence masses, whereas electric fields involve charges. Gravitational fields are always attractive. Electric fields can be attractive or repulsive, since they can exert forces in opposite directions depending on the charge of the object that is experiencing the field. 20. The direction of an electric field between a positive charge and a negative charge is from the positive charge toward the negative charge, since electric fields are always directed the way that a positive test charge would be forced. 21. The electric potential energy is greater between two like charges than between two unlike charges the same distance apart because of the differing sign of the electric potential energy. The calculation of the


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22.

23.

24.

25.

26.

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electric potential energy involves multiplying the two charges. The product of two like charges is positive and therefore greater than the product of two unlike charges, which is negative. A high-voltage wire falling onto a car produces a situation in which there is a highpotential source (the wire) very close to a low-potential region (the ground). The people in the car will be safe from electrocution as long as they do not complete a circuit between this high and low electric potential. They should not open the car door, for example, and step to the ground while maintaining contact with the car. Although opposite electric charges occur at the two plates of a parallel-plate apparatus when it is connected to a power supply, the overall charge on the apparatus remains zero. For every charge at one plate, there is an opposite charge at the other plate, which balances the overall charge to zero. a) If the distance between the plates is doubled then the field strength between the plates will be halved. b) If the charge on each plate is doubled then the field strength will double. c) If the plates are totally discharged and neutral then the field strength will drop to zero. Two point charges of like charge and equal magnitude should be placed side by side so that both the electric field strength and the electric potential will be zero at the midpoint between the charges. If one of the two like charges were doubled, the field strength and the potential would both be zero at a point two-thirds the separation distance away from the doubled charge. In the presence of electric fields, a field strength and a potential of zero would exist at a point where the sum of all electric forces was zero. In question 25, the sum of the repulsive forces from each of the two like charges is zero at some point between the two charges.

27. If a proton and an electron were released at a distance and accelerated toward one another, the electron would reach the greater speed just before impact. The reason is that both particles would be acted upon by the same force of attraction, but the electron has less mass. The acceleration of each particle is F, which described by the formula a m shows that for the same force, the smaller mass would have the greater acceleration over the same time period and therefore the greater final speed. 28. q

This type of motion is like upside-down projectile motion, since the charge moves in a parabolic path. This is the type of motion that an object would take if it were thrown horizontally in Earth’s gravitational field. The only difference here is that this charge appears to be “falling upward” instead of downward. 29. No, a parallel-plate capacitor does not have uniform electric potential. It does have uniform field strength between the two plates, but the potential varies in a linear fashion from one plate to the other. By definition, the electric potential is uniform along any equipotential line, which in this case is any line parallel to the two plates. 30. Charge Distribution Equipotential Lines (a) (b) (c)

(iii) (i) (ii)

31. a) The electrostatic interaction responsible for the large potential energy increase at very close distances is the repulsion between the two positive nuclei. b) This repulsion of the nuclei, and the associated increase in electric potential energy, is one of the main stumbling blocks for generating energy through nuclear fusion. This repulsion between nuclei means that


a very large amount of energy is required to begin the reaction process. c) The smaller increase in electric potential energy upon separation of the two atoms is caused by the attraction between the positively charged nucleus in each atom and the negatively charged electron in the other atom. d) A stable bond is formed when two hydrogen atoms are about 75 pm apart because this is the distance at which the electric potential energy is minimized — any closer and the repulsion between nuclei pushes the atoms apart, any farther away and the nucleus-electron attraction draws the atoms closer together. 32. A positive test charge moving along a line between two identical negative point charges would experience a topography similar to a vehicle moving up a hill (away from one charge), increasing the vehicle’s gravitational potential energy, and then rolling down the other side of the hill (toward the other charge). a) If the two identical point charges were both positive, the hill would change to a valley with the lowest part in the middle. b) If a negative test charge was placed between the two identical positive charges, the topography would still resemble a valley but now there would be a very deep crater at the lowest part of the valley.

3. A material that is attracted to a magnet or that can be magnetized is called ferromagnetic. Examples of ferromagnetic materials include materials made from iron, nickel, or cobalt. These materials are ferromagnetic because they have internal domains that can be readily aligned, due to the fact that these materials have unpaired electrons in their outermost electron energy level. 4. Magnets can lose their strength over time because their domains, which initially are aligned (pointing in the same direction), can become randomized and point in other directions. This randomizing of the domains reduces the overall strength of the entire magnet. 5. When a magnet is dropped or heated up, the domains of the magnet, which initially are aligned (pointing in the same direction), can be disrupted and forced to point in other, random directions. This randomizing of the domains reduces the overall strength of the entire magnet. 6. a) F

F T

Currents in the same direction— wires forced together

b) F

F T

Chapter 9 1. The law of magnetic forces states that like (similar) magnetic poles repel one another and different (dissimilar) poles attract one another, even at a distance. 2. A magnet can attract non-magnetic materials as long as they are ferromagnetic in nature. The magnet causes the internal domains (small magnets) of a ferromagnetic substance to line up in such a way that a new magnet is induced in the substance such that there are opposite magnetic poles which attract one another.

x

Currents in opposite directions— wires forced apart

7. The electrons in the beam that is illuminating your computer monitor’s screen are directed from the back of the monitor forward to the front of the screen, toward your face. Therefore, conventional (positive) current points in the opposite direction, away from your face and back into the computer monitor. This is the direction of the thumb of the right


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8.

9.

10.

11.

12.

13.

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hand when applying right-hand rule #1 for current flow. From your perspective, the magnetic field forms circular formations in the clockwise direction in and around the computer monitor. Relative to the direction of the electron beam, the magnetic field is directed in the counterclockwise direction around the beam. A wire possessing an eastbound conventional (positive) current has an associated circular magnetic field that points upward on the north side of the wire and downward on the south side. The magnetic field strength of a coil (an insulated spring) varies inversely with the length of the coil. Therefore, a reduction in the coil length to half its original length will cause a doubling of the magnetic field strength. This all depends on the assumption that the length of the coil is considerably larger than its diameter. a) For the force applied to a current-carrying conductor to be at a maximum, the magnetic field must cross the conductor at an angle of 90°. b) For the force applied to a current-carrying conductor to be at a minimum, the magnetic field must cross the conductor at an angle of 0°. According to right-hand rule #3 for the motor principle, the direction of the force on the conductor will be to the north. An electron moving vertically downward that enters a northbound magnetic field will be forced toward the west. A current-carrying solenoid produces a magnetic field coming directly out of one end of the coil and into the other end. An electron passing by either end of this coil experiences a force that is at right angles to its motion. As this force changes the direction of motion (a centripetal force), the electron takes on a curved path (circular motion). Application of the appropriate right-hand rules predicts that the electron’s motion will curve in the same

14.

15.

16.

17.

18.

19.

direction as the direction of conventional current flow through the coil. The cathode rays will be deflected away from the current-carrying wire, moving in a plane that contains the wire. Current passing through a helical spring will produce a situation very similar to having two parallel conductors with a current flowing in the same direction. Application of the appropriate right-hand rules predicts that the magnetic field interaction between each pair of the helical loops will force the spring to compress, reducing its length. Current passing through a highly flexible wire loop will tend to result in magnetic field interactions that will force apart nearby sections of the wire, so that the wire loop will most likely (if the proper conditions exist) straighten out. Faraday’s principle states that a magnetic field that is moving or changing in intensity in the region around a conductor causes or induces electrons to flow in the conductor. To improve the electromotive force induced in a conductor, we can increase the magnetic field strength, the length of the conductor, and the strength of the current flowing through the conductor. Current can be induced to flow in a conductor if the conductor is moving with respect to a magnetic field. The maximum induced current occurs if the conductor and the field cross each other at right angles. Lenz’s law states that the direction of the induced current creates an induced magnetic field that opposes the motion of the inducing magnetic field. Lenz derived this law by reasoning that a decrease in kinetic energy in the inducing magnetic field must compensate for the increase in the electric potential energy of the charges in the induced current, according to the law of conservation of energy. This decrease in kinetic energy is felt as an opposition to the inducing magnetic field by an induced magnetic field.


20. The induced current can only create a magnetic field that opposes the action of movement (conductor or field) in order to follow the law of conservation of energy and Lenz’s law. If the motion were not opposed and the induced magnetic field instead “boosted” the motion, this would increase the kinetic energy of the moving conductor or magnet, which would violate the law of conservation of energy. 21. a) Electromagnetic brakes might work by using the undesirable motion of the vehicle to provide the energy to induce current flow in a conductor. The resulting creation of electrical energy would be at the expense of the kinetic energy of the vehicle, which would slow down. This would be a case of energy being transformed from one form to another, following the law of conservation of energy. b) Electromagnetic induction brakes would be capable of recovering some of the kinetic energy of a vehicle that is normally lost as heat in conventional brakes, thereby saving money. The electrical energy generated could be used to recharge the battery for an electric vehicle/hybrid.

Chapter 10 1. The motion of a vibrating spring can be modelled mathematically by a sine wave, which resembles (visually) an electromagnetic wave. As well, both waves are periodic. 2. The magnetic field is induced by the electric field and thus they would both decrease. If one component vanishes then the electromagnetic radiation ceases to exist. 3. “Visible light” is relative to the human being perceiving it. Also, some other animals see in other regions such as the infrared and ultraviolet.

4.

θi

θr

Normal

5. When metallic objects are placed in a microwave oven, they can absorb electromagnetic microwaves, which dislocate loose electrons in the metal and allow charges to build up on the surfaces, until the cumulative charge is large enough to “jump” across an air gap to another conductive material in the oven, causing a spark. 6. Simple harmonic motion refers to a physical “state” where the restoring force, acting on an object when it is pulled away from some equilibrium position, is proportional to the displacement of the object from the equilibrium position. Since there is a net force acting on the object, it experiences an acceleration, and thus the speed cannot be constant. 7. If a circle is viewed edge-on, with a dot painted on the edge, and the circle is spun, the dot will seem to exhibit simple harmonic motion as it moves around the circle. From the edge it will seem as if the dot is moving back and forth, constantly passing the equilibrium position. 8. Electron oscillators absorb energy from the incoming wave, causing it to be retarded. When this secondary wave interferes with the incident wave, a phase lag is created retarding the wavefront, slowing it down. 9. Newton’s theory of refraction predicts that light speeds up as it changes direction. This is incorrect since light decreases its speed when bending toward the normal. You can show his theory by rolling a marble across a boundary between a flat area and an incline. As the marble crosses the boundary, it bends toward a line drawn perpendicular to the edge but it speeds up.


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10. One example of an invisible medium is a vacuum. The refractive index of a vacuum is 1.00. When the refractive index is 1.00, there is no component of an incoming light ray that is reflected. Since no light is reflected, the medium is invisible. Another possibility is that the medium is of the same refractive index as the environment. 11. Using a laser, which is a powerful coherent source of visible light, you can measure the refraction of the ray as it enters a medium, or the extent of polarization upon reflection and/or transmission, all of which can be combined to calculate the optical density of the medium. 12. Because the refractive index is wavelength dependent, when white light refracts through a material, each component of light bends slightly differently. This separates the light. If the separation is great enough, dispersion occurs. 13. As light passes through a prism, both refractions cause the light to refract in the same spatial direction. This accentuates the spreading of the colours. 14. No, sound waves cannot be polarized. Sound waves are mechanical waves and refer to compressions and rarefactions within a medium. Sound waves have only one component, not two like electromagnetic waves, and thus polarization is impossible. 15. A polarizer and an analyzer are both thin pieces of film. They are given different names based on the order in which a wave enters them. If two pieces of thin film are positioned side by side, the first one struck by the wave is known as the polarizer and the second one the analyzer. If the two are flipped, the analyzer will become the polarizer and the polarizer will become the analyzer. 16. The lenses in polarized sunglasses are normally oriented in such a way as to restrict the passage of plane-polarized light reflecting off the surface of the ground and water (glare). If the lenses are rotated, they will no longer block the glare. 78

17. Yes, the effectiveness of Polaroid sunglasses varies as the relative positions of the Sun and the horizon vary, since the distribution of scattered angles varies as well. The amount of polarization is angle dependent, hence the effectiveness of the glasses varies. 18. No, Polaroid sunglasses are not effective on circularly polarized light, which is composed of the two polarization directions combined in a specific phase relationship causing the direction of the electric field vector to rotate around. The linear polarizer cannot block out both components, hence light is transmitted. 19. With a powerful light source, you can easily notice that light reflects off dust particles in the air. Sometimes, depending on the size of the particles, certain frequencies of the electromagnetic spectrum are deflected/reflected more than others. This effect causes a certain colour to appear in the medium (for example, the blue colour of the sky). By noting the colour, you can determine the frequency and thus the wavelength of light associated with that colour. Once you know the wavelength, you can calculate the approximate size of the particles that would deflect waves of those frequencies. Also, you can use the intensity of the colour to estimate the density of the particles in the air. 20. Polarization: Electric fields of electromagnetic radiation behave sinusoidally. The direction of these fields is randomly oriented in any direction for unpolarized light. Two components are obtained by using plane polarizers. The two components can be combined using the wave equation ( sin t) to form circularly or elliptically polarized light. Scattering: The wavelength of light, , comparable to the size of particles in the air creates the maximum scattering. The extent of scattering of light by air molecules is proportional to 4. Refraction: Using wavefronts, Snell’s law of refraction is derived. Based on phase relationships between the incident wave and the


transmitted wave, light is bent and slowed down in different mediums.

Chapter 11 1. Refraction, polarization, interference, and diffraction 2. Refraction, diffraction, and interference can be demonstrated using water waves in ripple tanks. Polarization cannot. 3. The film on a soap bubble is thicker at the bottom than at the top, forming a wedge shape, since gravity pulls the soap down. As the film’s thickness changes, the interference changes (destroys some wavelengths) and the colours change. 4. As the gasoline evaporates, it becomes thinner, changing the interference pattern and the colours. 5. A camera lens has a thickness and material designed to block out certain colours, whereas a car windshield does not. These properties of a lens produce interference patterns and a colour change. Camera lenses are designed to correct chromatic aberration caused by different wavelengths bending at different angles while being refracted. 6. a) Newton believed that light was a particle. b) Changing people’s environments through innovation can leave people feeling not in control, especially in cases where a new technology has the possibility of replacing people in jobs. c) Accepting theories prematurely hinders progress, since it discourages research. 8. No, there are no interference patterns because the two car headlights are not coherent light sources and do not form a double slit. 9. Any imperfections are in the order of magnitude of the wavelengths of light used for the experiment. This washes out the effect with its own random interference patterns. 10. Sound waves are comparable in wavelength size to the openings, increasing the diffractive effect. Light waves have much smaller wavelengths and hence do not show these effects.

11. The resolving power of your eyes restricts your ability to distinguish between objects at great distances. This is because your pupils are circular, allowing diffraction to occur. 12. No, diffraction patterns place a limit on resolving power as well as the magnitude of the wavelength of light used. 13. Both spectroscopes separate white light into its colour components, but the prism spectroscope uses refraction and dispersion while the grating spectroscope uses diffraction. 14. Continuous spectra involve an extensive range of frequencies (example: sunlight spectrum). With line spectra, on the other hand, discrete frequencies are observed (example: molecular gas spectrum). 15. Each piece of a hologram contains the complete interference pattern of the object from which the hologram was created, whereas a piece of a normal photograph contains only local information and nothing about the complete photograph. 16. Diffraction gratings and interference gratings are really the same thing. Diffraction gratings actually use the interference superposition formula. Gratings show both effects—those due to the width of a single opening and the combination of all the openings. 17. Close spacing in a grating provides strong mutual coupling, increasing the effect of interference. The separation of the maxima increases. 18. Gratings with many slits have high resolving power. This means that the individual maxima become sharper. 19. Yes, because increasing the number of slits decreases the slit separation. If the slit separation is reduced beyond what is comparable to the wavelength of light, no light will get through. 20. A single slit has a double central maximum, with the intensities of the maxima dropping off dramatically with order number. A diffraction grating has a single central maximum and the intensities do not drop off as dramatically.


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21. Diffraction occurs as light enters the pupil. This places a limit on the eye’s resolving power. As you move away from the picture, sooner or later you cannot distinguish between the dots and they blend together to form a continuous picture. 22. Electrons have a smaller wavelength to that of visible light, and therefore have a higher resolution. This also minimizes diffraction. In fact, the beams of electrons have an effective wavelength that is 105 times that of visible light. This is a 100 000-fold increase in resolution.

Chapter 12 1. A photon is a unit particle (as opposed to wave) of electromagnetic radiation that moves at the speed of light. Its energy is proportional to the frequency of the radiation. 2. Ultraviolet radiation from the Sun is very energetic due to its high frequency. The photons that possess this energy are the cause of sunburn. These photons are energetic enough to remove electrons from our body cells, causing a change in our skin biology and in severe cases causing cancer. 3. Visual light is mostly in the infrared-visual spectrum. The energy of these photons is not sufficient to damage skin cells. 4. If h 0, quantization would not exist. There would be no energy levels in atoms. Electrons in atoms would therefore not attain any real value for energy, resulting in the absence of orbitals in atoms. 5. The electron volt (eV) corresponds to the energy of an electron at a potential of one volt. Hence, one electron volt is the energy equalling the charge of an electron multiplied by the potential of one volt: 1 eV qe 1 V. 6. Wien’s law relates the wavelength of photons to the temperature of the black body. 7. W0, the work function, is the amount of energy required to produce the photoelectric effect in a given metal. It is the minimum energy required to liberate electrons from a metal.

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8. Since the photons have detectable linear momentum, their mass equivalence can be computed. Momentum is an intrinsic property of matter, therefore we can assume that mass equivalency is correct. 9. An empirical relationship is a relationship that is determined experimentally. It is not backed up by theory. 10. Determinacy is a condition of a measurement being characterized definitely. An example of an everyday event could be a repetitive measurement of the length of a table. Each time the measurement is made, errors are encountered. If determinacy existed at the macroscopic level, we would get the same length every time. 11. The computation of uncertainties using Heisenberg’s uncertainty principle yields minute values for speed and position. The limitations of human perception prevent us from experiencing such minute variances at the macroscopic level. 12. Another device besides the STM that operates using the principle of quantum tunnelling is the electron tunnelling transistor, which is an on-off switch that uses the ability of an electron to pass through impenetrable energy obstacles. 13. The energy of an orbital varies as the inverse square of the radius. Hence, the spectral lines are closer together farther away from the nucleus. 14. a) The peak wavelength emitted by a mercury lamp lies in the visual spectrum. However, this implies that there is a tail in the ultraviolet spectrum. The ultraviolet photons are energetic enough to damage skin cells. b) An appropriate shielding that blocks ultraviolet light but allows photons in the visual spectrum to pass through could be used. 15. Consider two particles that have the same de Broglie wavelength and masses m1 and m2 such that m1 m2. According to de Broglie’s h h equation, and , where v1 m1v1 m2v2 and v2 are the velocities of the two particles.


Since is the same for both particles, the following equation can be written: h h m1v1 m2v2 This equation can be simplified: m1v1 m2v2 Since m1 m2, it follows that v2 v1. If the mass of the first particle is much greater than that of the second particle, the velocity of the second particle must be much greater than that of the first particle. 16. According to Planck, the energy is quantized. The angular momentum is certainly related to the energy. Hence, the angular momentum needs to be quantized as well. To quantize L, Bohr had to quantize both the velocity, v, and the radius, r. 17. Although the initial and the final speed and the scatter angles are known, the manner in which the actual collision occurs cannot be precisely predicted, and the exact position of the particles during the collision is not known. Hence, the uncertainty principle is not violated.

5.

6.

7.

8.

Chapter 13 1. Your car is in an inertial frame when it is stopped, or when it is moving at a steady speed in a straight line. Your car is in a non-inertial frame when it is accelerating, such as when you are braking, or when you are making a turn. 2. Donovan’s reference frame is inertial because the 100-m dash is in a straight line. Leah’s frame is non-inertial because the 400-m oval requires her to constantly change direction. 3. No, without reference to the outside world, it would be difficult to determine whether the cruise ship was at rest or moving with a constant velocity. 4. Suppose v swim speed and w water speed. To swim upstream and back down, it would take a total time of: d 2dv d . vw vw v2 w2

To swim straight across the stream (perpendicular to the current) and back, it would take a total time of: 2 d 2dv w2 . 2 2 2 2 v w2 v w 2 w2, so But v v 2 2dv 2dv w2

. 2 2 2 v w v w2 Therefore it would take longer to swim upstream and back down than to swim across the stream and back. The Michelson-Morley null result led to the development of special relativity, a tool needed in the understanding of high-energy physics. Analogous to the Doppler shift of sound, the constant speed of light in a vacuum, c, requires the wavelength of the approaching amber light to shorten or become more yellowish. In terms of Einstein’s first postulate involving relative motion, the two situations are equivalent. The same physics occurs whether a magnet is moved into a stationary coil or a coil is moved around a stationary magnet. Proper time is the time measured by one watch between the beginning and the end of the experiment. This is the time measured by a watch moving with the muon. The scientists of Earth would require at least two watches, one at the birth of the muon and the other at its disintegration. The relativity equation for length is v2 v2 . If v > c, then 1 is L L0 1 c2 c2 negative and L becomes imaginary, which is not physically reasonable. Since the electrons would have a greater relative velocity than the protons, the space between the electrons would be more contracted. As a result, the concentration of electrons would exceed that of the protons, and the wire would seem negatively charged. For this reason magnetism is a result of special relativity.

9.

10.


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11. No, you have not travelled faster than the speed of light. Instead, you have measured the Earthstar distance to be contracted and thus it took less time to travel at your speed of v 0.98c. 12. No, although time is dilated (events seem longer at relativistic speeds), it cannot slow to a complete standstill unless v c. 13. In both cases of the Doppler shift for sound, there is a shift to higher frequencies. However the physics of sound waves generated by a moving vibrating source colliding with air molecules and perceived by a stationary receiver is different from that of sound created by a stationary source and perceived by a moving source. For light, the frequency shift depends only on the relative speed of the source and receiver because the speed of light is always c, according to the second postulate. 14. Only Barb is correct in saying that Phillip’s clock ran slow, because his time was the proper time that was at the beginning and finish of his experiment. If Phillip observed stationary Barb doing a similar experiment beside the train, he would be correct in saying that her clock ran slow. 15. If the charge of an electron depended on its speed then the neutrality of atoms would be upset by the motion of electrons within the atoms. Experiments have shown that the charge on an electron is the same at all speeds. 16. The radius of the orbit becomes smaller as the magnetic field is increased because the radius mv is equal to , where B is the magnetic field qB strength. v2 , but 17. No, because mass dilates as 1 c2 mass density , and volume contracts as volume

1 vc .

v2 1 2 . Therefore, density dilates as c 2

2

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18. The starlight will pass you at a speed of c according to the second postulate of special relativity. 19. The occupants of the spacecraft would say that they observed the same things about us, due to relative motion. 20. No, according to the second postulate of special relativity, the light leaving the receding mirror travels with speed c relative to you. 21. Tachyons or particles that travel with a speed greater than c would seem to require infinite energy. Experiments do not support their existence. 22. Particle A would have the greater speed because its total energy due to mass dilation (mc2) is three times its rest energy, whereas particle B has a total energy dilated by a factor of only two. 23. Since the ice and the water have the same mass, they have the same total energy (m0c2 Ek). However, the kinetic energy, Ek, of the water is higher than that of the ice and for that reason the rest energy of the water is less than that of the ice. 24. If you consider that energy is equivalent to mass (E mc2), then electromagnetic energy in the form of light could be considered to have an equivalent mass. 25. A 100-eV electron has a dilated mass according to: mc2 m0c2 qV mc2 (0.511 0.000 100) MeV mc2 0.5111 MeV This means that its mass is less than 0.02% greater than its rest mass. A 100-MeV electron has a mass equivalent to (0.511 100) MeV of energy, which means that its mass is about 197 times its rest mass. 26. When we say that the rest mass of a muon is 106 MeV/c2, we mean that its rest energy is equivalent to the kinetic energy of an electron accelerated from rest through 106 million volts. 27. When a particle is travelling at an extremely high speed, say 90% of the speed of light, a lot of energy is needed to increase the particle’s velocity by a few percent. As a result, the


mass of the particle increases by a large amount. Therefore, it may be more accurate to say that a particle accelerator increases the mass of electrons rather than their speed.

Chapter 14 1. Every atom of the same element has the same number of protons, and the number of protons in the nucleus, Z, determines the chemical properties of the atom. However, atoms of different isotopes of the same element have different numbers of neutrons (and thus different A values), which results in different physical properties such as nuclear stability or decay. 2. Many elements are composed of several naturally occurring isotopes, each with a different atomic mass number, A. The weighted average of the isotopes’ mass numbers often results in a non-integral value for the atomic mass of that element. 3. Each nuclear isotope has a unique total binding energy determined by its nuclear structure. This binding energy is equivalent to the mass difference between the nucleus and its constituent nucleons (protons and neutrons) according to E mc2. 4. The missing mass was converted to energy of various forms such as gamma radiation emitted during the formation of the deuterium atom. 5. Your body, composed of many elements, likely has more neutrons than protons, since stable atoms with A > 20 have more neutrons than protons. 6. During a nuclear reaction, nucleons may be converted from one type to another, such as neutrons to protons in beta decay. However, the total nucleon number is conserved or remains constant. On the other hand, various forms of energy may be absorbed or emitted, resulting in an equivalent change in mass. 7. The average binding energy per nucleon is greater in the more stable isotopes because it is the “glue” holding the nucleons together, or the average amount of energy needed to break them apart.

8. During alpha decay of a uranium-238 nucleus, N for example, the ratio of the parent nucleus Z 146 is or about 1.59, and the ratio of the 92 144 N 2 daughter nucleus, , is or about 90 Z2 1.60. This leads to greater nuclear stability by reducing the electrical repulsion of the protons relative to the nuclear attraction of N nucleons. During beta decay, the ratio of Z 146 the parent nucleus, or about 1.59, is 92 greater than the ratio of the daughter nucleus, N1 145 , which is or about 1.56. 93 Z1 Although the greater ratio of protons to neutrons in the daughter tends to increase the electrical repulsive forces, the beta-decay process can lead to greater nuclear stability through the pairing of previously unpaired neutrons or protons in the nuclear shells. 9. During alpha decay, the daughter nucleus has a mass, M, that is much larger than the mass of the alpha particle, m. Since momentum is conserved, the velocity of the daughter nucleus, v, is much smaller than the velocity of the alpha particle, V (Mv mV). Therefore, the kinetic energy of the alpha particle, 0.5mV2, is much greater than that of the daughter nucleus, 0.5Mv2. 10. If an alpha particle had enough initial kinetic energy to contact a gold nucleus then a nuclear process such as fusion or fission could occur, because at that closeness the shortrange nuclear force would overpower the electrical force of proton repulsion that is responsible for scattering. 11. a) p b) c) d) e) 12. The strong nuclear force differs from the electrical force in that: (i) the strong nuclear force is very short-range, acting over distances of only a few femtometres (1015 m); (ii) the strong nuclear force is much stronger than the electrical force over nuclear distances of 1 or 2 fm; (iii) the strong nuclear force does


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13.

14.

15. 16.

17. 18.

19.

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1 not vary with distance r as 2 as does the r electrical force; and (iv) the strong nuclear force is attractive only, acting between all nucleons (protonproton, protonneutron, and neutronneutron). The rate of decay of radioactive isotopes was not affected by combining them in different molecules or by changing the temperature. These changes usually affect the rate of chemical reactions, thus radioactivity must be found deeper within the atom (in the nucleus). The nuclear force only binds nucleons that are neighbours. This short-range energy is proportional to the number of nucleons, A, in the nucleus. On the other hand, the electrical repulsion of protons is long-range and acts between all proton pairs in the nucleus. The electrical energy is therefore proportional to Z2. Repulsion would overcome attraction in a larger nucleus if there were not more neutrons than protons to keep the forces balanced and the nucleus stable. Alpha particles are ions, since they are helium atoms stripped of their electrons. If human life expectancy were a random process like radioactive decay then you would expect 25% of the population to live to 152 years. However, this is not the case. As humans age, their expected number of years left to live decreases. Carbon-14 undergoing beta decay results in the daughter isotope nitrogen-14. Industrialization and automobile emissions have effected changes in our atmosphere such as global warming and ozone-layer depletion. Such changes in the past 100 years may be altering the 14C:12C ratio in the air. Potassium salts are rapidly absorbed by brain tumours, making them detectable. The short half-life of potassium-42 means that the dosage decays to a safe, insignificant level quickly. The transmutation to a stable calcium salt by beta decay is not harmful to the body.

20. Aquatic creatures do not respire or breathe atmospheric gases directly. The 14C:12C ratio in the ocean is different than in the air. 21. Relics that are more than 60 000 years old have lasted more than 10.5 half-lives of carbon-14. The 14C:12C ratio in these relics is about 1500 times smaller now and is difficult to determine. 22. The more massive lead atoms scatter the radiation particles more effectively than do the less massive water molecules, and may also present a larger “target” for a high-speed electron or alpha particle. 23. Transmutation involves a change in the proton number, Z. This occurs during alpha and beta decay but does not occur during gamma decay, in which a nucleus merely becomes less energized. 24. Alpha particles are more massive than beta or gamma particles and transfer more energy to a molecule of the body during a collision. This has a much more disastrous effect upon the cells of the body. 25. Yes, 4.2 MeV of kinetic energy is sufficient for an alpha particle to overcome the electrical repulsion of the positively charged nuclei (see problem 72) and contact the nitrogen-14 nucleus, thus a nuclear interaction or process is possible. 26. The matches are as follows: gaseswind; liquidswater; plasmasfire; and solidsearth. 27. For fission to occur in naturally occurring deposits of uranium, a minimum concentration of uranium would be needed in order to sustain a source of slow neutrons necessary to maintain the fission process. This concentration is not present in uranium deposits. 28. The huge inward pull of the Sun’s gravitational field confines the solar plasma. Lacking this huge confining force on our less massive Earth, scientists instead use strong electromagnetic fields to confine plasmas. 29. In a fusion reactor, the major problem is to create the exact and difficult conditions of high temperatures and plasma concentrations


30.

31.

32.

33.

34.

35.

36.

needed to initiate a fusion process. The moment these conditions are not met, the process stops. The high temperature in fusion means that the ions have a very high speed, which allows them to approach one another very closely during collisions. If the ions’ kinetic energy is sufficient to overcome the electrical repulsion of the nuclei, and the nuclei touch, then fusion is possible. Critical mass in fission involves the existence of enough fuel so that the fast neutrons emitted during fission are slowed and absorbed within the fuel itself before they escape. In this way the reaction is sustained by a continual source of slow neutrons. Natural uranium is not concentrated enough (“it’s too wet”) to provide the critical mass needed to slow down any fast neutrons (“the spark needed”) and capture them to create a sustainable reaction. A bubble chamber is superheated almost to the point of instability. When a charged particle passes through, it triggers the formation of a fine stream of bubbles in its wake. Neutral particles such as neutrons carry no electric field and leave no visible tracks in the chamber. High-energy accelerators provide ions with enough kinetic energy that each ion’s total energy, E mc2, becomes many times greater than its rest mass. In a collision there is a probability that this energy could be converted to a massive elementary particle. In the high-energy accelerator at UBC, the strong nuclear force, acting over a very brief period of time (1023 s) during collisions, produces pi-mesons or pions. The weak nuclear force is usually masked by the stronger (by a factor of 103) electromagnetic force or (by a factor of 105) strong nuclear force, unless these forces are forbidden. Any process involving the neutrino, such as beta decay, involves the weak force. The neutrino reacts rarely, or weakly, with other elementary particles over a longer time span (108 s) compared with the shorter interaction

37.

38.

39. 40.

41.

42.

times (1023 s) of the strong nuclear force. Gravitational interactions are the weakest of the four forces. At elementary particle distances the gravitational force is 1040 times as great as the strong nuclear force. For this reason the graviton or messenger of the gravitational force is extremely difficult to detect. The weak force is 103 times as great as the electromagnetic force at elementary particle distances. The weak force is mainly involved in neutrino interactions or processes where the electric and strong forces are forbidden. The exchange bosons of the weak force are W and Z bosons of mass 80 GeV/c 2 and 91 GeV/c 2 respectively, as compared to the photons of the electric force. The range of the weak force is about 1018 m, compared to infinity for electromagnetism. The weak force acts on both leptons (particles not affected by the strong force, such as electrons) and hadrons (particles affected by the strong force), whereas electromagnetism acts only on charged particles. Strong nuclear processes are the fastest (or shortest), with a lifetime of about 1023 s. A high-energy particle travels close to the speed of light, c 3 108 m/s. Thus, in a strong nuclear interaction of 1023 s, the cloud-chamber track would be 3 1015 m, too small to measure. No, a heavier, unstable version of the electron, the tauon or tau, , has a mass of 1777 MeV/c 2, greater than the mass of the proton or neutron (931.5 MeV/c 2). Since gluons, the quanta of the quark force field, carry one colour and one anti-colour, there should be 32 9 possible combinations (rR, rB, rG, bR, bB, bG, gR, gB, and gG). However, the three colour-neutral gluons (rR, bB, and gG) must be handled differently because of what are known as symmetry laws. For this reason only two possible neutral couplings exist, not three, making a total of eight colour gluons to act as the source of quarkquark interactions.


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PART 3 Solutions to End-of-chapter Problems Chapter 1 16. a) Distance is a scalar, so your total distance travelled would be 10 20 m 200 m. b) Displacement is a vector, and since you end up 0 m from where you started, your displacement is 0 m. 17. a) Since distance is a scalar, his total distance travelled would be: d |15 m [E]| |6.0 m [W]| |2.0 m [E]| d 15 m 6.0 m 2.0 m d 23 m b) Since displacement is a vector, his total displacement would be: 15 m [E] 6.0 m [E] 2.0 m [E] d 11 m [E] d 9.8 m 100 cm 0.394 in 0.083 ft 18. g 1 s2 1m 1 cm 1 in 2 g 32 ft/s 10 nm 6080 ft 12 in 19. a) 10 knots 1h 1 nm 1 ft 5 1.0 10 km 2.54 cm 1 in 1 cm 10 knots 18.5 km/h b) from a), 18.5 km 1000 m 10 knots 1h 1 km 4 2.78 10 h 1s 10 knots 5.14 m/s 20. To find the number of centimetres in one light year, simply express the speed of light in centimetres per year: 3.0 108 m 100 cm 60 s 1m 1 min 1s 60 min 24 h 365 d 1h 1d 1y 17 9.5 10 cm/y Therefore, there are 9.5 1017 cm in one light year.

21. Catwoman: d vavg t 100 m vavg 15.4 s vavg 6.5 m/s Robin: d vavg t 200 m vavg 28.0 s vavg 7.1 m/s 22. a) The speed of the sweep second hand at the 6 o’clock position is the same as anywhere else on the clock: d vavg t 2r vavg t 2(0.02 m) vavg 60 s vavg 2.1 103 m/s b) The velocity of the second hand at the 6 o’clock position is 2.1 103 m/s [left] because the velocity is always tangent to the face and perpendicular to the hand. 23. a) The time it would take the shopper to walk up the moving escalator is: d t vt where vt is the sum of the velocity of the escalator and the woman: d d vt 15 s 8.0 s 23d vt 120 s d Therefore, t 23d 120 s t 5.2 s

Solutions to End-of-chapter Problems

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b) In this case, vt is equal to her walking speed minus the speed of the escalator: d d vt 8.0 s 15 s 7d vt 120 s Since this velocity is positive, she could walk down the escalator. Also, intuitively, if the escalator takes 15 s to go the same distance that the woman can in 8 s, then she is faster and will therefore make it down the escalator. To find how long it will take her, solve for time: d t 7d 120 s t 17 s 24. Because the rabbit accelerates at a constant rate, v2 v1 at v2 (0.5 m/s) (1.5 m/s2)(3.0 s) v2 5.0 m/s 25. Mach 1 332 m/s Mach 2 2(332 m/s) Mach 2 664 m/s Because the jet accelerates at a constant rate: v2 v1 at (v2 v1) t a (664 m/s 332 m/s) t 50 m/s2 t 6.6 s (v2 v1) 26. a t (25 m/s [E] 15 m/s [W]) a 0.10 s (25 m/s [E] 15 m/s [E]) a 0.10 s 2 a 400 m/s [E] 27. Let t be the time when the two friends meet. Let x be the distance travelled by the second friend to reach the first friend. For the first friend: d v t d vt

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and d 50 x Therefore: x 50 vt For the second friend: at2 x v1t 2 at2 x 2 Now we set these two expressions for x equal to each other and solve for time: at2 50 vt 2 2 t t 100 0 1 1 4( 100) t 2 t 9.5 s 28. a) v2 v1 at and v1 equals zero, so v2 at v t 2 a 60 km/h t 10 km/h/s t 6.0 s b) To find Batman’s distance travelled, we must first convert his acceleration into standard SI units: 1000 m 10 km 1h 1 h s 3600 s 1 km 2.78 m/s2 Now: at2 d 2 (2.78 m/s)(6.0 s)2 d 2 d 50 m c) Robin’s speed in SI units is: 60 km 1h 1000 m 1h 3600 s 1 km 16.7 m/s When Batman catches up with Robin, Robin will have travelled: d v1t d (16.7 m/s)t relative to Batman’s initial position.


Similarly, Batman will have travelled: at2 d 2 d (1.39 m/s2)t2 Setting these two expressions equal and solving for t gives: ( 16.7 m/s)t (1.39 m/s2)t2 0 t 12.0 s 29. If the child catches the truck, she will have travelled 20 m d, and the truck will have travelled d in the same amount of time, t. For the truck: at2 d 2 d (0.5 m/s2)t2 For the child: (d 20 m) vavg t d (4.0 m/s)t 20 m Setting these two equations equal and solving for t gives: (4.0 m/s)t 20 m (0.5 m/s2)t2 t2 8t 40 0 This expression has no real roots, therefore the child will not catch the truck. 30. a) After ten minutes, the runner has gone (4000 m 800 m) 3200 m, at a speed of: d vavg t 3200 m vavg 600 s vavg 5.33 m/s If she then accelerates at 0.40 m/s2 for the final 800 m, it will take: at2 d v1t 2 800 m (5.33 m/s)t (0.20 m/s2)t2 (5.33)2 4( 0.20)( 800) 5.33 t 2(0.20) t 51 s b) Since she had two minutes to go, she will finish under her desired time. 31. We can use the information given to find the speed of the flower pot at the top of the window, and then use the speed to find the height

above the window from which the pot must have been dropped. Since the pot accelerates at a constant rate of 9.8 m/s2, we can write: at2 d v1t 2 d at v1 t 2 (9.8 m/s2)(0.20 s) 19 m v1 0.20 s 2 v1 8.5 m/s Now we can find the distance above the window: v12 vo2 2ad (v12 vo2) d 2a ((8.5 m/s)2 (0 m/s)2) d 2(9.8 m/s2) d 3.7 m 32. a) The only force acting on the ball while it is falling is that of gravity, so its acceleration is 9.8 m/s2 downward. b) Since the ball is being constantly accelerated downward, it cannot slow down. at2 c) d v1t 2 d (8.0 m/s)(0.25 s) (9.8 m/s2)(0.25 s)2 2 d 2.3 m at2 33. d v1t 2 (4.0 m) (4.0 m/s)t (9.8 m/s2)t2 2 2 2 (4.9 m/s )t (4.0 m/s)t 4.0 m 0 16 4 (4.9)( 4) 4 t 9.8 t 1.4 s 34. For the first stone, the distance it falls before reaching the second stone is: at2 h d 2 d (4.9 m/s2)t2 h


89

For the second stone, its distance travelled is given by: at2 d vit 2 d vit 4.9t2 Setting these expressions equal to each other and solving for t gives: 4.9t2 h vit 4.9t2 h t vi 35. a) Because the jackrabbit’s distance vs. time is changing at a constant rate during segments B, C, and D, he is undergoing uniform motion at these times. b) Because the jackrabbit’s distance vs. time is not changing at a constant rate during segment A but is increasing exponentially, his velocity vs. time must be increasing at a constant rate, and he is undergoing uniform acceleration during this segment. c) The average velocity during segment B is: d vavg t 150 m 100 m vavg 20 s 10 s vavg 5 m/s During segment C: d vavg t 150 m 150 m vavg 30 s 20 s vavg 0 m/s During segment D: d vavg t 50 m 150 m vavg 50 s 30 s vavg 10 m/s d d) vavg t 150 m 0 m vavg 17.5 s 1.0 s vavg 9.1 m/s

90

e) Since the jackrabbit’s displacement is not changing between 20 s and 30 s, his velocity over this interval, and at 25 s, is 0 m/s. f) The jackrabbit is running in the opposite direction. g) The jackrabbit’s displacement is: d 100 m 50 m 0 m 120 m d 30 m 36. a) The car’s acceleration for each segment can be found by taking the slope of the graph during that segment: During segment A: v a1 t (5 m/s 0 m/s) a1 (5 s 0 s) a1 1 m/s2 During segment B: v a2 t (13 m/s 5 m/s) a2 (9 s 5 s) a2 2 m/s2 During segment C: v a3 t (1 m/s 13 m/s) a3 (15 s 9 s) a3 2 m/s2 b) The car is slowing down, or decelerating. c) To find the distance travelled by the car, we must find the area under the graph, which can be approximated by the sum of rectangles and triangles: vt A: d1 2 (5 m/s 0 m/s)(5 s 0 s) d1 2 d1 12.5 m vt B: d2 v1t 2 (12.5 m/s 5 m/s)(9 s 5 s) d2 2 (5 m/s)(9 s 5 s) d2 35 m


vt C: d3 v2t 2 (1 m/s 12.5 m/s)(15 s 9 s) d3 2 (1 m/s)(15 s 9 s) d3 40.5 m dtotal d1 d2 d3 dtotal 12.5 m 35 m 40.5 m dtotal 88 m NOTE: The solutions to problem 37 are based on the velocity axis of the graph reading 60, 40, 20, 0, 20, 40, 60. 37. a) Because the skateboarder has a positive velocity between 0 and 5 seconds, this portion of the graph must describe his upward motion. b) Since the skateboarder has a negative velocity from 5 to 10 seconds on the graph, he must be descending during this portion of the graph. c) The skateboarder is undergoing uniform acceleration. d) The skateboarder is at rest when his velocity equals zero, at t 5 s. When his velocity equals zero, he is at the top of the side of the swimming pool, or ground level. e) The skateboarder’s acceleration can be found from the slope of the graph. It should be equal to g: v a t (50 m/s 50 m/s) a (10 s 0 s) a 10 m/s2 38. a) At t 4.0 s, each Stooge’s acceleration is: Curly: v a t a 0 m/s2 Larry: v a t (10 m/s 0 m/s) a (4.0 s 0 s) a 2.5 m/s2 Moe:

v a t (20 m/s 0 m/s) a (4.0 s 0 s) a 5.0 m/s2 b) To find their distance travelled, we take the area under the graph for each Stooge: Curly: d vt d (25 m/s)(4.0 s) d 100 m Larry: vt d 2 (10 m/s)(4.0 s) d 2 d 20 m Moe: vt d 2 (20 m/s)(4.0 s) d 2 d 40 m c) Since Curly is travelling at a constant velocity: d v t d t v (600 m) t (25 m/s) t 24 s Larry accelerates for the first 18 s of the race. His distance travelled at this point is: vt d 2 (45 m/s)(18 s) d 2 d 405 m


91

Then, travelling at a constant velocity of 45 m/s, he will traverse the last 195 m in: d t v (195 m) t (45 m/s) t 4.3 s His total time is: 18 s 4.3 s 22.3 s Moe accelerates for the first 8 s of the race. His distance travelled in this time is: vt d 2 (40 m/s)(8.0 s) d 2 d 160 m Then, travelling at a constant velocity of 40 m/s, he will traverse the last 440 m in: d t v (440 m) t (40 m/s) t 11 s His total time is: 8 s 11 s 19.0 s Therefore, Moe wins with the fastest time of 19.0 s. 39.

Ff

Fg

41.

Fsupport

Baby

Fg

42.

Fn

Textbook

Ff

Fg

Fg

40.

F2,1

Box #2

43. a)

Fbat

Ball

Fn

Ftension

The gravitational force downward is equal in magnitude to the tension in the elevator cable.

Elevator Fn Fg

Ff F 1,2

Box #1

Fg

Fapplied

b)

Ftension

Elevator

Fg

92


The gravitational force downward is equal in magnitude to the tension in the elevator cable.

c)

The acceleration downward is 2 Elevator 9.8 m/s .

Fg

d)

Fn

The gravitational force downward is equal in magnitude to the normal force upward.

Car

Fg

e)

Fn

F-14

Fg

Fcatapult

The gravitational force is equal in magnitude to the normal force, and the force due to the catapult must be large in order to accelerate the jet.

44. The driver’s initial velocity is the same as that of the car: 1000 m 50 km 1h 1 min v1 1h 60 min 60 s km v1 13.9 m/s His final velocity is zero, and the distance he travels is 0.6 m: v22 v12 2ad (v22 v12) a 2d ((0 m/s)2 (13.9 m/s)2) a 2(0.6) a 161 m/s2 a 16.4 g

45. Fnet ma, and (v22 v12) a , so 2d m(v22 v12) Fnet 2d (10 000 kg)[(150 m/s)2 (100 ms)2] Fnet 2(1000 m) 4 Fnet 6.2 10 N 46. Fnet ma Fnet a m 400 N a 200 kg a 2.0 m/s2 We also know that: (v2 v1) t a (0 m/s 0.5 m/s) t 2.0 m/s2 t 0.25 s 47. Since F ma(6.0 m/s2) and F mb(8.0 m/s2), it follows that: ma(6.0 m/s2) mb(8.0 m/s2) mb 0.75ma If the same force were used to accelerate both masses together, we would have: F (ma mb)a F (ma 0.75ma)a F 1.75ama But we already know that F ma(6.0 m/s2), so we now have: ma(6.0 m/s2) 1.75ama a 3.4 m/s2 48. The force applied by the hammer is given by: Fnet ma But we also know that: at2 d 2 2d a t2


93

Therefore: 2md Fnet t2 2(1.8 kg)(0.013 m) Fnet (0.10 s)2 Fnet 4.7 N Therefore, the force applied to the nail by the hammer is 4.7 N and the force applied to the hammer by the nail is 4.7 N. 49. The force due to the cows on the plate is: Fnet (m1 m2 m3 m4 m5)g Fnet 5(200 kg)(9.8 m/s2) Fnet 9800 N From Newton’s third law, the steel plate exerts a force of 9800 N upward. 50. a) The acceleration of the water skiers can be found using: Fnet ma Fnet a (m1 m2 m3) 10 000 N a (75 kg 80 kg 100 kg) a 39.2 m/s2 b) The force applied by the first skier on the second two skiers is equal to the sum of their masses times their acceleration: Fnet mt a Fnet (75 kg 80 kg)(39.2 m/s2) Fnet 6.1 103 N The force applied by the third skier on the first two skiers is equal to his mass times his acceleration: Fnet mt a Fnet (75 kg)(39.2 m/s2) Fnet 2.9 103 N From Newton’s third law, the forces applied by the second skier on the first and third skiers are 6.1 103 N and 2.9 103 N, respectively.

94

51. The forces in the vertical direction are balanced: Fg mg Fg Fn, therefore: Ff Fn Ff mg Ff (0.16)(2.0 kg)(9.8 m/s2) Ff 3.1 N 52. Fnet Ff Fnet ma Fnet Fn (v22 v12) We also know that a , therefore: 2d m(v22 v12) Fn 2d m(v22 v12) mg 2d (v22 v12) d 2 g [(0 m/s)2 (2.0 m/s)2] d 2(3.0)(9.8 m/s2) d 6.8 102 m Fengine Ffriction 53. Fnet Fengine Fnet Ffriction Fnet ma Ffriction Fn Fengine ma Fn Fengine ma mg (v2 v1) a , therefore: t m(v2 v1) Fengine mg t (800 kg)(27.8 m/s 13.9 m/s) Fengine 6.0 s (0.3)(800 kg)(9.8 m/s2) Fengine 4.2 103 N Gm1m2 54. Fg r2 (6.67 1011 N m2/kg2)(300 000 kg)2 Fg (1000 m)2 Fg 6.0 106 N Their acceleration would be: F a g m (6.0 106 N) a (300 000 kg) a 2.0 1011 m/s2


55. If the mass of Earth where doubled, the acceleration due to gravity would be: Gm12mE Fg r2 Gm12mE m1 g r2 G2mE g r2 g

2(6.67 1011 N m2/kg2)(5.97 1024 kg) (6.38 106 m)2

g 19.6 m/s2 56. The net gravitational force on planet Z would be equal to the sum of the gravitational forces caused by each planet: net Fx Fy F Gmzmx Gmzmy Fnet rzx2 rzy2 Fnet (6.67 1011 N m2/kg2)(5.0 1024 kg) 3.0 1024 kg (6.0 1010 m 5.0 1010 m)2

4.0 1024 kg (5.0 1010 m)2 Fnet 6.16 1017 N 57. The astronaut’s weight, or his mass times the acceleration due to gravity, is: GmAmE W rAE2

W

(6.67 1011 N m2/kg2)(100 kg)(5.98 1024 kg) (6.38 106 m 3.0 105 m)2

W 894 N

Chapter 2 14. a) Horizontal: dx (25 km) cos 20° x 23 km [E] d Vertical: dy (25 km) sin 20° y 8.6 km [N] d b) Horizontal: Fx (10 N) sin 30° x 5.0 N [E] F Vertical: Fy (10 N) cos 30° y 8.7 N [S] F c) Horizontal: ax (30 m/s) cos 45° x 21 m/s2 [W] a Vertical: ay (30 m/s) sin 45° y 21 m/s2 [S] a d) Horizontal: px (42 kg·m/s) sin 3° x 2.2 kg·m/s [W] p Vertical: py (42 kg·m/s) cos 3° y 42 kg·m/s [N] p 15. a) l (10 m) cos 40° l 7.7 m b) h (10 m) sin 40° h 6.4 m 16. Horizontal: ax (4.0 m/s2) cos 35° ax 3.3 m/s2 Vertical: ay (4.0 m/s2) sin 35° ay 2.3 m/s2 17. Adding by components:

d (2.0 km) (3.0 km) cos 20° d 4.8 km [W] d (3.0 km) sin 20° d 1.0 km [N] (4.8 k m) (1.0 km ) d d 4.9 km x x y y

2

1.0 km tan 1 4.8 km

2

12° Therefore, d 4.9 km [W12°N]. 2 m/s) (20 m/s) 2 18. |vi| (10 |vi| 22 m/s 20 m/s tan 1 10 m/s 63° Therefore, vi 22 m/s inclined 63° to the horizontal.


95

19. Adding by components:

d d d

x

(50 cm) sin 35° (100 cm) cos 15°

x

67.9 cm

y

(20 cm) (50 cm) cos 35° (100 cm) sin 15°

d 46.8 cm cm) (46.8 cm) d (67.9 d 82 cm y

2

2

67.9 cm tan 1 46.8 cm 55° d 82 cm [S55°W]. Therefore,

20. v vf vi v 28 m/s [N30°W] 30 m/s [S] v 28 m/s [N30°W] 30 m/s [N] Adding the vectors by components, v 56 m/s [N15°W] 21. v vf vi v 1.8 m/s [N30°E] 2.0 m/s [S30°E] v 1.8 m/s [N30°E] 2.0 m/s [N30°W] Adding the vectors by components, v 3.3 m/s [N2°W] v a t 3.3 m/s [N2°W] a 0.10 s a 33 m/s2 [N2°W] 22. a) The current velocity has no effect on the vertical component of the swimmer’s velocity, which is needed for crossing the river. Therefore: d t vs 0.80 km t 1.8 km/h t 0.44 h b) The current velocity determines how far the swimmer travels downstream, therefore: dd (vc)(t) dd (0.50 km/h)(0.44 h) dd 0.22 km

96

2 2 c) vg v s vc vg (1.8 k m/h)2 (0.5 km/ h)2 vg 1.9 km/h v tan c vs

0.5 km/h tan1 1.8 km/h 16° The ground velocity is vg 1.9 km/h [N16°E]. 23. a) In order to go north, his ground velocity must be north. vc

vs

vg

θ

Since vs and vc are known, v sin c vs

0.5 km/h sin1 1.8 km/h 16° The swimmer must swim [N16°W] in order to go straight north. 2 vs2 v b) vg c vg (1.8 k m/h)2 (0.5 km/ h)2 vg 1.7 km/h His ground velocity is vg 1.7 km/h [N]. d c) t vg 0.8 km t 1.7 km/h t 0.46 h


24. The time it takes the sandwich to reach the road is: d t p vs 10 m t 2.0 m/s t 5.0 s The distance of the pick-up truck when the sandwich is released is: dt (vt)(t) dt (60 km/h)(5.0 s) dt (17 m/s)(5.0 s) dt 83 m 25. v g

vw

θ

vh

vw cos vh

20 km/h cos1 150 km/h 82° The pilot must fly [N82°E] or [E7.7°N]. 26.

γ vc β

vs

vg

θ

45°

Use the sine law to find : sin sin vc vs vc sin sin1 vs 6.8° Use the sum of the interior angles of a triangle to find : 180° 180° 180° 135° 6.8° 38° The ship’s required heading is [N38°E].

27. a) Since the velocities are given relative to the deck, they are the velocities relative to the ship. Walking towards stern: v 0.5 m/s [S]; walking towards port, v 0.5 m/s [W]. b) velocity of velocity of velocity of pass. relative to water

pass. relative to ship

ship relative to water

vps vsw vpw Walking towards stern: vpw vps vsw vpw 0.5 m/s [S] 10 km/h [N] vpw 0.5 m/s [N] 2.78 m/s [N] vpw 2.3 m/s [N] Walking towards port: vpw vps vsw vpw 0.5 m/s [W] 2.78 m/s [N] vps2 vsw2 vpw 2 vpw (0.5 m/s) (2.78 m/s)2 vpw 2.8 m/s vps tan vsw

0.5 m/s tan1 2.78 m/s 10° Walking towards stern, v 2.3 m/s [N]; walking towards port, v 2.8 m/s [N10°W]. d 28. a) vf g t dg t vf 6.0 m t 5.0 m/s t 1.2 s To reach the pail, the quarterback must be 1.2 s away from reaching the garbage pail, therefore: dqg (vq)(t) dqg (4.0 m/s)(1.2 s) dqg 4.8 m The quarterback must release the ball 4.8 m in advance. b) 1.2 s as calculated in part a. 2 2 c) vg v f vq 2 vg (5.0 m/s) (4.0 m/s) 2 vg 6.4 m/s


97

v tan f vg

5.0 m/s tan1 4.0 m/s 51° The ground velocity is vg 6.4 m/s [E51°N]. 29. a) In order for the football to reach the garbage pail, the football’s ground velocity must be pointing north at the time of release. v cos q vf 4.0 m/s cos1 5.0 m/s 37° The ball must be thrown [W37°N]. b) Calculate the magnitude of vg: vf2 vg2 vq2 vg2 vf2 vq2 2 2 vg v f vq 2 vg (5.0 m/s) (4.0 m/s) 2 vg 3.0 m/s d vg t d t vg 10 m t 3.0 m/s t 3.3 s c) The ball is thrown such that its direction is north. The ground velocity is vg 3.0 m/s [N]. 30. The time it takes the ball to reach the ground is: 1 h vi t ayt2 2 1 10 m (0 m/s)t (9.8 m/s2)t2 2 t 1.4 s The horizontal distance travelled in 1.4 s is: 1 dx vi t axt2 2 1 dx (3.0 m/s)(1.4 s) (0 m/s2)(1.4 s)2 2 dx 4.2 m The friend must be 4.2 m away to catch the ball at ground level.

y

x

98

31. a) Find the time it takes the rock to reach the ground: 1 dx vi t axt2 2 1 20.0 m (10.0 m/s)t (0 m/s2)t2 2 t 2.00 s Find the height of the water tower: 1 h vi t ayt2 2 1 h (0 m/s)(2.00 s) (9.8 m/s2)(2.00 s)2 2 h 19.6 m b) In the horizontal direction, vf vi axt vf 10.0 m/s (0 m/s2)(2.00 s) vf 10.0 m/s In the vertical direction, vf vi ayt vf 0 m/s (9.8 m/s2)(2.00 s) vf 19.6 m/s 2 2 vf v f v f 2 vf (10.0 m/s) (19.6 m/s )2 vf 22.0 m/s vf tan vf x

y

x

x

x

x

y

y

y

y

x

y

y

x

19.6 m/s tan1 10.0 m/s 63.0° The rock’s final velocity is 22.0 m/s, 63° below the horizontal. 32. Find the time it takes the mail to reach the second building: 1 dx vi t axt2 2 1 100 m [(20 m/s) cos 15°]t (0 m/s2)t2 2 t 5.2 s Find the drop in height during the 5.2 s: 1 h vi t ayt2 2 h [(20.0 m/s) sin 15°](5.2 s) 1 (9.8 m/s2)(5.2 s)2 2 h 26.9 m 132.5 m h 105 m


x

y

Find the height of the second building: h2nd building h1st building h h2nd building 200 m 105 m h2nd building 95 m The second building is 95 m high. 1 33. a) h vi t ayt2 2 1 1.3 m (0 m/s)t (9.8 m/s2)t2 2 t 0.52 s b) The cup lands at the tourist’s feet, since both the cup of coffee and tourist are not moving horizontally relative to the train. c) dx (vtrain)(t) dx (180 km/h)(0.52 s) dx (50 m/s)(0.52 s) dx 26 m The train is 26 m closer to Montreal. 34. Find the time it takes the Humvee to drop down to the other ramp: 1 h vi t ayt2 2 Since both ramps are the same height, h 0 m. 0 m [(30 m/s) sin 20°]t 1 (9.8 m/s2)t2 2 1 [(30 m/s) sin 20°]t (9.8 m/s2)t2 2 t 2.1 s Find the maximum horizontal distance the Humvee can travel in 2.1 s: 1 dx vi t axt2 2 dx [(30 m/s) cos 20°](2.1 s) 1 (0 m/s2)(2.1 s)2 2 dx 59 m The maximum width of the pool is 59 m. 35. Find the time required to reach maximum height: 1 h vi t ayt2 2 1 h (vi sin )t ayt2 (eq. 1) 2 y

y

x

y

Since the ball’s motion is symmetrical, it will take twice the time for the soccer ball to reach the ground: 1 0 m (vi sin )2t ay(2t)2 2 2 ay2t (vi sin )2t ayt vi sin vi sin ay (eq. 2) t For the range, R: 1 R vi 2t ax(2t)2 2 1 R vi 2t (0 m/s2)(2t)2 2 R vi 2t R (vi cos )2t R t (eq. 3) 2vi cos Substitute equation 1 into equation 2: 1 vi sin 2 h (vi sin )t t t 2 1 h (vi sin )t (vi sin )t 2 1 h (vi sin )t (eq. 4) 2 Substitute equation 3 into equation 4, 1 R h (vi sin ) 2vi cos 2 x

x

x

1 sin h R 4 cos 1 h (tan )R 4 h 0.25R tan 36. If the ball clears the 3.0-m wall 130 m from home plate, then the ball rises (3.0 m 1.3 m) 1.7 m during this time. Thus, for the vertical height: 1 h vi t ayt2 2 1 1.7 m vi (sin 45°)t (9.8 m/s2)t2 (eq. 1) 2 Find the time it takes the ball to clear the wall: 1 dx vi t axt2 2 130 m vi (cos 45°)t 130 m t (eq. 2) vi cos 45°

y

x


99

Substitute equation 2 into equation 1: 130 m 1.7 m (vi sin 45°) vi cos 45°

130 m 2 1 (9.8 m/s2) vi cos 45° 2 1.7 m (tan 45°)(130 m) 33 800 m (4.9 m/s2) vi2 vi 36 m/s The player strikes the ball at 36 m/s, 45° above the horizontal. N )2 (10 N)2 37. a) Fnet (30 Fnet 32 N 30 N tan1 10 N 72° net 32 N [N72°E] So F b) Horizontal components:

F F

x

(60 N) sin 40°

x

38.6 N [W]

Vertical components:

F F

y

(60 N) cos 40° 80 N

y

34.0 N [S]

net 51 N [S49°W] F c) Horizontal components:

F F

x

(50 N) cos 60° 10 N

x

15 N [E]

Vertical components:

F F

y

(50 N) sin 60° 60 N

y

16.7 N [S]

net 22 N [S42°E] F net Fa Ff 38. a) F The sum of the x components is: a F1 F2 F3 F Fa (100 N) cos 20° [W] (200 N) cos 40° [E] Fa 59 N [E] The sum of the y components is: a F1 F2 F3 F Fa (100 N) sin 20° [N] (200 N) sin 40° [S] 300 N [S] Fa 394 N [S] x

x

x

x

y

y

y

x

Fa 2 Fa 2 Fa Fa (59 N )2 (394 N)2 Fa 399 N Fa tan Fa x

x

y

x

Find the kinetic frictional force, Fk: Fk kFn Fk (0.30)(Fg Fa sin 50°) Fk (0.30)[(20 kg)(9.8 m/s2) (100 N) sin 50°] Fk 36 N Fnet (100 N) cos 50° 36 N Fnet 28 N Fnet ma 28 N ma 28 N a 20 kg a 1.4 m/s2

y

y

100

394 N tan1 59 N 8.5° Fa 399 N [S8.5°E] Ff kFn Ff kmg Ff (0.10)(300 kg)(9.8 m/s2) Ff 294 N f 294 N [N8.5°W] F Fa Ff Fnet net 399 N [S8.5°E] 294 N [N8.5°W] F Fnet 399 N [S8.5°E] 294 N [S8.5°E] Fnet 105 N [S8.5°E] net 105 N [S8.5°E] The net force is F net ma b) F Fnet a m 105 N [S8.5°E] a 300 kg a 0.35 m/s2 [S8.5°E] net kinetic friction Fapplied force in the x directionF 39. F Fnet Fa Fk

x

y

y


40. The hockey stick provides the only force on the puck, therefore it is the net force acting on the puck: Fs Fnet F net ma Fnet a m 300 N [N25°E] a 0.25 kg a 1200 m/s2 [N25°E] Find vf : vf vi a t at vf vi vf at vi vf (1200 m/s2 [N25°E])(0.20 s) 12 m/s [S] vf 240 m/s [N25°E] 12 m/s [S] The vector sum of the x components is: vfx (240 m/s) sin 25° [E] vfx 101 m/s [E] The vector sum of the y components is: vfy (240 m/s) cos 25° [N] 12 m/s [N] vfy 206 m/s [N] 2 2 vf F f F f 2 vf (101 m/s) (206 m/s) 2 vf 229 m/s vf tan vf x

y

x

y

206 m/s tan1 101 m/s 26° The final velocity is vf 229 m/s [N26°E]. 41. a) Fk kFn Fk (0.50)(100 kg)(9.8 m/s2) Fk 4.9 102 N b) The frictional force is the only force acting on the baseball player, therefore it is also the net force. Fnet Fk ma Fk 490 N a 100 kg a 4.9 m/s2

Find vi: vf vi a t at vi vi (4.9 m/s2)(1.3 s) vi 6.4 m/s 42. Fnet Fk ma k Fn ma k mg a k g a (0.3)(9.8 m/s2) a 2.94 m/s2 Find distance, d: vf2 vi2 2ad vi2 2ad vi2 d 2a (2.0 m/s)2 d 2(2.94 m/s2) d 0.68 m The key will slide 0.68 m across the dresser. 43. Fnet Fa Fk The horizontal acceleration of 1.0 m/s2 is the net acceleration of the mop, therefore: Fnet max max Fa kFn max (30 N) cos 45° [(0.1)(Fg Fa sin 45°)] max 21.2 N [(0.1)(mg 21.2 N)] max 19.09 N 0.1mg (1.0 m/s2)m 19.09 N 0.1mg m(1.0 m/s2 0.1g) 19.09 N m(1.98 m/s2) 19.09 N m 9.6 kg 44. Let be the angle of the inclined plane when the box starts to slide. At this angle, Fs sFn Fs (0.35)(mg cos ) (eq. 1) (eq. 2) Fx mg sin


x

101

Set equation 1 equal to equation 2: (0.35)(mg cos ) mg sin sin 0.35 cos tan 0.35 tan1 (0.35) 19° The minimum angle required is 19°. 45. a) The acceleration for child 1: Fnet Fx m1a1 m1 g sin a1 g sin a1 (9.8 m/s2) sin 30° a1 4.9 m/s2 The acceleration for child 2: Fnet Fx m2a2 m2 g sin a2 g sin a2 4.9 m/s2 Both children accelerate downhill at 4.9 m/s2. b) They reach the bottom at the same time. 46. a) Fnet Fx Fk ma mg sin kFn ma mg sin k(mg cos ) a g sin k g cos a (9.8 m/s2) sin 25° (0.45)(9.8 m/s2)cos 25° a 0.14 m/s2 The acceleration of the box is 0.14 m/s2. b) vf2 vi2 2ad vf2 2(0.14 m/s2)(200 m) vf 7.6 m/s The box reaches the bottom of the hill at 7.6 m/s2. vf vi c) a t vf t a 7.6 m/s t 2 0.14 m/s t 53 s It takes the box 53 s to reach the bottom of the hill.

102

47. Find his final speed, vf, at the bottom of the ramp by first finding his acceleration: Fnet Fx ma mg sin a g sin a (9.8 m/s2) sin 35° a 5.6 m/s2 His final speed at the bottom of the ramp is: vf2 vi2 2ad vf2 2(5.6 m/s2)(50 m) vf 23.6 m/s vf will be the initial speed, vi2, for the horizontal distance to the wall of snow. Find the deceleration caused by the snow: Fnet Fk ma kFn ma (0.50)(9.8 m/s2)m a 4.9 m/s2 Find the distance Boom-Boom will go into the wall of snow: vf2 vi2 2ad 0 vi2 2ad vi2 2ad vi2 d 2a (23.6 m/s)2 d 2(4.9 m/s2) d 57 m Boom-Boom will go 57 m into the wall of snow. 48. Find the net force on Spot, then solve for the net acceleration: Fnet Fr Fx Fnet 2000 N mg sin ma 2000 N (250 kg)(9.8 m/s2)(sin 20°) ma 2000 N 838 N ma 1162 N 1162 N a 250 kg a 4.6 m/s2


Find time, t: 1 d vit at2 2 1 2 d at 2 1 250 m (4.6 m/s2)t2 2 2 t 108 s2 t 10 s 49. a) (a) For m1: Fnet1 T (eq. 1) T m1a For m2: Fnet2 Fg T (eq. 2) m2a m2 g T Substitute equation 1 into equation 2: m2a m2 g m1a (m1 m2)a m2 g (40 kg)a (20 kg)(9.8 m/s2) 4.9 m/s2 [left] a For tension T, substitute acceleration into equation 1: T m1a T (20 kg)(4.9 m/s2) T 98 N (b) Assume the system moves towards m3: For m1: Fnet1 T1 F1g m1a T1 m1 g (eq. 1) For m2: Fnet2 T2 T1 m2a T2 T1 (eq. 2) For m3: Fnet3 F3g T2 m3a m3 g T2 (eq. 3) Add equations 1, 2, and 3: m1a T1 m1 g (eq. 1) (eq. 2) m2a T2 T1 m3a m3 g T2 (eq. 3) (m1 m2 m3)a m3 g m1 g (10 kg 10 kg 30 kg)a (30 kg)(9.8 m/s2) (10 kg)(9.8 m/s2) (50 kg)a 196 N 3.9 m/s2 [right] a

Find T1: (eq. 1) m1a T1 m1 g 2 T1 (10 kg)(3.9 m/s ) (10 kg)(9.8 m/s2) T1 137 N Find T2: (eq. 3) m3a m3 g T2 2 T2 (30 kg)(9.8 m/s ) (30 kg)(3.9 m/s2) T2 176 N (c) For m1: Fnet1 T Fx m1a T mg sin (eq. 1) For m2: Fnet2 F2g T (eq. 2) m2a m2 g T Add equations 1 and 2: (eq. 1) m1a T mg sin (eq. 2) m2a m2 g T (m1 m2)a m2 g m1 g sin (25 kg)a (15 kg)(9.8 m/s2) (10 kg)(9.8 m/s2) sin 25° 4.2 m/s2 [right] a For tension T, substitute acceleration into equation 2: m2a m2 g T T (15 kg)(9.8 m/s2) (15 kg)(4.2 m/s2) T 84 N b) (a) For m1: Fnet1 T Fk m1a T km1 g (eq. 1) For m2: Fnet2 Fg T (eq. 2) m2a m2 g T Add equations 1 and 2: m1a T km1 g (eq. 1) (eq. 2) m2a m2 g T m1a m2a m2 g km1 g a(m1 m2) g(m2 km1) (20 kg 20 kg)a 9.8 m/s2[20 kg 0.2(20 kg)] a 3.9 m/s2 [left]


103

For tension T, substitute acceleration into equation 2: m2a m2 g T T (20 kg)(9.8 m/s2) (20 kg)(3.9 m/s2) T 118 N (b) Assume the system moves towards m3: For m1: Fnet1 T1 F1g m1a T1 m1 g (eq. 1) For m2: Fnet2 T2 T1 Fk m2a T2 T1 km2 g (eq. 2) For m3: Fnet3 F3g T2 m3a m3 g T2 (eq. 3) Add equations 1, 2, and 3: m1a T1 m1 g (eq. 1) m2a T2 T1 (eq. 2) km2 g m3a m3 g T2 (eq. 3) (m1 m2 m3)a m3 g km2 g m1 g (10 kg 10 kg 30 kg)a 9.8 m/s2[30 kg 0.2(10 kg) 10 kg] a 3.5 m/s2 [right] Find T1: (eq. 1) m1a T1 m1 g 2 T1 (10 kg)(3.5 m/s ) (10 kg) (9.8 m/s2) T1 133 N Find T2: (eq. 3) m3a m3 g T2 2 T2 (30 kg)(9.8 m/s ) (30 kg)(3.5 m/s2) T2 188 N (c) For m1: Fnet1 T Fx Fk m1a T m1 g sin (eq. 1) km1 g cos For m2: Fnet2 F2g T (eq. 2) m2a m2 g T

104

Add equations 1 and 2: m1a T m1 g sin (eq. 1) km1 g cos (eq. 2) m2a m2 g T (m1 m2)a m2 g m1 g sin km1 g cos (25 kg)a (9.8 m/s2)[15 kg (10 kg) sin 25° 0.2(10 kg) cos 25°] 3.5 m/s2 [right] a For tension T, substitute acceleration into equation 2: m2a m2 g T T (15 kg)(9.8 m/s2) (15 kg)(3.5 m/s2) T 94 N 50. For m1: Fnet1 T Ff1 m1a T kFn m1a T km1 g

(eq. 1)

For m2: Fnet2 F2g T1 m2a m2 g T1 (eq. 2) Add equations 1 and 2: (eq. 1) m1a T km1 g (eq. 2) m2a m2 g T (eq. 3) (m1 m2)a m2 g km1 g 2 (9.0 kg)a (4.0 kg)(9.8 m/s ) (0.10)(5.0 kg)(9.8 m/s2) a 3.8 m/s2 [right] 51. For the system to be NOT moving, the acceleration of the whole system must be 0. Using equation 3: (eq. 3) (m1 m2)a m2 g km1 g 0 m2 g km1 g km1 g m2 g k(5.0 kg) 4.0 kg k 0.80 52. First find the system’s acceleration: For Tarzana: FnetTA T (eq. 1) mTAa T For Tarzan: FnetTZ FTZg T (eq. 2) mTZa mTZ g T


Add equations 1 and 2: (eq. 1) mTAa T (eq. 2) mTZa mTZ g T (mTA mTZ)a mTZ g (65 kg 80 kg)a (80 kg)(9.8 m/s2) (80 kg)(9.8 m/s2) a (65 kg 80 kg) a 5.4 m/s2 To find time t: 1 d vit at2 2 1 15 m at2 2 30 m t2 a t

5.4 m/s 30 m

2

t 2.4 s 42r 53. ac Assuming ac is a constant, T2

42r ac a) If the radius is doubled, the period 2. increases by a factor of b) If the radius is halved, the period decreases 2. by a factor of 42r 54. a) ac T2 42(0.35 m) ac (0.42 s)2 ac 78 m/s2 b) The clothes do not fly towards the centre because the wall of the drum applies the normal force that provides the centripetal force. When the clothes are not in contact with the wall, there is no force acting on them. The clothes have inertia and would continue moving at a constant velocity tangential to the drum. The centripetal force acts to constantly change the direction of this velocity. 42r 55. ac T2 T 365 days 3.15 104 s 42(1.5 1011 m) ac (3.15 107 s) ac 6.0 103 m/s2 T

Fc Ff mac Fn mv 2 mg r gr v v 21 m/s It is not necessary to know the mass. 57. Vertically: Fn cos mac mg Fn cos Horizontally: Fc Fn sin mac Fn sin mg sin mac cos v2 g tan r tan 25° v rg v 19 m/s 58. Fc Fg mac mg v2 g r v gr v 9.9 m/s 59. a) T mg T (0.5 kg)g T 4.9 N mv2 b) T mg r mv2 T mg r (0.5 kg)(2.4 m/s)2 T (0.6 m) (0.5 kg)(9.8 m/s2) T 9.7 N 60. Maximum tension occurs when the mass is at its lowest position. Tension acts upward, and gravity acts downward. The difference between these forces is the centripetal force: mv2 Tmax mg r 2 mv Tmax mg r (2.0 kg)(6.6 m/s)2 Tmax (2.0 kg)(9.8 m/s2) 3.0 m Tmax 49 N 56.


105

The tension is minimized when the mass is at the top of its arc. Tension and gravity both act downward, and their sum is the centripetal force: mv2 Tmin mg r mv2 Tmin mg r (2.0 kg)(6.6 m/s)2 Tmin 3.0 m (2.0 kg)(9.8 m/s2) Tmin 9.4 N 61. a) Fnet ma Fn mg m(9g) Fn 9mg mg Fn 10mg Fn 5.9 103 N v2 b) ac r v2 9g r v2 r 9g (91.67 m/s)2 r 9(9.8 m/s2) r 95 m 62. a) G 6.67 1011 N·m2/kg2, T 365 days 3.15 107 s GmEmS 4 2r mE 2 r T2 4 2r 3 mS GT 2

mS

63. On mass 2: 42r Fc m2 T2

4 r T m T 4 (L L ) T m T 2

2

2

2

2

2

1

2

2

On mass 1: 42r Fc m1 T2

42r T1 T2 m1 T2

42L1 42(L1 L2) T1 m1 m 2 T2 T2

42 T1 (m1L1 m2(L1 L2)) T2

42(1.5 1011 m)3 (6.67 1011 N·m2/kg2)(3.15 107 s)2

mS 2.0 1030 kg m b) Density of the Sun V 2.0 1030 kg 4 r 3 3 1.4 103 kg/m3 mEarth 5.98 1024 kg 5.98 1024 kg Density of Earth 4 r 3 3 5.5 103 kg/m3 1 The Sun is about as dense as Earth. 4 106

2


Chapter 3

24. 60°

60°

21.

Tcable Fstrut

30°

Fs = 2500 N

2500 N

T

Fg

T

m

mg

30°

Fg

Fs

flower pot

Fs = 2500 N

F sin 30° g T

30°

Fg T sin 30° mg T sin 30° (10 kg)(9.8 N/kg) T sin 30° T 196 N F 22. tan g Fs Fg Fs tan 98 N Fs tan 30° Fs 169.7 N Fs 170 N 23. 30°

30° 30°

T1

T

mg sin 30° Fs Fs sin 30° m g (2500 N) sin 30° m 9.8 N/kg m 128 kg 25.

Tcable 12°

12°

500 kg

Trope T1

T2 60°

Fg — 2

2 T T1 T

30°

Tcable

mg

Fg

Fg

60° mg

T2

2 cos 30° Fg T

2 T Fg

(cos 30°)

2 (100 kg)(9.8 N/kg)

Trope

mg cos 12° Tcable mg Tcable cos 12° (500 kg)(9.8 N/kg) Tcable cos 12° Tcable 5009.5 N Tcable 5.01 103 N Frope tan 12° mg Frope mg tan 12° Frope (500 kg)(9.8 N/kg) tan 12° Frope 1.04 103 N

T (cos 30°) T 566 N Solutions to End-of-chapter Problems

107

27.

26. a)

Fapp

Ff

100 kg

mg

θ

1.5 m tan 25.0 m 2

250 kg d

T θ

250 kg

Fapp

tan 0.12 6.8° Fapp sin T Fapp T sin 425 N T sin 6.8° T 3.59 103 N The rope pulls with a force of 3.59 103 N.

L = 10 m

mg

θ

car

b)

T

T

Fn

425 N θ

0.63 Fapp Ff 0 Fapp Ff Fapp Fn Fapp mg Fapp 0.63(100 kg)(9.8 N/kg) Fapp 617.4 N

10 m

25.0 m ——— 2

1.5 m

25.0 m

θ

Fapp

1.5 m

Fn

28. T

d

Fapp = 617 N

Bird

mg

Using similar triangles, find T first: T 2 (mg)2 Fapp2 2 T [(250 kg)(9. 8 N/kg)] ( 617.4 N) 2 T 2526.6 N T 2.53 103 N d Fapp L T FappL d T (617.4 N)(10 m) d 2526.6 N d 2.4 m

108

T

mBg 18.0 m

9.0 m θ 0.52 m

0.52 m tan 9.0 m tan 0.058 3.3°


30.

T

Th θ θ = 3.3°

mBg

Ff

Pulley

Tv T

T

2 sin mBg

T

T 2T sin mB g 2(90 N) sin 3.3° mB 9.8 N/kg mB 1.1 kg

29.

mg

Th

T

L – 2

T1 x – 2

80°

mLg

40° 40° L – 2

T2

Th Ff 0 With left taken to be the positive direction, Th Ff 0 Th Ff Th Fn mg Th 2 From Pythagoras’ theorem: mg 2 T2 Th2 2

T1

Leg T2

T (5.0 kg)(9.8 N/kg) T 49 N 40° T2

T1

mg mg T 2 2 mg T ( 1) 2 mg T 2 ( 1) 2

40°

2

Fapp

app T 1 T 2 F F app 2 cos 40° T Fapp 2(T cos 40°) Fapp 2mg cos 40° Fapp 2(5.0 kg)(9.8 N/kg) cos 40° Fapp 75 N [left]

2

2

2

2

2

2

From similar triangles: x 2 T h L T 2 x T h L T ThL x T


109

b)

Substituting for Th and T, mgL 2 x mg 2 1 2 L x 2 1 L x 2 1 31. a)

mT = m1 + m2

mTg +

P

2 1 kg

1 3 kg 2.0 m

centre of mass ? net 0 With clockwise as the positive rotation, 1 2 0 1 2 r1m1 g sin r2m2 g sin m2g r1 r2 m1g

m r r m 2

1

2

1

r r1 2 3 But r2 r1 rT 3r1 rT r1 4r1 rT rT r1 4 2.0 m r1 4 r1 0.5 m The centre of mass is 0.5 m from m1 and 1.5 m from m2.

110

T

T? If up is positive, T mTg T (4.0 kg)(9.8 N/kg) T 39.2 N [up] 32. T 0 The pivot is the left support. 1 0 2 Board Duck 0 2 B D 2 rBFgB rDFgD 2 rBmBg rDmDg 2 (2.0 m)(50 kg) (9.8 N/kg) (4.0 m) (8.5 kg)(9.8 N/kg) 2 1313.2 N/m 1313.2 Nm F2 0.8 m F2 1641.5 N 2 1.6 103 N [up] F For F1: FT 0 With down as positive, 0 F1 F2 FB FD F1 FB FD F2 F1 (mBg) (mDg) F2 F1 (50 kg)(9.8 N/kg) (8.5 kg)(9.8 N/kg) 1.6 103 N F1 1068.2 N 1 1.1 103 N [down] F and F2 1.6 103 N [up]


y2 = 0.5 m

y1 = 1.0 m

33.

X1 = 0.5 m X2 = 2.5 m

x1 x2 xcm 2 0.5 m 2.5 m xcm 2 xcm 1.5 m [right] y1 y2 ycm 2 0.5 m 1.0 m ycm 2 ycm 0.75 m [up] Centre of mass 1.5 m [right], 0.75 m [up]

x 5.0 m 3.75 m x 1.25 m 35. T 0 man L(left) L(right) rock 0 With clockwise as the positive direction of rotation, 0 man L(left) L(right) rock rock man L(left) L(right) rrockmrock g sin rmanmman g sin rL(left)mL(left) g sin rL(right)mL(right) g sin rrockmrock [(1.90 m)(86 kg)] 1.90 m 2

1.90 m (2.0 kg) 2.40 m 0.5 m 0.50 m (2.0 kg) 2 2.40 m

rrockmrock 163.4 kg·m 1.504 kg·m 0.104 kg·m rrockmrock 164.8 kg·m 164.8 kg·m mrock 0.50 m mrock 329.6 kg mrock 3.3 102 kg

34.

x 5.0 m

F23

F1

2.5 m

36. a)

17 kg 1

20 kg 3

27 kg 2 +

P

x

3.8 kg Fg

Let F1 be the pivot. T 0 2 3 L 0 With clockwise as positive, 223 L 0 2 r23 m g rLmg 3 3r r23 L 2 5.0 m 3 2 r23 2

15.0 m r23 4 r23 3.75 m

T 0 1 2 3 TL TR 0 With clockwise as the positive rotation, 1 2 3 TL TR 0 3 2 1 r3m3 g r2m2 g r1m1 g 3.8 m r3m3 (27 kg) 2

3.8 m (17 kg) 2

r3m3 51.3 kg·m 32.3 kg·m r3m3 19 kg·m 19 kg·m r3 20 kg r3 0.95 m


111

The third child of mass 20 kg must sit 0.95 m from the centre of the teeter-totter and on the same side as the 17.0-kg child. b) No, the mass of the teeter-totter does not matter. 37.

F1

F2

5.0 kg mp = 2.0 kg

+

P

1.5 m 2.5 m

Let F2 be pivot. net 0 1 B C 0 With clockwise as the positive rotation, 1 p c 0 1 p c r1F1 rpFgp rcFgc rpFgp rcFgc F1 r1

2(2.0 kg)(9.8 Nkg) [(2.5 m 1.5 m)(5.0 kg)(9.8 N/kg)] 2.5 m

F1

2.5 m

F1 29.4 N But Fnet 0 FgB FgC F2 0 F1 With up as the positive direction, 0 F1 FgB FgC F2 F2 FgB FgC F1 F2 mB g mC g F1 F2 (2.0 kg)(9.8 N/kg) (5.0 kg)(9.8 N/kg) 29.4 N F2 39.2 N The man farthest from the cement bag (F1) lifts with 29.4 N and the second man lifts with 39.2 N of force. 38. Take front two and back two legs as single supports. net 0 with front legs as pivot D Back 0

112

Let clockwise be positive. D Back 0 Back D rB FB rD FgD rB FB rDmD g rB FB (0.30 m)(30 kg)(9.8 N/kg) rB FB 88.2 N·m 88.2 N·m FB 1.0 m FB 88.2 N FB 8.8 101 N net 0 But F FD FB 0 FF Let up be positive. 0 F F FD FB FF FD FB FF mDg FB FF (30 kg)(9.8 N/kg) 88.2 N FF 205.8 N FF 2.1 102 N Front legs: 1.05 102 N each; back legs: 4.4 101 N each (each divided by 2). 39. a) P

C of m 2.4 m 20 kg

0.8 m

Fnet 0 FD 0 FT Taking up to be positive, 0 F T FD FT mD g FT (20 kg)(9.8 N/kg) T 196 N [up] F


b)

41. a)

P

1.2 m θD

1.0 m

rD 75 kg

1.6 m 0.4 m

C of m

C of m

P

Fnet 0 Ff 0 Fapp-h Taking the direction of force application to be positive, Fapp-h Ff Fapp-h Fn Fapp-h mg Fapp-h 0.42(75 kg)(9.8 N/kg) app-h 308.7 N [horizontally] F app-h 3.1 102 N [horizontally] F

0.4 m tan1 1.2 m 18.4° Assume the upper hinge is the pivot. B door 0 B door 0 B door rB FB sin B rDmD g sin D rDmD g sin D FB rB sin B (1.26 m)(20 kg)(9.8 N/kg) sin 18.4° FB (2.4 m) sin (90° 18.4°) FB 34.2 N [out horizontally]

b)

θbox θa

40.

P 1.2 m

+

72 kg θp

7.0 m

65°

P

p 90° 65° p 25° Choose bottom as pivot. net 0 wall p 0 Taking right (horizontally) as positive, wall p 0 wall p rw Fw sin w rpmp g sin p rpmp g sin p Fw rw sin w Fw

[(7.0 m 1.2 m)(72 kg)(9.8 N/kg)(sin 25°)] (7.0 m) sin 65°

Fw 272.6 N [horizontal] Fw 2.7 102 N But Fh(bottom) Fh(top) so 2.7 102 N is required to keep the ladder from sliding.

Just to the tip the box, net 0 a box 0 Taking the direction of force application to be positive, a box 0 a box Take bottom corner as pivot. 1.6 m 2 tan a 1.0 m 2 a 58° raFa sin a rboxmbox g sin box rboxmbox g sin box ra Fa sin a

ra

(0.8 m )2 (0 .5 m)2(75 kg)(9.8 N/kg) sin (90° 58°) (308.7 N) sin 58°

ra 1.40 m But: h ra sin 58° h 1.2 m


113

Let the contact point of F2 be the pivot P. 1 w 0 With clockwise being the positive torque direction, 1 w 0 1 w r1F1 rwmg rwmg F1 r1 (0.12 m)(65 kg)(9.8 N/kg) F1 0.04 m F1 1911 N 1 1.9 103 N [up] F Fnet 0 Fg 0 F1 F2 With up taken to be the positive direction, F2 F1 Fg F2 1911 N (65.0 kg)(9.8 N/kg) F2 2548 N 2 2.5 103 N [down] F

42.

+ 10 kg

P 5.0 cm 16 cm

35 cm

net 0 muscle arm water 0 With clockwise as the direction of positive rotation, m a w 0 m a w rmFm sin ramag sin rwmwg sin ramag rwmwg Fm rm Fm

(0.16 m)(3.0 kg)(9.8 N/kg) (0.35 m)(10 kg)(9.8 N/kg) 0.050 m

Fm 780.1 N Fm 7.8 102 N [up] 43.

1.9 kg

1.2 kg

45.

F = 0.5 N 0.01 m

P

0.4 kg

+

0.15 m 0.40 m

0.02 m 0.60 m

The total of all three torques must be equivalent to the total torque through the centre of mass. cm ua fa hand rcmm T g ruamua g rfamfa g rhandmhand g ruamua rfamfa rhandmhand rcm mT rcm

(0.15 m)(1.9 kg) (0.40 m)(1.2 kg) (0.60 m)(0.4 kg) 3.5 kg

rcm 0.29 m from shoulder 44.

F1 4.0 cm

114

W

P 12 cm

+

F2

Use pivot P as the point of contact of F1. net 0 F2 F 0 With clockwise taken to be the positive torque direction, F2 F 0 F2 F rF2F2 rFF rFF F2 rF2 (0.01 m)(0.5 N) F2 0.02 m F2 0.25 N


Fnet 0 F2 0 F F With right taken to be the positive direction, F1 F F2 F1 0.5 N 0.25 N F1 0.75 N 1 0.75 N [left], and F 2 0.25 N [right] F

48.

1

46.

h hcm

1.00 m

2 tan

28 cm 11 cm

L

7.25 kg

hcm

2.4 cm P

hcm

2

hcm

2

L

C of m

FT

θ

1.00 m

θ L — 2

+

tan

1.00 m

Set P at elbow joint. net 0 T arm sp 0 With clockwise taken to be the positive torque direction, T arm sp 0 T arm sp rTFT rarmFg(arm) rspFg(sp) rarmmarmg rspmspg FT rT FT

(0.11 m)(2.7 kg)(9.8 N/kg) (0.280 m)(7.25 kg)(9.8 N/kg) 0.024 m

tan 30° hcm 0.8660 m But: h 2hcm h 2(0.8660 m) h 1.73 m NOTE: The solution to problem 49 is based on the pivot point of the glass being at the corner of the base. 49.

d x

FT 9.5 102 N 47.

θ

0.6 m

θ

0.14 m

0.050 m

θ

0.3 m θ

0.020 m

0.3 m tan 0.6 m 26° The tipping angle is 26° from the horizontal.

θ

0.020 m tan 0.050 m 21.8° x sin h 0.050 m x (0.14 m 0.050 m) sin 21.8° x 0.033 m dxr d 0.033 m 0.020 m d 0.053 m


115

50.

Use hinge as pivot. net 0 s m 0 With clockwise taken to be the positive torque direction, s m 0 s m rs Fs sin s rm Fm sin 90° rmmm g Fs rs sin s (1.0 m)(10.0 kg)(9.8 N/kg) Fs (0.75 m) sin 10° Fs 752.5 N But: Fs kx F k s x 752.5 N k 4.0 102 m

θ 2.5 m

θ

2.5 m

2 tan base hcm

2 tan 2.5 m

2.5 m 26.6°

51.

Fs

k 1.88 104 N/m

nails

53.

x2

Fg

s Fg 0 F With up taken to be the positive direction, Fs mg But: Fs kx So: kx mg mg k x (3.0 kg)(9.8 N/kg) k 1.8 102 m k 1.6 103 N/m 52. +

x1

45° 0.50 m θ 0.50 m

0.50 m

2 m) (0.50 m)2 x1 (0.50 x1 0.7071 m 2 (0.50 m) (1.50 m)2 x2 x2 1.58 m 1.50 m tan 0.50 m 71.56° ∆x 1.58 m 0.7071 m ∆x 0.874 m

θ

T

Fs

T bar

10°

P 0.75 m 1.0 m

116

1.5 m

10 kg

Fgm


Fg

θ

T Fs T k∆x T (1.5 102 N/m)(0.874 m) T 1.311 102 N

T

θ

mg — 2

T

2 sin mg

T 2T sin m g 2(1.311 102 N) sin 71.56° m 9.8 N/kg m 25.4 kg 54. L 20 m r 2.0 103 m Limit FL 6.0 107 N/m2 F a) Stress A F A(Stress) F r2(Stress) F (2.0 103 m)2(6.0 107 N/m2) F 753.6 N F 7.5 102 N b) E for Al is: EAl 70 109 N/m2 Stress E Strain F A E L L

F L A L

E (20 m)(6.0 107 N/m2) L 70 109 N/m2 L 0.017 m L 1.7 102 m

55. a) A 0.1 m2 F Stress A (100 kg)(9.8 N/kg) Stress 0.1 m2 Stress 9.8 103 N/m2 Strain Eiron 100 109 N/m2 Stress E Strain Stress Strain E 9.8 103 N/m2 Strain 100 109 N/m2 Strain 9.8 108 b) ∆L ? L 2.0 m ∆L L(Strain) ∆L (2.0 m)(9.8 108) ∆L 1.96 107 m ∆L 2.0 107 m c) Maximum stress is 17 107 N/m2. Fmax Stress(A) Fmax (17 107 N/m2)(0.1 m2) Fmax 1.7 107 N mg Fmax 1.7 107 N m 9.8 N/kg m 1.7 106 kg 56. Maximum stress for femur is 13 107 N/m2. A 6.40 104 m2 F Stress A Fmax A(Stress) Fmax (6.40 104 m2)(13 107 N/m2) Fmax 8.32 104 N 57. Fc 200 N A 1 105 m2 L 0.38 m E 15 109 N/m2 F A E L L FL L AE (200 N)(0.38 m) L (1 103 m2)(15 109 N/m2)

L 5.067 106 m Solutions to End-of-chapter Problems

117

F k x 200 N k 5.067 106 m k 3.95 107 N/m 58. 2.0 m

L

R θ

R

τ

Gsteel 80 109 N/m2 rF C 2r C 2(20 m) C 125.6 m 2.0 m 125.6 m 360° (360°)(2 m) 125.6 m 5.73° R cos R L R R L cos 20 m R L cos 5.73° R L 20.1005 m L 20.1005 m R L 20.1005 m 20 m L 0.1005 m F A G L L

Arod (0.01 m)2 GLA F L (80 109 N/m2)(0.1005 m)[(0.01 m)2] F 2.0 m F 1 262 920 N F 1.26 106 N rod rF sin rod 2.0 m(1 262 920 N) sin 90° rod 2.52 106 N·m The torque on rod is 2.5 106 N·m.

118

59. Stress is 10% of Tmax. Stress 0.10(50 107 N/m2) Stress 5.0 107 N/m2 a) A ? F Stress A F A Stress mg A Stress (1.00 104 kg)(9.8 N/kg) A 5.0 107 N/m2 A 1.96 103 m2 r

A

1.96 103 m2 r 0.025 m r 2.5 102 m b) a 2.0 m/s2 Fnet Fapp mg Fnet ma mg Fnet m(a g) Esteel 200 109 N/m2 Stress E Strain Stress Strain E F A Strain E [m(a g)] Strain A E r

Strain

[(1.00 104 kg)(2.0 m/s2 9.8 m/s2)] 1.96 1 03 m2 200 10 9 N/m2

4

Strain 3.01 10 60. L ? Epine 10 109 N/m2 L 3.0 m A (10 102 m)(15 102 m) A 1.5 103 m2 Fg 1000 N


A E L L F

a)

Stress E Strain F Stress A 1000 N Stress 1.5 103 m2 Stress 6.67 105 N/m2 Stress Strain E 6.67 105 N/m2 Strain 10 109 N/m2 Strain 6.67 105 b) ∆L L(Strain) ∆L (3.0 m)(6.67 105) ∆L 2.0 104 m 61. m 2.5 104 kg Fapp (2.5 104 kg)(9.8 N/kg) Fapp 2.45 105 N A1 r2 1.00 m 2 A1 2 A1 0.785 m2 A2 r22 0.80 m 2 A2 2 A2 0.5024 m2 Emarble 50 109 N/m2 ∆L1 ? L Stress 1 L1 E

For column 2: Strain Stress E F Stress A2E 2.45 105 N Stress (0.5024 m2)(50 109 N/m2) Stress 9.75 106 L 2 9.75 106 L2 L2 L2 (9.75 106) But: L2 L2 21.999 863 m L2 L2(9.75 106 m) 21.999 863 m 21.999 863 m L2 1 9.75 106 m L2 22.000 0775 m The narrower column needs to be only 7.8 105 m longer than the wider column.

A L F

1

L1

1

E

FL1 L1 A1E (2.45 105 N)(22.0 m) L1 (0.785 m2)(50 109 N/m2) L1 1.37 104 m Column 1 final loaded: Loaded 22.0 m L1 Loaded 22.0 m 1.37 104 m Loaded 21.999863 m Solutions to End-of-chapter Problems

119

Chapter 4 16. p mv p (7500 kg)(120 m/s) p 9.0 105 kg·m/s 17. p mv p (0.025 kg)(3 m/s) p 0.075 kg·m/s 18. 90 km/h 25 m/s, m 25 g 0.025 kg p mv p (0.025 kg)(25 m/s) p 0.63 kg·m/s 19. v 500 km/h 138.89 m/s, p 23 000 kg·m/s p m v 23 000 kg·m/s m 138.89 m/s m 165.6 kg p 20. v m 1.00 kg·m/s v 1.6726 1027 kg

250 N [forward], 23. m 50 kg, F ∆t 3.0 s, v1 0 Ft mv

24.

25.

v 6.00 1026 m/s, which is much greater than the speed of light. mv 21. p (0.050 g)(10 m/s [down]) p p 0.5 kg·m/s [down] p = 0.5 kg·m/s [down]

1000 m 1h 22. v (300 km/h) 1 km 3600 s 83.3 m/s p mv (6000 kg)(83.3 m/s [NW]) p p 5 105 kg·m/s [NW]

26.

p = 5 x 105 kg·m/s [NW]

27. 45°

28.

120

(250 N [forward])(3.0 s) (50 kg)(v2 v1) 740 N [forward] v2 50 kg v2 15 m/s [forward] m 150 kg, v1 0, a 2.0 m/s2, ∆t 4.0 s a) v2 v1 a∆t v2 0 (2.0 m/s2)(4.0 s) v2 8.0 m/s p mv p (150 kg)(8.0 m/s) p 1200 kg·m/s b) J ∆p J m2v2 m1v1 J (150 kg)(8.0 m/s) (150 kg)(0) J 1200 kg·m/s m 1.5 kg, ∆h 17 m, v1 0, a 9.8 N/kg 1 a) h v1t at2 2 1 17 m 0 (9.8 m/s2)t2 2 t 1.86 s b) F ma F (1.5 kg)(9.8 N/kg) F 14.7 N c) J F∆t J (14.7 N)(1.86 s) J 27.3 kg·m/s F 700 N, ∆t 0.095 s a) J F∆t J (700 N)(0.095 s) J 66.5 kg·m/s b) J ∆p ∆p 66.5 kg·m/s m 0.20 kg, v1 25 m/s, v2 20 m/s ∆p m2v2 m1v1 ∆p (0.2 kg)(20 m/s) (0.2 kg)(25 m/s) ∆p 9.0 kg·m/s F∆t m∆v (N)(s) (kg)(m/s) (kg·m/s2)(s) (kg)(m/s) kg·m/s kg·m/s


p2

33°

p1

30. v1 0, v2 250 m/s, m 3.0 kg, F 2.0 104 N a) J ∆p J m2v2 m1v1 J (3.0 kg)(250 m/s) 0 J 750 kg·m/s b) J Ft J t F 750 kg·m/s t 2 104 N t 0.038 s 31. m 7000 kg, v1 110 km/h 30.56 m/s, ∆t 0.40 s, v2 0 p a) F t m2v2 m1v1 F t 0 (7000 kg)(30.56 m/s) F 0.40 s F 5.3 105 N p b) F t m2v2 m1v1 F t 0 (7000 kg)(30.56 m/s) F 8.0 s 4 F 2.7 10 N 32. m 30 g 0.03 kg, v1 360 m/s, ∆d 5 cm 0.05 m a) p mv p (0.03 kg)(360 m/s) p 11 kg·m/s b) v22 v12 2a∆d 02 (360 m/s)2 2a(0.05 m) a 1.3 106 m/s2

t (×10

F (×104 N)

p

c) F ma F (0.03 kg)(1.3 106 m/s2) F 3.9 104 N v2 v1 d) a t v2 v1 t a 0 360 m/s t 1.3 106 m/s2 t 2.8 104 s e) J p J m2v2 m1v1 J (0.03 kg)(0) (0.03 kg)(360 m/s) J 11 kg·m/s f) –4

33. a) F (×106 N )

29. p p2 p1

0 –1

s)

1 2 3

–2 –3 –4 8 7 6 5 4 3 2 1 0 5 10 15 t (s)

1 b) Area h(a b) 2 1 J (15 s)(5 106 N 8 106 N) 2 J 9.8 107 N·s 34. J area under the curve 1 J (90 N)(0.3 s) (120 N)(0.2 s) 2 1 (75 N)(0.4 s) 2 J (13.5 N·s) (24 N·s) (15 N·s) J 25.5 N·s 35. J area under the graph Counting roughly 56 squares, J 56(0.5 103 N)(0.05 s) J 1.4 103 N·s


121

36.

37.

38.

39.

J ∆p J mv2 mv1, where v1 0, 1.4 103 N·s (0.250 kg)(v2) v2 5.6 103 m/s pTf pTo m1v1o m2v2o (m1 m2)vf, where v2o 0 (5000 kg)(5 m/s [S]) (10 000 kg)(vf) vf 2.5 m/s [S] To p Tf p m1v1o m2v2o (m1 m2)vf, where v2o 0 (45 kg)(5 m/s) (47 kg)(vf) vf 4.8 m/s [in the same direction as v1o] pTf pTo m1v1o m2v2o m1v1f m2v2f (65 kg)(15 m/s) (100 kg)(5 m/s) 1 (65 kg) (15 m/s) (100 kg)(v2f) 3 (975 500 325) kg·m/s (100 kg)(v2f) v2f 1.5 m/s Tf pTo p m1v1o m2v2o (m1 m2)vf,

43.

44.

40.

where v2o 0 (0.5 kg)(20 m/s) 0 (30.5 kg)(vf) vf 0.33 m/s 41. pTo pTf m1v1o m2v2o m1v1f m2v2f (0.2 kg)(3 m/s) (0.2 kg)(1 m/s) (0.2 kg)(2 m/s) (0.2 kg)(v2f) 0.4 kg·m/s 0.4 kg·m/s (0.2 kg)(v2f) v2f 0 42. v1o 90 km/h 25 m/s, vf 80 km/h 22.2 m/s pTo pTf m1v1o m2v2o (m1 m2)vf, where v2o 0 m1(25 m/s) 0 (m1 6000 kg) (22.2 m/s) m1(25 m/s) m1(22.2 m/s) 133 333.3 kg·m/s m1(25 m/s 22.2 m/s) 133 333.3 kg·m/s m1

133 333.3 kg·m/s 2.8 m/s

m1 4.8 104 kg

122

F1 F2 ma1 ma2 v2f v2o v1f v1o m m t t m(v1f v1o) m(v2f v2o) mv1f mv1o mv2f mv2o mv1f mv2f mv1o mv2o pTf pTo pTf pTo 0 p 0 m1 1.67 1027 kg, m2 4m1, v1 2.2 107 m/s pTo pTf m1v1o m2v2o (m1 m2)vf, where v2o 0 m1(2.2 107 m/s) (5m1)vf 2.2 107 m/s vf 5 vf 4.4 106 m/s m1 3m, m2 4m, v1o v pTo pTf m1v1o m2v2o (m1 m2)vf, where v2o 0 (3m)v (7m)vf 3 vf v 7 m1 99.5 kg, m2 0.5 kg, v1f ?, v2f 20 m/s pTo pTf 0 (99.5 kg)(v1f) (0.5 kg)(20 m/s) 10 kg·m/s v1f 99.5 kg v1f 0.1 m/s d t v 200 m t 0.1 m/s t 2 103 s 1o 375 kg·m/s [E], p 2o 450 kg·m/s [N45°E] p p1o p2o a) pTo

45.

46.

47.

pTo θ p1o = 375 kg·m /s


p2o = 450 kg·m /s 45°

b) pTf pTo Using the cosine and sine laws, To2 (375 kg·m/s)2 (450 kg·m/s)2 p 2(375 kg·m/s)(450 kg·m/s) cos 135° pTo 762.7 kg·m/s Tf 762.7 kg·m/s p sin 135° sin 762.7 kg·m/s 450 kg·m/s 24.7° Therefore, pTf 763 kg·m/s [E24.7°N] 48. m1 3.2 kg, v1o 20 m/s [N], 1o 64 kg·m/s [N], m2 0.5 kg, p 2o 2.5 kg·m/s [W] v2o 5 m/s [W], p pTo pTf p1o p2o pTf m1v1o m2v2o (m1 m2)vf Using the diagram and Pythagoras’ theorem, p2o = 2.5 kg·m /s

pTf θ

p2o tan 60° 60 000 kg·m/s p2o (60 000 kg·m/s)(tan 60°) p2o 103 923 kg·m/s m2v2o 103 923 kg·m/s 103 923 kg·m/s v2o 5000 kg v2o 20.8 m/s [E] o 0, 50. mo 1.2 1024 kg, vo 0, p 25 m1 3.0 10 kg, v1 2.0 107 m/s [E], 1 6 1018 kg·m/s [E], p m2 2.3 1025 kg, v2 4.2 107 m/s [N], 2 9.66 1018 kg·m/s [N] p m3 1.2 1024 kg 3.0 1025 kg 2.3 1025 kg m3 6.7 1025 kg pTo 0 pTf pTo 0 p1 p2 p3 Drawing a momentum vector diagram and using Pythagoras’ theorem,

p1o = 64 kg·m /s

θ p3 = m3v3

p2o = 9.66 × 10–18 kg·m /s p1o = 6.0 × 10 –18 kg·m /s

2 Tf (2.5 p kg ·m/s) (6 4 kg· m/s) 2 Tf 64.05 kg·m/s p 2.5 kg·m/s tan 64 kg·m/s 2.2° pT (m1 m2)vf 64.05 kg·m/s [N2.2°W] (3.7 kg)vf vf 17 m/s [N2.2°W] 49. m1 3000 kg, v1o 20 m/s [N], 1o 60 000 kg·m/s [N], m2 5000 kg, p 2o ? [E], vf ? [E30°N], v2o ? [E], p f ? [E30°N] p To pTf p 1o p p2o pTf

Using the following momentum diagram, p2o = m2 v2o p1o = 60 000 kg·m /s

p Tf 30°

32 (9.66 1018 kg·m/s)2 p (6 1018 kg·m/s)2 3 1.1372 1017 kg·m/s p 3 p v3 m3 1.1372 1017 kg·m/s v3 6.7 1025 kg v3 1.7 107 m/s

6 1018 kg·m/s tan 9.66 1018 kg·m/s

31.8° Therefore, v3 1.7 107 m/s [S32°W] 60 m 51. m1 m2 m, v1o 12.5 m/s 4.8 s [R], v2o 0, v2f 1.5 m/s [R25°U] To pTf p m1v1o m2v2o m1v1f m2v2f Since m1 m2 and v2o 0, v1o v1f v2f


123

Drawing a vector diagram and using trigonometry, v1f

v2f = 1.5 m/s 25°

Using the vector diagram and trigonometry, v1f

θ

θ v1o = 6.0 m / s

v1o = 12.5 m /s

v1f2 (1.5 m/s)2 (12.5 m/s)2 2(1.5 m/s)(12.5 m/s) cos 25° v1f 11.16 m/s sin sin 25° 11.6 m/s 1.5 m/s 3.3° Therefore, the first stone is deflected 3.3° or [R3.3°D]. 52. m1 10 000 kg, v1 3000 km/h [E] 833.3 m/s [E], 1 8.333 106 kg·m/s [E], m2 ?, p v2 5000 km/h [S] 1388.9 m/s [S], 2 m2(1388.9 m/s) [S], p m3 10 000 kg m2, v3 ? [E10°N], 3 (10 000 kg m2)(v3) [E10°N] p p2 p3 p1 Drawing a momentum diagram and using trigonometry,

v2f = 4.0 m / s 25°

v1f2 (6.0 m/s)2 (4.0 m/s)2 2(6.0 m/s)(4.0 m/s) cos 65° v1f 5.63 m/s sin sin 65° 5.63 m/s 4.0 m/s 40° Therefore v1f 5.63 m/s [U40°R] 54. 2m1 m2, v1o 6.0 m/s [U], v2o 0, v2f 4 m/s [L25°U], v1f ? To pTf p m1v1o m2v2o m1v1f m2v2f, since 2m1 m2 and v2o 0 v1o v1f 2v2f Using the vector diagram and trigonometry, v1f

θ V1o = 6.0 m /s

p1 = 8.33 × 106 kg·m /s v2f = 8.0 m /s

p2 10°

p3

p2 tan 10° p1 2 (8.33 106 kg·m/s)(tan 10°) p 2 1.47 106 kg·m/s p p2 m2(1388.9 m/s) [S] 1.47 106 kg·m/s m2 1388.8 m/s m2 1057.6 kg The mass of the ejected object is 1.058 103 kg. 53. m1 m2 m, v1o 6.0 m/s [U], v2o 0, v2f 4 m/s [L25°U], v1f ? To pTf p m1v1o m2v2o m1v1f m2v2f Since m1 m2 and v2o 0, v1o v1f v2f

124

25°

v1f2 (6.0 m/s)2 (8.0 m/s)2 2(6.0 m/s)(8.0 m/s) cos 65° v1f 7.7 m/s sin 65° sin 7.7 m/s 8.0 m/s 70° Therefore, v1f 7.7 m/s [R20°U] 55. Counting ten dots for a one-second interval and measuring the distance with a ruler and the angle with a protractor gives: 33 mm a) v1o 1s v1o 0.033 m/s v2o 0 33 mm v1f 1s v1f 0.033 m/s 33 mm v2f 1s v2f 0.033 m/s


b) v1o 0.033 m/s [E] v2o 0 v1f 0.033 m/s [E45°S] v2f 0.033 m/s [E45°N] c) p1o (0.3 kg)(0.033 m/s [E]) p1o 9.9 103 kg·m/s [E] 2o (0.3 kg)(0) 0 p pTo 9.9 103 kg·m/s [E] 1f (0.3 kg)(0.033 m/s [E45°S]) p p1f 9.9 103 kg·m/s [E45°S] 2f (0.3 kg)(0.033 m/s [E45°N]) p p2f 9.9 103 kg·m/s [E45°N] The vector diagram for the final situation is shown below. 45°

p1f = 9.9 × 10–3 kg·m /s

p2f = 9.9 × 10–3 kg·m /s

45° p Tf

Using Pythagoras’ theorem, Tf2 (9.9 103 kg·m/s)2 p (9.9 103 kg·m/s)2 pTf 1.4 102 kg·m/s [E] d) p1oh 9.9 103 kg·m/s p1ov 0 p2oh 0 p2ov 0 p1fh (9.9 103 kg·m/s)(cos 45°) p1fh 7.0 103 kg·m/s p1fv (9.9 103 kg·m/s)(sin 45°) p1fv 7.0 103 kg·m/s p2fh (9.9 103 kg·m/s)(cos 45°) p2fh 7.0 103 kg·m/s p2fv (9.9 103 kg·m/s)(sin 45°) p2fv 7.0 103 kg·m/s e) Momentum is not conserved in this collision. The total final momentum is about 1.4 times the initial momentum. 56. m1 0.2 kg, v1 24 m/s [E], 1 4.8 kg·m/s [E], m2 0.3 kg, p 2 5.4 kg·m/s [N], v2 18 m/s [N], p m3 0.25 kg, v3 30 m/s [W], 3 7.5 kg·m/s [W], m4 0.25 kg, p 4 ? v4 ?, p To 0 p 1 p2 p3 p4 0 p

Drawing a vector diagram and using trigonometry, A

p4

B p3 = 7.5 kg·m /s

p2 = 5.4 kg·m /s

θ

C p = 4.80 kg·m /s 1

Using triangle ABC, 2.7 kg·m/s tan 5.4 kg·m/s 26.6° 2 p4 (2.7 kg·m/s)2 (5.4 kg·m/s)2 4 6.037 kg·m/s p 6.037 kg·m/s v4 0.25 kg v4 24.1 m/s Therefore, v4 24.1 m/s [S26.6°E] 57. a) Masstotal 5000 kg 10 000 kg Masstotal 15 000 kg 5000 kg 1 b) the distance to the larger 15 000 kg 3 1 truck, or (400 m) 133.3 m from the 3 larger truck. 58. a) m1 2000 kg, v1o 200 m/s [E], 1o 4 105 kg·m/s [E] p p1o = 4 × 105 kg·m /s [E]

b) m2 1000 kg, v2o 200 m/s [S30°E], 2o 2 105 kg·m/s [S30°E] p p2o = 2 × 105 kg·m /s [S30°E] 30°

p1o p2o c) pcmo p1o p2o pcmo

d) pcmf pcmo pcmf

59. a) pcmo pTo cmo 9.9 103 kg·m/s [E] (see 55c) p pTf b) pcmf pcmf 1.4 102 kg·m/s [E] (see vector diagram for 55c)


125

Chapter 5 11. a) W Fd W (4000 N)(5.0 m) W 2.0 104 J b) W (570 N)(0.08 m) W 46 J c) W Ek W Ek2 Ek1 1 W mv2 0 2 1 W (9.1 1031 kg)(1.6 108 m/s)2 2 W 1.2 1014 J 12. a) W Fd W (500 N)(5.3 m) W 2.7 103 J b) W Fd cos W (500 N)(5.3 m) cos 20° W 2.5 103 J c) W (500 N)(5.3 m) cos 70° W 9.1 102 J 13.

The force that the plow applies in 1 s is: Fapp Fg Fapp mg Fapp (3556 kg)(9.8 N/kg) Fapp 34 848.8 N This force is applied over a distance of 5 m: W Fd W (34 848.8 N)(5.0 m) W 174 244 J To find the number of seconds it takes to plow the road: d t v 8000 m t 10 m/s t 800 s WT (800 s)(174 244 J/s) WT 1.4 108 J 15. The two campers must overcome 84 N of friction, or 42 N each in the horizontal direction since both are at the same 45° angle. 45°

Fc

25.0 m h

θ

W Eg Fd mgh (350 N)(25.0 m) (50.0 kg)(9.8 N/kg)h h 18 m h sin 25.0 m 18 m sin 25.0 m 46° 14. Using the plow’s speed, in 1 s, the plow will push a “block” of snow that is 0.35 m deep, 4.0 m wide and 10.0 m long. This snow has a mass of: (0.35 m)(4.0 m)(10.0 m)(254 kg/m3) 3556 kg

126

45°

Fh

The horizontal component of Fc, called Fh, must be equal to 42 N. W Fh d W (42 N)(50 m) W 2.1 103 N·m Each camper must do 2.1 103 J of work to overcome friction. 16. W Fd ,where d for each revolution is zero. Therefore, W 0 J. 17. 350 J indicates that the force and the displacement are in the opposite direction. An example would be a car slowing down because of friction. Negative work represents a flow or transfer of energy out of an object or system.


18. dramp 5 m m 35 kg dheight 1.7 m a) F ma F (35 kg)(9.8 N/kg) F 343 N F 3.4 102 N b) W Fd W (3.4 102 N)(1.7 m) W 583.1 J W 5.8 102 N c) W Fdramp 583.1 J F(5 m) F 116.62 N F 1.2 102 N 19. W Area under the graph (20 m)(200 N) W (10 m)(200 N) 2 (20 m)(600 N) (20 m)(200 N) 2 (20 m)(400 N) (10 m)(800 N) 2 (20 m)(800 N) (10 m)(1200 N) W 5.4 104 J 20. a) W area under the graph (1 m)(100 N) (1 m)(200 N) W 2 2 (1 m)(100 N) (2 m)(300 N) W 8.5 102 J b) The wagon now has kinetic energy (and may also have gained gravitational potential energy). c) W Ek 1 Ek mv2 2 1 850 J (120 kg)v2 2 v 3.8 m/s 1 21. a) Ek mv2 2 1 Ek (45 kg)(10 m/s)2 2 Ek 2.3 103 J

2 r b) v t 2 (0.1 m) v 1s v 0.628 m/s 1 Ek mv2 2 1 Ek (0.002 kg)(0.628 m/s)2 2 Ek 3.9 104 J 100 km 1h 1000 m c) v 1h 3600 s 1 km v 27.7778 m/s 1 Ek mv2 2 1 Ek (15 000 kg)(27.7778 m/s)2 2 Ek 5.8 106 J d 22. v t 5.0 m v 2.0 s v 2.5 m/s 1 Ek mv2 2 1 450 J m(2.5 m/s)2 2 m 1.4 102 kg 1 23. Ek mv2 2 1 5.5 108 J (1.2 kg)v2 2 2(5.5 108 J) 1.2 kg v 3.0 104 m/s 24. Ek-gained Eg Ek-gained mgh2 mgh1 Ek-gained mg(h2 h1) Ek-gained (15 kg)(9.8 N/kg)(200 m 1 m) Ek-gained 2.9 104 J 2mEk 25. p kg·m/s (kg)(J) kg·m/s (kg)(N ·m) kg·m/s kg(kg m/s2 )m 2 2 2 kg·m/s kg m /s kg·m/s kg·m/s v


127

1000 eV 1.6 1019 J 26. 5 keV 1 keV 1 eV 8 1016 J 1 Ek mv2 2 1 8 1016 J (9.1 1031 kg)v2 2 v 4.2 107 m/s As a percentage of the speed of light: 4.2 107 m/s 100 14% 3 108 m/s (v22 v12) 27. a) a 2d 0 (350 m/s)2 a 2(0.0033 m) a 1.86 107 m/s2 F ma F (0.015 kg)(1.86 107 m/s2) F 2.8 105 N b) F force of bullet F 2.8 105 N 28. For 1 m: W (50 N)(1 m) W 50 J W Ek 1 Ek mv2 2 1 50 J (1.5 kg)v2 2 v 8 m/s For 2 m: 1 W 50 J (50 N)(1 m) (250 N)(1 m) 2 W 225 J 1 Ek mv2 2 1 225 J (1.5 kg)v2 2 v 17.3 m/s For 3 m: 1 1 W 225 J (50 N) m 2 6

5 1 1 (300 N) m (350 N) m 6 2 6 W 425 J

128

1 Ek mv2 2 1 425 J (1.5 kg)v2 2 v 23.8 m/s 2mEk 29. p p 2(5 kg )(3.0 102 J) p 55 N·s 30. m1 0.2 kg m2 1 kg v1o 125 m/s v1f 100 m/s v2o 0 v2f ? d2 3 m a) pTo pTf m1v1o m2v2o m1v1f m2v2f (0.2 kg)(125 m/s) 0 (0.2 kg)(100 m/s) (1 kg)v2f v2f 5 m/s 1 b) Ek mv2 2 1 Ek (1 kg)(5 m/s)2 2 Ek 12.5 J c) This collision is not elastic since some kinetic energy is not conserved. Some energy may be lost due to the deformation of the apple. d) v22 v12 2ad 0 (5 m/s)2 2a(3.0 m) a 4.1667 m/s2 F ma F (1.0 kg)(4.1667 m/s2) F 4.2 N 31. a) Eg mgh Eg (2.0 kg)(9.8 N/kg)(1.3 m) Eg 25 J b) Eg mgh Eg (0.05 kg)(9.8 N/kg)(3.0 m) Eg 1.5 J c) Eg mgh Eg (200 kg)(9.8 N/kg)(469 m) Eg 9.2 105 J d) Eg mgh Eg (5000 kg)(9.8 N/kg)(0) Eg 0 J


F 32. a) m a 4410 N m 9.8 N/kg m 4.5 102 kg b) W Fd W (4410 N)(3.5 m) W 1.5 104 J 33. Using conservation of energy: ETo ETf 1 1 mgh mvo2 mvf2 2 2 1 1 (9.8 m/s2)(1.8 m) (4.7 m/s)2 v2 2 2 1 17.64 m2s2 11.045 m2s2 v2 2 v 7.6 m/s 34. Ee Eg 1 kx2 mgh 2 1 (1200 N/m)x2 (3.0 kg)(9.8 N/kg)(0.80 m) 2 x 0.2 m x 20 cm 35. m 0.005 kg h 2.0 m Initial: E mgh E (0.005 kg)(9.8 N/kg)(2.0 m) E 0.098 J At half the height: E (0.005 kg)(9.8 N/kg)(1.0 m) E 0.049 J After the first bounce: E (0.80)(0.098 J) E 0.0784 J After the second bounce: E (0.80)(0.0784 J) E 0.062 72 J After the third bounce: E (0.80)(0.062 72 J) E 0.050 176 J After the fourth bounce: E (0.80)(0.050 176 J) E 0.040 140 9 J Therefore, after the fourth bounce, the ball loses over half of its original height.

36. a) The greatest potential energy is at point A (highest point) and point F represents the lowest amount of potential energy (lowest point). b) Maximum speed occurs at F when most of the potential energy has been converted to kinetic energy. Eg-lost Ek-gained 1 mgh mv2 2 1 (1000 kg)(9.8 N/kg)(75 m) (1000 kg)v2 2 v 38 m/s c) At point E, 18 m of Ep is converted to Ek. 1 mgh mv2 2 1 (1000 kg)(9.8 N/kg)(18 m) (1000 kg)v2 2 v 19 m/s d) Find the acceleration, then use F ma. v22 v12 2ad 0 (38 m/s)2 2a(5 m) a 144.4 m/s2 F ma F (1000 kg)(144.4 m/s2) F 1.4 105 N 37. Ee Ek 1 1 kx2 mv2 2 2 2 (890 N/m)x (10 005 kg)(5 m/s)2 x 16.8 m x 17 m v12 sin 2 38. dh g v12 sin 90° 15 m 9.8 m/s2 v1 12.1 m/s Ek Ee 1 1 mv2 kx2 2 2 (0.008 kg)(12.1 m/s) (350 N/m)x2 x 0.058 m x 5.8 cm


129

39. 85% of the original energy is left after the first bounce, therefore, (0.85)mghtree mghbounce (0.85)(2 m) hbounce h 1.7 m 40. Ee Ek 1 1 kx2 mv2 2 2 2 (35 000 N/m)(4.5 m) (65 kg)v2 v 104.4 m/s 2 v1 sin 2 dh g (104.4 m/s)2 sin 90° dh 9.8 m/s2 dh 1.1 103 m 41. k slope rise k run F k x 120 N k 0.225 m k 5.3 102 N/m 42. W area under the graph 1 a) W (0.05 m)(2 103 N) 2 W 50 J b) W 50 J (0.02 m)(2 103 N) 1 (0.02 m)(4.5 103 N) 2 W 135 J W 1.4 102 J 43. Ek Ee 1 2 1 2 mv kx 2 2 2 (0.05 kg)v (400 N/m)(0.03 m)2 v 2.7 m/s 44. Ek Ee 1 1 mv2 kx2 2 2 3 2 (2.5 10 kg)(95 m/s) k(35 m)2 k 1.8 104 N/m 45. Ee Ek 1 1 2 kx mv2 2 2 7 2 (5 10 N/m)(0.15 m) (1000 kg)v2 v 34 m/s 130

Ek Ee 1 2 1 mv kx2 2 2 2 (3 kg)v (125 N/m)(0.12 m)2 v 0.77 m/s b) Ff Fn Ff (0.1)(3 kg)(9.8 N/kg) Ff 2.94 N F ma F a m 2.94 N a 3 kg a 0.98 m/s2 v22 v12 2ad 0 (0.77 m/s)2 2(0.98 m/s2)d d 0.3 m d 30 cm Ek Ee 1 1 mv2 kx2 2 2 2 (3.0 kg)v (350 N/m)(0.1 m)2 v 1.1 m/s F kx F (4000 N/m)(0.15 m) F 600 N F ma F (100 kg)(9.8 N/kg) F 980 N Divided into 20 springs: 980 N F 20 F 49 N per spring F kx 49 N k(0.035 m) k 1.4 103 N/m F kx mg kx (10 kg)(9.8 N/kg) k(1.3 m) k 75.3846 N/m 1 2 Ee kx 2 1 2 106 J (75.3846 N/m)x2 2 x 2.3 102 m

46. a)

47.

48.

49.

50.


24 h 60 min 60 s 51. 3 d 259 200 s 1d 1h 1 min 60 min 60 s 8 h 28 800 s 1h 1 min 60 s 15 min 900 s 1 min t 259 200 s 28 800 s 900 s t 288 900 s E P t E Pt E (60 W)(288 900 s) E 1.7 107 J 1 kWh 1.7 104 kJ 4.8 kWh 3600 kJ 52. a) E Eg E mgh E (3500 kg)(9.8 m/s2)(13.4 m) E 459 620 J E P t 459 620 J P 23 s P 19 983 W 19 983 W PE 0.46 PE 4.3 104 W 1 hp b) 4.3 104 W 58 hp 746 W 54. a) P Fv P Fgv P mgv P (4400 kg 2200 kg)(9.8 m/s2)(2.4 m/s) P 1.6 105 W 55. Since the cyclist’s speed is 2.78 m/s, the cyclist travels 2.78 m up the hill per second. The cyclist’s change in height per second is: h d sin h (2.78 m) sin 7.2° h 0.348 m The increase in potenial energy is: Ep mgh Ep (75 kg)(9.8 m/s2)(0.348 m) Ep 255.78 J

In 1 s: 255.78 J P 1.0 s P 256 W 56. Using the conservation of momentum: m1v1o m2v2o m1v1f m2v2f m1v1o m1v1f m2v2f m1(v1o v1f) m2v2f (eq. 1) Using the conservation of kinetic energy: (eq. 2) m1(v1o2 v1f2) m2v2f2 Dividing equation 2 by equation 1: m1(v1o2 v1f2) m2v2f2 m1(v1o v1f) m2v2f v1o v1f v2f v1f v2f v1o (eq. 3) Substituting equation 3 into equation 1: m1(v1o v1f) m2v2f m1(v1o v2f v1o) m2v2f m1(2v1o v2f) m2(v2f) v1o(2m1) v2f(m1 m2) 2m1v1o v2f m1 m2 m1 m2 57. a) v1f v1o m1 m2 15 kg 3 kg v1f (3 m/s) 15 kg 3 kg v1f 2 m/s 2m1 v2f v1o m1 m2 2(15 kg) v2f (3 m/s) 15 kg 3 kg v2f 5 m/s 1 b) Ek mv2 2 1 Ek (3 kg)(5 m/s)2 2 Ek 37.5 J Ek 38 J 58. pTf pTo (m1 m2)vf m1v1o m2v2o (0.037 kg)vf (0.035 kg)(8 ms) (0.002 kg)(12 m/s) vf 6.9 m/s


131

59. a) pTo m1v1om2v2o pTo (3.2 kg)(2.2 m/s) (3.2 kg)(0) pTo 7.0 kg·m/s 1 Ek-To mv2 2 1 Ek-To (3.2 kg)(2.2 m/s)2 2 Ek-To 7.7 J b) Using the conservation of momentum and m1 m2: m1v1o m2v2o m1v1f m2v2f 2.2 m/s 0 1.1 m/s v2f v2f 1.1 m/s 1 1 c) Ek-Tf mv1f2 mv2f2 2 2 1 Ek-Tf (3.2 kg)(1.1 m/s)2 2 1 (3.2 kg)(1.1 m/s)2 2 Ek-Tf 3.8 J d) The collision is not elastic since there was a loss of kinetic energy from 7.7 J to 3.8 J. 60. pTo pTf m1v1o m2v2o m1v1f m2v2f (0.015 kg)(375 m/s) 0 (0.015 kg)(300 m/s) (2.5 kg)v2f v2f 0.45 m/s 61. m1 6m v1o 5 m/s m2 10m v2o 3 m/s Changing the frame of reference so that v2f 0: v1o 8 m/s m1 m2 v1f v1o m1 m2 6m 10m v1f (8 m/s) 6m 10m v1f 2 m/s 2m1 v2f v1o m1 m2 2(6m) v2f (8 m/s) 6m 10m v2f 6 m/s Returning to our original frame of reference (subtract 3 m/s): v1f 2 m/s 3 m/s 5 m/s, v2f 6 m/s 3 m/s 3 m/s 132

m1 m2 62. a) v1f v1o m1 m2 3m2 m2 v1f (5 m/s) 3m2 m2 v1f 2.5 m/s 2m1 b) v2f v1o m1 m2 2(3m2) v2f (5 m/s) 3m2 m2 v2f 7.5 m/s 63. mw 0.750 kg k 300 N/m mb 0.03 kg x 0.102 m a) Ee-gained Ek-lost 1 1 kx2 mv2 2 2 (300 N/m)(0.102 m)2 (0.78 kg)v2 v 2 m/s Using the conservation of momentum: pTo pTf mbvbo mwvwo m(bw)vf (0.03 kg)vbo 0 (0.78 kg)(2.0 m/s) vbo 52 m/s b) The collision is inelastic since: 1 Eko (0.03 kg)(52 m/s)2 2 Eko 40.56 J and Ekf 0 The kinetic energy is not conserved. 1 64. a) mgh mv2 2 1 (2.05 kg)(9.8 m/s)(0.15 m) (2.05 kg)v2 2 v 1.7 m/s b) m1v1 v2(m1 m2) (0.05 kg)v1 (1.71 m/s)(2.05 kg) v1 70 m/s 65. Using the conservation of momentum and m1 m2 m: pTo pTf mv1o mv2o mv1f mv2f v1o 0 v1f v1f (eq. 1)


Using the conservation of kinetic energy: EkTo EkTf 1 1 1 1 mv1o2 mv2o2 mv1f2 mv2f2 2 2 2 2 2 2 2 v1o 0 v1f v2f v1o2 v1f2 v2f2 (eq. 2) Equation 1 can be represented by the vector diagram: v1o

v1f

θ

v2f

The angle is the angle between the final velocity of the eight ball and the cue ball after the collision. Using the cosine law and equation 2: v1o2 v1f2 v2f2 2(v1f)(v2f) cos v1o2 v1o2 2(v1f)(v2f) cos 0 2(v1f)(v2f) cos 0 cos 90° Therefore, the angle between the two balls after collision is 90°.


133

Chapter 6 13. vi 4 km/s 4 103 m/s, vf 80 m/s 1 1 E mvi2 mvf2 2 2 1 E (100 000 kg)[(80 m/s)2 (4000 m/s)2] 2 E 7.9968 1011 J It has released 7.9968 1011 J of energy to the atmosphere. The shuttle’s initial height was 100 km, and it landed on Earth’s surface, therefore its change in height is 100 km. 14. mE 5.98 1024 kg, msat 920 kg, Ek 7.0 109 J a) At the start, the height is rE 6.38 106 m. Therefore, the total energy is ET Eki Epi GMm ET 7.0 109 J r 9 ET 7.0 10 J (6.67 1011 N·m2/kg2)(5.98 1024 kg)(920 kg) 6.38 106 m

ET 5.051 672 1010 J Since velocity is 0 at maximum height, total energy at maximum height Epf GMm ET rh ET

(6.67 1011 N·m2/kg2)(5.98 1024 kg)(920 kg) 6.38 106 m h

3.67 1017 N·m2 ET 6.38 106 m h The total energy is constant, therefore, 3.67 1017 N·m2 5.051 672 1010 J 6.38 106 m h h 884.1 km b) Escape velocity from Earth’s surface is given by: 2GM vesc r

2(6.67 1011 N·m2/kg2)(5.98 1024 kg) vesc (6.38 106 m)

vesc 1.12 104 m/s Therefore, the initial kinetic energy required for the escape should be greater than:

1 E mv2 2 1 E (920 kg)(1.12 104 m/s)2 2 E 5.75 1010 J c) The initial speed needed to keep going indefinitely should be greater than the escape speed, i.e., greater than 11.2 km/s. 15. ms 550 kg, mE 5.98 1024 kg, h 6000 km 6 106 m, rE 6.38 106 m GMm GMm a) Ep rh r

1 1 Ep GMm r rh 11 Ep (6.67 10 N·m2/kg2) (5.98 1024 kg)(550 kg) 1 6.38 106 m 6 106 m

1 6.38 106 m Ep 1.67 1010 J b) At the maximum height of 6000 km, the kinetic energy is 0 since the velocity is zero. Therefore, the change in Ep is the initial kinetic energy, i.e., Eki 1.67 1010 J. 16. mE 5.98 1024 kg, rE 6.38 106 m, mm 20 000 kg, vi 3.0 km/s 3.0 103 m/s, h 200 km 2.0 105 m Since the meteorite is headed from outer space, 1 Epi 0 and Eki mv2i 2 1 Therefore, ET mv2i 2 At 200 km, ET Ekf Epf 1 1 GMm mv2i mvf2 2 hr 2 GM 1 1 v2i vf2 2 hr 2 1 1 (3.0 103 m/s)2 vf2 2 2 (6.67 1011 N·m2/kg2)(5.98 1024 kg) 6.58 106 m

vf 11 412.1 m/s The meteorite’s speed 200 km above Earth’s surface is approximately 11.4 km/s. 134


17. vesc c 2.99 108 m/s, mE 5.98 1024 kg vesc

2GM r

2GM r 2 (vesc) r

v

2(6.67 1011 N·m2/kg2)(5.98 1024 kg) (2.99 108 m/s)2

GM rh

(6.67 1011 N·m2/kg2)(5.98 1024 kg) v 6.38 106 m 4.0 105 m

r 8.92 103 m 18. Given: dEM 3.82 108 m, mMoon 7.35 1022 kg, mEarth 5.98 1024 kg Equating the forces of gravity between Earth and the Moon, using the distance from Earth as r, GMMoonm GMEarthm (3.82 108 m r)2 r2 MEarth MMoon 8 2 (3.82 10 m r) r2 2 MMoonr MEarth(3.82 108 m r)2 0 MEarth(1.46 1017 m 7.64 108r r2) MMoonr2 0 8.73 1041 m 4.57 1033r 5.98 1024r2 7.36 1022r2 0 5.91 1024r2 4.57 1033r 8.73 1041 m r 4.29 108 m, 3.45 108 m The forces of gravity from Earth and the Moon are equal at both 4.43 108 m and 3.45 108 m from Earth’s centre. 19. mEarth 5.98 1024 kg, mMoon 7.35 1022 kg, rE 6.38 106 m, rM 1.738 106 m Let m be the mass of the payload. GMMoonm GM m Earth E r R E (6.67 1011 N·m2/kg2) 5.98 1024 kg 7.35 1022 kg 6 1.738 10 m 6.38 106 m E 5.97 107 J The energy required to move a payload from Earth’s surface to the Moon’s surface is 5.97 107 J/kg.

20. mE 5.98 1024 kg, rE 6.38 106 m, h 400 km 4.0 105 m Orbital speed is given by:

v 7.67 km/s The period of the orbit is the time required by the satellite to complete one rotation around Earth. Therefore, the distance travelled, d, is the circumference of the circular orbit. Therefore, d 2π(r h) d 2(3.14)(6.38 106 m 4.0 105 m) d 42 599 996 m Hence, speed is given by, d v T d T v 42 599 996 m T 7670 m/s T 5552 s The period of the orbit is 5552 s or 92.5 min. 21. mE 5.98 1024 kg, rE 6.37 106 m Since the orbit is geostationary, it has a period of 24 h 86 400 s. Using Kepler’s third law, GM r3 2 4 2 T GMT2 r 4 2

r

1 3

(6.67 1011 N·m2/kg2)(5.98 1024 kg)(86 400 s)2 4(3.14)2

1 3

r 4.22 10 m Subtracting Earth’s radius, r 4.22 107 m 6.37 106 m r 3.59 107 m The satellite has an altitude of 3.59 104 km. 22. mE 5.98 1024 kg, rE 6.37 106 m, r1 320 km 3.2 105 m, r2 350 km 3.5 105 m 7


135

Energy added to the station’s orbit is given by: GMm GMm E r2 rE r1 rE

1 1 E GMm r2 rE r1 rE E (6.67 1011 N·m2/kg2) (5.98 1024 kg)m 1 1 6.73 106 m 6.70 106 m E 2.65 105m J The shuttle has added 2.65 105m J of energy to the station’s orbit. 23. a) The total energy of a satellite in an orbit is the sum of its kinetic and potential energies. In all cases, total energy remains constant. Therefore, when r is increased, the gravitational potential energy increases as GMm Ep . As r increases, the energy r increases as it becomes less negative. Thus, when potential energy increases, kinetic energy decreases to maintain the total 1 energy a constant. Since Ek mv2, if 2 kinetic energy decreases, v also decreases and when r increases, v decreases. r3 b) In Kepler’s third law equation 2 K, T r is directly proportional to T. Therefore, as r increases, T also increases. 24. mE 5.98 1024 kg, mM 7.35 1022 kg, r 3.82 108 m The Moon’s total energy in its orbit around Earth is given by: 1 ET Ep 2 1 GMm ET 2 r 1 (6.67 10 N·m /kg )(5.98 10 kg)(7.35 10 kg) ET 3.82 10 m 2 28 ET 3.84 10 J

11

2

2

24

22

25. mSaturn 5.7 1026 kg, rSaturn 6.0 107 m Equating two equations for kinetic energy, 1 GMm mv2 2 2r v

GM r

(6.67 1011 N·m2/kg2)(5.7 1026 kg) v 6 107 m

v 2.5 10 m/s If an object is orbiting Saturn, it must have a minimum speed of 2.5 104 m/s. 26. mM 7.35 1022 kg, r rM 100 km r 1.738 106 m 1 105 m r 1.838 106 m 4

vesc

2Gm r Moon

2(6.67 1011 N·m2/kg2)(7.35 1022 kg) vesc 1.838 106 m

vesc 2.31 10 m/s The escape speed from the Moon at a height of 100 km is 2.31 km/s. 27. According to Kepler’s third law, GM r3 2 4 2 T 4 2r3 T2 GmMoon 3

4(3.14) (1.838 10 m) T 11 2 2 22 2

(6.67 10

3

N·m /kg )(7.35 10 kg)

T 7071 s It would take the Apollo spacecraft 7071 s or 1 h 58 min to complete one orbit around the Moon. 28. dMS 2.28 1011 m, rM 3.43 106 m, mM 6.37 1023 kg, mS 2.0 1030 kg a) Orbital speed is given by: v

8

GM r

(6.67 1011 N·m2/kg2)(2.0 1030 kg) v 2.28 1011 m

v 24.2 km/s

136

6


b) h 80 km 8 104 m v

Gm rh 11

(6.67 10 N·m /kg )(6.37 10 kg) v 6 4 2

2

23

3.43 10 m 8 10 m

v 3.48 km/s The speed required to orbit Mars at an altitude of 80 km is 3.48 km/s. 29. mM 7.35 1022 kg, rM 1.738 106 m Escape speed is given by: vesc

2GM r 11

2(6.67 10 N·m /kg )(7.35 10 kg) vesc 6 2

2

22

1.738 10 m

vesc 2.38 km/s 30. Three waves pass in every 12 s, with 2.4 m between wave crests. number of waves f time 3 f 12 s f 0.25 Hz 31. k 12 N/m, m 230 g 0.23 kg, A 26 cm 0.26 m At the maximum distance, i.e., A, v 0, therefore the total energy is: 1 E kA2 2 Also, at the equilibrium point, the displacement is zero, therefore the total energy is the kinetic energy: 1 E mv2 2 Hence, 1 1 kA2 mv2 2 2 v

v

(12 N/m)(0.26 m) 0.23 m

kA2 m

2

v 1.88 m/s The speed of the mass at the equilibrium point is 1.88 m/s.

32. m 2.0 kg, x 0.3 m, k 65 N/m 1 a) E kx2 2 1 E (65 N/m)(0.3 m)2 2 E 2.925 J Initial potential energy of the spring is 2.925 J. b) Maximum speed is achieved when the total energy is equal to kinetic energy only. Therefore, 1 E mv2 2 1 2.925 J (2.0 kg)v2 2 v 2.925 v 1.71 m/s The mass reaches a maximum speed of 1.71 m/s. c) x 0.20 m Total energy of the mass at this location is given by: 1 1 E mv2 kx2 2 2 1 2.925 J (2.0 kg)v2 2 1 (65 N/m)(0.2 m)2 2 v 1.625 v 1.275 m/s The speed of the mass when the displacement is 0.20 m is 1.275 m/s. 33. Given the information in problem 32, a) Maximum acceleration is achieved when the displacement is maximum since F kx and F ma Therefore, maximum displacement is x 0.30 m Hence, ma kx kx a m (65 N/m)(0.30 m) a (2.0 kg) a 9.75 m/s2 The mass’ maximum acceleration is 9.75 m/s2.


137

b) x 0.2 m kx a m (65 N/m)(0.2 m) a 2.0 kg a 6.5 m/s2 The mass’ acceleration when the displacement is 0.2 m is 6.5 m/s2. 34. dtide 15 m, mfloats m, spanfloats 10 km, Ttide 12 h 32 min 45 120 s a) Finding the work done by the upward movement of the floats, Wup Fg d Wup m(9.8 m/s2)(15 m) Wup 147m J Since there is a downward movement as well, Wup, down 2Wup Wup, down 294m J Since the linkages are only 29% efficient, Wactual 0.29(294m J) Wactual 85.26m J To find power: W P t 85.26m J P 45 120 s P 1.89 103m W P 1.89m mW The power produced would be 1.89m mW. b) 1.89m mW from the hydroelectric linkages is not even comparable to 900 MW from a reactor at Darlington Nuclear Power Station. In order for the linkages to produce the same power, the total mass of the floats would have to be 4.76 1011 kg, or 476 million tonnes. 35. m 100 kg, d 12 m, x 0.64 cm 0.0064 m First, we must find the speed at which the mass first makes contact with the spring. Using kinematics, v2 vo2 2ad vo2 2 ad v 2 v 0 2( 9.8 m/s )(12 m) v 15.34 m/s 138

Finding the maximum kinetic energy of the mass (instant before compression of spring), 1 Ekmax mv2 2 1 Ekmax (100 kg)(15.34 m/s)2 2 Ekmax 11 760 J Since kinetic energy is fully converted to elastic potential energy when the spring is fully compressed, 1 Epmax kx2 2 2Epmax k x2 2(11 760 J) k 2 (0.0064 m) k 5.7 108 N/m The spring constant is 5.7 108 N/m. 36. k 16 N/m, A 3.7 cm Since total energy is equal to maximum potential energy, maximum amplitude x at the point of maximum potential energy: Ep Etotal 1 Ep kx2 2 1 Ep (16 N/m)(0.037 m)2 2 Ep 0.011 J The total energy of the system is 0.011 J. 37. mbullet 5 g 0.005 kg, mmass 10 kg, k 150 N/m, vo bullet 350 m/s To find the final velocity, use the law of conservation of linear momentum: p o pf (0.005 kg)(350 m/s) 0 (10.005 kg)v v 0.175 m/s Therefore, the mass and bullet’s kinetic energy is: 1 Ek mv2 2 1 Ek (10.005 kg)(0.175 m/s)2 2 Ek 0.153 J


Since all of this energy is transferred to elastic potential energy, Ep Ek 1 kx20.153 J 2 x

2(0.153 J) 150 N/m

x 0.045 m 38. b 0.080 kg/s, m 0.30 kg, xo 8.5 cm, bt 2m

x xoe a) t 0.1 s x (8.5 cm)e x 8.39 cm b) t 1.5 s

(0.080 kg/s)(0.1 s) 2(0.30 kg)

x (8.5 cm)e x 6.96 cm c) t 15.5 s

(0.080 kg/s)(1.5 s) 2(0.30 kg)

(0.080 kg/s)(180 s) 2(0.30 kg)

(0.080 kg/s)(18 720 s) 2(0.30 kg)

(0.080 kg/s)(0.1 s) 0.30 kg

bt 2m

(0.080 kg/s)(22.3 s) 0.30 kg

1 xo xoe 2

bt 2m

(0.080 kg/s)t 2(0.30 kg)

(0.080 kg/s)t ln 0.5 0.30 kg t 5.2 s Therefore, the time required for the amplitude to reach one-half its initial value is 5.2 s. 40. k 100 N/m 1 E kxo2e 2 bt m

bt m

x (8.5 cm)e x 3.21 1010 cm e) t 5.2 h 18 720 s

0.5 e

bt m

(0.080 kg/s)(15.5 s) 2(0.30 kg)

x xoe

bt m

bt m

x (8.5 cm)e x 1.076 cm d) t 3.0 min 180 s

x (8.5 cm)e x 0 cm 1 39. x xo 2 Hence,

a) Initial energy: 1 Eo kxo2e (t 0 s) 2 1 Eo kxo2e0 2 1 Eo kxo2 2 One-half of the initial energy is: 1 1 1 kxo2 kxo2 2 2 4 Therefore, the time required for the energy to reach this value is: 1 Ef kxo2e 2 1 1 kxo2 kxo2e 4 2 1 e 2 1 bt ln 2 m (0.080 kg/s)t 1 ln 0.3 kg 2 t 2.6 s Therefore, it takes 2.6 s for the mechanical energy to drop to one-half of its initial value. b) i) t 0.1 s 1 E (100 N/m)(0.085 m)2e 2 E 0.352 J ii) t 22.3 s 1 E (100 N/m)(0.085 m)2e 2 E 9.45 104 J iii) t 2.5 min 150 s 1 E (100 N/m)(0.085 m)2e 2 E 1.53 1018 J iv) t 5.6 a 176 601 600 s 1 E (100 N/m)(0.085 m)2e 2 E0J


(0.080 kg/s)(150 s) 0.30 kg

(0.080 kg/s)(176 601 600 s) 0.30 kg

139

Chapter 7 1° 17. a) 0.0175 rad 57.3°/rad 1 b) 90° 2 rad 4

90° rad 2 220° c) 3.84 rad 57.3°/rad 459° d) 8.01 rad 57.3°/rad 1200° e) 20.9 rad 57.3°/rad 18. a) (15.3 rev)(2 rad/rev) 96.1 rad 3 2 rad 3

b) turn rad 4 turn 2 2 rad c) 4.4 h 2.3 rad 12 h 2 rad d) 28.5 h 7.46 rad 24 h 19. a) 0 rad 0° 2

b) rad (57.3°/rad) 120° 3 c) (20 rad)(57.3°/rad) 3600° d) (466.6 rad)(57.3°/rad) 2.67 104° 3.5 rad 20. a) 0.56 cycles 2 rad/cycle 1 b) rad cycle 2 1 cycle c) 50° 0.14 cycle 360° 1 cycle d) 450° 1.25 cycles 360° 21. a) s r s (40 m)(2 rad) s 80 m b) s r s (40 m)(6.7 rad) s 268 m 124° c) 2.16 rad 57.3°/rad s r s (40 m)(2.16 rad) s 86 m

140

22.

23.

24.

25.

26.

560° d) 9.77 rad 57.3°/rad s r s (40 m)(9.77 rad) s 3.9 102 m a) (15 cycles)(2 rad/cycle) 30 rad b) t 3.5 s t 30 rad 0 3.5 s 27 rad/s c) and become negative. t 26 s (4 cycles)(2 rad/cycle) 8 rad t 8 rad 0 26 s 0.97 rad/s 2 rad 1700 rev 1 min a) 1 min 60 s rev 178.0 rad/s b) t t (178.0 rad/s)(0.56 s) 1.0 102 rad a) 1 0 2 2.55 rad/s t 115 s (2 1) t 2.55 rad/s 0 115 s 0.0222 rad/s2 2.55 rad/s b) fmax 2 rad/cycle fmax 0.406 Hz 2 rad 45 rev 60 s 1 1 min 1 min 1 rev 1 4.7 rad/s 2 0 t 22.5 s


(2 1) t 0 4.7 rad/s 22.5 s 0.21 rad/s2 27. 1 18.0 rad/s 2 0 t 22.0 s (2 1) a) t (0 18.0 rad/s) 22.0 s 0.818 rad/s2 b) 22 12 2 12 2 (18.0 rad/s)2 2(0.818 rad/s2) 198 rad 198 rad c) number of cycles 2 rad/cycle number of cycles 31.5 d) 2 1 t 2 18.0 rad/s (0.818 rad/s2)(8.7 s) 2 11 rad/s 28. 0.95 rad/s2 1 1.2 rad/s a) t 0.30 s 2 1 t 2 1.2 rad/s (0.95 rad/s2)(0.30 s) 2 0.92 rad/s b) t 1.26 s 2 1 t 2 1.2 rad/s (0.95 rad/s2)(1.26 s) 2 3.0 103 rad/s c) t 13.5 s 2 1 t 2 1.2 rad/s (0.95 rad/s2)(13.5 s) 2 12 rad/s 29. r 0.028 m v 0.12 m/s v r v r 0.12 m/s 0.028 m 4.3 rad/s

Therefore, the angular speed of the reel is approximately 4.3 rad/s. 30. r 0.50 m (3.5 rev/s)(2 rad/rev) 22 rad/s ac r2 ac (0.50 m)(22 rad/s)2 ac 2.4 102 m/s2 31. ac 7.98 m/s2 2.50 103 m r 2 r 1.25 103 m v2 a) ac r acr v v (7.98 m/s2)( 1.25 103 m) v 99.9 m/s b) 0 v c) r 99.9 m/s 1.25 103 m 0.0799 rad/s number of revolutions 0.0799 rad/s 3600 s 24 h 2 rad/rev 1h number of revolutions 1.10 103 60 s 0.0799 rad d) 45 min 1s 1 min 216 rad s r s (1.25 103 m)(216 rad) s 2.70 105 m 32. c 3.0 108 m/s r 0.80 m d 2(10.0 km) d 20 000 m d a) t c 20 000 m t 3.0 108 m/s t 6.7 105 s 1° 57.3°/rad 0.0174 rad


141

t 0.0174 rad 6.7 105 s 2.6 102 rad/s b) v r v (0.80 m)(2.6 102 rad/s) v 2.1 102 m/s 33. Both people are travelling at the same angular speed but in the opposite direction. Therefore, they will meet halfway, after each

person has travelled rad. 2 t t

rad 2 t 1.3 rad/s t 1.2 s 34. A 1.3 rad/s B 1.6(1.3 rad/s) B 2.08 rad/s A B

B A tA tB A B A B A

A 1.3 rad/s 2.08 rad/s (2.08 rad/s)A (1.3 rad/s) (1.3 rad/s)A A 1.208 rad t tA A t A 1.208 rad t 1.3 rad/s t 0.93 s 35. 1 4.2 rad/s 1.80 rad/s2 t 2.8 s a) 2 1 t 2 4.2 rad/s (1.80 rad/s2)(2.8 s) 2 9.2 rad/s

142

1 b) 1t t2 2 (4.2 rad/s)(2.8 s) 1 (1.8 rad/s2)(2.8 s)2 2 19 rad 36. 1 190 rad/s 2 80 rad/s t 6.4 s 2 1 a) t 80 rad/s 190 rad/s 6.4 s 42 rad/s2 (2 1) b) t 2 (80 rad/s 190 rad/s) (6.4 s) 2 2 3.5 10 rad c) (3.5 102 rad)(57.3°/rad) 2.0 104° d) 2 0 42 rad/s2 2 1 t 0 190 rad/s t 42 rad/s2 t 4.5 s 37. 3.8 rad/s2 t 3.5 s 110 rad 1 a) 1t t2 2 1 t2 2 1 t 1 110 rad (3.8 rad/s2)(3.5 s)2 2 1 3.5 s 1 24.77 rad/s 1 25 rad/s b) 22 12 2 2 rad/s) 2(3.8 rad/s2) (110 ra d) 2 (25 2 38.22 rad/s 2 38 rad/s


2 1 c) t 38.22 rad/s 24.77 rad/s 3.5 s 2 3.8 rad/s (2 1) d) t 2 (38.22 rad/s 24.77 rad/s) (3.5 s) 2 110 rad 1 2t t2 2 (38.22 rad/s)(3.5 s) 1 (3.8 rad/s2)(3.5 s)2 2 110 rad 1 min 2 rad 400 rev 38. 1 1 min 60 s rev 1 41.9 rad/s t 1.2 s a) (10 turns)(2 rad/turn) 20 rad (1 2) b) t 2 2 2 1 t 2(20 rad) 2 41.9 rad/s 1.2 s 2 62.8 rad/s 2 63 rad/s 2 1 c) t 62.8 rad/s 41.9 rad/s 1.2 s 2 17 rad/s 39. 2.0 104 rad 1 3.5 103 rad/s 2 2.5 104 rad/s (1 2) a) t 2 2 t 1 2 t

2(2.0 104 rad) (3.5 103 rad/s) (2.5 104 rad/s)

t 1.4 s

2 1 b) t (2.5 104 rad/s) (3.5 103 rad/s) 1.4 s 4 2 1.5 10 rad/s 40. 1.5 104 rad/s2 (from 39b) 1 0 2 3.5 104 rad/s 2 1 t 3.5 104 rad/s 0 t 1.5 104 rad/s2 t 2.3 s 41. 2 15 rad/s t 3.4 s 2.3 rad/s2 1 a) 2t t2 2 1 (15 rad/s)(3.4 s) (2.3 rad/s2)(3.4 s)2 2 38 rad 2 1 b) t 1 2 t 1 15 rad/s (2.3 rad/s2)(3.4 s) 1 7.2 rad/s 42. TM 5.94 107 s TE 3.16 107 s headstart 30°

headstart rad 6

rad Mt Et 6

rad 6 t M E

rad 6 t 2 rad 2 rad 5.94 107 s 3.16 107 s

t 5.63 106 s 43. A 0.380 rad/s B 0.400 rad/s A 0.080 rad/s2 B 0


143

25° headstart 57.3°/rad headstart 0.436 rad A 0.436 rad B 1 At At2 0.436 rad Bt 2 1 0 At2 At Bt 0.436 rad 2 1 0 (0.080 rad/s2)t2 (0.380 rad/s)t 2 (0.400 rad/s)t 0.436 rad 0 (0.040 rad/s2)t2 (0.020 rad/s)t 0.436 rad Use the quadratic formula: t

2 0.020 rad/s (0.0 20 rad/s) 4(0.040 rad /s2)(0.436 rad) 2(0.040 rad/s2)

t 3.56 s 1 44. I mr2 2 Ia 14 kg·m2 Ib 4.8 kg·m2 Ic 6.8 kg·m2 The order is a, c, b. 45. Ia mr2 1 Ib mr2 2 1 2 Ic mr 2 1 Id (3m)l2 12 1 Id (3m)(2r)2 12 Id mr2 2 Ie mr2 5 2 1 2 Ie (2m) r 5 2 1 2 Ie mr 5 The order is a and d, b and c, e. 46. m 4200 kg r 0.3 m 1 I mr2 2 1 I (4200 kg)(0.3 m)2 2 I 189 kg·m2

144

47. m 3.5 kg a) I mr2 I (3.5 kg)(0.21 m)2 I 0.15 kg·m2 1 b) I mr2 2 1 I (3.5 kg)(0.21 m)2 2 I 0.077 kg·m2 2 c) I mr2 5 2 I (3.5 kg)(0.25 m)2 5 I 0.088 kg·m2 1 d) I mr2 2 1 I (3.5 kg)(0.50 m)2 2 I 0.44 kg·m2 48. m 1.4 kg r 0.12 m 1 a) I mr2 2 1 I (1.4 kg)(0.12 m)2 2 I 0.010 kg·m2 b) (60 rev/s)(2 rad/rev) 120 rad/s 377 rad/s 49. m 10.0 kg 1 ri (0.54 m) 2 ri 0.27 m 1 re (0.54 m)(1.4) 2 re 0.378 m 1 I m(ri2 re2) 2 1 I (10.0 kg)[(0.27 m)2 (0.378 m)2] 2 I 1.08 kg·m2 50. m 2.0 kg r 1.5 m 2 a) I mr2 3 2 I (2.0 kg)(1.5 m)2 3 I 3.0 kg·m2


1 min 2 rad 200 rev b) 1 min 60 s 1 rev 20.9 rad/s 2 c) I mr2 5 2 I (2.0 kg)(1.5 m)2 5 I 1.8 kg·m2 51. m 20 kg r 0.9 m 2 0 1 (12.3 rev/s)(2 rad/rev) 1 77.3 rad/s WR WR I (1 2) WR (mr2) t t 2 1 WR mr2(0 1)(1 0) 2 1 WR (20 kg)(0.9 m)2(77.3 rad/s)2 2 WR 4.8 104 J 52. m 1450 kg 1.35 m r 2 r 0.675 m 1.40 rad/s 1 a) I mr2 2 1 I (1450 kg)(0.675 m)2 2 I 330 kg·m2 1 b) Erot I2 2 1 Erot (330 kg·m2)(1.40 rad/s)2 2 Erot 3.24 102 J c) v r v (0.675 m)(1.40 rad/s) v 0.945 m/s d) t t (1.40 rad/s)(6.5 s) 9.1 rad

9.1 rad number of turns 2 rad/turn number of turns 1.4 53. m 5.98 1024 kg r 6.38 106 m t 3.16 107 s 2 rad 1 a) Erot I2 2 1 2 Erot mr 2 2 2 5 1 2 Erot mr2 t 5 1 Erot (5.98 1024 kg)(6.38 106 m)2 5 2 2 rad 7 3.16 10 s Erot 1.92 1024 J b) v r 2 rad v (6.38 106 m) 3.16 107 s v 1.27 m/s 54. m 8.30 1025 kg r 3.5 m a) Ie mr2 Ie (8.30 1025 kg)(3.5 m)2 Ie 1.0 1023 kg·m2 (1000 cycles)(2 rad/cycle) b) 1.0 s 3 6.3 10 rad/s 1 c) Ek mv2 2 1 Ek m(r)2 2 1 Ek (8.30 1025 kg)(3.5 m)2 2 (6.3 103 rad/s)2 Ek 2.0 1016 J 55. me 9.11 1031 kg mn 1.67 1027 kg r 5.0 1011 m L 1.05 1034 kg·m2/s a) I mer2 I (9.11 1031 kg)(5.0 1011 m)2 I 2.3 1051 kg·m2


145

b) L I L I 1.05 1034 kg·m2s 2.3 1051 kg·m2 4.6 1016 rad/s 1 c) Erot I2 2 1 Erot (2.3 1051 kg·m2) 2 (4.6 1016 rad/s)2 Erot 2.4 1018 J 56. r 0.20 m h1 2.5 m h2 0 v1 0 1 0 1 1 a) mgh1 mv12 I12 2 2 1 1 mgh2 mv22 I22 2 2 1 1 1 mgh1 mv22 mr2 22 2 2 2 v 2 1 1 gh1 v22 r2 2 r 2 4 1 2 1 2 gh1 v2 v2 2 4 3 gh1 v22 4

3 gh 4 (9.8 m/s )(2.5 m) 3

v2 v2

4

1

v2 5.7 m/s v b) 2 2 r 5.7 m/s 2 0.20 m 2 29 rad/s 57. r 0.20 m h1 2.5 m h2 0 v1 0 1 0

2

1 1 a) mgh1 mv12 I12 2 2 1 1 mgh2 mv22 I22 2 2 1 1 mgh1 mv22 (mr2)22 2 2 1 2 1 2 v2 2 gh1 v2 r r 2 2 1 2 1 2 gh1 v2 v2 2 2 gh1 v22 v2 gh1 2 v2 (9.8 m/s )(2.5 m) v2 4.9 m/s v b) 2 2 r 4.9 m/s 2 0.20 m 2 25 rad/s 1 1 58. mgh1 mv12 I12 2 2 1 1 mgh2 mv22 I22 2 2 1 1 2 2 v2 2 2 mgh1 mv2 mr r 2 2 5 1 1 gh1 v22 v22 2 5 10 v2 gh1 7 59. l 2.8 m r 2.8 m h1 2.8 m h2 0 v1 0 1 0 1 1 mgh1 mv12 I12 2 2 1 1 mgh2 mv22 I22 2 2 1 1 1 v 2 mgh1 mv22 ml2 2 r 2 2 3 1 2 1 2 gh1 v2 v2 2 6 2 2 gh1 v2 3 3 v2 gh1 2

v2

3 (9.8 m/s )(2.8 m) 2

v2 6.4 m/s 146


2

60. m 3.9 kg r 0.13 m 150 rad/s 2 a) I mr2 5 2 I (3.9 kg)(0.13 m)2 5 I 0.0264 kg·m2 b) L I L (0.0264 kg·m2)(150 rad/s) L 4.0 kg·m2/s 61. m 2.4 kg r 0.30 m 1 0 2 250 rad/s t 3.5 s 1 a) I mr2 2 1 I (2.4 kg)(0.30 m)2 2 I 0.108 kg·m2 b) 2 1 250 rad/s 0 250 rad/s c) L L2 L1 L I L (0.108 kg·m2)(250 rad/s) L 27 kg·m2/s d) t 250 rad/s 3.5 s 71.4 rad/s2 e) I (0.108 kg·m2)(71.4 rad/s2) 7.7 N·m 62. I 1.50 103 kg·m2 d 4.5 m (3.0 turns)(2 rad/turn) 6.0 rad a) v 17.0 m/s d t v 4.5 m t 17.0 m/s t 0.2647 s t 0.26 s

b) t 6.0 rad 0.2647 s 71 rad/s c) L I L (1.50 103 kg·m2)(71 rad/s) L 0.11 kg·m2/s 63. l 2.5 m m 3.2 kg t 13 s r 0.010 m (28 turns)(2 rad/turn) 56 rad 1 a) I mr2 2 1 I (3.2 kg)(0.010 m)2 2 I 1.6 104 kg·m2 b) L I 56 rad L (1.6 104 kg·m2) 13 s L 2.2 103 kg·m2s 64. l 2.5 m m 3.2 kg t 13 s (28 turns)(2 rad/turn) 56 rad 1 a) I ml2 12 1 I (3.2 kg)(2.5 m)2 12 I 1.667 kg·m2 I 1.7 kg·m2 b) L I 56 rad L (1.667 kg·m2) 13 s 2 L 22 kg·m s 65. m 3.2 kg l1 2.5 m 2.5 m l2 0.5 m 2 l2 0.75 m


147

I Icm ml22 1 I ml12 ml22 12 I 1.7 kg·m2 (3.2 kg)(0.75 m)2 I 3.5 kg·m2 66. 1 6.85 rad/s 2 4.40 rad/s where x is the factor by which I1 xI2 the moment of inertia changes. 1 x 2 6.85 rad/s x 4.40 rad/s x 1.56 67. I11 I22 1 I11 I1 2 2 1 1 2 2 21 2 Therefore, the angular speed will increase by a factor of 2. 68. Im 1.5 103 kg·m2 Is 8.5 kg·m2 s 10 rad/s a) Iss Imm Iss m Im

(8.5 kg·m2)(10 rad/s) m (1.5 103 kg·m2) m 5.7 104 rad/s b) (10 rad)(57.3°/rad) 573° c) (5.7 104 rad)(57.3°/rad) 3.3 106°

d) 45° rad 4 t s s

rad 4 t 10 rad/s t 0.0785 s m mt m (5.7 104 rad/s)(0.0785 s) m 4.45 103 rad 4.45 103 rad number of rotations 2 rad/rotation number of rotations 7.1 102 148

69. rp 4.3 m rt 4.3 m mp 600 kg p 6.4 rad/s mt 35 kg a) Ipp (Ip It)f Ipp f (Ip It) 1 mprp2p 2 f 1 mprp2 mtrt2 2 1 (600 kg)(4.3 m)2(6.4 rad/s) 2 f 1 (600 kg)(4.3 m)2 (35 kg)(4.3 m)2 2 f 5.7 rad/s b) t 3.1 rad/s Ipp Itt (Ip It)f Ipp Itt f (Ip It)

1 mprp2p mtrt2t 2 f 1 mprp2 + mtrt2 2

f

1 (600 kg)(4.3 m)2(6.4 rad/s) + (35 kg)(4.3 m)2(3.1 rad/s) 2 1 (600 kg)(4.3 m)2 (35 kg)(4.3 m)2 2

f 6.0 rad/s c) t 6.4 rad/s Ipp Itt f (Ip It) 1 mprp2p mtrt2t 2 f 1 mprp2 mtrt2 2

f

1 (600 kg)(4.3 m)2(6.4 rad/s) (35 kg)(4.3 m)2(6.4 rad/s) 2 1 (600 kg)(4.3 m)2 (35 kg)(4.3 m)2 2

f 5.0 rad/s 70. m1 30 kg r1 1.5 m m2 20 kg r2 1.0 m 1 12 rad/s


a) I11 (I1 I2)f I11 f (I1 I2) m1r121 f (m1r12 m2r22) (30 kg)(1.5 m)2(12 rad/s) f ((30 kg)(1.5 m)2 (20 kg)(1.0 m)2) f 9.2 rad/s b) I11-i I22-i (I1 I2)f I11-i I22-i f (I1 I2) m1r121-i m2r222-i f (m1r12 m2r22) f

(30 kg)(1.5 m)2(12 rad/s) (20 kg)(1.0 m)2(12 rad/s) ((30 kg)(1.5 m)2 (20 kg)(1.0 m)2)

f 12 rad/s I11-i I22-i c) f (I1 I2)

73. m 0.135 kg I 8.50 105 kg·m2 r 0.0030 m d 1.10 m 1 0 v1 0 g a) a I 1 2 mr

a

t

f 6.5 rad/s d) I11-i I22-i (I1 I2)f I11-i 2-i I2

(30 kg)(1.5 m)2(12 rad/s) (20 kg)(1.0 m)2

2-i 40 rad/s 71. I 250 kg·m2 r1 2.5 m r2 1.5 m t-1 2.0 rad/s md 40 kg I11 I22 I11 2 I2 (I mdr12)1 2 (I mdr22)

e)

f)

[250 kg·m2 (40 kg)(2.5 m)2](2.0 rad/s) (250 kg·m2 (40 kg)(1.5 m)2)

2 2.9 rad/s

c)

d)

m1r121-i 2-i m2r22

2

(30 kg)(1.5 m)2(12 rad/s) (20 kg)(1.0 m)2(12 rad/s) ((30 kg)(1.5 m)2 (20 kg)(1.0 m)2)

2-i

9.8 m/s2 8.50 105 kg·m2 1 2 (0.135 kg)(0.0030 m)

a 0.138 m/s2 1 b) d v1t at2 2 2d t a

m1r121-i m2r222-i f (m1r12 m2r22) f

g)

2(1.10 m) 0.138 m/s 2

t 3.99 s (v2 v1) a t v2 v1 at v2 0 (0.138 m/s2)(3.99 s) v2 0.551 m/s v2 r2 v 2 2 r 0.551 m/s 2 0.0030 m 2 184 rad/s 1 Ek(final) mv22 2 1 Ek(final) (0.135 kg)(0.551 m/s)2 2 Ek(final) 0.0205 J 1 Erot(final) I22 2 1 Erot(final) (8.50 105 kg·m2)(184 rad/s)2 2 Erot(final) 1.43 J ETotal(initial) mgh1 ETotal(initial) (0.135 kg)(9.80 m/s2)(1.10 m) ETotal(initial) 1.46 J


149

74. m 0.135 kg I 8.50 105 kg·m2 r 0.0030 m d 1.10 m v1 1.0 m/s g a) a I 1 2 mr

a

2

9.8 m/s 8.50 105 kg·m2 1 2 (0.135 kg)(0.0030 m)

1 1 g) ETotal(initial) mgh1 mv12 I12 2 2 ETotal(initial) (0.135 kg)(9.80 m/s2)(1.10 m) 1 (0.135 kg)(1.0 m/s)2 2 1 (8.50 105 kg·m2) 2 1.0 m/s 2 0.0030 m ETotal(initial) 6.24 J

a 0.138 m/s2 1 b) d v1t at2 2 1 0 at2 v1t d 2 1 0 (0.138 m/s2)t2 2 (1.0 m/s)t (1.10 m) 0 (0.0690 m/s2)t2 (1.0 m/s)t (1.10 m) Use the quadratic formula: t

c)

d)

e)

f)

150

1.0 m/s (1.0 m /s)2 4(0.0690 m /s2)( 1.10 m) 2(0.0690 m/s2)

t 1.03 s (v2 v1) a t v2 v1 at v2 1.0 m/s (0.138 m/s2)(1.03 s) v2 1.14 m/s v2 r2 v 2 2 r 1.14 m/s 2 0.0030 m 2 380 rad/s 1 Ek(final) mv22 2 1 Ek(final) (0.135 kg)(1.14 m/s)2 2 Ek(final) 0.088 J 1 Erot(final) I22 2 1 Erot(final) (8.50 105 kg·m2)(380 rad/s)2 2 Erot(final) 6.16 J Solutions to End-of-chapter Problems

Chapter 8 33. Positive signs: protons Negative signs: electrons 34. a) No charge b) Negative c) Positive d) No charge e) Positive 35. a) Negative b) Positive c) Negative d) Positive 36. a) Negative b) Electrons 37. a) Glass: positive; silk: negative b) Since they have opposite charges, they will be attracted 38. a) Insulator (non-metallic) b) Conductor (conducts lightning to ground) c) Insulator (non-metallic) d) Insulator (non-metallic) e) Insulator (non-metallic) f) Insulator (non-metallic) 39. Dog hair is positive since a silk shirt rubbed with wool socks would have a negative charge. 40. a) The electroscope becomes positive because it gives up some electrons to the glass rod to reduce the rod’s deficit of electrons. This is called charging by contact. b) The leaves become positively charged as well. In charging by contact, the charged object receives the same charge as the charging rod. c) Negative charges will enter the leaves if the system is grounded. 41. 1 C 6.25 1018 e, q 15 C q (15 C)(6.25 1018 e/C) q 9.38 1019 e 42. q 1.1 C q 1.1 106 C q (1.1 106 C)(6.25 1018 e/C ) q 6.9 1012 e

43. The electroscope has an overall positive charge: q 4.0 1011 e q (4.0 1011 e)(1.602 1019 C/e) q 6.4 108 C 1 44. q (5.4 108 e) 2 1 q (5.4 108 e)(1.602 1019 C/e) 2 q 4.3 1011 C 45. qn 2.4 1012 C (2.4 1012 C)(6.25 1018 e/C) 1.5 107 elementary charges This means that there are 1.5 107 protons in the nucleus, so the neutral atom must have an equal number of electrons: 1.5 107. kqq 46. Fe r2 kqq a) Fe1 2 (4r) kqq Fe1 2 16r 1 Fe1 Fe 16 k(2q)(2q) b) Fe2 r2 4kqq Fe2 r2 Fe2 4Fe 4 c) Fe3 Fe 16 1 Fe2 Fe 4 47. Each sphere loses half of its charge to balance with its identical neutral sphere. 1 1 q1 q1, q2 q2 2 2 kq1q2 Fe1 r21 kq1q2 Fe2 r22

1 1 k q1 q2 2 2 Fe2 2 r2 kq1q2 Fe2 4r22


151

But Fe2 Fe1 kq1q2 kq1q2 2 4r2 r21 1 1 2 2 4r2 r1 r2 r22 1 4

1 Therefore, r2 r1 2 The spheres should be placed one-half their original distance apart to regain their original repulsion. 48. r 100 pm 100 1012 m 1.00 1010 m, q1 q2 1.602 1019 C kq1q2 Fe r2 (9.0 109 N·m2/C2)(1.602 1019 C)2 Fe (1.00 1010 m)2

Fe2 q2 2 Fe1 3q Fe2 1 Fe1 3 1 The magnitude of Fe2 is Fe1, and in the 3 opposite direction of Fe1. 51. a) p

Fe e

Fe 2.3 108 N

Fg

49. r 25.0 cm 0.250 m, Fe 1.29 104 N, 2 q 1 q2 q qo 3 kqq a) Fe r2 q

kq q2 1 r2 Fe2 kq q2 Fe1 1 r2 Fe2 (q)(q) Fe1 (q)(3q)

Fer2 k

(1.29 104 N)(0.25 m)2 q (9.0 109 N·m2/C2) q 3.00 108 C 2 b) q is the original charge on each sphere. 3 3 qo q 2 3 qo (3.00 108 C) 2 qo 4.5 108 C The type of charge, positive or negative, does not matter as long as they are both the same. (Like charges repel.) 50. q1 q, q2 3q qT q (3q) 2q 2q So q1 q2 q 2

b) q1 1.602 1019 C, q2 1.602 1019 C, m 9.1 1031 kg, g 9.8 m/s2 Fg Fe kq1q2 mg r2 r

kq q mg 1 2

(9.0 10 N·m /C )(1.602 10 C) r 31 2 9

(9.1 10

2

2

2

r 5.1 m 52. q1 2.0 106 C, q2 3.8 106 C, q3 2.3 106 C a) r1 0.10 m, r2 0.30 m kq1q3 1Fe3 r21 Fe3

1

(9.0 109 N·m2/C2)( 2.0 106 C)(2.3 106 C) (0.10 m)2

Fe3 4.14 N (attraction) 1Fe3 4.14 N [right] kq2q3 2Fe3 r22 1

Fe3

2

(9.0 109 N·m2/C2)(3.8 106 C)(2.3 106 C) (0.30 m)2

Fe3 0.87 N (repulsion) 2Fe3 0.87 N [left] 2

152

19

kg)(9.8 m/s )


FeT 4.14 N [right] 0.87 N [left] eT 3.3 N [right] F b) r1 0.30 m, r2 0.10 m kq1q3 1Fe3 r21 Fe3

1

(9.0 109 N·m2/C2)( 2.0 106 C)(2.3 106 C) (0.30 m)2

Fe3 0.46 N (attraction) 1Fe3 0.46 N [left] 1

kq2q3 Fe3 r22

2

Fe3

2

(9.0 109 N·m2/C2)(3.8 106 C)(2.3 106 C) (0.10 m)2

Fe3 7.86 N (repulsion) 2Fe3 7.86 N [right] FeT 0.46 N [left] 7.86 N [right] eT 7.4 N [right] F 2

c) 1Fe3 4.14 N (attraction) 1Fe3 4.14 N [left] Fe3 7.86 N (repulsion) 2Fe3 7.86 N [left] eT 4.14 N [left] 7.86 N [left] F eT 12 N [left] F d) The third charge could only be placed to the left or to the right of the two basic charges for the forces to balance and give a force of 0. For the charge to be placed a distance of rx metres to the left of the first charge: 1Fe3 2Fe3 kq1q3 kq2q3 2 rx (0.20 m rx)2 6 ( 2.0 10 C) (3.8 106 C) rx2 (2.0 101 m rx)2 (3.8 106)rx2 (2.0 106)(4.0 102 4.0 101rx rx2) 2 6 (3.8 10 )rx 8.0 108 8.0 107rx 2.0 106rx2 Rearranging: 1.8 106rx2 8.0 107rx 8.0 108 0 Solve for rx using the quadratic formula. 2

rx

(8.0 107) (8.0 107 )2 4 (1.8 106)(8.0 108) 2(1.8 106)

Therefore, the charge must be placed 0.53 m to the left of the first charge. The other answer, 0.084 m, would place the charge between the two base charges and therefore is an inappropriate answer. For a charge placement to the right of the two charges, two inappropriate answers are calculated, meaning that the only possible placement for the charge is at 0.53 m to the left of the first charge. 53. The forces on the test charge from the repulsion by the other two charges must equal one another for the test charge to come to rest there. The force of charge 1 on the test charge (1Fqt) must equal the force of charge 2 on the test charge (2Fqt). 1Fqt 2Fqt kqqt k4qq 2 2t 1 2 r r 3 3 2 4r2 r 9 4(9) 2 r r2 Therefore, the net force on the charge would 1 be 0 if it was placed of the distance 3 between the two charges. 54. q2 q1 q3 1.0 104 C, r1 r2 r3 0.40 m

1

1

q

q

2

3

q

q

30° 120° 30°

40 cm

For q1: The force is the vector sum of two e1 and 3F e1. These two magnitudes forces, 2F must have the same value. kqq 2Fe1 r2 (9.0 109 N·m2/C2)(1.0 104 C)2 2Fe1 (0.40 m)2 2 2Fe1 5.6 10 N 3Fe1 F2eT 2F2e1 3F2e1 2(2Fe1)(2Fe1)(cos 120°) 102 C)2 2(5.6 102 C)2(co s 120°) FeT 2(5.6 2 FeT 9.7 10 N

So rx 0.53 m or 0.084 m.


153

From the isosceles triangle with angles of 30°, the total angle is 30° 60° 90°. eT1 9.7 102 N [up] F eT2 9.7 102 N [left 30° down] F eT3 9.7 102 N [right 30° down] F Each force is 9.7 102 N [at 90° from the line connecting the other two charges]. 55. a) l 2.0 102 m, q1 q2 q3 q4 1.0 106 C 2.0 cm

q

q

q

q

56.

57. The field is similar to the one above, but is now asymmetrical and has its inflection points pushed farther to the right.

2.0 cm

kqq Fe1 r22

2

Fe1

2

(9.0 109 N·m2/C2)( 1.0 106 C)2 (2.0 102 m)2

Fe1 22.5 N [left] 4Fe1 22.5 N [up] 2

58. Parallel plates:

kqq 3Fe1 r23 Fe1

3

+

Coaxial cable:

–

(9.0 109 N·m2/C2)( 1.0 106 C)2 2(2.0 102 m)2

Fe1 11.25 N [left 45° up] From Pythagoras’ theorem: 2(22.5 N)2 2Fe1 4Fe1 2Fe1 4Fe1 31.82 N [left 45° up] Therefore, FeT1 (31.82 N 11.25 N) [left 45° up] FeT1 43.1 N [left 45° up] eT2 43.1 N [right 45° up] F eT3 43.1 N [right 45° down] F eT4 43.1 N [left 45° down] F Each force is 43.1 N [symmetrically outward from the centre of the square]. b) The force on the fifth charge is 0 N because the forces from each charge are balanced. c) Sign has no effect. If the new fifth charge were either positive or negative, the attractive/repulsive forces would still balance one another. 3

154

59. q 2.2 106 C, Fe 0.40 N F e q 0.40 N 2.2 106 C 1.8 105 N/C 60. Fe 3.71 N, 170 N/C F q e 3.71 N q 170 N/C q 2.2 102 C


61. q1 4.0 106 C, q2 8.0 106 C, r 2.0 m kq2 kq1 1 2 1 2 r r 2 2

(9.0 109 N·m2/C2)(8.0 106 C) 2.0 m 2 2

(9.0 109 N·m2/C2)(4.0 106 C) 2.0 m 2 2 3.6 104 N/C Therefore, the field strength is 3.6 104 N/C towards the smaller charge. e 7.5 N [left] 62. a) q 2.0 106 C, F F e q 7.5 N [left] 2.0 106 C 3.8 106 N/C [left]

b) q2 4.9 105 C Take right to be positive. e q F

0.5 m

q

q

0.5 m

q

Fe (4.9 105 C)(3.8 106 N/C) Fe 1.86 102 N The force would be 1.86 102 N [left]. 63. r 0.5 m, q 1.0 102 C kq r2 (9.0 109 N·m2/C2)(1.0 102 C) (0.5 m)2 8 3.6 10 N/C [left] 64. q1 4.0 106 C, q2 1.0 106 C Take right to be positive. 2 1 (9.0 109 N·m2/C2)(4.0 106 C) (0.40 m)2 (9.0 109 N·m2/C2)( 1.0 106 C) (0.30 m)2 3.25 105 N/C [right] 65. r 5.3 1011 m, q 1.602 1019 C kq r2 (9.0 109 N·m2/C2)(1.602 1019 C) (5.3 1011 m)2

66. rT 0.20 m, q1 1.5 106 C, q2 3.0 106 C r22 (0.20 r1)2 1 2 kq1 kq2 r21 r22 1.5 106 C 3.0 106 C 2 r1 r22 r22 2r21 Substitute for r22 and rearrange: 0 r12 0.4r1 4.0 102 0.4 (0.4)2 4(4.0 102) r1 2 2 r1 8.3 10 m, therefore, r2 1.17 101 m 1.2 101 m 0 at 1.2 101 m from the larger charge, or 8.3 102 m from the smaller charge. 67. q1 q2 q3 q4 1.0 106 C, r 0.5 m

q

Since the magnitudes of all four forces are equal, and they are paired with forces in the e2 F e4 and opposite direction (F Fe1 Fe3), there is no net force. Therefore, there is no net field strength. 0 N/C 68. q1 q2 2.0 105 C, r 0.50 m 1 P

q 2

P

q 0.50 m

kq 1 r2 (9.0 109 N·m2/C2)(2.0 105 C) 1 (0.50 m)2 1 7.2 105 N/C 1 2 1 2 and T 2 Therefore, T 2( 2(1 )2(cos 120°) 1) 6 T 1.2 10 N/C [at 90° from the line connecting the other two charges]

5.1 1011 N/C Solutions to End-of-chapter Problems

155

69. q 0.50 C, ∆V 12 V W q∆V W (0.50 C)(12 V) W 6.0 J 70. W 7.0 102 J, ∆V 6.0 V W q V 7.0 102 J q 6.0 V q 1.2 102 C 71. q 1.5 102 C, Fe 7.5 103 N, ∆d 4.50 cm 4.50 102 m W V q Fed V q (7.5 103 N)(4.5 102 m) V 1.5 102 C V 2.3 104 V 72. 130 N/C, Fe 65 N, ∆V 450 V V W q VF W e (450 V)(65 N) W (130 N/C) W 2.3 102 J 73. d 0.30 m, q 6.4 106 C kq V d (9.0 109 N·m2/C2)(6.4 106 C) V 0.30 m 5 V 1.9 10 V 74. a) q1 1.0 106 C, q2 5.0 106 C, r 0.25 m kq1q2 Ee r Ee

(9.0 109 N·m2/C2)( 1.0 106 C)( 5.0 106 C) 0.25 m

Ee 0.18 J (repulsion) kq1q2 b) Ee1 r Ee1

W Ee2 Ee1 W 0.18 J 0.045 J W 0.14 J 75. Position in the field has no bearing on the field strength. 5.0 103 N/C, d 5.0 cm 5.0 102 m V d V (5.0 102 m)(5.0 103 N/C) V 2.5 102 V 76. a) q 1 105 C, 50 N/C Fe q Fe (1 105 C)(50 N/C) Fe 5.0 104 N b) ∆d 1.0 m ∆Ek W ∆Ek Fe∆d ∆Ek (5.0 104 N)(1.0 m) ∆Ek 5.0 104 J c) v 2.5 104 m/s 1 Ek mv2 2 2Ek m v2 2(5.0 104 J) m (2.5 104 m/s)2 m 1.6 1012 kg 77. d1 1.0 109 m, d2 1.0 108 m, q1 q2 1.602 1019 C, m1 m2 9.11 1031 kg Ee E2 E1 kq1q2 kq1q2 Ee d2 d1

(9.0 109 N·m2/C2)( 1.0 106 C)( 5.0 106 C) 1.00 m

Ee1 0.045 J (repulsion) W ∆Ee

156

1 1 Ee kq1q2 d2 d1 19 Ee 2.08 10 J Therefore, the electric potential energy was reduced by 2.08 1019 J, which was transferred to kinetic energy. The energy is spread over both electrons, so the energy for each electron is 1.04 1019 J.


For one electron: 1 Ek mv2 2 v v

qV ma d qV a md (1.602 1019 C)(7.5 103 V) a (3.3 1026 kg)(1.2 m)

2E m k

2(1.04 1019 J) 9.11 1031 kg

v 4.78 105 m/s 78. V2 2V1 and Ek ∆Ee qV With the same charge on each electron, the kinetic energy is also doubled, i.e., Ek2 2Ek1 Ek2 2Ek1 Ek1 Ek1 1 mv22 2 2 1 mv21 2 v22 2v21 v2 2 v1 Therefore, the speed is 1.41 times greater. 79. a) V 15 kV 1.5 104 V, P 27 W, 1 C 6.25 1018 e number of electrons/s

P(6.25 1018 e/C) V

number of electrons/s (27 J/s)

1C 1.5 104 J

(6.25 1018 e/C) number of electrons/s 1.1 1016 b) q 1.602 1019 C, m 9.11 1031 kg Accelerating each electron from rest, Ek Ee 1 mv2 Vq 2 2Vq v m

19

2(1.5 10 V)(1.602 10 C) v 31 4

9.11 10

kg

v 7.3 10 m/s 80. a) d 1.2 m, V 7.5 103 V, m 3.3 1026 kg Fe q 7

a 3.0 1010 m/s2 b) E Vq E (1.602 1019 C)(7.5 103 V) E 1.202 1015 J c) At this speed and energy, relativistic effects may be witnessed. Although the speed may not be what is predicted by simple mechanics, the total energy should be the same but may be partly contributing to a mass increase of the ion. 81. q1 q2 1.602 1019 C, m1 m2 1.67 1027 kg, v1 v2 2.7 106 m/s ∆Ek ∆Ee The total energy for both ions is: kq1q2 1 (2)mv2 r 2 kq1q2 r mv2 r

(9.0 109 N·m2/C2)(1.602 1019 C)2 (1.67 1027 kg)(2.7 106 m/s)2

r 1.9 1014 m 82. a) q 2e, m 6.696 1027 kg, v1v 0 m/s, v1h 6.0 106 m/s, V 500 V, dv 0.03 m, dh 0.15 m Acceleration is toward the negative plate: F a e m q a m qV a mdv 2(1.602 1019 C)(500 V) a (6.696 1027 kg)(3.0 102 m) a 7.97 1011 m/s2


157

Time between the plates is: d t h vh 0.15 m t 6.0 106 m/s t 2.5 108 s Therefore, 1 dh at2 2 1 dh (7.97 1011 m/s2)(2.5 108 s)2 2 dh 2.5 104 m dh 0.025 cm The alpha particle is 3.0 cm 0.025 cm 2.975 cm from the negative plate if it enters at the positive plate or 1.475 cm from the negative plate if it enters directly between the two plates. b) v2v v1v at v2v 0 (7.97 1011 m/s2)(2.5 108 s) v2v 2.0 104 m/s From Pythagoras’ theorem, 2 106 m/s) (2.0 104 m/s) 2 v2 (6.0 v2 6.0 106 m/s 83. d 0.050 m, V 39.0 V V d 39.0 V 0.050 m 7.80 102 N/C 84. 2.85 104 N/C, d 6.35 cm 6.35 102 m V d V (6.35 102 m)(2.85 104 N/C) V 1.81 103 V 85. a) m 2mP 2mn 4(1.67 1027 kg), g 9.80 N/kg, q 2e Fe Fg mg q 4(1.67 1027 kg)(9.80 N/kg) 2(1.602 1019 C) 2.04 107 N/C

158

86.

87.

88.

89.

90.

b) d 3.0 cm 3.0 102 m V d V (3.0 102 m)(2.04 107 N/C) V 6.1 109 V d 0.12 m, V 92 V V d 92 V 0.12 m 7.7 102 N/C 3 106 N/C, d 1.0 103 m V d V (1.0 103 m)(3 106 N/C) V 3 103 V Therefore, 3.0 103 V is the maximum potential difference that can be applied. Exceeding it would cause a spark to occur between the plates. V 50 V, 1 104 N/C V d 50 V d 1 104 N/C d 5.0 103 m V 120 V, 450 N/C V d 120 V d 450 N/C d 2.67 101 m a) m 2.2 1015 kg, d 5.5 103 m, V 280 V, g 9.80 N/kg Fe Fg q mg qV mg d mgd q V q

(2.2 1015 kg)(9.80 N/kg)(5.5 103 m) 280 V

q 4.2 1019 C 4.2 1019 C b) N 1.602 1019 e/C N 2.63 e 3 e The droplet has three excess electrons.


91. V 450 V, me 9.11 1031 kg, e 1.602 1019 C a) ∆Ee qV ∆Ee (1.602 1019 C)(450 V) ∆Ee 7.21 1017 J Therefore, Ek Ee 1 mv2 Ee 2 2Ee v m

v

2(7.21 1017 J) 9.11 1031 kg

v 1.26 107 m/s 1 1 b) mv2 Ee 2 3 2Ee v 3m

v

2(7.21 1017 J) 3(9.11 1031 kg)

v 7.26 106 m/s 92. k 6.0 103 N/m, d 0.10 m, V 450 V, x 0.01 m V a) d 450 V 0.10 m 4.5 103 N/C b) The force to deform one spring is: F kx F (6.0 103 N/m)(0.01 m) F 6.0 105 N The force to deform both springs is: 2(6.0 105 N) 1.2 104 N c) The force on the pith ball must also be 1.2 104 N d) Fspring Fe Fspring q Fspring q 1.2 104 N q 4.5 103 N/C q 2.7 108 C


159

B 1.8 102 T NOTE: The solutions to problem 27 are based on a distance between the two conductors of 1 cm. 27. a)

Chapter 9 22. I 12.5 A B 3.1 105 T I B 2 r I r 2 B (4 107 T·m/A)(12.5 A) r 2 (3.1 105 T)

F

r 8.1 102 m 23. r 12 m I 4.50 103 A I B 2 r (4 107 T·m/A)(4.50 103 A) B 2 (12 m) 5 B 7.5 10 T 24. I 8.0 A B 1.2 103 T N1 NI B 2r NI r 2B (4 107 T·m/A)(1)(8.0 A) r 2(1.2 103 T) r 4.2 103 m 25. N 12 r 0.025 m I 0.52 A NI B 2r (4 107 T·m/A)(12)(0.52 A) B 2(0.025 m) B 1.6 104 T N 35 turns 100 cm 26. L 1 cm 1m N 3500 turns/m L I 4.0 A NI B L N B I L

3500 turns B (4 107 T·m/A) (4.0 A) 1m 160

F

Currents in the same direction— wires forced together

Referring to the above diagram, the magnetic fields will cancel each other out because the field from each wire is of the same magnitude but is in the opposite direction. b) F

F x

Currents in opposite directions— wires forced apart

I 10 A r 1.0 102 m I B 2 r (4 107 T·m/A)(10 A) B 2 (1.0 102 m) B 2.0 104 T But this field strength (2.0 104 T) is for each of the two wires. Referring to the above diagram, the two fields flow in the same direction when the current in the two wires moves in the opposite direction. The result is that the two fields will add to produce one field with double the strength (4.0 104 T). 28. Coil 1: N 400 L 0.1 m I 0.1 A Coil 2: N 200 L 0.1 m I 0.1 A


BTotal Bcoil1Bcoil2 NI NI BTotal L L 7 (4 10 T·m/A)(400)(0.1 A) BTotal (0.1 m) 7 (4 10 T·m/A)(200)(0.1 A) (0.1 m) 4 BTotal 2.5 10 T 29. The single loop: NI Bsingle 2r (4 107 T·m/A)(1)I Bsingle 2(0.02 m) Solenoid: L 2 rsingle loop L 2 (0.02 m) L 0.04

100 cm 15 turns N L 1 cm 1m N (1500 turns/m)(0.04 m) N 188 NI Bsol L (4 107 T·m/A)(188)(0.4 A) Bsol 0.04

4 Bsol 7.5 10 T To cancel the field, the magnitude of the two fields must be equal but opposite in direction. Bsol Bsingle (4 107 T·m/A)(1)I 7.5 104 T 2(0.02 m) (7.5 104 T)2(0.02 m) I 4 107 T·m/A I 24 A 30. a) 45° L 6.0 m B 0.03 T I 4.5 A F BIL sin F (0.03 T)(4.5 A)(6.0 m) sin 45° F 0.57 N The direction of this force is at 90° to the plane described by the direction of the current vector and that of the magnetic field, i.e., upwards.

b) If the current through the wire was to be reversed, the magnitude and direction of the resultant force would be 0.57 N [downwards]. 31. a) d(linear density) 0.010 kg/m B 2.0 105 T 90° F dg L F (0.010 kg/m)(9.8 N/kg) L F 9.8 102 N/m (linear weight) L F BIL sin F I BL sin F L I B sin 9.8 102 N/m I (2.0 105 T) sin 90° I 4900 A b) This current would most likely melt the wire. 32. a) N 60 I 2.2 A B 0.12 T NI B L NI L B (4 107 T·m/A)(60)(2.2 A) L 0.12 T 3 L 1.38 10 m F BIL sin F (0.12 T)(2.2 A)(1.38 103 m) (sin 90°) F 3.64 104 N b) F ma F a m 3.64 104 N a 0.025 kg a 1.46 102 m/s2 33. B 0.02 T v 1.5 107 m/s 90°


161

q 1.602 1019 C m 9.11 1031 kg Fc FB mv2 qvB sin r mv r qB (9.11 1031 kg)(1.5 107 m/s) r (1.602 1019 C)(0.02 T) r 4.3 103 m 34. qalpha 2(1.602 1019 C) qalpha 3.204 1019 C v 2 106 m/s B 2.9 105 T malpha 2(protons) 2(neutrons) malpha 4(1.67 1027 kg) malpha 6.68 1027 kg mv r qB (6.68 1027 kg)(2 106 m/s) r (3.204 1019 C)(2.9 105 T) r 1.4 103 m 35. Fg mg Fg (9.11 1031 kg)(9.8 N/kg) Fg 8.9 1030 N Fmag Bqv sin Fmag (5.0 105 T)(1.602 1019 C) (2.8 107 m/s) Fmag 2.24 1016 N The magnetic force has considerably more influence on the electron. 36. q 1.5 106 C v 450 m/s r 0.15 m I 1.5 A 90° F Bqv sin I B 2 r Iqv sin F 2 r F

(4 107 T·m/A)(1.5 A)(1.5 106 C)(450 m/s)sin 90° 2 (0.15 m)

F 1.3 109 N

162

According to the right-hand rules #1 and #3, this charge would always be forced towards the wire. 37. a) v 5 107 m/s r 0.05 m I 35 A q 1.602 1019 C F Bqv sin I B 2 r Iqv sin F 2 r F

(4 107 T·m/A)(35 A)(1.602 1019 C)(5 107 m/s) sin 90° 2 (0.05)

F 1.12 1015 N According to the right-hand rules #1 and #3, this charge would always be forced toward the wire. b) If the electron moved in the same direction as the current, then it would be forced away from the wire. 38. a) v 2.2 106 m/s r 5.3 1011 m q 1.602 1019 C m 9.11 1031 kg At any given instant, the electron can be considered to be moving in a straight line tangentially around the proton. Fmag Fc mv2 qvB sin r mv B qr (9.11 1031 kg)(2.2 106 m/s) B (1.602 1019 C)(5.3 1011 m) B 2.36 105 T But this field would always be met by a field of the same magnitude but opposite direction when the electron was on the other side of its orbit. Therefore, the net field strength at the proton is zero. b) To keep an electron moving in a circular artificially simulated orbit, the scientist must apply a field strength of 2.36 105 T.


39. 475 V/m B 0.1 T The electron experiences no net force because the forces from both the electric and magnetic fields are equal in magnitude but opposite in direction. If all the directions are mutually perpendicular, both the electric and magnetic fields will move the electron in the same direction (based on the right-hand rule #3). Therefore, Fmag Fe qvB q v B (475 V/m) v (0.1 T) v 4750 m/s 40. B 5.0 102 T d 0.01 m v 5 106 m/s q 1.602 1019 C Fmag Fe qvB q V qvB q d V dvB V (0.01 m)(5 106 m/s)(5.0 102 T) V 2500 V 41. r 3.5 m I 1.5 104 A I1I2L F 2 r F

(4 107 T·m/A)(1.5 104 A)2(190 m) 2 (3.5 m)

F 2.44 103 N 42. L 0.65 m I 12 A B 0.20 T F BIL sin F (0.20 T)(12 A)(0.65 m)(sin 90°) F 1.56 N [perpendicular to wire] At the angle shown, the force is: (1.56 N)sin 30° 0.78 N

43. a) v 5.0 106 m/s r 0.001 m q 1.602 1019 C m 9.11 1031 kg Fc Fmag mv2 qvB r vm B qr (5.0 106 m/s)(9.11 1031 kg) B ( 1.602 1019 C)(0.001 m) B 2.8 102 T b) Fc mac Fc qvB mac qvB qvB ac m ac

(1.602 1019 C)(5.0 106 m/s)(0.028 T) 9.11 1031 kg

ac 2.5 1016 m/s2 44. a) r 0.22 m B 0.35 T q 1.602 1019 C m 1.67 1027 kg q v m Br qBr v m (1.602 1019 C)(0.35 T)(0.22 m) v (1.67 1027 kg) v 7.4 106 m/s mv2 b) Fc r (1.67 1027 kg)(7.4 106 m/s)2 Fc 0.22 m 13 Fc 4.2 10 N


163

d) No, the results would be exactly the same. The induced current flow would be in the opposite direction if the poles of the magnet were reversed, but the reduction in speed would be the same. 49. The copper conductor is cutting through the magnetic field lines as it moves, and therefore experiences a force that opposes its motion. The induced and external magnetic fields are in opposite directions, which causes the opposition. Aluminum wire would make no difference as long as it conducts electricity.

e 45. 5.7 108 C/kg m B 0.75 T q v m Br mv r Bq d v t 2 r v T 2 r T v mv 2 Bq T v

2

T q B m

2

T (0.75 T)(5.7 108 C/kg) T 1.5 108 s 46. m 6.0 108 kg q 7.2 106 C B 3.0 T 1 t T 2 1 2 m t 2 Bq

(6.0 108 kg) t (3.0 T)(7.2 106 C) t 8.7 103 s 47. Falling through the top of the loop, the current is clockwise. Falling out of the bottom, the current is counterclockwise. 48. a) The conventional current flow is clockwise (looking down from top). b) The induced magnetic field is linear (down at the south end). c) Yes, the falling magnet would experience a magnetic force that is opposing its motion, as described by Lenz’s law.

164


1 3

x(m)

180°

θ 360°

180°

θ 360°

–A v(m/s)

21. a) 4 m b) A 5 cm c) T 8 s 1 d) f T 1 f 8s f 0.1 s1 e) v f v (4 m)(0.1 s1) v 0.4 m/s cycles 22. f s 10 cycles f 3.2 s f 3.125 cycles/s 1 T f 1 T 3.125 cycles/s T 0.32 s/cycle cycles 23. f s 72 cycles f 60 s f 1.2 cycles/s 1 T f 1 T 1.2 cycles/s T 0.83 s/cycle 24. f 60 Hz 1 T f T 0.017 s/cycle 150 cycles 25. a) f 60 s f 2.5 Hz 1 b) T f T 0.4 s/cycle 26. For 78 rpm: 78 cycles f 60 s f 1.3 Hz

1 T f T 0.77 s/cycle For 45 rpm: 45 cycles f 60 s f 0.75 Hz 1 T f T 1.33 s/cycle For 33 rpm: 100 f rpm 3 100 cycles f 180 s f 0.56 Hz 1 T f T 1.8 s/cycle 27. a) x A cos x 1 cos (10°) x 0.98 m b) x A cos x 1 cos (95°) x 0.087 m c) x A cos 3

x 1 cos rad 4 x 0.71 m d) x A cos x 1 cos (2π rad) x1m 28. A

At equilibrium (x 0), v is a maximum (sin 90° 1). When x A, v is a minimum (sin 0° 0). 29.

a(m/s2)

Chapter 10

180°

θ 360°

The object always accelerates toward equilibrium and slows down as it moves away from equilibrium. Solutions to End-of-chapter Problems

165

a is a maximum when x A (cos 0°); a is a minimum at equilibrium (cos 90°). The a vector is always directed toward the equilibrium position. 30. a) T 2

L g

0.30 kg 23.4 N/m

T 0.711 s/cycle m k

b) T 2

80 m 9.8 m/s

T 2

2

0.40 kg 20 N/m

T 2

T 18 s/cycle L g

T 0.889 s/cycle

c) T 2

m k

c) T 2

0.15 m T 2 9.8 m/s2 T 0.78 s/cycle

0.21 kg 200 N/m

T 2

g L

Moon

2.1 m 2 1.6 m/s T 7.2 s/cycle T 2

g

ii) T 2

L

Moon

80 m 1.6 m/s

T 2

2

T 44 s/cycle

g

iii) T 2

L

Moon

0.15 m 1.6 m/s

T 2

2

T 1.9 s/cycle

g

T 2

L

Jupiter

2.1 m 24.6 m/s

T 2

2

T 1.8 s/cycle

g

ii) T 2

2

T 2

L g

L

Jupiter

80 m 24.6 m/s

T 2

T 11 s/cycle 166

0.15 m 24.6 m/s

T 2

m k

b) T 2

b) i)

Jupiter

32. a) T 2

T 2

L

T 0.49 s/cycle

2.1 m T 2 9.8 m/s2 T 2.9 s/cycle

31. a) i)

g

iii) T 2

2

T 0.204 s/cycle 1 k 33. a) f 2 m k 4 2f 2m k 4 2(12 Hz)2(0.402 kg) k 2.3 103 N/m b) F kx F (2.3 103 N/m)(0.35 m) F 8.0 102 N 34. a) v f v f 3.00 108 m/s f 6.50 107 m

f 4.62 1014 Hz v b) f 3.00 108 m/s f 6.00 107 m f 5.00 1014 Hz v c) f 3.00 108 m/s f 5.80 107 m f 5.17 1014 Hz


v d) f 3.00 108 m/s f 5.20 107 m f 5.77 1014 Hz v e) f 3.00 108 m/s f 4.75 107 m f 6.32 1014 Hz v f) f 3.00 108 m/s f 4.00 107 m f 7.50 10 Hz d 35. a) t v 1.49 1011 m t 3.00 108 m/s t 497 s t 8.28 min t 0.138 h d b) t v 3.8 108 m t 3.00 108 m/s t 1.3 s t 2.1 102 min t 3.5 104 h d c) t v 5.8 1012 m t 3.00 108 m/s t 1.9 104 s t 3.2 102 min t 5.4 h d d) t v 9.1 1010 m t 3.00 108 m/s t 3.03 102 s t 5.1 min t 8.4 102 h 36. 1 a (3600 s/h)(24 h/d)(365 d/a) 3.1536 107 s d v∆t

37.

38.

39.

14

40.

d (3.00 108 m/s)(3.1536 107 s) d 9.46 1015 m d 100 light years 9.46 1017 m, v 3.00 108 m/s d t v t 3.15 109 s t 100 a d t v 160 m t 3.00 108 m/s t 5.33 107 s rEarth 6.38 106 m, cEarth 2π(6.38 106 m) 4.01 107 m d t v 4.01 107 m t 3.00 108 m/s t 0.134 s For the minimum frequency, 4 107 m v f 3.00 108 m/s f 4 107 m f 8 1014 Hz For the maximum frequency, 8 108 m v f 3.00 108 m/s f 8 108 m

f 4 1015 Hz Thus, the range is 8 1014 Hz to 4 1015 Hz. 41. For the car: ∆tcar (50 h)(3600 s/h) ∆tcar 180 000 s For light: d tlight v 4.00 106 m tlight 3.00 108 m/s tlight 0.01 s Comparing the two, tcar 180 000 s 1.8 107 times tlight 0.01 s


167

sin 30° 0.5 sin 60° 0.866 sin 45° 0.707 sin 12.6° 0.218 sin 74.4° 0.963 sin 0° 0 sin 90° 1 sin1 (0.342) 20° sin1 (0.643) 40° sin1 (0.700) 44.4° sin1 (0.333) 19.5° sin1 (1.00) 90° c 44. v n 3.00 108 m/s v 0.90 v 3.3 108 m/s This speed is impossible, since it is greater than the speed of light. 45. n1 1.00, n2 1.98, 1 22 n1 sin 1 n2 sin 2 sin 22 1.98 sin 2 2 sin 2 cos 2 1.98 sin 2 1.98 cos 2 2 2 8.1° 46. 1.5 sin 30° n2 sin 50° n2 0.98 As in problem 44, this value is impossible. 47. 42. a) b) c) d) e) f) g) 43. a) b) c) d) e)

light ray

wavefronts glass

c 48. a) v n 3.00 108 m/s v 2.42 v 1.24 108 m/s c b) v n 3.00 108 m/s v 1.52 v 1.97 108 m/s

168

c c) v n 3.00 108 m/s v 1.33 v 2.26 108 m/s c d) v n 3.00 108 m/s v 1.30 v 2.31 108 m/s n 49. Use 1 with n1 1.00: n2 a) 0.413 b) 0.658 c) 0.752 d) 0.769 c 50. v n 3.00 108 m/s v 1.33 v 2.26 108 m/s d t v 1200 m t 2.26 108 m/s t 5.31 105 s 51. a) tan 1 tan B n tan 1 2 n1 tan 1 1.42 1 54.8° b) n1 sin 1 n2 sin 2

sin 54.8° 2 sin1 1.00· 1.42 2 35.2° c) 54.8° (i r) 52. Polaroid glasses are most effective when the light is most polarized. The light is 100% polarized at Brewster’s angle, B. n tan B 2 n1 1.33 tan B 1.00 B 53.1° elevation 90° 53.1° elevation 36.9°


53. a) I2 0.5Io cos2 I2 0.5Io cos2 30° I2 0.375Io I 2 37.5% Io I b) 2 20.7% Io I2 c) 5.85% Io 54. When viewing something through a doubly refracting crystal, two images are seen, since two rays of polarized light are produced. If another crystal was laid over-top and rotated, nothing would be seen, since the two crystals now act as a polarizeranalyzer pair, with an angle of 90° between their axes. 55. Use an analyzer (another polarizer, rotated). n 56. tan B 2 n1

1.33 B tan1 1.00 B 53° n 57. a) tan B 2 n1

1.33 B tan1 1.00 B 53.1° n b) tan B 2 n1

cos1

1.50 B tan1 1.00 B 56.3° n c) tan B 2 n1 1.33 B tan1 1.50 B 41.6° n d) tan B 2 n1 1.33 B tan1 1.30 B 45.7° n 58. tan B 2 n1 n2 tan 60° 1.00

n2 tan 60° n2 1.73 59. The first Polaroid will remove exactly one component (50%) of the incident light. The third Polaroid, having been placed at any angle but 90° to the first one, will remove a fraction of the remaining light, allowing one component of the light to pass through. The second Polaroid will then remove only a single component of the residual light. Thus, a fraction of the incident light passes through all three Polaroids. 60. a) I2 0.5Io cos2 I2 0.5Io cos2 10° I2 0.485Io I 2 48.5% Io I b) 2 37.5% Io I2 c) 5.85% Io I d) 2 0.380% Io 61. I2 0.4Io I2 0.5Io cos2 0.4Io 0.5Io cos2 0.4 0.5

26.6° 62. I2 0.5Io cos2 1 and I3 I2 cos2 2 I3 0.5Io cos2 1 cos2 2 I3 0.5Io cos2 60° cos2 10° I3 0.121Io I 3 12.1% Io


169

Chapter 11 23. a) b) c) d) 24.

1.0 m b) d 10 500 slits d 9.52 105 m m sin m d (2)(5.50 107m) sin 2 9.52 105 m

constructive constructive partial destructive 1

sin 2 0.01155 2 0.662° 27. L 1.0 m dxm m L mL xm d 2(5.50 107 m)(1.0 m) x2 2.0 106 m

S1 0 S1

1

25.

x2 0.55 m

2

28.

1

1

0

2

S1

1

S2

1

2

3

3

Minima numbers

2

26. m 2 550 nm a) d 2.0 106 m m sin m d (2)(5.50 107 m) sin 2 2.0 106 m sin 2 0.55 2 33.4°

2

2 1

1

29. 560 nm d 4.5 106 m a) m 1 m sin m d (1)(5.60 107 m) sin 1 4.5 106 m sin 1 0.12444 1 7.14°

170

Maxima numbers


1 m 2 b) sin d (5.60 10 m) 2 sin 3

7

4.5 106 m

sin 0.18667 10.8° m c) sin m d (3)(5.60 107 m) sin 3 4.5 106 m sin 3 0.37333 3 21.9° 1 m 2 d) sin m d

(5.60 10 m) 2 sin 7

3

7

4.5 106 m

sin 3 0.43556 3 25.8° 30. 610 nm m2 2 23o m d sin m (2)(6.10 107 m) d sin 23° d 3.12 106 m d 3.12 m 31. d 0.15 mm m2 x2 7.7 m L 1.2 m dxm mL (1.5 104 m)(7.7 m) (2)(1.2 m) 4.81 104 m 481 m

32. 585 nm L 1.25 m m9 x 3.0 cm 1 m L 2 d x

(5.85 10 m)(1.25 m) 2 d 19

7

3.0 102 m

d 2.3 104 m d 0.23 mm 33. 630 nm d 3.0 105 m d sin m For maximization, sin 1. d m 3.0 105 m m 6.30 107 m m 4.76 1011

34. a) The light now travels an extra twice 4 between the original and the second positions. This produces an extra shift of . The observer therefore sees a dark 2 band and the fringe pattern moves by half a band. b) The light now travels an extra twice 2 between the original and the second positions. This produces an extra shift of . The observer therefore sees a bright band and the fringe pattern moves by a full band. 3 c) The light now travels an extra twice 4 between the original and the second positions. This produces an extra shift 3 of . The observer therefore sees neither 2 a bright nor a dark band and the fringe 3 pattern moves by of a band. 2


171

d) The light now travels an extra twice between the original and the second positions. This produces an extra shift of 2. The observer therefore sees a bright band and the fringe pattern moves by two full bands. 35. ∆PD 4 n 1.42 600 nm

PD t 2(n 1) (4)(6.00 107 m) t 2(0.42) t 2.857 106 m t 2.86 m 36. ∆PD 12 t 3.60 microns 640 nm

PD n 1 2t (12)(6.40 107 m) n 1 2(3.60 106 m) n 2.07 37. ∆PD 10 vm 1.54 108 m/s t 2.80 microns c nm vm 3.0 108 m/s nm 1.54 108 m/s nm 1.948 nm 1.95 2t(n 1)

PD 2(2.80 106 m)(0.948) 10 7 5.309 10 m 531 nm 38. t 364 nm 510 nm ng 1.40 g ng

Because there is a half-phase shift between air and gas, 1 2t g 2 m g 1 2(3.64 107 m) (3.64 107 m) 2 m 3.64 107 m m 2.5 The interference is destructive. 1 3 39. a) 2 2 2 2 destructive 1 b) 2 2 4 constructive 5 1 c) 2 3 2 4 constructive 7 1 15 d) 2 2 2 2 destructive 40. ng 1.40 560 nm t 4.80 106 m g ng

5.60 107 m g 1.40 g 4.00 107 m Because there is a half-phase shift between air and gas, 1 2t g 2 m g 1 2(4.80 106 m) (4.00 107 m) 2 m 4.00 107 m m 24.5 The interference is destructive and a dark area will result.

5.10 107 m g 1.40 g 3.64 107 m 172


41. 500 nm a) nf 1.44 f nf

b) Because of the phase shift and constructive interference, 1 m s 2 t 2

5.00 107 m f 1.44 f 3.47 107 m Because there is a half-phase shift, 1 2t f 2 m f m 2 t 1

f

1 (4.36 10 m) 2 t 1

7

2

7

t 1.09 10 t 109 nm v 43. a) f

m

350 m/s 250 Hz

2

1.40 m v b) f

1 (3.47 10 m) 2 t 1

7

2

7

t 8.675 10 m t 86.8 nm b) nf 1.23 f nf 5.00 107 m f 1.23 f 4.07 107 m Because the shifts cancel, m t f 2 (1)(4.07 107 m) t 2 t 2.03 107 m t 203 nm 42. 580 nm ns 1.33 s ns 5.80 107 m s 1.33 s 4.36 107 m a) Because of the phase shift and destructive interference, m t s 2 (1)(4.36 107 m) t 2 t 2.18 107 m t 218 nm

2.50 108 m/s 4.81 1014 Hz 5.20 107 m 520 nm v c) f 3.0 108 m/s 1.2 108 Hz 2.5 m v 44. a) f 3.0 108 m/s f 2.0 1012 m f 1.5 1020 Hz 14 km 1h 1000 m b) v 1h 3600 s 1 km v 3.889 m/s v f 3.889 m/s f 1.2 m f 3.2 Hz


173

46. a) m 2 580 nm w 2.2 105 m 1 m 2 sin w

2 (5.80 10 m) 2 sin 1

7

2.2 105 m

sin 0.0659 3.78° b) m 2 550 nm w 2.2 105 m m sin m w (2)(5.50 107 m) sin 2 (2.2 105 m) sin 2 0.05 2 2.87° 47. w 1.2 102 mm m1 1 4° w sin m m (1.2 105 m) sin 4° 1 8.37 107 m 837 nm 48. L 1.0 m m2 837 nm w 1.2 102 mm Minimum: mL xm w (2)(8.37 107 m)(1.0 m) x2 1.2 105 m x2 0.1395 m x2 140 mm

Maximum: 1 m L 2 x w 1 2 (8.37 107 m)(1.0 m) 2 x 1.2 105 m

x 0.174 m x 174 mm 49. w 1.1 105 m 620 nm m2 a) Minimum: m sin m w (2)(6.20 107 m) sin 2 1.1 105 m sin 2 0.113 2 6.47° b) Maximum: 1 m 2 sin w

2 (6.20 10 m) 2 sin 1

7

1.1 105 m

sin 0.141 8.10° 50. Width of central maximum 6.6° 400 nm w sin 4.00 107 m w sin 3.3° w 6.949 106 m w 6.95 m 51. 585 nm w 1.23 103 cm L 1.2 m a) m 3 mL xm w (3)(5.85 107 m)(1.2 m) x3 1.23 105 m x3 0.171 m x3 171 mm

174


b) m 2 1 m L 2 x w

2 (5.85 10 m)(1.2 m) 2 1

x

7

1.23 105 m

x 0.1426 m x 143 mm 52. w 1.10 103 cm 470 nm m1 m sin 2 w

(1)(4.70 10 m) sin 2 1.10 10 m sin 0.0427 2 7

5

4.90° 53. 493 nm w 5.65 104 m L 3.5 m m1 1 mL a) xm 2 w (1)(4.93 107 m)(3.5 m) x 2 (5.65 104 m) x 6.1 103 m x 6.1 mm m b) sin 2 w

(1)(4.93 10 m) sin 2 (5.65 10 m) sin 8.73 10 2 7

4

4

0.10° 54. 450 nm 55. 530 nm N 10 000 slits w 1 cm m1

w d N 1 102 m d 10 000 d 1 106 m m sin m d (1)(5.30 107 m) sin 1 1 106 m sin 1 0.53 1 32° 56. 650 nm N 2000 slits w 1 cm m 11.25° w d N 1 102 m d 2000 d 5.00 106 m d sin m 1 m 2 6 1 (5.00 10 m) sin 11.25° m 2 (6.50 107 m) m1 57. m 2 1 d 4 2.3 10 slits/mm d 4.35 105 mm L 0.95 m 610 nm mL xm d (2)(6.10 107 m)(0.95 m) x2 (4.35 108 m) x2 27 m 58. N 10 000 slits w 1.2 cm w d N 1.2 102 m d 10 000 d 1.2 106 m


175

a) 600 nm d m 1.2 106 m m 6.00 107 m m2 b) 440 nm d m 1.2 106 m m 4.40 107 m m 2.7 m2 1 102 m 59. d 1000 d 1 105 m d m400 1 105 m m400 4.00 107 m m400 25 d m700 1 105 m m700 7.00 107 m m700 14.3 m700 14 Orders needed: 25 14 11 60. d 1.0 microns a) 610 nm d m 1.0 106 m m 6.10 107 m m 1.6 m1 b) 575 nm d m 1.0 106 m m 5.75 107 m m 1.7 m1

c) 430 nm d m 1.0 106 m m 4.30 107 m m 2.3 m2 61. 1 589 nm 2 589.59 nm w 2.5 cm N 104 w d N 2.5 102 m d 104 d 2.5 106 m m m sin1 2 sin1 1 d d sin1

5.8959 10 m 2.5 10 m 7

6

5.89 107 m sin1 2.5 106 m

13.641° 13.627° 1.39 102° 62. ∆ 2 1 ∆ 5.8959 107 m 5.89 107 m ∆ 5.9 1010 m 1 2 avg 2 (5.89 107 m) (5.8959 107 m) avg 2 7 avg 5.8930 10 m m2 avg N

m (5.8930 107 m) N (5.9 1010 m)(2) N 500 63. N 106 w 2.5 cm 520 nm w d N 2.5 102 m d 106 d 2.5 108 m

176


d m 2.5 108 m m 5.20 107 m m 0.0481 m0 64. N 4000 m1 ∆ (6.5648 107 m) (6.5630 107 m) ∆ 1.8 1010 m avg

(6.5648 107 m) (6.5630 107 m) 2

avg 6.5639 107 m avg R

6.5639 107 m R 1.8 1010 m R 3647 Nm (4000)(1) Nm 4000 R Nm, therefore it will not be resolved. 65. 0.55 nm 1m d 6 2.5 10 d 4.0 107 m m1 m sin d (1)(5.5 1010 m) sin 4.0 107 m sin 1.375 103 7.9 102° Diffraction is not apparent. 66. d 0.40 nm 0.20 nm m3 m sin 2d (3)(2.0 1010 m) sin 2(4.0 1010 m) sin 0.75 49°


177

Chapter 12 7

19. max 597 nm 5.97 10 m The temperature can be found using Wien’s law: 2.898 103 max T 2.898 103 T max 2.898 103 T 5.97 107 m T 4854.27 K T 4854.27 273°C T 4581.27°C 20. T 2.7 K max can be found using Wien’s law: 2.898 103 max T 2.898 103 max 2.7 K max 1.07 103 m 21. T 125 K max can be found using Wien’s law: 2.898 103 max T 2.898 103 max 125 K max 2.32 105 m The peak wavelength of Jupiter’s cloud is 2.32 105 m. It belongs to the infrared part of the electromagnetic spectrum. 22. P 2 W, 632.4 nm 6.324 107 m We are to find the number of photons leaving the laser tube per second. Let us symbolize this quantity by N . Using Planck’s equation, we can express the energy for a single photon: hc E The number of photons leaving the tube can be found as follows: P N E P N hc (2 W)(6.324 107 m) N (6.626 1034 J·s)(3.0 108 m/s)

23. E 4.5 eV, W0(gold) 5.37 eV E W0. The gold will absorb all of the energy of the incident photons, hence there will be no photoelectric effect observed (see Figure 12.13). 24. 440 nm 4.4 107 m, W0(nickel) 5.15 eV First, we shall calculate the energy of the incident photons. Using Planck’s equation: hc E (6.626 1034 J·s)(3.0 108 m/s) E 4.4 107 m E 4.52 1019 J 4.52 1019 J E 1.6 1019 C E 2.82 eV Since E W0, the photoelectric effect will not be exhibited (see Figure 12.13). 25. P 30 W, 540 nm 5.4 107 m We are to find the number of photons radiated by the headlight per second. Let us symbolize this quantity by N . Using Planck’s equation, we can express the energy for a single photon: hc E The number of photons radiated by the headlight can be found as follows: P N E P N hc (30 W)(5.4 107 m) N (6.626 1034 J·s)(3.0 108 m/s) N 8.15 1019 photons/s 26. W0 3 eV 4.8 1019 J, 219 nm 2.19 107 m a) The energy of photons with cut-off frequency is equal to the work function of the metal. Hence, E W0 4.8 1019 J

N 6.36 1018 photons/s

178


The frequency can be found using Planck’s equation: E hf E f h 4.8 1019 J f 6.626 1034 J·s f 7.24 1014 Hz b) The maximum energy of the ejected photons can be found using the equation: Ekmax E W0 hc Ekmax W0 Ekmax

(6.626 1034 J·s)(3.0 108 m/s) 2.19 107 m

4.8 1019 J Ekmax 4.28 1019 J 27. a) To avoid unwanted electrical currents and change in bonding structure of the material of the satellite, the number of electrons ejected from the material should be minimal. The greater the work function of the metal, the more photon energy it will absorb and the fewer electrons will be ejected. Hence, the material selected should have a relatively high work function. b) The longest wavelength of the photons that could affect this satellite would have an energy equal to the work function of the material, i.e., E W0 hc Using Planck’s equation E , hc max (if W0 is in Joules) W0 hc max (if W0 is in eV) W0e 28. W0(platinum) 5.65 eV 9.04 1019 J From problem 27, we know that: hc max W0 (6.626 1034 J·s)(3.0 108 m/s) max 9.04 1019 J max 2.2 107 m The maximum wavelength of the photon that could generate the photoelectric effect on the platinum surface is 2.2 107 m.

29. a) For a material with a work function greater than zero, the typical photoelectric effect graph has a positive x intercept. If the graph passes through the origin, the work function of the material is zero, which means that the photoelectric effect would be observed with incident photons having any wavelength. b) If the graph has a positive y intercept, we would observe the photoelectric effect without the presence of incident photons. 30. 400 pm 4.0 1010 m a) The frequency of the photon can be found using the wave equation: c f 3.0 108 m/s f 4.0 1010 m f 7.5 1017 Hz b) The momentum of the photon can be computed using de Broglie’s equation: h p 6.626 1034 J·s p 4.0 1010 m p 1.66 1024 N·s c) The mass equivalence can be found using de Broglie’s equation: p mv p m c 1.66 1024 N·s m 3.0 108 m/s m 5.53 1033 kg 31. mproton 1.673 1027 kg First, we have to express the rest energy of the proton. It can be found using: Eproton mc 2 The energy of the photon, which is equal to the rest energy of the proton, can be expressed using Planck’s equation: hc E


179

Then, Eproton E hc mc 2 h mc Using de Broglie’s equation: h p Hence, p mc p (1.673 1027 kg)(3.0 108 m/s) p 5.02 1019 N·s 32. 10 m 1 105 m Using de Broglie’s equation: h p 6.626 1034 J·s p 1 105 m p 6.63 1029 N·s 33. f 1 nm 1 109 m Consider the following diagram: y

To find the Compton shift,

f i

1 109 m 9.9552 1010 m

4.48 1012 m The Compton shift is 4.48 1012 m. 34. 180°, vf 7.12 105 m/s From the conservation of energy, Ei Ef Ek hc 1 hc mvf2 (eq. 1) 2 i f From the conservation of momentum, pi pf pe h h mvf (eq. 2) i f (The negative sign signifies a scatter angle equal to 180°.) Multiplying equation 2 by c and adding the result to equation 1, 2hc 1 mvf2 cmvf 2 i 2hc i 1 2 m vf cvf 2

i e xf

θ 43°

C

x

xi

From the conservation of energy, Ei Ef Ek hc 1 hc mvf2 (eq. 1) 2 i f From the conservation of momentum, p i pf pe In the direction of the x axis: h h cos 43° mvf cos (eq. 2) i f In the direction of the y axis: h sin 43° mvf sin (eq. 3) i Using math software to solve the system of equations that consists of equations 1, 2, and 3, the value for i 9.9552 1010 m. 180

2(6.626 1034 J·s)(3.0 108 m/s) (9.11 1031 kg) 1(7.12 105 m/s)2 (3.0 108 m/s)(7.12 105 m/s) 2

i 2.04 109 m 35. i 18 pm 1.8 1011 m, energy loss is 67% The initial energy of the photon can be computed using Planck’s equation: hc Ei i (6.626 1034 J·s)(3.0 108 m/s) Ei 1.8 1011 m Ei 1.1 1014 J Since 67% of the energy is lost, the final energy of the photon is: Ef 0.33Ei Ef 0.33(1.1 1014 J) Ef 3.64 1015 J The final wavelength can be calculated using Planck’s equation: hc f Ef (6.626 1034 J·s)(3.0 108 m/s) f 3.64 1015 J f 5.45 1011 m


The Compton shift as a percentage is: 5.45 1011 m f 100% i 1.8 1011 m f 302% i The wavelength of a photon increases by 302%. 36. m 45 g 0.045 kg, v 50 m/s Using de Broglie’s equation: h mv 6.626 1034 J·s (0.045 kg)(50 m/s) 2.9 1034 m The wavelength associated with this ball is 2.9 1034 m. 37. mn 1.68 1027 kg, 0.117 nm 1.17 1010 m Using de Broglie’s equation: h mv h v m 6.626 1034 J·s v (1.68 1027 kg)(1.17 1010 m) v 3371 m/s The velocity of the neutron is 3371 m/s. 38. mp 1.67 1027 kg, 2.9 1034 m Using de Broglie’s equation: h mv h v m 6.626 1034 J·s v (1.67 1027 kg)(2.9 1034 m) v 1.37 1027 m/s The speed of the proton would have to be 1.37 1027 m/s. Since v is much greater than c, this speed is impossible. 39. Ek 50 eV 8 1018 J, me 9.11 1031 kg a) We shall first compute the velocity using the kinetic energy value: 1 Ek mv2 2

v v

2E m k

2(8 1018 J) 9.11 1031 kg

v 4.19 106 m/s Now can be found using de Broglie’s equation: h mv 6.626 1034 J·s (9.11 1031 kg)(4.19 106 m/s) 1.73 1010 m b) The Bohr radius is 5.29 1011 m. The wavelength associated with an electron is longer than a hydrogen atom. 40. The photon transfers from n 5 to n 2. The energy at level n is given by: 13.6 eV En n2 The energy released when the photon transfers from n 5 to n 2 is:

E E5 E2 13.6 eV 13.6 eV

E 2 22 5

E 2.86 eV

E 4.58 1019 J To compute the wavelength: hc E (6.626 1034 J·s)(3.0 108 m/s) 4.58 1019 J 4.34 107 m The wavelength released when the photon transfers from n 5 to n 2 is 4.34 107 m. It is in the visual spectrum and it would appear as violet. 41. a) The electron transfers from n 1 to n 4. The energy of the electron is given by: 13.6 eV En n2 The energy needed to transfer the electron from n 1 to n 4 is:

E E4 E1 13.6 eV 13.6 eV

E 12 42

E 12.75 eV


181

b) The electron transfers from n 2 to n 4. Similarly, the energy needed to transfer the electron from n 2 to n 4 is:

E E4 E2 13.6 eV 13.6 eV

E 22 42

E 2.55 eV 42. We need to find the difference in the radius between the second and third energy levels. The radius at a level n is given by rn (5.29 1011 m)n2 The difference in radii is: ∆r r3 r2 ∆r (5.29 1011 m)(3)2 (5.29 1011 m)(2)2 ∆r 2.64 1010 m 43. n 1 The radius of the first energy level can be found using: rn (5.29 1011 m)n2 rn (5.29 1011 m)(1)2 rn 5.29 1011 m The centripetal force is equal to the electrostatic force of attraction: ke2 F r2 (8.99 109 N·m2/C2)(1.6 1019 C)2 F (5.29 1011 m)2 F 8.22 108 N The centripetal force acting on the electron to keep it in the first energy level is 8.22 108 N. 44. F 8.22 108 N, r 5.29 1011 m F m4 2rf 2 1 F f 2 mr

1 f 2

8.22 108 N (9.11 1031 kg)(5.29 1011 m)

f 6.56 1015 Hz The electron is orbiting the nucleus 6.56 1015 times per second.

182

45. Consider an electron transferring from n 4 to n 1. As computed in problem 41, the energy released is equal to 12.75 eV 2.04 1018 J. The frequency is then equal to: E f h 2.04 1018 J f 6.626 1034 J·s f 3.08 1015 Hz The frequency of the photon is 3.08 1015 Hz, or one-half the number of cycles per second completed by the electron in problem 44. 46. Bohr predicted a certain value for energy at a given energy level. From the quantization of energy, there can be only specific values for velocity, v, and radius, r. Thus, the path of the orbiting electron can attain a specific path (orbit) around the nucleus, which is an orbital. 48. v 1000 m/s, m 9.11 1031 kg ∆py ∆y ≥ –h ∆p m∆v –h

y m v 1.0546 1034 J·s

y (9.11 1031 kg)(1000 m/s)

y 1.16 107 m Hence, the position is uncertain to 1.16 107 m. 49. ∆y 1 104 m The molecular mass of oxygen is 32 mol. The mass of one oxygen molecule is 32 mol 5.32 1026 kg 6.02 1023 mol/g From ∆py∆y ≥ –h and ∆p m∆v, the maximum speed is: –h v m v 1.0546 1034 J·s v (5.32 1026 kg)(1 104 m) v 1.98 105 m/s


Chapter 13 28. a)

b)

c) d)

e) 29. a)

b)

c)

3 cm 1m 1a 3 cm/a 1a 100 cm 365.25 d 1d 86400 s 9.5 1010 m/s 9.5 1010 m/s 3.16 1018 3.0 108 m/s 0.1 mm 1m 0.1 mm/s 1s 1000 mm 4 1.0 10 m/s 1.0 104 m/s 3.3 1013 3.0 108 m/s 10.8 m/s 3.6 108 3.0 108 m/s Mach 6.54 6.54 332 m/s Mach 6.54 2171.28 m/s 2171.28 m/s 7.24 107 3.0 108 m/s 2.2 106 m/s 7.33 103 3 108 m/s Snoopy must fly 50 km/h [N then E]. Let y resultant ground speed (130 k m/h)2 (50 km/h )2 y y 120 km/h The Baron going west has a ground speed of: bvg bvw wvg bvg 130 km/h 50 km/h bvg 180 km/h [W] While going east, bvg 130 km/h 50 km/h bvg 80 km/h [E] The time for Snoopy: 200 km 3600 s 6000 s 120 km/h 1h Time for the Baron: 100 km 100 km 3600 s 180 km/h 80 km/h 1h

6500 s Therefore, Snoopy wins the race by 500 s or 0.139 h.

200 v2 w2 tS d) 100 100 tB v w vw

200 )(v (v w w) tS 200v tB (v w)(v w)

(v w)(v w) t S tB v (v w )(v w) t (v w )(v w) S tB v 2 t v w2 S tB v

v2 w2

v t w 1 t v t S tB

2

2

S

2

B

30. In our rest frame, we observe the contracted length: v2 L L0 1 c2 L (1.0 m)1 (0 .080)2 L 0.6 m 1 31. L L0 3

v2 L L0 1 c2

1 v2 1 3 c2 1 v2 1 2 9 c v2 8 2 9 c

9

vc

8

v 0.943c v 2.83 108 m/s


183

32. Length is contracted for the moving stopwatch. The time it measures is: L t v v2 L0 1 2 c t v

(180 m)1 (0 .7)2 t 8 0.7(3 10 m/s) t 6.12 107 s 33. We observe the dilated half-life of the muon: t0 t v2 1 2 c 2.6 108 s t 2 1 (0 .998) t 4.11 107 s The distance travelled is: d vt d (0.998c)(4.11 107 s) d 123 m 34. Katrina measures a contracted distance: v2 L L0 1 c2 L (7.83 1010 m)1 (0 .25)2 10 L 7.58 10 m 35. The time the girlfriend measures is: L0 tf 35 m/s The time Henry measures is: d th v v2 L0 1 2 c th 35 m/s Their time difference is: v2 L0 1 2 L0 c

t 35 m/s 35 m/s

Using the low-speed approximation when v

c: v2 v2 1 1 2 2c2 c

v L 1 1 2c

t 2

0

2

35 m/s

(35 m/s)2 35 000 m 2(3.0 108 m/s)2

t 35 m/s 12

t 6.81 10 s 36. Given the muon’s dilated half-life: t 2.8 106 s and its rest half-life: t0 2.2 106 s t0 t v2 1 2 c 2.2 v2 1 2.8 c2 2 v 2.2 2 1 2 2.8 c

2.2 c1 v 2.8 2

1.856 108 m/s v d circumference d 2r d vt vt r 2 (1.856 108 m/s)(2.8 106 s) r 2 r 82.7 m 37. Only the component of L0 in the direction of travel is contracted: Lx L0 cos 30° The contracted length seen by Tanya in the direction of travel (x) is: v2 Lx Lx 1 c2

v2 Lx L0 cos 30° 1 c2 The perpendicular length, Ly, is L0 sin 30° for both Katrina and Tanya.

184


Ly tan 45° Lx Ly 1 Lx Ly L0 sin 30° v2 Lx L0 cos 30° 1 2 c Therefore: L0 sin 30° 1 v2 L0 cos 30° 1 2 c

v2 1 2 tan 30° c 1 v2 1 2 3 c

v2 1 1 2 3 c 2 v 2 2 3 c v 0.816c v 2.45 108 m/s 38. The time to travel a circumference is: 2r t v 2(6.38 106 m) t0 300 m/s t0 1.336 105 s For the clocks on Earth, use the low-speed approximation for v

c: t0 t v2 1 2 c v2 t t0 1 2 2c The difference in the flying clocks compared to the ones on Earth is:

t t t0 v2

t t0 1 2 t0 c

v2 2r

t 1 2 1 2c v

(300 m/s)2

t (1.336 105 s) 2(3.0 108 m/s)2

t 6.68 108 s

39. 1 ca vt 1 ca (3.0 108 m/s) (365.25 24 60 60 s) 1 ca 9.47 1015 m 40. Using spacetime invariance: (∆s2) c2(∆tJ)2 (∆xJ)2 and: (∆s2) c2(∆tT)2 (∆xT)2 For Ted, the distance between events is: c 2(1.0 106 s)2 (600 m)2 0 ( xT)2 9 104 m2 3.6 105 m2 ( xT)2 ( xT)2 2.7 105 m2

xT 5.20 102 m 41. Ted’s length, L, has contracted relative to Jane’s length, L0: v2 L L0 1 c2

v2 520 m (600 m) 1 c2 169 v2 1 2 225 c 2 56 v 2 225 c v 0.499c v 1.50 108 m/s 42. The dilated time of the stationary observer is: t0 t v2 1 2 c 4.0 s v2 1 5.0 s c2 2 16 v 1 2 c 25 3 v c 5 The distance travelled in the 5.0 s is: d vt 3 d (3.0 108 m/s)(5.0 s) 5 d 9.0 108 m 43. See problem 42: 3 v c 5 v 1.8 108 m/s


185

46. The centripetal force is provided by the electrical coulomb force: mv2 kQq r2 r ke2 r 2 mv v2 ke2 1 2 c r 2 m0v

44. Trevor’s time is: d t0 v v2 2L0 1 2 c t0 v His sister’s time is: 2L t 0 v The time difference is:

t 1 a

r

v2 2L0 1 2 c 2L

t 0 v v

v 1 c

2L

t 0 1 v

2

2

tv L0 v2 2 1 1 2 c (1 a)(0.95c) L0 2(1 1 (0 .95c)2) L0 0.691 ca 45. q 1.6 1019 C v 0.8c B 1.5 T m0 m v2 1 2 c mv r qB m0v r v2 qB 1 2 c [9.11 1031 kg][0.8(3.0 108 m/s)] r (1.602 1019 C)(1.5 T)1 (0 .8)2 r 1.52 103 m

(9.0 109 N·m2C2)(1.602 1019 C)2 1 (0 .6)2 31 8 2 (9.11 10 kg)[0.6(3.0 10 m/s)]

r 6.26 1015 m 47. The difference between the dilated and rest masses is: ∆m m m0 Use the low-speed binomial approximation when v

c: 1 v2 2 1 v 2c2 1 2 c v2

m m0 1 2 m0 2c

v

m m 1 1 2c m v

m 2 c (60 kg) (3.0 10 m/s)

m 2 (3.0 10 m/s) 2

0

2

2

0

2

2

8

2

m 3.0 107 kg 48. Use the high-speed approximation: v2 v 1 2 2 1 c c

m0 m v2 1 2 c m0 m v 2 1 c 9.11 1031 kg m 2(1 0.999 999 9 99 67) 31 9.11 10 kg m 2(3.3 1010 ) 26 m 3.55 10 kg

186

4


49. For a charge moving perpendicular to a magnetic field, the centripetal force equals the magnetic force: mv2 Bqv r Due to mass dilation, the magnetic field is: m0v B v2 qr 1 2 c

B

(9.1 1031 kg)(3.0 108 m/s)(0.999 999 986) (1.602 1019 C)(450 m) 1 (0 .999 9 99 986 )2

B 2.26 102 T mass 50. density volume m density xyz where x, y, and z are the rectangular dimensions. Contraction occurs only in the direction of motion, so density is: m0 v2 1 2 c v2 x0 1 2 yz c 0 v2 1 2 c When the density of an object is dilated twice as much as its density at rest, 20 : 0 20 v2 1 2 c v2 1 1 2 2 c v2 1 2 2 c v 0.7071c v 2.1 108 m/s

51. Using the relativistic equation of velocity addition, the velocity of the light relative to the duck is: lvc cvd lvd lvccvd 1 c2 c 0.2c lvd (c)(0.2c) 1 c2 1.2c lvd 1.2 lvd c 52. Using the relativistic equation of velocity addition, the velocity of star A relative to star B is: avE Evb avb avEEvb 1 c2 0.2c 0.3c avb 1 (0.2)(0.3) avb 0.472c 8 avb 1.42 10 m/s 53. The speed of rocket A relative to Earth is: avb bvE avE avbbvE 1 c2 0.8c 0.7c avE 1 (0.8)(0.7) avE 0.962c 8 avE 2.88 10 m/s 54. The speed of the positron relative to the electron is: pvg gve pve pvggve 1 c2 0.95c 0.85c v 1 (0.95)(0.85) pve 0.996c 8 pve 2.988 10 m/s 55. Bob’s velocity relative to Earth, bvE 0.3c; Nicole’s velocity relative to Earth, nvE 0.9c pvE, the phaser bullet’s velocity relative to Earth. p e


187

The velocity of the phaser bullet relative to Bob, pvb, is: pvE Evb pvb pvEEvb 1 c2 0.9c 0.3c v 1 (0.9)(0.3) pvb 0.822c 8 pvb 2.47 10 m/s 56. Kirk’s velocity relative to Earth: kvE X the module’s velocity relative to Kirk: mvk X the module’s velocity relative to Earth: mvE 0.8c mvk kvE mvE mvkkvE 1 c2 X X 0.8c X2 1 c2 2 0.8X 0.8c 2X c 2 0.8X 2cX 0.8c2 0 2X2 5cX 2c2 0 (2X c)(X 2c) 0 The speed of the Enterprise is: c 1.5 108 m/s 2 57. The mass, m, equivalent to the chemical energy released is: E mc2 3.2 104 J m c2 m 3.56 1013 kg 58. The mass, m, equivalent to the chemical energy released is: E mc2 9.2 1010 J m c2 m 1.02 106 kg 59.To find the energy equivalent of 1.0 kg of bananas: E mc2 E (1.0 kg)(3.0 108 m/s)2 E 9.0 1016 J 9.0 1016 J E 3.6 106 J/kWh E 2.5 1010 kWh p b

188

At a typical consumer rate of $0.08/kWh, 1.0 kg of bananas is equivalent to: (2.5 1010 kWh)($0.08) $2 109 or $2 billion Conversely, the rate of relativistic “banana” power is: $1.29 $0.000 000 000 052/kWh 2.5 1010 kWh 60. E mc2 E (m0c2 Ek) The work done in increasing an electron’s speed is:

Ek Ek Ek

Ek (mc2 m0c2) (mc2 m0c2)

Ek (mc2) mc2 1 1

Ek m0c2 2 v v2 1 2 1 2 c c For v 0.5c to v 0.9c: 1 1

Ek m0c 2 2 2 1 (0 .9) 1 (0 .5) 2

Ek 1.139m0c For v 0.9c to v 0.95c: 1 1

Ek m0c 2 2 2 1 (0 .95) 1 (0 .9) 2

Ek 0.908m0c It takes more work to increase from 0.5c to 0.9c. 61. To find the equivalent mass of the particle: E mc2 8.19 1014 J m (3.0 108 m/s)2 m 9.1 1031 kg m the mass of an electron 62. To find the difference between the dilated relativistic and the classical momentum,

p p p0

p mv m0v 1

p m0v 1 v2 1 2 c


Since v 75 103 m/s, or v

c, use the low-speed binomial approximation: v2 1

1 2c2 v2 1 2 c v2

p m0v 1 2 1 2c

m0v3

p 2c2 (125 kg)(75 000 m/s)3

p 2(3.0 108 m/s)2

p 0.29 kg·m/s 63. Since v

c, use the low-speed approximation: v2 1

1 2c2 v2 1 2 c The work done, ∆Ek, in speeding Mercury from rest is given by:

Ek mc2 m0c2 1

Ek m0c2 1 v2 1 2 c 2 v

Ek m0c2 1 2 1 2c

m0v2

Ek 2 (3.28 1023 kg)(4.78 104 m/s)2

Ek 2 32

Ek 3.75 10 J The mass equivalent, ∆m, to this amount of energy is:

Ek

m c2 3.75 1032 J

m (3.0 108 m/s)2

m 4.16 1015 kg E 64. m 2 c qV m c2 (1.602 1019 C)(1.35 108 V) m (3.0 108 m/s)2 m 2.4 1028 kg

65. Accelerating the electron of mass, m0 9.1 1031 kg, and charge, q 1.6 1019 C, from rest, through a potential of V results in a new total energy: E m0c2 Vq E mpc2 mpc2 m0c2 Vq mpc2 m0c2 V q V

(3.0 108 m/s)2[(1.67 1027 kg) (9.1 1031 kg)] 1.60 1019 C

V 9.38 108 V V 938 MV 66. Using the energy triangle, E2 (mvc)2 (m0c2)2 E2 (m0c2 Ek)2 For particle A: (21 J 8 J)2 (21 J)2 (mvc)2 (mvc)2 841 J2 441 J2 (mvc)2 400 J2 mvc 20 J To find the velocity of A, v mvc (where E mc2 m0c2 + Ek) c mc2 20 J v c 29 J v 0.69c For particle B: (22 J 7 J)2 (22 J)2 (mvc)2 (mvc)2 841 J2 484 J2 (mvc)2 357 J2 mvc 18.9 J To find the velocity of B, v mvc c mc2 18.9 J v c 29 J v 0.65c Particle A has the greater speed. 67. E2 (mvc)2 (m0c2)2 E2 (mvc)2 (938.3 MeV)2 E2 (0.996mc2)2 (938.3 MeV)2 E2(1 0.9962) 8.804 105 MeV2 E 1.05 104 MeV


189

68. E mc2 m0c2 Ek mc2 (0.511 MeV) (3.1 103 MeV) mc2 3100.511 MeV From the energy triangle: m0c cos E 0.511 MeV cos 3100.5 MeV 89.990557° mvc sin mc2 v sin c v c sin v (3.0 108 m/s) sin 89.990557° v 2.999 999 96 108 m/s 69. Using the energy triangle: mvc sin mc2 v mvc mc2 c mvc 2 tan m0c For particle A: (4 108 N·s)(3 108 m/s) mvc 2 20 J m0c mvc 2 0.60 m0c tan 0.60 30.96° v sin (30.96°) c v 0.514c For particle B: (5 108 N·s)(3 108 m/s) mvc 2 mc 30 J mvc 0.50 mc2 v 0.50c Particle A is faster.

190


Chapter 14 43. a) Cl b) Rn c) Be d) U e) Md 44. For AZ X, Z is the number of protons and A Z is the number of neutrons: a) 17 protons, 18 neutrons b) 86 protons, 136 neutrons c) 4 protons, 5 neutrons d) 92 protons, 146 neutrons e) 101 protons, 155 neutrons 45. Since 1 u 931.5 MeV/c 2, then 18.998 u 931.5 MeV/c 2/u 17 697 MeV/c 2. 106 MeV/c2 46. Conversely, 0.114 u. 931.5 MeV/c2/u 47. To find the weighted average of the two isotopes: 0.69(62.9296 u) 0.31(64.9278 u) 63.55 u This is closest to the mean atomic mass of Cu. 48. B [Zm(1H) Nmn m(146 C )]c 2 B [6(938.78) 8(939.57) (14.003 242 u)(931.5)] MeV B 105.22 MeV B 105.22 MeV 7.5 MeV/nucleon A 14 nucleons N 49. Since 146C → 147N 10 e v, the ratio Z 8 7 4 changes from to or from to the 6 7 3 1 more stable . 1 50. The binding energy is: B [m(3He) mn m(4He)]c 2 B [3.0160 u 1.008 665 u 4.002 60 u]c 2 931.5 MeV/c 2/u B 20.55 MeV 228 4 51. Since 232 92U → 90Th 2He Ek, Ek [mU mTh m]c 2 Ek [232.037 131 u 228.028 716 u 4.002 603 u]c2 931.5 MeV/c2/u Ek 5.41 MeV

52. Assuming the uranium nucleus is fixed at rest and the kinetic energy of the alpha particle becomes electrical potential, kq1q2 Ek r kq1q2 r Ek r

53. 54.

55.

56.

57.

(8.99 109 J·m/C2)(1.6 1019 C)2(2)(92) (5.3 106 eV)(1.6 1019 J/eV)

r 5.0 1014 m 0 231 231 v 90Th → 91Pa 1e 235 231 4 92U → 90Th 2He The mass difference is: ∆m mn (mp me) ∆m [939.57 938.27 0.511] MeV/c 2 ∆m 0.789 MeV/c2 From problem 54, the energy equivalent of 0.789 MeV/c2 is 0.789 MeV. 2 Thus (0.789 MeV) 0.526 MeV. 3 Since the total momentum before decay is equal to the total momentum after decay, and p 0 p, the three momentum vectors must form a right-angle triangle. From Pythagoras’ theorem: pC2 pe2 p2 21 2 pC (2.64 1021 )2 (4.76 10 ) pC 5.44 1021 N·s p2 Using Ek , the recoiling carbon nucleus 2m will have (5.44 1021 N·s)2 Ek 2(12.011 u)(1.6605 1027 kg/u)

Ek 7.42 1016 J 58. For a fixed gold nucleus at rest, the kinetic energy of the 449-MeV alpha particle is converted to electrical potential. Thus, for the radius, kq1q2 Ek r kq1q2 r Ek r

(8.99 109 J·m/C2)(1.6 1019 C)2(2)(79) (449 106 eV)(1.6 1019 J/eV)

r 5.07 1016 m


191

59. 100

62. If the amount of radioactive material is 23% of the original amount after 30 d, then, 1 t N N0 T 2 30 d 1 0.23N0 N0 T 2 30 d 1 log (0.23) log T 2

% of Original Dose still Radioactive vs. Time

% radioactive

80 60

40 20 0

4

8

12 t (h)

16

60. For carbon-14, T 5730 a. Comparing the relative amount, NR, of a 2000-a relic with the amount, NS, in a shroud suspected of being 2002 a 1350 a 650 a, yields: t 1 T 2 NR t 1 NS T 2 1 2

N 1 N 2 R

R

1 2

1 2

5.12 63. The molar amount of 235U is 0.0218 235 3.4 2 207 and of Pb is 0.0165. The 207 original molar amount of 235U was 0.0218 0.0165 0.0383. Using the decay formula where T 7.1 108 a, 1 2

1 N N0 2

t T 1 2

1 t 0.0218 0.0383 T 2 0.0218 t 1 log log 8 0.0383 7.1 10 a 2

2000 a 650 a 5730 a

S

NR 0.85 NS 61. The half-life of Po-210 is: T 138 d 198 720 min The half-life of Po-218 is T 3.1 min After 7.0 min, there will be: 1 t 1 210 Po: N N0 T 2 2 1 log N (3.5 105)log 2 N 100% 1 t 1 218 Po: N N0 T 2 2 1 log N (2.26)log 2 N 20.9% There will be a total of: 1(1 g) 0.209(1 g) 1.21 g Therefore, 1.21 106 g of radioactive Po remains.

1 2

7.0 min 198 720 min

7.0 min 3.1 min

1 2

1 2

1 2

0.0218 log (7.1 108 a) 0.0383 t 1 log 2 8 t 5.78 10 a

1 2

192

T 14 d

1 2

1 2

R

1 2

1 (30 d) log 2 T log (0.23)

20

When t 8 h, 39.7% of the original dose is still radioactive.

1 2

1 2

64. Using the activity decay formula where T 5730 a for 14C decay, 1 2

1 t N N0 T 2 1 750 900 2 750 t 1 log log 900 5730 a 2 5 log (5730 a) 6 t 1 log 2 t 1507 a

1 2

t 5730 a


65. For an isotope to be doubly stable, its values for both Z and N A Z must be “magic” nuclear shell numbers, where the numbers are 2, 8, 20, 28, 50, 82, and 126. The other 48 doubly stable isotopes are 42He, 168O, 40 20Ca, 20Ca, 78 132 28Ni, and 50 Sn. 0 137 66. 137 55Cs → 56Ba 1 e v Ek. To determine the maximum Ek available per disintegration, find the mass difference of the parent nucleon and the daughter plus the electron. Ek [136.9071 u (136.9058 u 0.000 549 u)]c 2 931.5 MeV/c 2/u Ek 0.6996 MeV 67. Dose Dose

activity time energy percentage absorbed mass 19

(3700 Bq)(365 24 60 60 s)(1.0 10 eV)(1.6 10 J/eV)(5%) 70 kg 6

Dose 0.013 mGy 68. The pilots fly for 52 weeks 20 h/week 1040 h per year. Thus, their exposure is: (7.0 106 Sv/h)(1040 h/a) 7.28 103 Sv/a. Compared with the average of 2 mSv/a, this 7.28 value is about 3.64 times greater. 2 238 206 69. Since 92U → 82Pb, 238 206 32 nucleons are lost through alpha decay in groups of 4 nucleons per decay. 32 Thus, there are 8 alpha particles 4 emitted. The number of beta decays is equal to the number of neutrons changed into protons. N protons in Pb protons left after alpha decay N 82 (92 8 2) N 6 beta particles emitted 70. Four beta decays means that four neutrons 208 were changed into protons, or 208 82Pb 824 X. Six alpha decays means that 208 78X came from 208 64 232 78 62Y 90Y. From the periodic table, this element is thorium-232 or 232Th.

71. The separation distance of an alpha particle (A 4) and a nitrogen nucleus (AN 14) is given by: r s r rN 3 3 rs 1.2A 1.2 AN 3 3 rs 1.2 4 1.2 14 rs 4.8 fm 72. Considering the nitrogen nuclei to be fixed at rest, the Ek of the incoming alpha particle is converted to electrical potential, or kq1q2 , where q1 2e and q2 7e Ek r Ek

(9.0 109 J·m/C2)(1.6 1019 C)2(2)(7) (4.8 1015 m)(1.6 1013 J/MeV)

Ek 4.2 MeV 73. The half-life of hassium-269 is T 9.3 s. The original amount of hassium is 1 2

N

mass of hassium Avogadro’s number mass per mole

Using the activity equation: 0.693N Activity T 1 2

0.693 Activity

(1.0 103 g)(6.022 1023 mol1) 269 g/mol

9.3 s Activity 1.67 1017 Bq Using the decay formula for a time of 1 s: 1 t N N0 T 2 1 N 2 N 92.82% If 92.82% remains after 1 s, then 100% 92.82% 7.18% has decayed. This activity equals:

1 2

1s 9.3 s

Activity (7.18%)

(1.0 103 g)(6.022 1023 mol1) 269 g/mol

Activity 1.61 10 Bq 74. The energy released is equivalent to the energy of the mass difference: E [m(1H) m(2H) m(3He)]c 2 E [1.007 825 u 2.014 102 u 3.016 029 u]c 2 931.5 MeV/c 2/u E 5.49 MeV


17

193

75. One mole of 235U releases 23 500 GJ of energy. mo 2(fuel used) mo 2(moles of U used)(mass/mol) power time mo 2 (0.235 kg/mol) energymol

(0.7 GW)(2)(3.1536 10 s) m 2 23 500 GJ/mol 7

o

(0.235 kg/mol) mo 883 kg electrical energy produced 76. %E fission energy released (electrical power) time %E mass of U (energy/mol) molar mass (0.7 GW)(86400 s) %E 2.5 kg (23 500 GJ/mol) 0.235 kg/mol %E 0.242 About 24.2% of the fission energy is transformed into electrical energy. 77. Since a mole of 235U releases 23 500 GJ of energy, the 50 kg releases (50 kg)(23 500 GJ/mol) 5 106 GJ 0.235 kg/mol 5 1015 J 78. Since the electron keeps only 10% of its kinetic energy with each collision, the energy remaining after x collisions is given by: Ex Eo(0.1)x 0.05 eV (5.0 106 eV)(0.1)x 0.05 eV log x log (0.1) 5.0 106 eV log (108) x log (101)

x 8 collisions 79. The incoming speed of a neutron with 3.5 MeV of kinetic energy is: 2E v k m 2(3.5 106 eV)(1.602 1019 J/eV) v (1.008 665 u)(1.6605 1027 kg/u)

1.008 665 u 1.007 276 u v1 1.008 665 u 1.007 276 u (2.5876 107 m/s) v1 1.782 104 m/s A 1 141 1 80. For the reaction 235 92U 0n → 56Ba Z Y 30n, conservation of atomic mass number for the reaction yields 235 1 141 A 3(1), or A 92. Conservation of atomic number yields 92 0 56 Z 3(0), or Z 36. The daughter isotope, from the periodic table, is 92 36Kr. 81. Working in MeVs, assume the rest mass of lead-207 is: m0 (207 u)(931.5 MeV/c 2/u) m0 1.928 105 MeV/c 2 Its total energy is: E m0c2 Ek E 1.928 105 MeV 7.000 106 MeV E 7.1928 TeV At relativistic speeds, use Einstein’s energy triangle: (mvc)2 E2 (m0c2)2 (mvc)2 (7.1928 1012 eV)2 (1.928 1011 eV)2 mvc 7.1902 1012 eV Rearranging for v, v 7.1902 1012 eV c mc2 v 7.1902 1012 eV c 7.1928 1012 eV v 0.999639c v 2.9989 108 m/s 82. The de Broglie wavelength is: h mv hc mvc (6.626 1034 J·s)(3.0 108 m/s) (7.19 1012 eV)(1.6 1019 J/eV)

1.73 1019 m

v 2.5876 107 m/s For head-on elastic collisions, mn mx v, where v is the recoil v mn mx velocity of the neutron.

194


83. At relativistic speeds, the mass becomes dilated: m0 m v2 1 2 c 1.673 53 1027 kg m 1 0. 752 m 2.53 1027 kg The de Broglie wavelength is: h mv

qB The cyclotron frequency, f , yields: 2m 2mf B q 2(2.3856 1027 kg)(20 106 Hz) B 1.6 1019 C

87.

6.626 1034 J·s (2.53 1027 kg)(0.75c) 1.16 1015 m 1.16 fm qB 84. f 2m 2mf B q 2(2.53 1027 kg)(23 106 Hz) B (1.6 1019 C) B 2.28 T 85. Electrons and protons with the same de Broglie wavelength have the same h momentum . Using Einstein’s mv energy triangle and MeV units, for the electron: (mvc)2 (m0c2 Ek)2 (m0c2)2 (mvc)2 (0.511 MeV 9 103 MeV)2 (0.511 MeV)2 mvc 9.0005 GeV The proton has an equal mvc, so (9000.5 MeV)2 (938.27 MeV Ek)2 (938.27 MeV)2 938.27 MeV Ek 2 (9000.5 MeV) (938.27 MeV) 2 Ek 8951.47 MeV 938.27 MeV Ek 8.1 GeV 86. Using the energy equation mc2 m0c2 Ek to find the dilated mass of the proton, Ek m m0 c2 400 MeV m 938.27 MeV/c 2 c2 m 1338.27 MeV/c 2 m 2.3856 1027 kg

88.

89. 90.

91.

B 1.87 T 2 1 1 a) uds 0 3 3 3 2 1 b) ud 1 3 3 1 1 c) db 0 3 3 2 2 d) cc 0 3 3 a) lambda (baryon) b) pion or rho (mesons) c) b-zero (meson) d) eta-c (meson) A neutron consists of udd, therefore an anti d . neutron is ud The mass of the top quark is 176 103 MeV/c2 188.94 u. The element 931.5 MeV/c2/u with the closest atomic mass is osmium (Os), with an atomic mass of 190.2 u. The pion has a quark combination of ud and a charge of e. Conversely, a pion has the combination u d, and its charge is 2 1 e e e. 3 3

d

t v 2.4 1015 m

t 3 108 m/s

t 8 1024 s a) Two protons approach and exchange a virtual meson, then recoil from each other. b) An atom sits at rest, then one of its electrons drops to a lower energy level and emits a photon, so the atom is pushed in the opposite direction. c) A pion decays into a muon and a muon neutrino.

92.

93.


195

94. Antiproton decay: p n W → n 10 e v → t e

n

v

W–

p

x

95. For a neutron and a proton, the interaction is: p n→p n

1 98. The charge of the strange quark, s, is and 3 2 the charge of the anticharm quark, c , is . 3 1 2 The charge of the meson is: 1 3 3 99. The charge of the baryon is 1 2 2 ttb 1 3 3 3 100. The blue quark could either emit a blue-antigreen gluon or absorb a greenantiblue gluon. t Blue

t udu Proton

ddu Neutron

u u π0 udu Proton

ddu Neutron

Green

Blueantigreen gluon

Green

Blue

x x

96. In the reaction p → n π , the energy associated with the mass difference is: E (mp mn)c2 From Einstein’s energy relationship, E2 [(mp mn)c2]2 E2 p2c2 m02c4 p2c2 E2 (m0c2)2 p2c2 (939.6 MeV 938.3 MeV)2 (139.6 MeV)2 2 2 p c 19 486.5 MeV p2 2.165 1013 N2·s2 This result does not have a solution in the real numbers, so the momentum, p, is imaginary (or virtual). 97. For a strange, s, quark and an antistrange, s , quark, the two new quarks created at the bro , accordken ends could be u and u or d and d sd. These particles ing to ss → su su or sd are known as mesons.

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Physics Concepts and Connections, Book Two - Solutions Manual

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