Physics Concepts and Connections, Book Two - Solutions Manual[1]

8/2/2019 Physics Concepts and Connections, Book Two - Solutions Manual[1]

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BOOK TWO

Brian Heimbecker

Igor Nowikow

Christopher T. Howes

Jacques Mantha

Brian P. Smith

Henri M. van Bemmel

Solutions Manual


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Physics: Concepts and ConnectionsBook Two Solutions Manual

AuthorsBrian Heimbecker

Igor NowikowChristopher T. Howes Jacques ManthaBrian P. SmithHenri M. van Bemmel

NELSONDirector of PublishingDavid Steele

PublisherKevin Martindale

Project EditorLina Mockus-OBrien

EditorKevin Linder

First Folio Resource GroupProject ManagementRobert Templeton

CompositionTom Dart

Proofreading and Copy EditingChristine SzentgyorgiPatricia Trudell

IllustrationsGreg DuhaneyClaire Milne

COPYRIGHT 2003 by Nelson, adivision 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

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Every effort has been made to traceownership of all copyrighted material andto secure permission from copyrightholders. In the event of any questionarising as to the use of any material, wewill be pleased to make the necessarycorrections in future printings.


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Table of Contents iii

Table of Contents

Chapter 1Section 1.3 1

1.4 11.6 11.7 31.8 31.11 41.12 51.13 61.14 61.15 7


2.2 92.3 102.4 112.5 122.6 132.7 142.8 15


3.4 183.5 183.6 203.7 213.8 223.9 22


4.3 24

4.4 254.5 254.6 26


5.3 285.4 295.5 295.6 305.7 31


6.2 336.3 34


7.3 367.4 367.5 37 7.6 387.7 387.8 387.9 397.10 39

7.11 40Chapter 8Section 8.4 41

8.5 418.6 428.7 438.8 448.9 44



10.3 47 10.4 4810.5 48


11.5 4911.6 4911.8 4911.9 5011.10 51


12.3 5212.4 5212.5 5312.6 5312.8 54


13.2 5513.3 5513.4 5613.5 5613.6 57

13.7 57 13.8 58


14.2 5914.3 5914.4 5914.5 5914.6 6014.7 60

14.8 60

I Solutions to Applying the Concepts Questions II Answers toEnd-of-chapterConceptualQuestions

Chapter 1 61Chapter 2 63Chapter 3 65Chapter 4 66Chapter 5 67 Chapter 6 68Chapter 7 69Chapter 8 71Chapter 9 75

Chapter 10 77 Chapter 11 79Chapter 12 80Chapter 13 81Chapter 14 83

III Solutions to End-of-chapterProblems

Chapter 1 87 Chapter 2 95Chapter 3 107 Chapter 4 120Chapter 5 126Chapter 6 134Chapter 7 140Chapter 8 151Chapter 9 160Chapter 10 165Chapter 11 170Chapter 12 178Chapter 13 183Chapter 14 191


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Solutions to Applying the Concepts 1

Section 1.3

1. (30 days) 2.6 106 s (units cancel to give answer in

seconds)

2. (7 furlongs) 1.4 km(units cancel to give answer in kilometres)

3. (1 quart) 5.5 102 mL (units cancel to give answer

in millilitres)

Section 1.41. Since the question is asking for velocity, the

answer must include a direction. Since thedirection in which the train travels isconstant,

v

avg

v

avg

v

avg 14 m/s [N]2. a) Since the question is asking for average

speed, direction is not required.vavg

vavg

vavg 1.6 km/h b) Since the question is asking for average

velocity, direction is required.

v

avg

v

avg

v

avg

v

avg

v

avg 0.40 km/h [E]

3. a) Since the question asks for the carsvelocity, direction is important. Since thedirection is constant,

v

avg

v

avg

v

avg 1.1 m/s [E] b) The cars instantaneous velocity at 5 s can

be approximated by the difference betweenthe distance travelled after 6 s and thedistance travelled after 5 s, divided by thetime during that interval:

v

avg

v

avg

v

avg 0 m/s

Section 1.61. v22 v12 2a d

d

d

d 9.4 103 m2. 10 cm 1.0 10 1 m

d

v1 v2

v1 0.05 m/s

v1 1.7 10 2 m/s

3. a) Igor: dI vI t Brian: dB

12

a B t 2

If they meet, dI dB 8.0 m

vI t 8.0 m12

a B t 2

0 12

(2.8 m/s 2) t 2 (7.0 m/s) t 8.0 m

0 (1.4 m/s 2) t 2 (7.0 m/s) t 8.0 m

2(1.0 10 1 m)3.0 s

2 dt

(v1 v2) t 2

(600 m/s) 2 (350 m/s) 2

2(12.6 m/s 2)

v22 v12

2a

8.0 m [E] 8.0 m [E]6.0 s 5.0 s

d

t

9.0 m [E] 0 m [E]8.0 s

d

t

2.0 km [E]5.0 h

3.0 km [E] 5.0 km [E]5.0 h

3.0 km [W] 5.0 km [E]5.0 h

d

t

8.0 km5.0 h

dt

2.5 104 m [N]1.8 103 s

d

t

27.5 mL1 oz

20 oz1 quart

1 km0.63 mile

1 mile8 furlong

60 s1 min

60 min1 h

24 h1 day

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


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t

t

t

t 3.2 s or t 1.8 sWe will take the lower value: t 1.8 s.

b) dB12

(2.8 m/s 2)(1.8 s) 2

dB 4.4 m4. 8.0 cm 8.0 10 2 m

v22 v12 2a d

a

a

a 7.7 105 m/s 25. t total t 1 t 2 t 3

t total 3.0 s 6.0 s 10 st total 19.0 s

In the first 3.0 s, the truck travels a distanceof:

d112

(v1 v2) t 1

d112

(0 m/s 8.0 m/s)(3.0 s)

d1 12 mSince the truck travels at a constant speedover the second interval,

d2 v2 t 2d2 (8.0 m/s)(6.0 s)d2 48 m

For the final interval,

d3 v1 t 312

a t 32

d3 (8.0 m/s)(10 s)12

(2.5 m/s 2)(10 s) 2

d3 2.1 102

mdtotal d1 d2 d3dtotal 12 m 48 m 2.1 102 mdtotal 2.7 102 m

vavg

vavg

vavg 14 m/s

6. 100 km/h 27.8 m/s

d v1 t 12

a t 2

500 m (27.8 m/s) t 12

(30 m/s 2) t 2

0 (15 m/s 2) t 2 (27.8 m/s) t 500 m

t

t 4.9 s

7. a) d v1 t 12

a t 2

80 m (17 m/s) t 12

(9.8 m/s 2) t 2

0 (4.9 m/s 2) t 2 (17 m/s) t 80 m

t

t 2.7 s b) v22 v12 2a d

v2 v12 2 a d v2 (17 m /s) 2 2(9.8 m/s 2) (80 m ) v2 43 m/s

8. a) a

t (eq.1)

d t (eq. 2)Substituting equation 1 into equation 2,

d 2a d v22 v12 v2v1 v1v2v22 v12 2a d

b) a

v1 v2 a t (eq. 1)

d t (eq. 2)Substituting equation 1 into equation 2,

d

t

d v2 t 12

a t 2

v2 v2 a t

2

v2 v12

v2 v1t

v2 v1a

v2 v12

v2 v12

v2 v1a

v2 v1t

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

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

2.7 102 m19.0 s

d totalt total

(0 m/s) 2 (350 m/s) 2

2(8.0 10 2 m)

v22 v12

2 d

7.0 m/s 2.05 m/s2.8 m/s 2

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

b b2 4 ac 2a

2 Solutions to Applying the Concepts


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Section 1.71. a) v22 v12 2a d

Assuming up is positive,

d

d

d 330 m b) v2 v1 a t

t

t

t 8.16 sc) 2(8.16 s) 16.3 s

2. a) d v1 t 12

a t 2

Assuming down is positive,30.0 m (4.0 m/s) t 1

2(9.8 m/s 2) t 2

0 (4.9 m/s 2) t 2 (4.0 m/s) t 30.0 m

t

t

t

t 2.1 s

b) d v1 t 12a t 2

Assuming down is positive,

30.0 m ( 4.0 m/s) t 12

(9.8 m/s 2) t 2

0 (4.9 m/s 2) t 2 ( 4.0 m/s) t 30.0 m

t

t

t

t 2.9 s

3. d v1 t 12

a t 2

Assuming down is positive,

35 m v1(3.5 s)12

(9.8 m/s 2)(3.5 s) 2

v1 7.2 m/s or 7.2 m/s [up]

Section 1.8

1. a) a t 7.0s

a t 7.0s

a t 7.0s 2.0 m/s 2

a t 12s

a t 12s 0 m/s 2

a t 3.0s

a t 3.0s 12 m/s 2

b) The distance travelled by Puddles fromt 5.0 s to t 13 s can be found byfinding the area under the curve betweenthose times. We must consider two

separate intervals: between 5.0 s and 10 s,and between 10 s and 13 s. The area underthe graph in the first interval can beexpressed as the sum of the areas of atriangle and a rectangle:

d1 t 1 v1

d1

(10 s 5.0 s)(50 m/s)

d1 275 mThe area under the graph in the secondinterval can be expressed as a rectangle:d2 t 2 v2d2 (13 s 10 s)(60 m/s 0 m/s)d2 180 mdT d1 d2dT 275 m 180 mdT 455 m

2. a) For Super Dave, Sr.,

vavg

t

t

t 5.0 s

50 m10 m/s

dvavg

dt

(10 s 5.0 s)(60 m/s 50 m/s)2

t 1 v12

32.0 m/s 8.0 m/s4.0 s 2.0 s

60 m/s 60 m/s13 s 11 s

55.0 m/s 51.0 m/s8.0 s 6.0 s

vt2 vt1t 2 t 1

4.0 m/s 24.6 m/s9.8 m/s 2

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

b b2 4 ac 2a

4.0 m/s 24.6 m/s9.8 m/s 2

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

b b2 4 ac 2a

0 80.0 m/s9.8 m/s 2

v2 v1a

0 (80.0 m/s) 2

2( 9.8 m/s 2)

v22 v12

2a

Solutions to Ap ply in g the Con ce pts3


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We can find the acceleration of SuperDave, Jr. from the slope of his v

t graph:

a

a

a 3 m/s 2

d v1 t

But v1 0 m/s, so

d

t t t 6 s

b) Super Dave, Sr. wins the race by 1 s.c) Super Dave, Sr.:

vavg

t

t

t 10 sSuper Dave, Jr.:

d v1 t , where v1 0 m/s, so

d

t t t 8 s

Super Dave, Jr. wins.3. a) For segment 1,

d1 2.0 m 0.5 md1 1.5 mt 1 0.6 s 0.0 st 1 0.6 s

vavg1

vavg1

vavg1 2.5 m/s

For segment 2,d2 2.0 m 2.0 md2 0 m

vavg2 0 m/sFor segment 3,

d3 1.0 m 2.0 md3 1.0 mt 3 1.8 s 1.0 st 3 0.8 s

vavg3

vavg3

vavg3 1.25 m/sFor segment 4,

d4 2.2 m 1.0 md4 1.2 mt 4 2.6 s 1.8 st 4 0.8 s

vavg4

vavg4

vavg4 1.5 m/s

b) vavg

vavg

vavg 0.65 m/s

Section 1.111. a) Forces are unbalanced as

the force provided by thekicker, F k, will cause the ball to accelerate.

Ball

F n

F k

F g

2.2 m 0.5 m

2.2 s 0.0 s

d totalt total

1.2 m0.8 s

d4t 4

1.0 m0.8 s

d3t 3

1.5 m0.6 s

d1t 1

2(100 m)3 m/s 2

2 da

a t 2

2

a t 2

2

100 m10 m/s

dvavg

dt

2(50 m)3 m/s 2

2 da

a t 2

2

a t 22

6 m/s2 s

vt



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b) The forces are balanced. Theforce he provideson the gun, F m,will balance theforce of the bullet.

c) The forces are not balanced, asthe penny still acceleratesdownward, but at a slower rate.

d) These forces are balanced, andthe soldier falls downward at aconstant speed.

Section 1.121. a) F 1 m1a 1

a 1

a 1

a 1 5.0 m/s 2

b) F 1 2m1a 2

a 2

a1

a 2

a 2

a 2 2.5 m/s 2

c) m1a 3

F 1 2m1a 3

a 3

a 3

a 3 2.5 m/s 22. F ma

F g F f ma F f m( g a) F f (90 kg)(9.8 m/s 2 6.8 m/s 2) F f 270 N

3. v22 v12 2a d

a

a

a 2.5 104 m/s 2

F ma F (8.0 10 2 kg)( 2.5 104 m/s 2) F 2000 N

4. For the first kilometre,

d v1 t 12

a 1 t 2

d 12

a 1 t 2

a 1

a 1

a 1 4.54 m/s 2

v22 v12 2a dv2 2a d v2 2(4.54 m/s 2) (1000 m) v2 95.3 m/s

For the last 1.4 km, the cars acceleration is:v22 v12 2a 2 d

a 2

a 2

a 2 3.24 m/s 2

F f ma 2 F f (600 kg)( 3.24 m/s 2) F f 1.94 103 N

(0 m/s) 2 (9.53 m/s) 2

2(1.40 103 m)

v22 v12

2 d

2(1000 m)(21.0 s) 2

2 dt 2

(0 m/s) 2 (15 m/s) 2

2(4.5 103

m)

v22 v12

2 d

5.0 m/s 2

2

a 12

F 12

5.0 m/s 2

2

a 12

F 1m1

F 12m1

10 N2.0 kg

F 1m1

Soldier

F parachute

F g

Penny

F buoyant

F g

Gun

F support

F m F B

F g



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b) F f k F n F n mg F f kmg

k

k

k 0.172. a) Since the toy duck is travelling at a

constant velocity, it is not being acted upon by an unbalanced force. Therefore, theforces must have equal magnitudes andopposite directions.

b) From a), we know that the applied force, F app, is equal in magnitude to the force dueto friction, F f . F n mg

F app F f F app k F n F app kmg

a

a k g a (0.15)(9.8 m/s 2)a 1.5 m/s 2

3. F f mak F n ma

kmg maa k g

v22 v12 2a d

d

d

d 42 m

Section 1.15

1. F g

F g

F g 5.5 10 67 N

2. F g

F g

F g

F g 2.1 1020 N

3. a) F g1

F g2 F g2 ( F g1)

b) F g2

F g2 F g2 ( F g1)

c) F g2

F g2

F g2 F g1

4. ( F gEarth) F g2

r 22 2r Earth r 2 r Earth 2 0

r 2

r 2 2.6 106 m

5. F g

myou g Jupiter

g Jupiter

g Jupiter

g Jupiter 24 m/s 2

(6.67 10 11 N m2 kg2)(1.9 1027 kg)(7.2 107 m)2

Gm Jupiterr Jupiter 2

Gm youm Jupiterr Jupiter 2

Gm youm Jupiterr Jupiter 2

2r Earth (2r Earth )2 4 (1)( r Earth 2) 2

1r Earth 2 2r Earth r 2 r 22

12(r Earth 2)

Gm youmEarth(r

Earthr

2)2

Gm youmEarthr

Earth

212

12

Gm 1m2r 2

G 4m1m2(2r )2

29

Gm 1m2r 2

29

G (2m1)m2(3r )2

18

Gm 1m2r 2

18

Gm 1m2r 2

(6.67 10 11 N m2 kg2)(0.013)(5.97 1024 kg)2

(3.82 108 m)2

G (0.013) mEarth 2

r 2

Gm Earth mMoonr 2

(6.67 10 11 N)(9.11 10 31 kg)2

(0.01 m) 2

Gm 1m2r 2

(0 m/s) 2 (22.2 m/s) 2

2(0.60)( 9.8 m/s 2)

v22 v12

2 k g

F appm

6.7 103 N(4000 kg)(9.8 m/s 2)

F f mg



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Section 2.11. dA [E35N] or [N55E]

dB [S12E] or [E78S]dC [S45W] or [W45S]dD [W80N] or [N10W]

2. a) In the N-S direction,dy d cosd

y (50 m) cos 14 [S]d

y 49 m [S]In the E-W direction,

dx d sind

x (50 m) sin 14 [E]d

x 12 m [E] b) In the N-S direction

vy v sinv

y (200 m/s) sin 30 [S]v

y 100 m/s [S]In the E-W direction,vx v cosv

x (200 m/s) cos 30 [W]v

x 173 m/s [W]c) In the N-S direction,

a y a sina

y (15 m/s 2) sin 56 [N]a

y 12 m/s 2 [N]In the E-W direction,a x a cosa

x (15 m/s 2) cos 56 [E]a

x 8.4 m/s 2 [E]3. Horizontally,

vx v cosvx (5.0 m/s) cos 25vx 4.5 m/sVertically,vy v sinvy (5.0 m/s) sin 25vy 2.1 m/s

4. v

g v

w v

b

v

g 4.0 m/s [forward] 3.0 m/s [upward]Since vw and v b are perpendicular,

vg vw2 v b2 vg (4.0 m /s) 2 (3.0 m /s) 2 vg 5.0 m/s

tan

tan 1 53

v

g 5.0 m/s [up 53 forward]5. a) Component Method:

v

f v

1 v

2

For the x components,v

fx v

1x v

2x

v

fx (50 m/s) cos 36 [W](70 m/s) cos 20 [E]

vfx ( 50 m/s) cos 36(70 m/s) cos 20

v

fx 25.3 m/s [E]For the y components,v

fy v

1y v

2y

v

fy (50 m/s) sin 36 [N](70 m/s) sin 20 [S]

vfy (50 m/s) sin 36(70 m/s) sin 20

v

fy 5.45 m/s [N]vf vf x

2 vf y2 vf (25.3 m/s) 2 (5.4 5 m/s) 2 vf 26 m/s

tan

tan 1 78

v

f 26 m/s [N78E]Sine/Cosine Method:

54 90 54 20 16

vf 2 v12 v22 2v1v2 cosvf 2 (50 m/s) 2 (70 m/s) 2

2(50 m/s)(70 m/s) cos 16vf 26 m/s

To find direction,

32

25.9 m/ssin 16

50 m/ssin

vf sin

v1sin

25.3 m/s5.45 m/s

vf xvf y

4.0 m/s3.0 m/s

vwv b



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To find , 180 54 78

v

f 26 m/s [N78E]

b) 37 (parallel line theorem) 180 53 37 (supplementary

angles theorem) 90

Sine/cosine Method:df 2 d12 d22 2 d1 d2 cosdf 2 (28 m) 2 (40 m) 2

2(28 m)(40 m) cos 90df 49 m

To find direction,

sin

35

To find , 180 37 18d

f 49 m [W18N]

c) Component Method: F

net F

1 F

2 F

3

For the x components, F

netx F

1x F

2x F

3x

F

netx 140 N [W] (200 N) cos 30 [E](100 N) sin 35 [W]

F netx 140 N (200 N) cos 30(100 N) sin 35

F netx 24.15 N F

netx 24.15 N [W]

For the y components, F

nety F

1y F

2y F

3y

F

nety (200 N) sin 30 [N](100 N) cos 35 [S]

F nety (200 N) sin 30(100 N) cos 35

F

nety 18.08 N [N] F net F net x2 F net y2 F net (24.15 N)2 (18.0 8 N) 2 F net 30.1 N

tan

tan 1 53

F

net 30.1 N [N53W]6. vx v2 sin 40 v1 sin 15

v

x 25.8 m/s [W]vy v2 cos 40 ( v1 cos 15)v

y 1.17 m/s [N]v (25.8 m/s) 2 (1.1 7 m/s) 2 v 26 m/s

tan 1 87v

26 m/s [N87W]

Section 2.21. a) v

og v

mg v

om

cos

cos

76The ships heading is [S76E].

b) v2og v2om v2mgvog (20 km /h) 2 (5.0 k m/h) 2 v

og 19 km/h [E]

c) t dv

t

t 5.2 h

100 km19 km/h

5.0 km/h20 km/h

vmgvom

25.8 m/s1.17 m/s

24.15 N18.08 N

F net x F net y

d

f

d

2

d

1

53 37

28 m49 m

28 msin

49 msin 90

d1sin

df sin

20

36

v

2

v

1

v

f



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2. a) sin

sin 1 9.6

The girls heading is [N9.6E]. b) The girl:

vg (3.0 m /s) 2 (0.50 m/s) 2 v

g 2.96 m/s [N]

t

t

t 169 sThe boy:

t

t

t 167 sc) The boy travels an extra distance west of

the girls landing point, caused by the horizontal component of his velocity (equalto the rivers current).d vt d (0.50 m/s)(167 s)d 83 m

d) The time required for the boy to run theextra 83 m at 5.0 m/s is 17 s. The boystotal time is 167 s 17 s 184 s. Thegirls time was 169 s. She wins the race.

3. v

pw v

sw v

ps

v2pw v2sw v2psv2pw (10 km /h) 2 (6.0 k m/h) 2 vpw 12 km/h

tan

59v

pw 12 km/h [N59E]4. a) v

og v

om v

mg

cos

cos 1 76

Terry must throw at [S76E].

b) vom (2.0 m /s) 2 (0.50 m/s) 2 v

om 1.9 m/s [E]

t

t

t 2.6 s

Section 2.3

1. a) dy viy t a y t 2

15 m (0 m/s) t (9.8 m/s 2) t 2

t 2

t 1.7 s

b) dx vix t a x t 2

dx (25 m/s)(1.7 s) (0 m/s 2) t 2

dx 43 m

2. a) a y

t

t

t 2.3 s b) Since the curve Blasto travels is

symmetrical (a parabola), the time he takesto reach maximum height is the same asthe time he takes to reach the ground.t total 2(2.3 s)t total 4.6 sSolving for horizontal distance,

dx vix t a x t 2

dx (35 m/s) cos 40(4.6 s)

dx 120 m3. a) To find the time required for the bomb toreach the ground,

dy viy t a y t 2

200 m (97.2 m/s) cos 25 t

(9.8 m/s 2) t 2

200 m (88.1 m/s) t (4.9 m/s 2) t 2

12

12

12

0 m/s (35 m/s) sin 409.8 m/s 2

v2y v1ya

v2y v1yt

12

1

2

30 m9.8 m/s 2

12

12

5.0 m1.9 m/s

dvom

0.50 m/s2.0 m/s

vmgvog

10 km/h6.0 km/h

500 m3.0 m/s

dv

500 m2.96 m/s

dv

0.50 m/s3.0 m/s

vmgvom



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0 (88.1 m/s) t (4.9 m/s 2) t 2

200 m

t

t 2.0 sTo calculate the horizontal distance,

dx vix t a x t 2

Since there is no horizontal acceleration,dx vix t dx (97.2 m/s) sin 25(2.0 s)dx 82 m

b) The y component of the final velocity, vfy, isvf y

2 viy2 2a d

vf y2 [(97.2 m/s) cos 25] 2

2(9.8 m/s 2)(200 m)vf y 108 m/svf x (97.2 m/s) sin 25vf x 41.1 m/svf (108 m /s) 2 (41.1 m/s) 2 vf 115.6 m/s

tan

21v

f 116 m/s inclined at 21 to the vertical4. Since the time it takes for the ball to hit the

green is not given, we can find two time-

related equations (one for the horizontalcomponent and one for the verticalcomponent), for the golf balls velocity, equate both equations, and solve for horizontalvelocity. For the vertical component,

dy viy t 12

a y t 2

Since the change in height is 0 m,

0 (vig sin ) t 12

( 9.8 m/s 2) t 2

(4.9 m/s 2) t vig sin

t (eq. 1)

For the horizontal component,

dx vix t 12

a x t 2

250 m (vig cos ) t

t (eq. 2)

Equating equations 1 and 2,

vig2

vig2

vig 66 m/sv

ig 66 m/s, 17 above the horizontal

Section 2.41. F

p F

1 F

2

F

p 200 N [N] 300 N [W] F p F 12 F 22 F p (200 N )2 (3 00 N) 2 F p 361 N

tan

tan

56

F

p 361 N [N56W]For the frictional force, F f kmg F f 0.23(200 kg)(9.8 m/s 2) F f 451 N

This is the maximum force of friction betweenthe stove and the floor. However, friction onlyacts to oppose motion, so F

f 361 N [S56E]. F

net F

p F

f

F

net 361 N [N56W] 361 N [S56E] F

net 361 N[N56W] 361 N [N56W] F net 0 N F net ma

a

a20

0

0

N

kga 0 m/s 2

Since the frictional force is stronger than theforce provided by the peoples pushing, thestove does not move.

F netm

300 N200 N

F

2

F

f

F

1

F

p

F 2

F 1

1225 m2/s 2

sin 17cos 17

(250 m)(4.9 m/s 2)sin cos

250 mvig cos

vig sin4.9 m/s 2

250 mvigcos

vig sin4.9 m/s 2

41.1 m/s108 m/s

12

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



16/200

2. a) F

net F

1 F

2 F

3

F

net 25 N [S16E] 35 N [N40E]45 N [W]

Adding the x components, F

netx F

1x F

2x F

3x

F

netx (25 N) sin 16 [E](35 N) sin 40 [E] 45 N [W]

F netx (25 N) sin 16(35 N) sin 40 45 N

F netx 15.6 N F

netx 15.6 N [W]Adding the y components, F

nety F

1y F

2y F

3y

F

nety (25 N) cos 16 [S](35 N) cos 40 [N]

F nety ( 25 N) cos 16(35 N) cos 40

F

nety 2.78 N [N] F net F net x

2 F net y2 F net (15.6 N)2 (2.78 N)2 F net 15.8 N

tan

tan

80 F

net 15.8 N [N80W]

b) F

net ma

a

a

a

0.20 m/s 2 [N80W]3. F

net ma

F

net (0.250 kg)(200 m/s 2 [W15S]) F

net 50.0 N [W15S] F

net F

1 F

2

F

2F

netF

1 F

2 50.0 N [W15S] 100 N [N25W] F

2 50.0 N [W15S] 100 N [S25E]Adding the x components, F

2x (50.0 N) cos 15 [W](100 N) sin 25 [E]

F

2x 6.03 N [W]

Adding the y components, F

2y (50.0 N) sin 15 [S] (100 N) cos 25 [S] F

2y 103.6 N [S] F 2 F 2x

2 F 2y2 F 2 (6.03 N)2 (103.6 N)2 F 2 104 N

tan

tan

86.7 90 86.7 3.3

F

2 104 N [S3.3W]4. The only two forces in the x direction are F x

and F f . F

net F

x F

f

F x F cos 45 F x (250 N) cos 45 F x 177 N F f k F n F

n F

g F

y

F n mg F sin 45 F n (20 kg)( 9.8 m/s 2) 177 N F n 372 N F f (0.40)( 372 N) F f 149 N

F net 177 N 149 N F net 27.9 N F net ma

a

a

a 1.38 m/s 2

Section 2.51. The only two unbalanced forces are F || and F f .

F net F || F f (eq. 1) F || F g sin 25 (eq. 2) F f F n F f F g cos 25 (eq. 3)

27.9 N20 kg

F netm

103.6 N6.03 N

F 2y F 2x

16 N [N80W]80 kg

F

net

m

15.6 N2.78 N

F net x F net y



17/200

Substituting equations 2 and 3 into equation 1, F net F g sin 25 F g cos 25 F net F g(sin 25 cos 25) F net (2.0 kg)(9.8 m/s 2)(sin 25 cos 25) F net (19.6 N)(sin 25 cos 25) F net 6.51 N F net ma6.51 N (2.0 kg)a

a 3.26 m/s 2

d vi t a t 2

4.0 m (3.26 m/s 2) t 2

t t 1.6 s

2. Since there is no friction, the only force that

prevents the CD case from going upward isthe deceleration due to gravity, F || . F net F || F net F g sin 20Since F net ma ,ma mg sin 20

a g sin 20a 3.35 m/s 2

a

t

t

t 1.2 s3. To find the distance the skateboarder travels

up the ramp, we need to find the velocity of the skateboarder entering the second ramp atv1. Since there is no change in velocity on the horizontal floor, v1 v2.For the acceleration on ramp 1, F net F ||ma mg sin 30

a g sin 30a 4.9 m/s 2

v22 v12 2adv22 0 m/s 2(4.9 m/s 2)(10 m)v2 9.9 m/s

For the deceleration on ramp 2, F net F || F nma mg sin 25 (0.1)mg cos 25

a 5.02 m/s 2

For d,v32 v22 2a d

(0 m/s) 2 (9.9 m/s) 2 2( 5.02 m/s 2) dd 9.8 m

4. F net m(0.60 g ) F net also equals the sum of all forces in theramp surface direction:

F

net F

|| F

f F

engine

m(0.60 g ) mg sin 30 F n F enginem(0.60 g ) mg sin 30 (0.28)mg cos 30

F engine F engine (0.60)mg mg sin 30

(0.28)mg cos 30 F engine mg (0.60 sin 30

(0.28) cos 30) F engine 3.36m N

Section 2.61. a) For m1,

F net m1aT m1 g m1a (eq. 1)

For m2, F net m2a

m2 g T m2a (eq. 2)Adding equations 1 and 2,m2 g m1 g a(m1 m2)

a

a

a

5.1 m/s 2 [right]Substitute a into equation 2:T m2 g m2aT 71 N

b) For m1, F net m1a

T m1 g sin 35 m1 g cos 35 m1a(eq. 1)For m2,

F net m2am2 g T m2a (eq. 2)

(15 kg)(9.8 m/s 2) 0.20(10 kg)(9.8 m/s 2)25 kg

m2 g m1 g m1 m2

4.0 m/s3.35 m/s 2

v2 v1a

v2 v1t

8.0 m3.26 m/s 2

12

12



18/200

Adding equations 1 and 2,m2 g m1 g sin 35 m1 g cos 35

a (m1 m2)

a

a

a

3.5 m/s 2 [right]Substitute a into equation 2:T m2 g m2aT (5.0 kg)(9.8 m/s 2) (5.0 kg)(3.5 m/s 2)T 32 N

c) For m1, F net m1a

T m1 g sin 40 1m1 g cos 40 m1a(eq. 1)For m2,

F net m2am2 g sin 60 T 2m2 g cos 60 m2a(eq. 2)Adding equations 1 and 2,m2 g sin 60 2m2 g cos 60m1 g sin 40 1m1 g cos 40

a (m1 m2)

a

a

a

1.1 m/s 2 [right]Substitute a into equation 1:T m1a m1 g sin 40 1m1 g cos 40T (20 kg)(1.1 m/s 2) (20 kg)(9.8 m/s 2)

sin 40 (0.20)(20 kg)(9.8 m/s 2)cos 40

T 1.8 102 Nd) For m1,

F net m1am1 g sin 30 T 1 m1a (eq. 1)For m2,

F net m2aT 1 T 2 m2a (eq. 2)

For m3, F net m3a

T 2 m3 g m3a (eq. 3)

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

a

a

a

0.82 m/s 2 [left]Substitute a into equation 3:T 2 m3a m3 g T 2 (10 kg)(0.82 m/s 2) (10 kg)(9.8 m/s 2)T 2 106 NSubstitute a into equation 2:T 1 m2a T 2T 1 106 N (20 kg)( 0.82 m/s 2)T 1 122 N

Section 2.7

1. a c

a c

a c 21 m/s 2

2. v dt

v 25 a c

a c

a c

a c 8.9 m/s 2

3. a c

a) If v is doubled, a c increases by a factor of 4. b) If the radius is doubled, a c is halved.c) If the radius is halved, a

cis doubled.

4. a) v 2T

r , where

r 3.8 105 kmr 3.8 108 mT 27.3 daysT 2.36 106 s

v2

r

2500 2(1.3 m)(60 s) 2

2500 2r t 2

v2

r

2 r t

(25 m/s) 2

30 m

v2

r

(30 kg)(9.8 m/s 2)sin 30 (10 kg)(9.8 m/s 2)60 kg

m1 g sin 30 m3 g m1 m2 m3

(9.8 m/s 2)[(30 kg) sin 60 0.30(30 kg) cos 60 (20 kg) sin 40 0.20(20 kg) cos 40]50 kg

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

(9.8 m/s 2)[5.0 kg (3.0 kg) sin 35 0.18(3.0 kg) cos 35]8.0 kg

g (m2 m1 sin 35 m1 cos 35)m1 m2



19/200

a c

a c

a c

a c 2.7 10 3 m/s 2

b) The Moon is accelerating toward Earth.c) The centripetal acceleration is caused by

the gravitational attraction between Earthand the Moon.

5. r 60 mmr 0.06 ma c 1.6 m/s 2

a c

v a cr

v 0.31 m/s6. Since d 500 m, r 250 m

v

f T 1

v 2 rf a c g

a c

g 4 2rf 2

f f f 0.0315 rotations/s f (0.0315 rotations/s)

f 2724 rotations/day

Section 2.81. a) v d

t

v

v 3.5 m/s

b) F c ma c

F c (10 kg)

F c 24 Nc) Friction holds the child to the merry-go-

round and causes the child to undergocircular motion.

2. Tension acts upward and the gravitationalforce (mg ) acts downward. F c F net andcauses Tarzan to accelerate toward the pointof rotation (at this instant, the acceleration isstraight upward).

F c ma c

T mg

T m g

T (60 kg) 9.8 m/s 2

T 9.7 102 N3. Both tension and gravity act downward.

F c ma c

T mg

When T 0,

mg

v gr

v (9.8 m /s 2)(1. 2 m) v 3.4 m/s

4. a)

b) F c mg tan 20

mg tan 20

v rg tan 20 v (100 m )(9.8 m /s 2) ta n 20 v 19 m/s

c) The horizontal component of the normalforce provides the centre-seeking force.

mv2r

N

N

cos 20

N

sin 20

20

mg

mv2

r

mv2

r

(4 m/s)2

2.5 m

v2

r

mv2

r

v2

r

20(2 r )180 s

24 h1 day

60 min1 h

60 s1 min

9.8 m/s 2

4 2(250 m)

g 4 2r

v2

r

2 r T

v2

r

4 2(3.8 108 m)(2.36 106 s)2

4 2r T 2

v2

r



20/200

d) If the velocity were greater (and the radiusremained the same), the car would slide upthe bank unless there was a frictional forceto provide an extra centre-seeking force.The normal force would not be sufficientto hold the car along its path.

e) Friction also provides a centre-seekingforce.

5. G 6.67 10 11 Nm 2/kg2, mE 5.98 1024

kg F c mMa c

Gm E v2r

Gm E , where v

T

T T 1.97 106 sT 22.8 days

6. G 6.67 10 11 Nm 2/kg2,mE 5.98 1024 kg, r E 6.37 106 m

F c mHa c

Gm E v2

r v r height of orbit r Er 6.00 105 m 6.37 106 mr 6.97 106 m

v v v 7.57 103 m/s

7. G 6.67 10 11 Nm 2/kg2

mM (0.013) mEmM 7.77 1022 kgr M 1.74 106 m

F c mApolloa c

Gm M v2r

Gm M , where v

r height of orbit r Mr 1.9 105 m 1.74 106 mr 1.93 106 m

T T T 7.4 104 s

400 2(1.93 106 m)3

(6.67 10 11 N m2/kg2)(7.77 1022 kg)

400 2r 3

Gm M

10(2 r )T

400 2r 3

T 2

mApollov2

r Gm MmApollo

r 2

(6.67 10 11 N m2/kg2)(5.98 1024 kg)6.97 106 m

Gm Er

Gm Er

mHv2

r Gm EmH

r 2

4 2(3.4 108 m)3

(6.67 10 11 N m2/kg2)(5.98 1024 kg)

4 2r 3

Gm E

2 r T

4 2r 3

T 2

mMv2

r Gm EmM

r 2



21/200

Section 3.31.

Horizontal:T h T cos 60T h (1.0 104 N) cos 60T h 5.0 103 NVertical:T v T sin 60T v (1.0 104 N) sin 60T v 8.7 103 N

2.

F net T v T A T A F net ma F net 0T v T A T AT v 2T AT v 2( 100.0 N) cos 70T v 68.4 N

3. a)

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

c) F net 2T v F g F net ma F net 0 F g 2T vm

m

m 0.45 kg4. a)

tan 71.1

F net mg 2FBv F net ma F net 0

0 mg 2 F B sin

F B

F B

F B 20.7 N b) F h F B cos

F h (20.7 N) cos 71.1 F h 6.71 N

c) F v F B sin F v (20.7 N) sin 71.1 F

v 19.6 N [down] (not including theweight of the beams)

6.

F || mg sin F f mg cos F net T F f F || F net ma

F fF n

F ||F

T

boat

(4.0 kg)( 9.8 N/kg)2 sin 71.1

mg 2 sin

1.90 m0.650 m

pail

F BF B

mg

+

F g

1.90 m

0.65 m

=

2(85 N) sin 1.59.8 N/kg

2T sin g

bag

F g mg

+

55

=

T T = 85 N = 85 N

T v

T a T a

+

70 70 = 100.0 N= 100.0 N

60

T v

T h

T h = 1.0 10 4 N



22/200

F net 0T F || F f T mg sin mg cosT mg (sin cos )T (400.0 kg)(9.8 N/kg)

(sin 30 (0.25) cos 30)T 1.11 103 N

Section 3.41. a)

b) rF sin rmg sin (1.50 m)(45.0 kg)(9.8 N/kg) sin 40 425 Nm

2. a) 2.0 103 Nmr 1.5 m

90 F ?

rF sin

F

F

F 1.3 103 N3.

a) V w 10.0 L

w 1000 kg/m 3

V w (10.0 L)

V w 0.0100 m 3

mw w V wmw (1000 kg/m 3)(0.0100 m 3)mw 10.0 kg F g mg F g (10.0 kg)(9.8 N/kg) F g 98.0 N

b) Position B provides the greatest torque because the weight is directed at 90 to thewheels rotation.

c) A rF sinA (2.5 m)(10.0 kg)(9.8 N/kg) sin 45A 1.7 102 NmB rF sinB (2.5 m)(10.0 kg)(9.8 N/kg) sin 90B 2.4 102 NmC AC 1.7 102 Nm

d) A larger-radius wheel or more and largercompartments would increase the torque.

Section 3.5

1.

90r 1 ?m1 45.0 kg

m2 20.0 kg m2 5.0 kgr 2

r 2 0.375 mm3 20.0 kg m2m3 15.0 kg

0.75 m2

0.753.0

20.0 kg P

0.75 m3.0 m

1 m3

1.00 106 cm3

1000 cm3

1 L

10.0 L

2.5 m

B

C

A

2.0 103 Nm(1.5 m) sin 90

r sin

1.50 m

50

45.0 kg

F g mg =



23/200

r 3

r 3 1.12 m0 1 2 30 r 1 F 1 sin 1 r 2 F 2 sin 2 r 3 F 3 sin 3

r 1

r 1

r 1 0.332 m2. a)

t-t rF sin

t-t

t-t 147 NmThis torque applies to both sides of theteeter-totter, so the torques balance eachother.

b)

H L 0L H

r L

r L

r L 2.63 mc)

cos

75.5At the horizontal position:

H (1.75 m)(45 kg)(9.8 N/kg)H 7.7 102 Nm

At maximum height:H (1.75 m)(45 kg)(9.8 N/kg) sin 75.5H 7.5 102 Nm

% 100

% 2.6%3.

a)

1

2 0r 1 F 1 sin r cm F g sin

F 1

F 1

F

1 24.5 N b) F rv F v2 0

F rv F v2 F rv (5.00 kg)( 9.8 N/kg) F

rv 49 N [up] F rh F h1 0

F rh F h1 F rh 24.5 N F

rh 24.5 N [left]The vertical reaction force is 49 N [up]and the horizontal reaction force is 24.5 N[left].

4.

1

2

3 0r 1 F 1 r 2 F 2 r 3 F R3 0

r 1

r 1 0.375 m

0.75 m2

F 4 F 3

F 2

P

0.4 m1.6 m

F 1

(0.375 m)(5.00 kg)(9.8 N/kg)0.75 m

r cm F g sinr 1 sin

40

50

40

P

F gF 1

(7.7 102 Nm 7.5 102 Nm)7.7 102 Nm

0.5 m2.0 m

2.0 m

0.50 m

(1.75 m)(45.0 kg)30.0 kg

r HmH g mL g

r h = 1.75 m r l = ?

(1.0 m)(30.0 kg)(9.8 N/kg)2

1.7 m

4.0 m

(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)

r 3 F 3 r 2 F 2 F 1

3.0 m 0.75 m2



24/200

r 2

r 2 1.0 mr3 1.60 m

90sin 1

F 3

F 3

F

3 306 N [up] F 4 F 1 F 2 F 3 F 4 (120.0 kg)(9.8 N/kg)

(5.0 kg)(9.8 N/kg) 306 N F

RP 919 N [up]Left saw horse: 919 N [up]Right saw horse: 306 N [up]

Section 3.61.

45r w 48.0 cmr w 0.480 mmw 10.0 kg

r L

r L 24.0 cmr L 0.240 mmL 5.00 kg

w

L 0 w L ( r w F w sin 45)

( r L F L 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 Nm [clockwise]

2.

1

2 02 12 1

r 2 F 2 sin 2 r 1 F 1 sin 1

F 2

F 2

F 2 529.2 N F 2 5.3 102 N

The angle makes no difference it cancelsout.

3.

a)

m

b

s 0m b s 0

m b sr m F m sin m r b F b sin b r s F s sin s

F m

F m

F m

F m

F m 5.57 103 N (tension)

(75 102 m)(9.8 N/kg) sin 75[(0.57)85 kg 19.0 kg](45 102 m) sin 11

rg sin (m b m s)

r m sin m

r bm b g sin b r sm s g sin sr m sin m

r b F b sin b r s F s sin sr m sin m

15 11

30 cm

45 cm

P F gs

F m F gb

(8.0 10 2 m)(27 kg)(9.8 N/kg) sin(4.0 10 2 m) sin

r 1 F 1 sin 1r 2 sin 2

2

1

F 1 F n

F 2

P

+8 cm

4 cm=

48.0 cm2

F w

F L

+45 48 m

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

r 1 F 1 r 2 F 2r 3

2.0 m2



25/200

Reaction forces:0 F

py F

my F

by F

sy

F py F my F by F sy F py ( 5.57 103 N)(sin 4)

(19.0 kg)( 9.8 N/kg)(0.57)(85 kg)( 9.8 N/kg)

F py 1049.6 N F

py 1.05 103 N [up]0 F

px F

mx F

bx F

sx

F px F mx F bx F sx F px (5.57 103 N)(cos 4) 0 0 F

px 5.55 103 N [right]Horizontal force: 1.49 103 N [right]; verticalforce: 7.65 102 N [up]

Section 3.7

1. a) sin 43

htipped

htipped 49.8 cm

b) tan 43

hstraight

hstraight 36.5 cm2. Four-wheeled ATV:

tan T

T 31.0

Three-wheeled ATV:

tan

25.64

sin

x (0.55 m)(sin 25.64) x 0.237 m

tan T

tan T 13.3

0.237 m1.00 m

x0.55 m

0.60 m1.25 m

1.25 m

0.6 m

0.6 m

0.7 m

0.55 m

1.0 m

Back View

x

x

T

T

Top View

0.60 m1.0 m

1 . 0

m

0.6 m

34.0 cmtan 43

34.0 cmhstraight

34.0 cmsin 43

34.0 cm

htipped



26/200

Section 3.81. k 16.0 N/m

x 30.0 cm x 30.0 10 2 ma) F k x

F (16.0 N/m)(30.0 10 2 m) F 4.80 N

b) F ma

am F

a

a 1.78 103 m/s 2

2. F g (67.5 kg)(9.8 N/m) F g 661.5 N F kx

k

k

k 66150 N/mk 6.61 104 N/m F g-truck mg F g-truck (2.15 103 kg)(9.8 N/kg) F g-truck 2.1 104 NThis weight is distributed equally over foursprings.

F s

F s 5267.5 N/spring

x

x

x 0.0796 m x 8.0 10 2 m

3. F kx F (120 N/m)(30.0 10 2 m) F 36 N

Section 3.91. d 0.29 mm

L 0.90 m L 0.22 mm Esteel 200 109 N/m 2

A

E

E

F

F

F

F 0.807 NFor nylon, Enylon 5 109 N/m 2

d 2 d 2 d 1.83 mmd 1.83 10 3 m

2. Emarble 50 109 N/m 2

A 3.0 m2m 3.0 104 kg

a) Stress A F

Stress

Stress 9.8 104 N/m 2

b) E

Strain

Strain

Strain 2.0 10 6

9.8 104 N/m 2

50 109 N/m 2

Stress

E

StressStrain

(3.0 104 kg)(9.8 N/kg)3.0 m2

4(0.807 N)(0.90 m)(5 109 N/m 2)(0.22 10 3 m)

4 FL E L

0.29210 3 m 2(200 109 N/m 2)(0.22 10 3 m)

4(0.90 m)

d2

2 E L

4 L

AE L L

FL A L

A F

L L

d2

2

4

5267.5 N/spring6.6150 104 N/m

F sk

2.1 104

N4 springs

661.5 N1.0 10 2 m

F

x

4.80 N2.7 10 3 kg



27/200

c) L 15 m L ?

Strain L L

L L(Strain) L (15 m)(2.0 10 6) L 3.0 10 5 m

3. a) Compressive strength of bone17 107 N/m 2

d bone 4.0 10 2 mBone cross-sectional area is: A r2

A (2.0 10 2 m)2

A 1.26 10 3 m2

F b

F b

Breakage occurs if Strength

Strength

m

m

m 4.4 104 kg

2(17 107 N/m 2)(1.26 10 3 m2)9.8 N/kg

2(Strength) A g

m2 g

A

m2 g

A F b A

F b A

mg 2

F g2

F g F g

200 kg

2

2



28/200

Section 4.21. p

mv

p

(8 kg)(16 m/s [W20N])p

128 kgm/s [W20N]p

1.3 102 kgm/s [W20N]2. p

9.0 104 kgm/s [E]

v

(72 km/h [E]) v 20 m/s [E]m

m

m 4.5 103 kg3. a) p

mv

p

(0.5 kg)(32 m/s [S])p

16 kgm/s [S]Using a scale factor of 1 mm 1 kgm/s,

b) p

mv

p

(0.5 kg)(45 m/s [N])p

22.5 kgm/s [N]

c) p

p

2 p

1

p

22.5 kgm/s [N] 16 kgm/s [S]p

22.5 kgm/s [N] 16 kgm/s [N]p

38.5 kgm/s [N]

Section 4.31. a) J

F

t J

(3257 N [forward])(1.3 s) J

4234.1 Ns [forward] J

4.2 103 Ns [forward] b) J

F

t J

ma

t

J

m t J

m(v

2 v

1) J

(0.030 kg)(200 m/s 0 m/s) J

6.0 Ns [out of gun]c) J

F

t J

ma

t J

(0.500 kg)(9.8 N/kg [down])(3.0 s) J

14.7 Ns [down] J

15 Ns [down]2. p

p

2 p

1

p

mv

2 mv

1

p

m(v

2 v

1)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 Ns [up]

3. a) F

F

F 1.3 104 N b) v1 0

v2 120 km/hv2 33.3 m/s

a

a

a 166.7 m/s 2

d v1 t 12 a t 2

d (0 m/s)(0.2 s) 12

(166.7 m/s 2)(0.2 s) 2

d 3.3 m

33.3 m/s 0 m/s0.2 s

v2 v1t

2.5 103 Ns0.2 s

J t

v

2 v

1

t

p = 38.5 kg m/s [N]

p = 22.5 kg m/s [N]

p = 16 kg m/s [S]

9.0 104 kgm/s20 m/s

pv

1 h3600 s

1000 m1 km



29/200

4. a) J 12 bh

J 1

2(5 s)(25 N [S])

J

62.5 Ns [S] b) J Area under triangle rectangle

J 1

2(500 250 N [W])(3 s)

(250 N [W])(6 s) J

1875 Ns [W]c) J Area above area below (counting the

squares: approximately) J (13 squares above) (4 squares

below) J 9 squaresMultiplying 9 by the length and width of each square, J

9(0.05 s)(100 N [E]) J

45 Ns [E]

Section 4.41. 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 v

2f 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

v2f

v2f

v

2f 11.8 m/s [back]3. m1 400 g 0.400 kg,

v

1o 3.0 m/s [forward],v

1f 1.0 m/s [forward],m2 0.400 kg, v

2o 0, v

2f ?p

o p

f

m1v

1o m2v

2o m1v

1f m2v

2f

v

2f

v

2f

v

2f 2.0 m/s [forward]4. m1 m , m2 80m , m (1 2) 81m, v(1 2)o ?,

v1f 1.5 106 m/s, v2f 4.5 103 m/s,po 7.9 10 17 kgm/s

po pf po m1v1f m2v2f

7.9 10 17 kgm/s m( 1.5 106 m/s)80m(4.5 103 m/s)

7.9 10 17 kgm/s m[ 1.5 106 m/s80(4.5 103 m/s)]

m

m 6.9 10 23 kg5. m1 5m , v1o v, v(1 2)f ?, m2 4m, v2o 0

po pf m1v1 m2v2 (m1 m2)v(1 2)f

(5m)(v) (4m)(0) (5m 4m)v(1 2)f 5mv 9mv (1 2)f

v(1 2)f 59

v

Section 4.51. m1 m2 2.0 kg, v

1o 5.0 m/s [W], v

2o 0,v

1f 3.0 m/s [N35W], v

2f ?p

1o (2.0 kg)(5.0 m/s [W])p

1o 10 kgm/s [W]p

1f (2.0 kg)(3.0 m/s [N35W])p

1f 6.0 kgm/s [N35W]p

o p

f

p

1o p

2o p

1f p

2f , where p

2o 0p

1o p

1f p

2f

Using the cosine law,p2f 2 (10 kgm/s) 2 (6.0 kgm/s) 2

2(10 kgm/s)(6.0 kgm/s) cos 55p2f 8.2 kgm/s

p mv

v2f

v2f 4.1 m/s

8.2 kgm/s2 kg

35

p 1f = 6.0 kg m/s

p 1o = 10 kg m/s

p 2f

7.9 10 17 kgm/s1.5 106 m/s 80(4.5 103 m/s)

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

m1v

1o m2v

2o m1v

1f

m2

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

m1v1o m2v2o m1v1f m2



30/200

Using the sine law to find direction,

37v

2f 4.1 m/s [W37S]2. m1 85 kg, v

1o 15 m/s [N],p

1o 1275 kgm/s [N], m2 70 kg,v

2o 5 m/s [E], p

2o 350 kgm/s [E]p

o p

f

p

1o p

2o p

f

Using Pythagoras theorem to solve for pf ,

pf 2

(1275 kgm/s)2

(350 kgm/s)2

pf 1322 kgm/s

tan

15.4p

f m f v

f

v

f

v

f 8.5 m/s [N15E]3. m1 0.10 kg, v

1f 10 m/s [N],p

1f 1.0 kgm/s [N], m2 0.20 kg,v

2f 5.0 m/s [S10E],p

2f 1.0 kgm/s [S10E], m3 0.20 kg,v

3f ?p

To 0p

To p

Tf

0 p

1f p

2f p

3f

Using the cosine law,

p3f 2 (1.0 kgm/s) 2 (1.0 kgm/s) 2

2(1.0 kgm/s)(1.0 kgm/s)(cos 10)p3f 0.1743 kgm/s

v3f

v3f 0.87 m/sUsing the sine law to find direction,

85v

3f 0.87 m/s [S85W] or 0.87 m/s [W5S]4. m1 0.5 kg, v

1o 2.0 m/s [R],p

1o 1.0 kgm/s [R], m2 0.30 kg, v

2o 0,p

2o 0, v

1f 1.5 m/s [R30U],p

1f 0.75 kgm/s [R30U], v

2f ?, p

2f ?p

To p

Tf

p

1o p

2o p

1f p

2f , where p

2o 0p

1o p

1f p

2f

Using the cosine law,

p2f 2 (1.0 kgm/s) 2 (0.75 kgm/s) 2

2(1.0 kgm/s)(0.75 kgm/s)cos 30p2f 0.513 kgm/sp mv

v2f

v2f 1.7 m/sUsing the sine law to find direction,

47v

2f 1.7 m/s [R47D] or 1.7 m/s [D43R]

Section 4.6

1. a) 1.5 m from both objects

b)

(60 cm)

17.1 cm from the larger mass

c) (20 km)6.67 km from the larger satellite

200600

2.0 kg

5.0 kg 2.0 kg

3.0 m2

sin 300.513 kgm/s

sin0.75 kgm/s

0.513 kgm/s0.30 kg

p 1f

= 0.75 kg m/s

p 1o = 1.0 kg m/s

30

p 2f

sin 100.1743 kgm/s

sin1.0 kgm/s

0.17 kgm/s0.2 kg

p 3f

p 1f = 1.0 kg m/s

p 2f = 1.0 kg m/s

10

1322 kgm/s [N15E]85 kg 70 kg

350 kgm/s1275 kgm/s

p 2o = 350 kg m/s

p 1o = 1275 kg m/s p f

sin 558.2 kgm/s

sin6.0 kgm/s



31/200

2. a) p1o (2.0 kg) p

1o 0.22 kgm/s [S20E]

p2o (1.0 kg) p

2o 0.17 kgm/s [S10W]

p1f (2.0 kg) p 1f 0.26 kgm/s [S5W]p2f (1.0 kg) p

2f 0.15 kgm/s [S30E]

pcm (3.0 kg) p

cm 0.39 kgm/s [S8E] b) i)

ii)

c) The total momentum before and aftercollision is the same as the momentum of the centre of mass. The total momentumvectors have the same length and directionas the momentum of the centre of mass.

5

30

p 1f

p 2f

p Tf

10

70

p 1o

p 2o

p To

0.013 m0.1 s

0.015 m0.1 s

0.013 m0.1 s

0.017 m0.1 s

0.011 m0.1 s



32/200

Section 5.21. a) W F d

W (40 N)(0.15 m)W 6.0 J

b) W F dW mg dW (50 kg)(9.8 N/kg)(1.95 m)W 9.6 102 J

c) W F d cosW (120 N)(4 m)(cos 25)W 4.4 102 J

2. 45 km/h 12.5 m/sTo find d,v22 v12 2a d

d

dd 31.25 m

W F dW (5000 N)(31.25 m)W 1.6 105 J

3. W F d cosW (78 N)(10 m)(cos 55)W 4.5 102 J

4. a

a

a 2.2 m/s 2

F ma F (52 000 kg)( 2.2 m/s 2) F 114 400 N

d

d

d 97.5 mW F dW ( 114 400 N)(97.5 m)W 1.1 107 J

5. a) W F dW (175 N)(55 m)W 9625 J

b) The triangular areas above and below theaxis 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

d

d

d 16 m

Section 5.3

1. a) Ek12

mv2

Ek1

2

(20 000 kg)(7500 m/s) 2

Ek 5.6 1011 J b) 20 km/h 5.6 m/s

Ek12

mv2

Ek12

(1.0 kg)(5.6 m/s) 2

Ek 15.4 J

c) Ek12

mv2

Ek12 (0.030 kg)(400 m/s)

2

Ek 2.4 103 J

2. Ek12

mv2

3900 J 12

(245 kg) v2

v v 5.6 m/s

3. Ek1

2mv2

m

m

m 6.5 kg4. p 2mE k

p 2(9.11 10 31 kg)( 6000 e V )(1.6 10 19 J/eV ) p 4.2 10 23 Ns

2(729 J)(15 m/s) 2

2 Ekv2

(3900 J)(2)245 kg

480 J29.4 N

W F

[(14 m/s) 2 (25 m/s) 2]2( 2.2 m/s 2)

(v22 v12)2a

(14 m/s 25 m/s)5.0 s

(v2 v1)t

(12.5 m/s) 2 02(2.5 m/s 2)

(v22 v12)2a



33/200

5. Ek Ek2 Ek1

Ek12

(60 kg)(5.0 m/s) 2 12

(60 kg)(14 m/s) 2

Ek 5.1 103 J

6. a) Ek12

mv2

Ek12 (0.350 kg)(25.0 m/s)

2

Ek 1.1 102 J

b) a

a

a 1.3 104 m/s 2

F ma F (0.350 kg)( 1.3 104 m/s 2) F 4557 N

W F dW ( 4557 N)(0.024 m)W 1.1 102 J

c) F avg ma F avg 4557 N F avg 4.6 103 N

Section 5.41. 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 2a d (or use the conservationof energy)

v22 (0) 2(9.8 m/s 2)(27 m)v2 23 m/s

b) E final E initial Ekf Ego Eko1

2(65 kg) vf 2 (65 kg)(9.8 N/kg)(27 m)

12

(65 kg)(3.0 m/s) 2

vf 23 m/s

3. a) Using the law of conservation of energy, E total 5460 J

12

mv2 mgh 5460 J

12

(3.0 kg) v2 (3.0 kg)(9.8 N/kg)(5.0 m) 5460 J

v 60 m/s b) Eg mgh

5460 J (3.0 kg)(9.8 N/kg) hh 185.7 m from the groundh 180.7 m from the pad

c) v2 v1 a t v2 (60 m/s) (9.8 N/kg)(2.0 s)v2 40.4 m/s

Ek12

mv2

Ek1

2(3.0 kg)(40.4 m/s) 2

Ek 2.4 103 J Ep E total Ek Ep 5460 J 2448.24 J Ep 3.0 103 J

4. F kx

k

k

k

k 1.2 106 N/mFor only one spring:

k

k 3.0 105 N/m

Section 5.5

1. a) k

k

k 200 N/mk 2.0 102 N/m

20 N0.1 m

riserun

1 225 000 N/m4

(5000 kg)(9.8 N/kg)0.04 m

mg x

F x

0 (25.0 m/s) 2

2(0.024 m)

(v22 v12)2 d



34/200

b) Maximum elastic potential energy occursat x 0.1 m.

Ep12

kx2

Ep12

(200 N/m)(0.1 m) 2

Ep 1.0 Jc) Ee Ee2 Ee1

E 12

(200 N/m)(0.04 m) 2

12

(200 N/m)(0.03 m) 2

Ee 7.0 10 2 J2. F g F e

mg kx(0.500 kg)(9.8 N/kg) k(0.04 m)

k 122.5 N/m

3. a) W EW E2 E1 where E1 0W E2

W 12

kx2

W 12

(55 N/m)( 0.04 m) 2

W 4.4 10 2 J b) W E

W E2 E1 where E1 0W E2

W 12

kx2

W 12

(85 N/m)(0.08 m) 2

W 2.7 10 1 J4. Ee Ek

12

kx2 12

mv2

(200 N/m)(0.08 m) 2 (0.02 kg) v2

v 8.0 m/s5. Ee Ek

12

kx2 12

mv2

(5 106 N/m) x2 (2000 kg)(4.5 m/s) 2

x 9 cm

6. The loss in elastic potential energy is equal tothe gain in kinetic energy.

Ee EkLet the subscript 1 represent the initialcompressed spring and subscript 2 represent the moment after the spring has been releasedwhen the cart has a velocity of 0.42 m/s.

( Ee2 Ee1) Ek2 Ek1

kx12 kx22 mv22 0

x2 x2

x2 0.056 m x2 5.6 cm

Section 5.61. The energy required to heat the water is

Ew (4.2 103 J/C/L)(65C 10C)(2.3 L) Ew 5.31 105 JThe energy expended by the stove is

P

Es P t Es (1000 W)(600 s) Es 6.0 105 J

The energy lost to the environment is E E s 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

P

P

P 590 W b) Ep 1.057 104 J

Ep 10 600 J3. Once the radiation of the Sun reaches Earth,

it has spread out into a sphere surroundingthe Sun. This sphere has a surface area of:SA 4 r 2

SA 4 (1.49 1011 m)2

SA 2.79 1023 m2

1.057 104 J18.0 s

Et

Et

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

65 N/m

kx12 mv22

k

12

12

12



35/200

The ratio of this area to the area of Earthexposed to the radiation will be equal to theratio of the power radiated by the Sun to thepower absorbed by Earth.

x

x 2 1017 WTherefore, Earth intercepts 2 1017 J of energy from the Sun each second.

4. The total time is 3(20 min)(60 s/min) 3600 sThe time the player spends on ice is(3600 s)(0.25) 900 s

P

E P t E (215 W)(900 s) E 1.935 105 JWhile sitting on the bench, the player expends100 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 5.73. a) m1 3000 kg

v

1o 20 m/s [W]v

1f 10 m/s [W]m2 1000 kg

v2o 0v2f ?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 Ek truck

W 12

(3000 kg)(10 m/s) 2

12

(3000 kg)(20 m/s) 2

W 4.5 105 J4. mp 0.5 kg

mg 75 kgdp 0.03 m

vpo 33.0 m/svgo 0vgf 0.30 m/sa) pgo mv

pgo (75 kg)(0)pgo 0

Ekgo 0ppo mvppo (0.5 kg)(33.0 m/s)ppo 16.5 kgm/s

Ekpo12

(0.5 kg)(33.0 m/s) 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

c) Ek p12

mpvpf 2

Ek p12

(0.500 kg)(12 m/s) 2

Ek p 36 J

Ek g12

mgvgf 2

Ek g12 (75 kg)(0.30 m/s)

2

Ek g 3.4 Jd) The collision is inelastic due to the loss of

kinetic energy.

Et

(3.9 1026 W)(1.48 1014 m2)2.79 1023 m2

3.9 1026 W x

2.79 1023 m2

(6.87 106 m)2

3.9 1026 W x

2.79 1023 m2

( dE2arth )2

Suns radiationabsorbed radiation

SASun AEarth



36/200

5. m1 10 gm2 50 gv1o 5 m/sv2o 0

v1f v1o

v1f (5 m/s)v1f 3.3 m/s

v2f v1o

v2f (5 m/s)

v2f 1.7 m/s6. m1 0.2 kg

m2 0.3 kgv1o 0.32 m/s

v2o 0.52 m/sChanging the frame of reference,v1o 0.84 m/sv2o 0 m/s

v1f (0.84 m/s)

v1f 0.168 m/s

v2f (0.84 m/s)

v2f 0.672 m/s

Returning to the original frame of reference,v1f 0.168 m/s 0.52 m/sv1f 0.69 m/sv2f 0.672 m/s 0.52 m/sv2f 0.15 m/s

8. a) E stored12 bh

Estored12

(0.06 m 0.02 m)(50 N)

Estored 1.0 J

b) E lost 1.0 J12 (0.005 m)(30 N)

(0.005 m)(20 N)12

(0.035 m)(20 N)

E lost 1.0 J 0.075 J 0.1 J 0.35 J E lost 0.475 J

9. a) Counting the squares below the top curve,there are about 16.5 squares, each with anarea of (0.01 m)(166.7 N) 1.6667 J. Theamount of energy going into the shockabsorber is (16.5)(1.6667 J) 27.5 J.

b) There are roughly 6 squares below the lower curve. The energy returned to theshock absorber is (6)(1.6667 J) 10 J

c) % energy lost 100

% energy lost 64%

27.5 J 10 J27.5 J

2(0.2 kg)0.2 kg 0.3 kg

0.2 kg 0.3 kg0.2 kg 0.3 kg

2(10 g)10 g 50 g

2m1m1 m2

10 g 50 g10 g 50 g

m1 m2m1 m2



37/200

Section 6.11. mE 5.98 1024 kg, mS 1.99 1030 kg,

r 1.50 1011 m

a) Ek

Ek

Ek 2.65 1033 J

b) Ep

Ep

Ep 5.29 1033 Jc) ET Ek Ep

ET 2.65 1033 J ( 5.29 1033 J) ET 2.65 1033 J

2. a g

a g

a g 7.32 m/s 2

3. v1000 km 6.0 km/s 6.0 103 m/s,h 1000 km 1 106 m

a) vesc vesc vesc 10 397 m/sSince the rocket has only achieved6000 m/s, it will not escape Earth.

b) Ek 1000 km12

mv2

Ek 1000 km12

m(6000 m/s) 2

Ek 1000 km 1.8 107m JSince all kinetic energy is converted togravitational potential energy at maximum

height, Ek Ep Ek E2 E1

1.8 107m J

r 2

r 2(6.67 10 11 Nm 2/kg 2)(5.98 1024 kg)

1.8 107 J

r 2 1.1 107 m

hmax r 2 r Ehmax 1.1 107 m 6.38 106 mhmax 4.7 106 m

Section 6.21. a) M Sun 1.99 1030 kg,

T 76.1 a 2.4 109 sT 2 ka 3

a 13

a 13

a 2.7 1012 m b) 0.97

c) v

v

v 7031 m/s2. r altitude 10 000 km 1 107 m,

r Jupiter 7.15 107 m, m Jupiter 1.9 1027 kg

vesc vesc vesc 56 000 m/s

3. mMoon 7.36 1022 kg,mEarth 5.98 1024 kg, r 3.82 108 m

a) vesc vesc vesc 1445 m/s

2(6.67 10 11 Nm 2/kg 2)(5.98 1024 kg)3.82 108 m

2GM r

2(6.67 10 11 Nm 2/kg 2)(1.9 1027 kg)7.15 107 m 1 107 m

2GM r

2 (2.69 1012 m)2.4 109 s

dt

(2.4 109 s)2

4 2

(6.67 10 11 Nm 2/kg 2)(1.99 1030 kg)

(2.4 109 s)2

G 4 M

2

(6.67 10 11 Nm 2/kg 2)(5.98 1024 kg)6.38 106 m 1 106 m

GM

1.8 107 J G r M

1

GMmr 1

GMmr 2

2(6.67 10 11 Nm 2/kg 2)(5.98 1024 kg)(6.38 106 m 1 106 m)

2GM r

(6.67 10 11 Nm 2/kg 2)(5.98 1024 kg)(6.38 106 m 1 106 m)2

GM

r 2

(6.67 10 11 Nm 2/kg 2)(1.99 1030 kg)(5.98 1024 kg)(1.50 1011 m)

GMmr

(6.67 10 11 Nm 2/kg 2)(1.99 1030 kg)(5.98 1024 kg)2(1.50 1011 m)

GMm2r



38/200

To find the current speed of the Moon,

mv2

v v v 1022 m/s

To find the additional speed required forescape,vadd esc 1445 m/s 1022 m/svadd esc 423 m/s

b) Ek mvesc2 mv 2

Ek (7.36 1022 kg)[(1445 m/s) 2

(1022 m/s) 2] Ek 3.84 1028 J

c) This value is comparable to a 900-MWnuclear power plant (e.g., Darlington)running for 2.35 1011 years!

4. Geostationary Earth satellites orbit constantlyabove the same point on Earth because theirperiod is the same as that of Earth.

5. M 5.98 1024 kg, r 6.378 106 m,v 25 m/sTo find the semimajor axis,

ET Ep Ek

mv2

v2

a

a

a 3.19 106 m

To find the period,

T 2 ka 3, where k

T T 1792 s

Section 6.31. a) At the equilibrium point, the bobs kinetic

energy accounts for all the energy in thesystem. This total energy is the same as themaximum elastic potential energy. Ek equil ET Ek equil Epmax

Ek equil12

kx2

Ek equil1

2(33 N/m)(0.23 m) 2

Ek equil 0.87 J b) 0

c) Ek12

mv2

v v v 1.9 m/s

2. a) To find the period of an object in simple harmonic motion,

T 2 T 2 T 0.76 s

b) At 0.16 m, the elastic potential energy of the bob is

Ep 0.16m1

2kx2

Ep 0.16m12

(33 N/m)(0.16 m) 2

Ep 0.16m 0.42 J ET Ek Ep Ek ET Ep Ek 0.87 J 0.42 J Ek 0.45 J

0.485 kg33 N/m

mk

2(0.87 J)0.485 kg

2 Ekm

4 2(3.19 106 m)3

(6.67 10 11 Nm 2/kg 2)(5.98 1024 kg)

4 2

GM

(6.67 10 11 Nm 2/kg 2)(5.98 1024 kg)(6.378 106 m)2(6.67 10 11 Nm 2/kg 2)(5.98 1024 kg) (25 m/s) 2(6.378 106 m)

GMr 2GM v2r

2GM v2r GMr

1a

v2

GM 2r

1a

2GM r

GM a

12

GMmr

GMm2a

12

12

12

(6.67 10 11 Nm 2/kg 2)(5.98 1024 kg)3.82 108 m

GM r

GMm2r

12



39/200

Ek12

mv2

v v v 1.36 m/s

c) Ek 0.45 J, from part b3.

D i s p

l a c e m e n

t ( m )

Position vs. Time

Time (s)

0.6 0.4 0.2

00.20.4

0.20 0.4 0.6 0.8 1.0 1.2 1.4

2(0.45 J)0.485 kg

2 Ekm



40/200

Section 7.2

1. a) 0.17 rad

b) 1.0 rad

c) 1.6 rad

d) 3.07 rad

e) 4.47 rad

2. a) ( rad)(57.3/rad) 180

b) rad (57.3/rad) 45

c) (3.75 rad)(57.3/rad) 675d) (11.15 rad)(57.3/rad) 639e) (40 rad)(57.3/rad) 2.3 103

3. a) Earth rotates 2 radians every 24 h.6.0 h 1.57 rad

b) Earth moves 2 rad every 365 days.

265 d 4.56 rad

c) The second hand moves 2 rad every 60 s.

25 s 2.62 rad

d) A runner moves 2 rad for every lap.

25.6 laps 161 rad

Section 7.3

2. a) a c

v a cr v (9.8 m /s 2)(12 00 m) v 108 m/sv 1.1 102 m/s

b)

0.090 rad/sThe angular acceleration is zero becausethe angular velocity is constant.

3. a)

0.12566 rad/s0.13 rad/s

b) r 1500 ma c r 2

a c (1500 m)(0.12566 rad/s) 2

a c 24 m/s 2

c) The angular acceleration is zero becausethe angular velocity is constant.

d) a c-space-station 24 m/s 2

ac-Earth 9.8 m/s 2

2.4

Section 7.41. a) (3.35 rev/s)(2 rad/rev)

21.0 rad/st 2 min (50 sec)

t 170 s

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

b)

44 rad/s 2

2. a) t

t

t 8.3 s

b) t

(8.3 s)

7.3 radc) There are 2 radians in one cycle.

number of cycles

number of cycles 1.16number of cycles 1.2

7.3 rad2 rad/cycle

(1.75 rad/s 0)2

( 1 2)2

(0 1.75 rad/s)0.21 rad/s 2

(22.0 rad/s 0)0.5 s

t

t

60 s1 min

24 m/s 2

9.8 m/s 2

1 min60 s

2 rad1 rev

1.2 rev1 min

108 m/s1200 m

v

r

v2

r

2 rad1 lap

2 rad60 s

2 rad365 d

2 rad24 h

4

25657.3/rad

17657.3/rad

9057.3/rad

6057.3/rad

1057.3/rad



41/200

d) 0.58 cycles

(0.58 cycles)(2 rad/cycle)3.6 rad

2 t 12

t 2

3.6 rad 012 ( 0.21 rad/s

2

) t 2

t 5.9 s

3. a) t

t 2

t 6.026 st 6.03 s

b)

0.266 rad/s 2

Section 7.52. a) I

(0.045 kgm 2)( 1.90 rad/s 2)0.086 Nm

b) For 78 rpm:1 0

2

2 8.2 rad/s2

21

2 2

17.69 rad18 rad

number of turns

number of turns 2.8For 45 rpm:

1 0

2

2 4.7 rad/s

22

12 2

5.813 rad5.8 rad

number of turns

number of turns 0.93

For 33 13

rpm:

1 0

2 2 3.5 rad/s2

21

2 2

3.223 rad3.2 rad

number of turns

number of turns 0.51

3. I

I

I 0.693 kgm 2

4. a) I 12

mr 2 (moment of inertia for a disk)

I 12

(5.55 kg)(1.22 m) 2

I 4.13 kgm 2

b) rF (1.22 m)(15.1 N)18.4 Nm

c) I

4.46 rad/s 2

18.4 Nm4.13 kgm 2

8.45 Nm12.2 rad/s 2

3.223 rad2 rad/turn

(3.5 rad/s) 2 02( 1.90 rad/s 2)

( 22 12)2

1 min60 s

2 rad1 rev

1300 rev

1 min

5.813 rad2 rad/turn

(4.7 rad/s) 2 02( 1.90 rad/s 2)

( 22 12)2

1 min60 s

2 rad1 rev

45 rev1 min

17.69 rad2 rad/turn

(8.2 rad/s) 2 02( 1.90 rad/s 2)

( 22 12)2

1 min60 s

2 rad1 rev

78 rev1 min

14.5 rad/s 16.1 rad/s6.026 s

t

92.2 rad(16.1 rad/s 14.5 rad/s)

( 12

2)

1.16 cycles2



42/200

Section 7.61. a) rF

(0.20 m)(23.1 N)4.62 Nm4.6 Nm

W RW R (4.62 Nm)(2 rad)W R 29 J

b) W RW R (4.62 Nm)(1.5 rad)W R 6.9 J

c) 951.66 rad

W RW R (4.62 Nm)(1.66 rad)W R 7.7 J

2. a) 45

4rad

W RW R rF

W R (0.556 m)(12.2 N) 4rad

W R 5.3 J b) The work done does not change.

Section 7.7

1. I 25 mr 2

I 25

(0.0350 kg)(0.035 m) 2

I 1.7 10 5 kgm 2

Erot 12 I 2

Erot 12

(1.7 10 5 kgm 2)(165 rad/s) 2

Erot 0.23 J2. a) (5.3 rev/s)(2 rad/rev)

33.3 rad/s

Erot 412 I 2

Erot 412

(0.900 kgm 2)(33.3 rad/s) 2

Erot 2.0 103 J

b) v r v (0.320 m)(33.3 rad/s)v 10.7 m/s

Ek12

mv2

Ek12

(1000 kg)(10.7 m/s) 2

Ek 5.7 104 J

Section 7.8

1. a) vr

78 rad/s

Erot 412 I 2

Erot 2(0.900 kgm2

)(78 rad/s)2

Erot 1.1 104 J

b) Ek12

mv2

Ek12

(1300 kg)(25 m/s) 2

Ek 4.1 105 Jc) ET Ek E rot

ET (4.1 105 J) (1.1 104 J) ET 4.2 105 J

2. v1 01 0

h1 12.0 mm 2.2 kgr 0.056 m I mr 2 (moment of inertia for a hollowcylinder)a) ET mgh 1

ET (2.2 kg)(9.8 m/s 2)(12.0 m) ET 2.6 102 J

b) To find the gravitational potential energy halfway down: Eg mgh 2

Eg mg Eg (2.2 kg)(9.8 m/s 2) Eg 1.29 102 J

12.0 m2

h12

25 m/s0.320 m



43/200

To find the velocity halfway down: ET1 ET2

mgh 112

mv2212 I 2 mgh 2

mgh 112

mv2212

mr 2 2 mgh 2mgh 1 mv22 mgh 2mv22 mgh 1 mgh 2

v22 gh1 g 2v22 2 gh1 gh12v22 gh1

v2 v2 v2 7.67 m/s

c) vr

1.9 102 rad/s

Section 7.9

1.

1.99 10 7 rad/s

I 25 mr 2 (moment of inertia for a sphere)

L I

L 25

mr 2

L 25

(5.98 1024 kg)(6.38 106 m)2

(1.99 10 7 rad/s) L 1.94 1031 kgm 2/s

2.

25.7 rad/s

r

r 0.9 m

I 25

mr 2 (moment of inertia for a sphere)

L I

L 25

mr 2

L 25

(85 kg)(0.9 m) 2(25.7 rad/s)

L 7.1 102 kgm 2/s3. At perihelion,

v 5472.3 m/sr 4.4630 1012 mm 1.027 1026 kg

vr

1.2261 10 9 rad/s L I

L 2

5mr 2

L 25

(1.027 1026 kg)(4.4630 1012 m)2

(1.2261 10 9 rad/s) L 1.003 1042 kgm 2/sAt aphelion:v 5383.3 m/sr 4.5368 1012 mm 1.027 1026 kg

vr

1.1866 10 9 rad/s L I

L 25

mr 2

L 25

(1.027 1026 kg)(4.5368 1012 m)2

(1.1866 10 9 rad/s) L 1.003 1042 kgm 2/s

Section 7.10

2. 1

1

1 2.94 10 6 rad/s

2 rad2.14 106 s

t

5383.3 m/s4.5368 1012 m

5472.3 m/s4.4630 1012 m

1.8 m2

2 rad1 cycle

4.5 cycles1.1 s

2 rad1 rev

1 h3600 s

1 d24 h

1 rev365 d

10.8 m/s0.056 m

(9.8 m/s 2)(12.0 m)2

gh12

h12

v2r



44/200

I 1 1 I 2 2

mr 12 1 mr 22 2

r 12 1 r 22 2

2

2

2 4.69 104 rad/s

T 2

T 2

T 2 1.34 10 4 s3. r a 1.52 1011 m

r p 1.47 1011 mvp 30 272 m/s

I a a I p p

mr a2 mr p2 r ava r pvp

va

va

va 2.93 104 m/sva 29.3 km/s

Section 7.113. R 0.040 m

r 0.0070 m

a

a g

1

a 9.8 m/s2

1

a 0.64 m/s 2

12

(0.040 m) 2

(0.0070 m) 2

12

mR 2

mr 2

g

m I r 2

1

(1.47 1011 m)(30 272 m/s)1.52 1011 m

r pvpr a

vpr p

var a

2 rad4.69 104 rad/s

2 rad2

(6.95 108 m)2(2.94 10 6 rad/s)

(5500 m) 2

r 12 1r 22

25

25



45/200

Section 8.41. q1 3.7 10 6 C, q2 3.7 10 6 C,

d 5.0 10 2 m, k 9.0 109 Nm 2/C 2

F

F

F 49 N F 49 N (attraction)

2. F 2( 49 N) F 98 N

r r r 3.5 10 2 m

3. a)

b)

c) How close do the dust balls get and what isthe charge on the tethered dust ball?m 2.0 10 10 kg , l 0.42 m,dwall-1 0.35 m, q 3.0 10 6 C, 21dwall-2 0.35 m 0.42 m(sin 21)

dwall-2 0.35 m 0.15 mdwall-2 0.20 m

From the force vector diagram we see that,

tan

F e mg tan

mg tan

q1

q1

q1 1.1 10 15 CThe dust balls are 0.20 m apart, and thecharge on the tethered dust ball is1.1 10 15 C.

Section 8.51. a)

b)

c)

+

(0.20 m) 2(2.0 10 10 kg)(9.8 N/kg)(tan 21)(9.0 109 Nm 2/C2)(3.0 10 6 C)

r 2mg tankq2

kq1q2r 2

F emg

mg T

F e

F e

T

mg

(9.0 109 Nm 2/C 2)(3.7 10 6 C)( 3.7 10 6 C)98 N

kq1q2 F

(9.0 109 Nm 2/C 2)(3.7 10 6 C)( 3.7 10 6 C)(5.0 10 2 m)2

kq1q2d2



46/200

Section 8.61. a) q 1.0 10 6 C,

1.7 106 N/C [right]Let right be the positive direction. F

e q

F e ( 1.0 10 6 C)(1.7 106 N/C) F e 1.7 N F

e 1.7 N [left]

b) q 1.0 10 6 C, 2(1.7 106 N/C) [right]If right is still the positive direction, F

e q

F e (1.0 10 6 C)[2(1.7 106 N/C)] F e 3.4 N F

e 3.4 N [right]2.

q 1.0 10 6 C,

1.7 106 N/C [right] F e mg tan F

e 1.7 N [left]3. a)

The field lines radiate outward, away fromthe charge.

b) k 9.0 109 Nm 2/C2, q 3.0 10 6 CAt 2 cm away from the charge:

6.8 107 N/C

At 4 cm away:

1.7 107 N/CAt 6 cm away:

7.5 106 N/Cc) Doubling the distance,

1

114

Tripling the distance,2

219

1 decreases to14

and 2 decreases to19

of

the original strength.

d) . The field strength varies as the

inverse square of the distance away from

the charge.e) q1 1.0 10 6 C, q2 3.0 10 6 C,

r 8.0 10 2 m

4.22 106 N/C F

e q

F e (1.0 10 6 C)(4.22 106 N/C)

F

e 4.22 N [right]4. a) q1 q2 1.0 10 6 C, r 0.20 m

Let the positive direction be left.At point A:r 1 0.05 m, r 2 0.25 m

(9.0 109 Nm 2/C 2)(3.0 10 6 C)(8.0 10 2 m)2

kq1r 2

1r 2

kq(3r )2

kq(2r )2

(9.0 109 Nm 2/C 2)(3.0 10 6 C)(6.0 10 2 m)2

kqr 2

(9.0 109 Nm 2/C 2)(3.0 10 6 C)(4.0 10 2 m)2

kqr 2

(9.0 109 Nm 2/C2)(3.0 10 6 C)(2.0 10 2 m)2

kqr 2

+

mg

F e

T

Stationary chargecreating a field



47/200

TA

1

2

TA

TA (9.0 109 Nm 2/C 2)(1.0 10 6 C)

TA 3.7 106 N/C [left]

At point B:r 1 0.10 m, r 2 0.10 mThe addition of these two distances as wasdone in the previous question will yield azero quantity.

TB 0 N/CAt point C:r 1 0.15 m, r 2 0.05 m

TC

1

2

TC

TC (9.0 109 Nm 2/C 2)(1.0 10 6 C)

TC 3.2 106 N/C [left]

b) At the centre point, 1 is equal inmagnitude but opposite in direction to 2,therefore there is no net field strength asthe fields cancel out.

c) For all field strengths to cancel out, the

magnitudes of the ratio of must beequal and pointing in opposite directions.

Section 8.7

1. a) Ee

Ee

Ee 6.8 10 1 J

b) V

V

V 4.5 105 V

c) V

V

V 9.0 105 V

V V 2 V 1 V 9.0 105 V ( 4.5 105 V) V 4.5 105 V

2. a) m1 m2 5.0 10 9 g 5.0 10 12 kg,q1 4.0 10 10 C, q2 1.0 10 10 COn particle 1:W 1 qV W 1 (4.0 10 10 C)(50 V)W 1 2.0 10 8 JOn particle 2:W 2 qV W 2 (1.0 10 10 C)(50 V)W 2 5.0 10 9 J

b) W Ek

W mv2

v The similar masses cancel.

2

3. a) Extensive: electric force, potential energyIntensive: field strength, electric potential

b) Electric force Charge and the fieldstrengthPotential energy Charge and the electricpotential

c) Extensive propertiesProduct cost (per package)

MassVolumeLengthForce of gravityEtc.Intensive propertiesUnit product cost (per unit weight or measure)DensityHeat capacity

v1v2

2.0 10 8 J5.0 10 9 J

v1v2

W 1W 2

v1v2

2mW

1

1 2mW

2

2 v1v2

2W m

12

(9.0 109 Nm 2/C 2)( 5.0 10 6 C)5.0 10 2 m

kqr

6.8 10 1 J1.5 10 6 C

Ee

q

(9.0 109 Nm 2/C2)( 5.0 10 6 C)(1.5 10 6 C)10 10 2 m

kq1q2r

qr 2

1(0.15 m) 2

1(0.05 m) 2

kq1

r 21

kq2

r 22

1(0.25 m) 2

1(0.05 m) 2

kq2r 22

kq1r 21



48/200

Indices of refractionGravitational field strengthEtc.

Section 8.81. qA 2e, qB 79e,

Ek 7.7 MeV

(7.7 106 eV)(1.602 10 19 J) Ee Ek

Ee

r

r

r 2.96 10 14 mr 3.0 10 14 m

3. q 1.5 105

C12

mv2 q(V 2 V 1)

v v v

6.0 m/s [left]4. a) V 1.5 103 V, m 6.68 10 27 kg,

q 2e 3.204 10 19 C Ek Ee

12

mv2 Vq

v v v 3.8 105 m/s

b) 12

mv2 12

Vq

v v v 2.7 105 m/s

5. a) V 20 kV 2.0 104 V,q 1.602 10 19 C, m 9.11 10 31 kg Ek E e Ek Vq Ek (2.0 104 V)(1.602 10 19 C) Ek 3.2 10 15 J

b) Ek12 mv

2

v v v 8.4 107 m/s

Section 8.91. W 2.4 10 4 J, q 6.5 10 7 C

V

V

V 3.7 102 V2. d 7.5 10 3 m, V 350 V,

4.7 104 N/C3. m 2.166 10 15 kg, V 530 V,

d 1.2 10 2 m F e F g

mg

q

q

q 4.8 10 19 C

(2.166 10 15 kg)(9.8 N/kg)(1.2 10 2 m)530 V

mgdV

qV d

350 V7.5 10 3 m

V d

2.4 10 4 J6.5 10 7 C

W q

2(3.2 10 15 J)9.11 10 31 kg

2 Ekm

(1.5 103 V)(3.204 10 19 C)6.68 10 27 kg

Vqm

2(1.5 103 V)(3.204 10 19 C)6.68 10 27 kg

2Vqm

2( 1.5 10 5 C)( 12 V)(1.0 10 5 kg)

2q(V 2 V 1)m

(9.0 109 Nm 2/C2)(1.602 10 19 C)2(2)(79)(7.7 106 eV)(1.602 10 19 J)

kqAqB Ee

kqAqBr



49/200

Section 9.51. 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 10 2 N B 3.5 10 2 T

90

I

I

I 18 A3. a) L 50 m

I 100 A F 0.25 N

45

B

B

B 7.1 10 5 T b)

4. B 3.0 105

T L 0.20 m N 200

4 10 7 Tm/A

I

I

I 2.4 10 2 A

5. a) I 100 A L 50 m B 3.0 10 5 T

45

r

r r 0.67 m

b) Referring to the diagram in question 3,Earths field lies in a line that is crossingthe wire at 45 below the horizontal. Themagnetic field would form a circular ringin the clockwise direction (rising on thesouth side of the wire, descending on thenorth with a radius of 0.67 m). Therefore,the field will cancel that of Earth on thesouth side below the wire, as shown in thediagram.

2 x2

(0.67 m)2

x 0.47 mThe fields will cancel 4.7 10 1 m southand 4.7 10 1 m below the wire.

6. a) r 2.4 10 3 m I 13.0 A L 1 m

F

F

F 1.4 10 2 N/m7. q 20 C

B 4.5 10 5 Tv 400 m/s

90 F qvB sin F (20 C)(400 m/s)(4.5 10 5 T) sin 90 F 0.36 N

(4 10 7 Tm/A)(13.0 A) 2(1 m)2 (2.4 10 3 m)

I 2 L2 r

N

45 x

x

0.67 m

(4 10 7 Tm/A)(100 A)2 (3.0 10 5 T)

I 2 B

(3.0 10 5 T)(0.20 m)(4 10 7 Tm/A)(200)

BL N

Tower

Directionof Force

Wire

(cross-section) Earth'sMagneticField45

S N

45

(0.25 N)(100 A)(50 m) sin 45

F IL sin

(9.2 10 2 N)(3.5 10 2 T)(0.15 m) sin 90

F BL sin



50/200

8. q 1.602 10 19 Cv 4.3 104 m/s B 1.5 T

90 F qvB sin F (1.602 10 19 C)(4.3 104 m/s)(1.5 T) sin 90 F

1.0 10 14 N [south]



51/200

Section 10.2

1. a) T

T

T 75 min

b) T T 0.67 s

c) T

T 1.80 s

d) T

T 0.838 s

2. a) f

f

f 60 Hz

b) f

f 0.75 Hz

c) f

f 0.009 26 Hz

d) f

f 1.35 Hz

3. a) i) f

f

f 2.22 10 4 Hz

ii) f

f 1.49 Hz

iii) f

f 0.556 Hz

iv) f

f 1.19 Hz

b) i) T

T

T 0.0167 s

ii) T

T 1.33 s

iii) T

T 108 s

iv) T

T 0.74 s5. a) x (30 cm) cos

x (30 cm) cos 30 x 26 cm

b) x (30 cm) cos 180 x 30 cm

c) x (30 cm) cos 270 x 0 cm (equilibrium)

d) x (30 cm) cos 360 x 30 cm

e) x (30 cm) cos4

x 21 cm

Section 10.34. a) v f

f

f

f 4.7 1014 Hz

b) f

f 2.5 108 Hz

c) f

f 1.5 1017 Hz5. a) v f

2.0 10 5 m

3.0 108 m/s1.5 1013 Hz

v f

3.0 108 m/s2 10 9 m

3.0 108 m/s1.2 m

3.0 108 m/s640 10 9 m

v

11.35 Hz

10.009 26 Hz

10.75 Hz

160 Hz

1 f

10.838 s

1

1.80 s

10.67 s

175 60 s

1T

6548 s

401.2 60 60 s

4560 s

1202.0 s

cyclest

57 s68

60 s33.3

6.7 s10

375 min5

t cycles



52/200

b)

0.15 m

c)

1.0 10 14 m

Section 10.44. a) n

v

v

v 2.26 108 m/s

b) v

v 1.24 108

m/sc) v

v 1.99 108 m/s

5. a) n

n

n 1.43

b) n

n 2.0

c) n

n 1.276. a) n1 sin 1 n2 sin 2

2 sin 1 2 sin 1 2 18.5

b) 2 sin 1 2 10.1

c) 2 sin 1 2 16.3

Section 10.5

5. a) n

vo ray

vo ray

vo ray 1.81 108 m/s

ve ray

ve ray

ve ray 2.02 108 m/s

b) 100%

111.6%

Therefore, the speed of the e ray is 11.6%greater than the speed of the o ray.

ve rayvo ray

2.02 108 m/s1.81 108 m/s

ve rayvo ray

3.0 108 m/s1.486

c ne ray

3.0 108 m/s1.658

cno ray

c v

sin 251.51

sin 252.42

sin 251.33

n1 sin 1n2

3.0 108 m/s0.79(3.0 108 m/s)

3.0 108 m/s1.5 108 m/s

3.0 108 m/s2.1 108 m/s

c v

3.0 108 m/s1.51

3.0 108 m/s2.42

3.0 108 m/s1.33

c n

c v

3.0 108 m/s3.0 1022 Hz

3.0 108 m/s2.0 109 Hz



53/200

Section 11.42. d 5.6 m

x2 28 cm L 1.1 mm 2

7.13 10 7 m713 nm

3. 510 nmd 5.6 m L 1.1 m

x

x

x 0.10 m x 10 cm

4. m 3d 5.6 m L 1.1 m

713 nm

xm

x3

x3 0.42 m x3 42 cm

Section 11.52. PD 3

ng 1.52624 nm

t

t

t 1.8 10 6 mt 1.8 m

Section 11.62. m 22

625 nm

t

t

t 6.87 10 6 mt 6.9 m

3. t 1.75 10 5 m625 nm

2t m

m

m

m 55.5m 55

Section 11.81. w 5.5 10 6 m

550 nm L 1.10 mm 2

a) sin m

sin 2

sin 2 0.22 11.5

b) xm L sin m xm (1.10 m)(0.2) xm 0.22 m xm 22 cm

2. a) x

x

x 0.22 m x 22 cm

2(5.50 10 7 m)(1.10 m)(5.5 10 6 m)

2 Lw

(2)(5.50 107

m)5.5 10 6 m

mw

12

2(1.75 10 5 m)(6.25 10 7 m)

12

2t

12

(22)(6.25 10 7 m)2

m2

(6.24 107

m)(3)2(0.52)

PD2(ng 1)

(3)(7.13 10 7 m)(1.1 m)

(5.6 106

m)

m Ld

(5.10 10 7 m)(1.1 m)

5.6 10 6 m

Ld

(5.6 10 6 m)(0.28 m)(2)(1.1 m)

dxmmL



54/200

b) sin 12

sin 12

sin 12

0.1

11.53. x

x

x 0.11 m x 11 cm

6. R 1 10 7 radd 2.4 m

a)

1.97 10 7 m197 nm

b) sin = L x

L

L 5000 m L 5 km

Section 11.91. N 8500

w 2.2 cm530 nm

d

d

d 2.59 10 6 m

sin m

sin 1

sin 1 0.2051 12

sin m

sin 2

sin 2 0.4102 24

sin m

sin 3

sin 3 0.6143 38

2. a) m

m

m 4

b) m

m

m 4.7m 4

c) m

m

m 5.7m 5

3. m 22 8.41o

614 nm

a) d

d

d 8.396 10 6 m

d 8.40 m b) w 1.96 cm

N

N

N 2334 slits

1.96 10 2 m8.396 10 6 m

wd

(2)(6.14 10 7 m)sin 8.41

msin m

2.59 10 6 m4.50 10 7 m

d

2.59 10 6 m5.50 10 7 m

d

2.59 10 6 m6.50 10 7 m

d

3(5.30 10 7 m)2.59 10 6 m

md

2(5.30 10 7 m)2.59 10 6 m

md

5.30 10 7 m2.59 10 6 m

md

2.2 10 2 m

8500

w N

12

(1.0 10 3 m)

sin(1 10 7 rad)

(1 10 7 rad)(2.4 m)1.22

Rd

1.22

(5.50 10 7 m)(1.10 m)(5.5 10 6 m)

Lw

5.50 10 7 m5.5 10 6 m

w



55/200

Section 11.10

1. 300 000 lines/m

100 000 lines/m

Therefore, 3000 lines/cm produces the best

resolution.3 . sin Red

sin Red

sin Red 0.386Red 22.7

sin Violet

sin Violet

sin Violet 0.211Violet 12.2

sin Green

sin Green

sin Green 0.269Green 15.6

This can be similarly proven for the next 3orders using the appropriate m .The sequence is violet, green, red.At the fourth order, green and red maxima areno longer visible.

5. d 2.5 10 10 m12o

m 2

5.198 10 11 m52 pm

6.

sin

sin

sin 0.208168, 192

(5.2 10 11 m)(2)2(2.5 10 10 m)

m2d

2d sinm

2(2.5 10 10 m) sin 122

2d sinm

(1)(5.10 10 7 m)1.89 10 6 m

md

(1)(4.00 10 7 m)1.89 10 6 m

md

(1)(7.30 10 7 m)1.89 10 6 m

md

100 cm1 m

20 000 lines20 cm

100 cm1 m

3000 lines1 cm



56/200

Section 12.21. T 12 000 K

a) The maximum wavelength can be foundusing Wiens law:

max

max

max 2.4 10 7 mThe peak wavelength of Rigel is2.4 10 7 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 foundusing Wiens law:

max

max

max 3.2 10 6 mThe peak wavelength of the light is3.2 10 6 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 thevisual spectrum.

Section 12.3

1. V f 0eV hf 0 W 0Choosing two pairs of values from the tableand subtracting,

(1.6 10 19 C)(0.95 V) h(7.7 1014 Hz) W 0(1.6 10 19 C)(0.7 V) h(7.2 1014 Hz) W 0

(1.6 10 19 C)(0.25 V) h(0.5 1014 Hz)h 8 10 34 Js

W 0 4.64 10 19 JW 0 2.9 eV

2. a) Increasing the work function by 1.5 wouldcause a vertical shift of the line. Hence,potential would have to be greater, but thefrequencies would not change.

b) The term he

is constant and hence the

slope would not change.3. 230 nm 2.3 10 7 m

The energy can be found as follows:

E W 0

E

4.64 10 19 J E 5.79 10 19 J

Section 12.4

2. E 85 eV, 214 nm 2.14 10 7 ma) Momentum of the original electron can be

found using:

p

p

p 4.53 10 26 Ns b) Momentum of the resultant electron can be

found using:

p

p

p 3.1 10 27 Nsc) The energy imparted can be found by:

E E hc

6.626 10 34 Js2.14 10 7 m

h

(85 eV)(1.6 10 19 C)3.0 108 m/s

Ec

(8 10 34 Js)(3.0 108 m/s)2.3 10 7 m

hc

3

2

1

0 7 8

V s

t o p

( V )

f 0 ( 10 14 Hz)

V stop vs. f 0

9 10 11 12 13

W 0e

he

2.898 10 3

900 K

2.898 10 3

T

2.898 10 3

Physics Concepts and Connections, Book Two - Solutions Manual[1]

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