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
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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
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Illustrations
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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
7.0 m/s (7.0
m/s) 4(1
.4 m/s
)(8.0 m)
t 2(1.4 m/s2)
2
2
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
27.8 m/s (27.8
m/s) 4(1
5 m/s )(50
0 m)
t 2(15 m/s2)
2
2
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
17 m/s (17 m
/s) 4(4.9
m/s )(80
m)
t 2(4.9 m/s2)
2
2
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
Solutions to Applying the Concepts
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
2
4.0 m/s (4.0 m
/s) 4(4.9
m/s )(30.
0 m)
t 2(4.9 m/s2)
2
2
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
4.0 m/s (4.0
m/s) 4(4
.9 m/s
)(30
.0 m)
t 2(4.9 m/s2)
2
2
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
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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 kg
a (0.15)(9.8 m/s2)
a 1.5 m/s2
3.
Ff ma
k Fn ma
kmg ma
a kg
v22 v12 2ad
v22 v12
d 2 kg
(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
11
(6.67 10 N m kg )(0.013)(5.97 10 kg)
Fg (3.82 108 m)2
2
2
24
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
11
(6.67 10 N m kg )(1.9 10 kg)
gJupiter (7.2 107 m)2
2
2
27
gJupiter 24 m/s2
Solutions to Applying the Concepts
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
tan
vw
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°
Solutions to Applying the Concepts
To find ,
180° 54°
78°
vf 26 m/s [N78°E]
ϕ
θ
20°
v2
γ
v1
36°
b)
For the y components,
nety F
1y F
2y F
3y
F
Fnety (200 N) sin 30° [N] β
vf
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
Solutions to Applying the Concepts
9
2. a) sin
vmg
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
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
b) dx vi t axt2
2
1
dx (25 m/s)(1.7 s) (0 m/s2)t2
2
dx 43 m
v2 v1
2. a) ay t
v2 v1
t a
0 m/s (35 m/s) sin 40°
t 9.8 m/s2
t 2.3 s
b) 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
3. a) 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
1. a)
Solutions to Applying the Concepts
y
x
y
y
y
y
x
y
0 (88.1 m/s)t (4.9 m/s2)t2 200 m
88.1 m/s (88.1
m/s) 4(4
.9 m/s
)(20
0 m)
t 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
Solutions to Applying the Concepts
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)
Solutions to Applying the Concepts
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
(15 kg)(9.8 m/s ) 0.20(10 kg)(9.8 m/s )
a 25 kg
2
2
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)
Solutions to Applying the Concepts
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
m/s )[5.0 kg (3.0 kg) sin 35° 0.18(3.0 kg) cos 35°]
a (9.8
8.0 kg
2
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)
g(m2 sin 60° 2m2 cos 60° m1 sin 40° 1m1 cos 40°)
a m m
1
a
2
(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
(30 kg)(9.8 m/s )sin 30° (10 kg)(9.8 m/s )
a 60 kg
2
2
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
Solutions to Applying the Concepts
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)
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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
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)
tan
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
+
θ
2.
Ta = 100.0 N
FB
θ
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
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
Solutions to Applying the Concepts
3.0 m 0.75 m
r3 2
r3 1.12 m
0 1 2 3
0 r1F1 sin 1 r2F2 sin
r3F3 r2F2
r1 F1
2
r3F3 sin
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
3.
P
(1.12 m)(15.0 kg)(9.8 N/kg) (0.375 m)(5.0 kg)(9.8 N/kg)
r1 (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)
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
cos
50°
40°
F1
4.
F4
F3
P
F1
θ
0.5 m
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
Solutions to Applying the Concepts
19
2.
2.0 m
r2 2
r2 1.0 m
r3 1.60 m
90°
sin 1
r1F1 r2F2
F3 r3
F2
θ2
F1 = Fn
4 cm
P
+
8 cm
(0.375 m)(120.0 kg)(9.8 N/kg) (1.0 m)(5.0 kg)(9.8 N/kg)
F3 1.60 m
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
2
(8.0 10 m)(27 kg)(9.8 N/kg) sin
F2 (4.0 102 m) sin
Section 3.6
1.
τ
48 m
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°
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
Fg–b
15°
a)
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
Solutions to Applying the Concepts
rbmb g sin b rsms g sin s
Fm r sin
m
m
rg sin (mb ms)
Fm rm sin m
(75 10 m)(9.8 N/kg) sin 75°[(0.57)85 kg19.0 kg]
Fm (45 10 m) sin 11°
–2
–2
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
x
θT
θ
0.6 m
0.60 m
tan T 1.0 m
T 31.0°
0.60 m
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°
tan
Solutions to Applying the Concepts
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
0.29 103 m 2
(200 109 N/m2)(0.22 103 m)
2
F 4(0.90 m)
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
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
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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 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°
(0.03 kg)(0) (1.9 kg)(0) (0.03 kg)(750 m/s)
v2f 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
(0.400 kg)(3.0 m/s [forward]) (0.400 kg)(0) (0.400 kg)(1.0 m/s [forward])
v2f 0.400 kg
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
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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.
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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.
Solutions to Applying the Concepts
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%
Solutions to Applying the Concepts
GM
r2 GM
1.8 107 J r1
Section 6.1
1. mE 5.98 10 kg, mS 1.99 10 kg,
r 1.50 1011 m
GMm
a) Ek 2r
24
11
30
r2 (6.67 10 N·m /kg )(1.99 10 kg)(5.98 10 kg)
Ek 2(1.50 10 m)
2
2
30
24
11
Ek 2.65 1033 J
GMm
b) Ep r
11
(6.67 10 N·m /kg )(1.99 10 kg)(5.98 10 kg)
Ep (1.50 10 m)
2
2
30
24
11
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
(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
Solutions to Applying the Concepts
33
To find the period,
To find the current speed of the Moon,
1
GMm
mv2 2
2r
v
4 2
T2 ka3, where k GM
GM
r
4 (3.19 10 m)
T 11
2
2
24
2
(6.67 10
11
(6.67 10 N·m /kg )(5.98 10 kg)
v 8
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
11
(6.67 10 N·m /kg )(5.98 10 kg)(6.378 10 m)
a 2(6.67 10 N·m /kg )(5.98 10 kg) (25 m/s) (6.378 10 m)
11
2
2
2
2
24
24
a 3.19 106 m
34
6
2
6
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
T2
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
Solutions to Applying the Concepts
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)
Solutions to Applying the Concepts
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
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
3. a)
Section 7.4
(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 1. a)
(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
Solutions to Applying the Concepts
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
0
1
100
rev
1 min
2 rad
3
2 60 s
1 min
1 rev
2 3.5 rad/s
2
2
2 1 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
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
2
2
1
Section 7.5
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
2
2
2 1 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
2. a)
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
Solutions to Applying the Concepts
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
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°
1. a)
Section 7.8
v
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
1. a)
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 7.7
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
1
E 4(0.900 kg·m )(33.3 rad/s)
2
2
rot
Erot 2.0 10 J
3
38
2
12.0 m
Eg (2.2 kg)(9.8 m/s2) 2
2
Eg 1.29 10 J
Solutions to Applying the Concepts
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
c)
v2 gh
2
v2 (9.8 m/s )(12.0 m)
2
1
2
v2 7.67 m/s
v
r
10.8 m/s
0.056 m
1.9 102 rad/s
Section 7.9
1 rev
1d
2 rad
1h
365 d
24 h
3600 s
1 rev
7
1.99 10 rad/s
2
I mr2 (moment of inertia for a sphere)
5
LI
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
1.
LI
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
LI
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
LI
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.
t
2 rad
1 2.14 106 s
6
rad/s
1 2.94 10
1
Solutions to Applying the Concepts
39
I1 1 I2 2
2
2
mr12 1 mr22 2
5
5
2
r1 1 r22 2
r12 1
2 r22
6
(6.95 10 m) (2.94 10 rad/s)
(5500 m)2
8
2
2
4.69 104 rad/s
2 rad
T2 2
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
Solutions to Applying the Concepts
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
6
6
(9.0 10 N·m /C )(3.7 10 C)(3.7 10 C)
F (5.0 102 m)2
9
2
2
F 49 N
F 49 N (attraction)
2. F 2(49 N)
F 98 N
r
2
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
10
(0.20 m) (2.0 10 kg)(9.8 N/kg)(tan 21°)
q1 (9.0 109 N·m2/C2)(3.0 106 C)
2
98 N
r 3.5 102 m
3. a) T
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)
+
Solutions to Applying the Concepts
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 (2r)2
1
1 4
Tripling the distance,
kq
2 (3r)2
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
Solutions to Applying the Concepts
TA 1 2
kq1
kq2
TA 2 r1
r 22
9
2
2
6
C)
TA (9.0 10 N·m /C )(1.0 10
1
1
2 2
(0.05 m)
(0.25 m)
6
TA 3.7 10 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
TC 1 2
kq2
kq1
TC r 22
r 21
9
2
2
6
C)
TC (9.0 10 N·m /C )(1.0 10
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
6
6
(9.0 10 N·m /C )(5.0 10 C)(1.5 10 C)
Ee 10 10 m
9
2
2
2
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
Solutions to Applying the Concepts
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
(9.0 109 N·m2/C2)(1.602 1019 C)2(2)(79)
r (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
15
2
(2.166 10 kg)(9.8 N/kg)(1.2 10 m)
q 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
Solutions to Applying the Concepts
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
Solutions to Applying the Concepts
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)
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
5. 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) 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
Solutions to Applying the Concepts
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
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°
sin 1.33
1
2
2 18.5°
sin 25°
b) 2 sin1 2.42
2 10.1°
sin 25°
c) 2 sin1 1.51
2 16.3°
48
c
n v
Solutions to Applying the Concepts
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)
2(0.52)
t 1.8 106 m
t 1.8 m
t
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
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)
sin
2
x 0.22 m
x 22 cm
Solutions to Applying the Concepts
49
2 w
5.50 10 m
1
sin
2 5.5 10 m
b) sin
1
7
6
0.1
11.5°
L
w
(5.50 107 m)(1.10 m)
x (5.5 106 m)
3. x x 0.11 m
x 11 cm
6. R 1 107 rad
d 2.4 m
d
a) R
1.22
(1 107 rad)(2.4 m)
1.22
1.97 107 m
197 nm
=
x
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
50
0.205
1 12°
1
m
0.410
2 24°
m
sin m d
3(5.30 107 m)
sin 3 2.59 106 m
sin
1
sin
2
b) sin
m
d
2(5.30 107 m)
sin 2 2.59 106 m
sin
2
0.614
38°
3
d
2. a) m 2.59 106 m
m
6.50 107 m
sin
3
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
o
2 8.41
614 nm
m
a) d sin m
d
(2)(6.14 107 m)
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
Solutions to Applying the Concepts
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
sin
Red
2d sin
m
m
2d
(5.2 1011 m)(2)
2(2.5 1010 m)
6.
sin
sin
sin
0.208
168°, 192°
(1)(7.30 107 m)
1.89 106 m
0.386
22.7°
Red
m
sin Violet d
(1)(4.00 107 m)
sin Violet 1.89 106 m
sin
Red
0.211
Violet 12.2°
m
sin Green d
(1)(5.10 107 m)
sin Green 1.89 106 m
sin
Violet
0.269
15.6°
Green
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
sin
Green
2(2.5 1010 m) sin 12°
2
5.198 1011 m
52 pm
Solutions to Applying the Concepts
51
Section 12.2
Vstop vs. f0
Section 12.3
1.
V
e f e
h
W0
0
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
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
W0
E (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 Solutions to Applying the Concepts
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
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
(6.626 1034 J·s)(3.0 108 m/s)
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
72 v 5.27 106 m/s
The speed increase of the electron is
5.27 106 m/s.
81 (6.26 1034 J·s)(3.0 108 m/s)
5 1019 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 2
12
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
Solutions to Applying the Concepts
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 2
22
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 2
42
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:
v 2Ek
mp
v
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.
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)
54
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)
Section 12.8
y
2(1.92 1016 J)
1.673 1027 kg
6.3 102 m
Solutions to Applying the Concepts
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 2
mv Vq
2
v 2Vq
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
(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
2
L L0 1 v2
c
t
2
0.5L0 L0 1 v2
c
2
v
0.25 1 2
c
v
0.75
c
v 0.866c
v 2.60 108 m/s
Solutions to Applying the Concepts
55
3. L0 200 ca
v 0.9986c
6. L0 5.75 1012 m
The time you take:
distance measured
t0 velocity
v2
c2
L0 1 t0 t0 v
v2
c2
(vt0)2 L02 1 L v c v
ct0 2 2
2
2
0
v
v
c2
1
ct0
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
L0
v
7 ca
7a3a
v
7 ca
v
10 a
v 0.7c
2. The age or time difference for the twins is:
5 a tS tT
d
d
5 a S T
v
v
v2
2L0 1 2
c
2L0
5 a
v
v
v2
10 ca 1 2
c
10 ca
5 a
v
v
t
2
v 2c 2c 1 v2
c
2
v 2c 2c 1 v2
c
2
2
2
2
v 4vc 4c 4c 4v
v(5v 4c) 0
v 0.8c
since v ≠ 0
56
v2
c2
L0 1 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
(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
1
(0
.9)2
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.
m
Solutions to Applying the Concepts
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,
C2
E2 3
C1
E1
C2 ($100 million)
5000 MeV 3
500 MeV
C2 $100 billion
5. The radius for charges moving at right angles
mv
. The ratio of the
to a magnetic field is r Bq
r
mv
radius of a fast to slow electron is f 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:
rp
mv
p p and since vp ve:
re
meve
mpre
me
(1.67 1027 kg)re
rp 9.11 1031 kg
rp 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 vv
1 b c 2c E
c
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)
(1.6 10 C)(5.0 105 T)
1 (0
.996)2
19
r 6.98 105 m
Section 13.6
2. Using the relativistic formula for velocity
addition:
vN NvL
vL v v
1 N 2N L
c
c 0.999c
vL (c)(0.999c)
1
c2
vL c
Solutions to Applying the Concepts
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 2
c
(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 2
v
c
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.
Solutions to Applying the Concepts
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
1
0.0044
t(1.196 10 a ) log log 0.030
2
235
235
238
Section 14.3
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
.
9
1
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
8
Section 14.5
2. T 1.28 109 a, N0 5 mg, N 1 mg
1
2
N N0
log
1
1
4.45 109 a
7.04 108 a
0.8337
t (0.3010)(1.196
109 a1)
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
1
2
t
T238
0
t
Section 14.2
or
238
t
T235
0
1
2
1u3u
v v
1 u 3 u
t
T
1
2
N T log 2 N
t
0
1
1
2
N
log
N0
tT
1
log
2
1
2
5 mg t (1.28 10 a)
1
log 2
log
9
t 2.97 109 a
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
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
Solutions to Applying the Concepts
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,
m c2
cos 0 2
mc
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
2f
4.35 107 m/s
r
2(32 106 Hz)
r 0.216 m
b) r Section 14.7
2
2
1
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
3 3 1
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
2. a) uud 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.
Solutions to Applying the Concepts
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.
Answers to End-of-chapter Conceptual Questions
Fn
Fn
Puck
on
carpet
Ff
Puck
on
ice
Fg
Fg
Fn
Fn
Puck
on
air table
Fg
Ff
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
Answers to End-of-chapter Conceptual Questions
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
Answers to End-of-chapter Conceptual Questions
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.
Answers to End-of-chapter Conceptual Questions
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.
Answers to End-of-chapter Conceptual Questions
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.
Answers to End-of-chapter Conceptual Questions
67
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.
68
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
Answers to End-of-chapter Conceptual Questions
(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.
Answers to End-of-chapter Conceptual Questions
69
3.
4.
5.
6.
7.
70
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
not visually apparent as the diver then enters
the water out of sight of the spectators and
judges.
8. 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.
9. 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.
10. 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.
11. 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.
Answers to End-of-chapter Conceptual Questions
Without the spin, uneven airflow over the surface of the object would make it tumble, experience greater air resistance, and travel a
shorter distance.
12. 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.”
13. 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.
14. 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.
15. 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.
16. 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
Answers to End-of-chapter Conceptual Questions
71
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
of universal gravitation
of electrostatic forces
Gm1m2
F 2
r
kq q
F 122
r
Constant of
G 6.67 1011 N·m2/kg2
proportionality
Type of
Attraction only
force(s)
Conditions
Acts between any
for use
two masses
k 9.0
Equation
109 N·m2/C2
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.
Attraction and repulsion
Acts between any two
electrostatic charges
–
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.
Answers to End-of-chapter Conceptual Questions
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
Answers to End-of-chapter Conceptual Questions
73
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.
22. 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.
23. 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.
24. 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.
25. 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.
26. 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.
74
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
Answers to End-of-chapter Conceptual Questions
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
Answers to End-of-chapter Conceptual Questions
75
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.
8. 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.
9. 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.
10. 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°.
11. According to right-hand rule #3 for the motor
principle, the direction of the force on the conductor will be to the north.
12. An electron moving vertically downward that
enters a northbound magnetic field will be
forced toward the west.
13. 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
76
direction as the direction of conventional current flow through the coil.
14. The cathode rays will be deflected away from
the current-carrying wire, moving in a plane
that contains the wire.
15. 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.
16. 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.
17. 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.
18. 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.
19. 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.
Answers to End-of-chapter Conceptual Questions
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.
Answers to End-of-chapter Conceptual Questions
77
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
Answers to End-of-chapter Conceptual Questions
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.
Answers to End-of-chapter Conceptual Questions
79
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.
Answers to End-of-chapter Conceptual Questions
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.
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.
5. The Michelson-Morley null result led to the
development of special relativity, a tool needed
in the understanding of high-energy physics.
6. 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.
7. 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.
8. 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.
9. 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.
10. 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.
Answers to End-of-chapter Conceptual Questions
81
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
82
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
Answers to End-of-chapter Conceptual Questions
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
Answers to End-of-chapter Conceptual Questions
83
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).
13. 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).
14. 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.
15. Alpha particles are ions, since they are helium
atoms stripped of their electrons.
16. 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.
17. Carbon-14 undergoing beta decay results in
the daughter isotope nitrogen-14.
18. 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.
19. 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.
84
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
Answers to End-of-chapter Conceptual Questions
needed to initiate a fusion process. The
moment these conditions are not met, the
process stops.
30. 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.
31. 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.
32. 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.
33. 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.
34. 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.
35. 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.
36. 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
times (1023 s) of the strong nuclear force.
37. 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.
38. 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.
39. Strong nuclear processes are the fastest (or
shortest), with a lifetime of about 1023 s.
40. 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.
41. 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).
42. 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.
Answers to End-of-chapter Conceptual Questions
85
86
Answers to End-of-chapter Conceptual Questions
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
87
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
88
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
1hs
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.
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
The gravitational
force is equal in
magnitude to the
normal force, and
Fcatapult 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 m s)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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
11
2(6.67 10 N m /kg )(5.97 10 kg)
g (6.38 106 m)2
2
2
24
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
11
(6.67 10 N m /kg )(100 kg)(5.98 10 kg)
W (6.38 106 m 3.0 105 m)2
W 894 N
2
2
24
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.
Solutions to End-of-chapter Problems
95
19. Adding by components:
d (50 cm) sin 35° (100 cm) cos 15°
d 67.9 cm
d (20 cm) (50 cm) cos 35° x
x
y
(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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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 (60 N) sin 40°
F 38.6 N [W]
x
x
Vertical components:
F (60 N) cos 40° 80 N
F 34.0 N [S]
y
y
net 51 N [S49°W]
F
c) Horizontal components:
F (50 N) cos 60° 10 N
F 15 N [E]
x
x
Vertical components:
F (50 N) sin 60° 60 N
F 16.7 N [S]
y
y
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
Solutions to End-of-chapter Problems
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 kg
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 Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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]
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
30 m
2
5.4 m/s
t 2.4 s
42r
53. ac Assuming ac is a constant,
T2
t
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.
Solutions to End-of-chapter Problems
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
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
42r
T1 T2 m1 T2
42L1
42(L1 L2)
T1 m1 m
2
T2
T2
42
T1 (m1L1 m2(L1 L2))
T2
3
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
2
4 (1.5 10 m)
mS (6.67 1011 N·m2/kg2)(3.15 107 s)2
11
1
On mass 1:
42r
Fc m1 T2
2
2
Solutions to End-of-chapter Problems
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°
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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]
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
(2.0 kg)(9.8 N kg) [(2.5 m 1.5 m)(5.0 kg)(9.8 N/kg)]
F1 2 2.5 m
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
Solutions to End-of-chapter Problems
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]
40.
b)
θbox
θa
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
[(7.0 m 1.2 m)(72 kg)(9.8 N/kg)(sin 25°)]
Fw (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
(0.8 m
)2 (0
.5 m)2(75 kg)(9.8 N/kg) sin (90° 58°)
ra (308.7 N) sin 58°
ra 1.40 m
But:
h ra sin 58°
h 1.2 m
Solutions to End-of-chapter Problems
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
(0.16 m)(3.0 kg)(9.8 N/kg) (0.35 m)(10 kg)(9.8 N/kg)
Fm 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
(0.15 m)(1.9 kg) (0.40 m)(1.2 kg) (0.60 m)(0.4 kg)
rcm 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
Solutions to End-of-chapter Problems
Fnet 0
F2 0
F F 48.
1
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
46.
h
hcm
1.00 m
2
tan 28 cm
11 cm
L
7.25 kg
hcm
2.4 cm P
2
h L
C of m
FT
θ
1.00 m
θ
L
—
2
+
cm
tan 2 h 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
(0.11 m)(2.7 kg)(9.8 N/kg) (0.280 m)(7.25 kg)(9.8 N/kg)
FT 0.024 m
cm
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
[(1.00 104 kg)(2.0 m/s2 9.8 m/s2)]
1.96 1
0 m
200 10 9 N/m2
Strain 2
3
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
Solutions to End-of-chapter Problems
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
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]
p = 5 x 105 kg·m/s [NW]
45°
120
250 N [forward],
23. m 50 kg, F
∆t 3.0 s, v1 0
Ft mv
(250 N [forward])(3.0 s) (50 kg)(v2 v1)
740 N [forward]
v2
50 kg
v2 15 m/s [forward]
24. 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
25. 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
26. 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
27. 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
28.
F∆t m∆v
(N)(s) (kg)(m/s)
(kg·m/s2)(s) (kg)(m/s)
kg·m/s kg·m/s
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
121
J ∆p
J mv2 mv1, where v1 0,
1.4 103 N·s (0.250 kg)(v2)
v2 5.6 103 m/s
pTf
37.
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
38.
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
39.
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
40.
pTo p
m1v1o m2v2o (m1 m2)vf,
36.
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
133 333.3 kg·m/s
m1 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
44. 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
45. m1 3m, m2 4m, v1o v
pTo pTf
m1v1o m2v2o (m1 m2)vf, where v2o 0
(3m)v (7m)vf
3
vf v
7
46. 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],
47. p
2o 450 kg·m/s [N45°E]
p
p1o p2o
a) pTo 43.
pTo
θ
p1o = 375 kg·m /s
Solutions to End-of-chapter Problems
p2o = 450 kg·m /s
45°
b) pTf pTo
Using the cosine and sine laws,
pTo 2 (375 kg·m/s)2 (450 kg·m/s)2 2(375 kg·m/s)(450 kg·m/s)
cos 135°
pTo 762.7 kg·m/s
pTf 762.7 kg·m/s
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
pTf (2.5
kg
·m/s)
(6
4 kg· m/s)
2
pTf 64.05 kg·m/s
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°
p3 2 (9.66 1018 kg·m/s)2 (6 1018 kg·m/s)2
p3 1.1372 1017 kg·m/s
p3
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
Solutions to End-of-chapter Problems
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
v1f 2 (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°
v1f 2 (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
p2 (8.33 106 kg·m/s)(tan 10°)
p2 1.47 106 kg·m/s
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°
v1f 2 (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
Solutions to End-of-chapter Problems
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,
pTf 2 (9.9 103 kg·m/s)2 (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
p4 6.037 kg·m/s
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)
Solutions to End-of-chapter Problems
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.
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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 m2 s2 11.045 m2 s2 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
Solutions to End-of-chapter Problems
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
47.
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
48. F kx
F (4000 N/m)(0.15 m)
F 600 N
49. 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
50.
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)
Solutions to End-of-chapter Problems
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 m s) (0.002 kg)(12 m/s)
vf 6.9 m/s
Solutions to End-of-chapter Problems
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)
Solutions to End-of-chapter Problems
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°.
Solutions to End-of-chapter Problems
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
11
(6.67 10 N·m /kg )(5.98 10 kg)(920 kg)
ET 6.38 106 m h
2
2
24
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
Solutions to End-of-chapter Problems
17. vesc c 2.99 108 m/s, mE 5.98 1024 kg
vesc 2GM
r
2GM
r 2
(vesc)
20. mE 5.98 1024 kg, rE 6.38 106 m,
h 400 km 4.0 105 m
Orbital speed is given by:
v
11
2(6.67 10 N·m /kg )(5.98 10 kg)
r (2.99 108 m/s)2
2
2
GM
rh
(6.67 1011 N·m2/kg2)(5.98 1024 kg)
v 24
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.
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
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.
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
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
7
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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.
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
(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
560°
d) 9.77 rad
57.3°/rad
s r
s (40 m)(9.77 rad)
s 3.9 102 m
22. 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.
23. 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
24. a) 1 min
60 s
rev
178.0 rad/s
b) t
t
(178.0 rad/s)(0.56 s)
1.0 102 rad
25. 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
26. 1 1 min
1 min
1 rev
1 4.7 rad/s
2 0
t 22.5 s
Solutions to End-of-chapter Problems
(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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
2(2.0 10 rad)
t (3.5 103 rad/s) (2.5 104 rad/s)
4
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
Solutions to End-of-chapter Problems
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:
0.020 rad/s (0.0
20 rad/s)
4(0.040
rad
/s )(0.436
rad)
t 2(0.040 rad/s )
2
2
2
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
145
b) L I
L
I
1.05 1034 kg·m2 s
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
v m/s )(2.5 m)
3 v2 4
1
2
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
3
(9.8
v m/s )(2.8 m)
2 2
v2 6.4 m/s
146
Solutions to End-of-chapter Problems
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·m2 s
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
Solutions to End-of-chapter Problems
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
1
mprp2p mtrt2t
2
f 1
mprp2 + mtrt2
2
1
(600 kg)(4.3 m)2(6.4 rad/s) + (35 kg)(4.3 m)2(3.1 rad/s)
2
f 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
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
(600 kg)(4.3 m)2(6.4 rad/s) (35 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 5.0 rad/s
70. m1 30 kg
r1 1.5 m
m2 20 kg
r2 1.0 m
1 12 rad/s
Solutions to End-of-chapter Problems
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)
(30 kg)(1.5 m)2(12 rad/s) (20 kg)(1.0 m)2(12 rad/s)
f ((30 kg)(1.5 m)2 (20 kg)(1.0 m)2)
f 12 rad/s
I11-i I22-i
c) f (I1 I2)
a
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
t (30 kg)(1.5 m) (12 rad/s) (20 kg)(1.0 m) (12 rad/s)
f ((30 kg)(1.5 m)2 (20 kg)(1.0 m)2)
2
2
f 6.5 rad/s
d) I11-i I22-i (I1 I2)f
I11-i
2-i I2
m1r121-i
2-i m2r22
(30 kg)(1.5 m) (12 rad/s)
2-i (20 kg)(1.0 m)2
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)
[250 kg·m (40 kg)(2.5 m) ](2.0 rad/s)
2 (250 kg·m2 (40 kg)(1.5 m)2)
2 2.9 rad/s
m1r121-i m2r222-i
f (m1r12 m2r22)
2
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
2
2(1.10 m)
0.138 m/s
2
t 3.99 s
(v2 v1)
c) a t
v2 v1 at
v2 0 (0.138 m/s2)(3.99 s)
v2 0.551 m/s
d) v2 r2
v
2 2
r
0.551 m/s
2 0.0030 m
2 184 rad/s
1
e) Ek(final) mv22
2
1
Ek(final) (0.135 kg)(0.551 m/s)2
2
Ek(final) 0.0205 J
1
f) Erot(final) I22
2
1
Erot(final) (8.50 105 kg·m2)(184 rad/s)2
2
Erot(final) 1.43 J
g) ETotal(initial) mgh1
ETotal(initial) (0.135 kg)(9.80 m/s2)(1.10 m)
ETotal(initial) 1.46 J
Solutions to End-of-chapter Problems
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:
1.0 m/s (1.0
m
/s) 4(0.0690
m
/s )( 1.10 m)
t 2(0.0690 m/s2)
2
2
t 1.03 s
(v2 v1)
c) a t
v2 v1 at
v2 1.0 m/s (0.138 m/s2)(1.03 s)
v2 1.14 m/s
d) v2 r2
v
2 2
r
1.14 m/s
2 0.0030 m
2 380 rad/s
1
e) Ek(final) mv22
2
1
Ek(final) (0.135 kg)(1.14 m/s)2
2
Ek(final) 0.088 J
1
f) Erot(final) I22
2
1
Erot(final) (8.50 105 kg·m2)(380 rad/s)2
2
Erot(final) 6.16 J
150
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
Solutions to End-of-chapter Problems
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
2
(9.1 10
19
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
6
6
(9.0 10 N·m /C )( 2.0 10 C)(2.3 10 C)
Fe3 (0.10 m)
9
2
2
1
2
Fe3 4.14 N (attraction)
1Fe3 4.14 N [right]
kq2q3
2Fe3 r22
1
6
6
(9.0 10 N·m /C )(3.8 10 C)(2.3 10 C)
Fe3 (0.30 m)
9
2
2
2
2
Fe3 0.87 N (repulsion)
2Fe3 0.87 N [left]
2
152
2
kg)(9.8 m/s )
Solutions to End-of-chapter Problems
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
6
6
(9.0 10 N·m /C )( 2.0 10 C)(2.3 10 C)
Fe3 (0.30 m)
9
2
2
1
2
Fe3 0.46 N (attraction)
1Fe3 0.46 N [left]
1
kq2q3
Fe3 r22
2
6
6
(9.0 10 N·m /C )(3.8 10 C)(2.3 10 C)
Fe3 (0.10 m)
9
2
2
2
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]
2Fe3 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.
(8.0 10 ) (8.0
10 ) 4
(1.8 10 )(8.0
10 )
rx 2(1.8 10 )
7
7 2
6
6
8
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.
Solutions to End-of-chapter Problems
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
6
(9.0 10 N·m /C )( 1.0 10 C)
Fe1 (2.0 102 m)2
9
2
2
2
2
Fe1 22.5 N [left]
4Fe1 22.5 N [up]
2
58. Parallel plates:
kqq
3Fe1 r23
+
6
Coaxial cable:
–
(9.0 10 N·m /C )( 1.0 10 C)
Fe1 2
2
9
3
2
2
2
2(2.0 10 m)
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
Solutions to End-of-chapter Problems
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
6
6
(9.0 10 N·m /C )( 1.0 10 C)( 5.0 10 C)
Ee 0.25 m
9
2
2
Ee 0.18 J (repulsion)
kq1q2
b) Ee1 r
6
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
6
(9.0 10 N·m /C )( 1.0 10 C)( 5.0 10 C)
Ee1 1.00 m
9
2
2
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.
Solutions to End-of-chapter Problems
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
P(6.25 10 e/C)
number of electrons/s V
18
1C
number of electrons/s (27 J/s) 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
19
(9.0 10 N·m /C )(1.602 10 C)
r (1.67 1027 kg)(2.7 106 m/s)2
9
2
2
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
Solutions to End-of-chapter Problems
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
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
86. d 0.12 m, V 92 V
V
d
92 V
0.12 m
7.7 102 N/C
87. 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.
88. V 50 V, 1 104 N/C
V
d 50 V
d 1 104 N/C
d 5.0 103 m
89. V 120 V, 450 N/C
V
d 120 V
d 450 N/C
d 2.67 101 m
90. 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
15
3
(2.2 10 kg)(9.80 N/kg)(5.5 10 m)
q 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.
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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°
Solutions to End-of-chapter Problems
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
7
6
(4 10 T·m/A)(1.5 A)(1.5 10 C)(450 m/s)sin 90°
F 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
7
19
(4 10 T·m/A)(35 A)(1.602 10 C)(5 10 m/s) sin 90°
F 2 (0.05)
7
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.
Solutions to End-of-chapter Problems
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
7
(4 10 T·m/A)(1.5 10 A) (190 m)
F 2 (3.5 m)
4
2
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
19
(1.602 10 C)(5.0 10 m/s)(0.028 T)
ac 9.11 1031 kg
6
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
L
g
b) T 2
80 m
9.8 m/s
2
T2
L
g
c) T 2
T2
g
L
Moon
2.1 m
2
1.6 m/s
T 7.2 s/cycle
T2
ii) T 2
g
L
Moon
T2
80 m
1.6 m/s
2
T 44 s/cycle
iii) T 2
g
L
Moon
T2
0.15 m
1.6 m/s
2
T 1.9 s/cycle
b) i)
T2
g
L
Jupiter
T2
2.1 m
24.6 m/s
2
T 1.8 s/cycle
ii) T 2
g
L
Jupiter
T2
80 m
24.6 m/s
T 11 s/cycle
166
2
m
k
0.30 kg
23.4 N/m
m
k
0.40 kg
20 N/m
T 0.889 s/cycle
T2
0.15 m
24.6 m/s
T 0.711 s/cycle
0.15 m
T2
2
9.8 m/s
T 0.78 s/cycle
31. a) i)
T2
T2
T 18 s/cycle
c) T 2
L
Jupiter
32. a) T 2
T2
g
T 0.49 s/cycle
2.1 m
T2
2
9.8 m/s
T 2.9 s/cycle
b) T 2
iii) T 2
2
m
k
0.21 kg
200 N/m
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
Solutions to End-of-chapter Problems
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 1014 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
d (3.00 108 m/s)(3.1536 107 s)
d 9.46 1015 m
37. 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
38. t v
160 m
t 3.00 108 m/s
t 5.33 107 s
39. 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
40. 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
Solutions to End-of-chapter Problems
167
42. a) sin 30° 0.5
b) sin 60° 0.866
c) sin 45° 0.707
d) sin 12.6° 0.218
e) sin 74.4° 0.963
f) sin 0° 0
g) sin 90° 1
43. a) sin1 (0.342) 20°
b) sin1 (0.643) 40°
c) sin1 (0.700) 44.4°
d) sin1 (0.333) 19.5°
e) 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.
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°
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
169
Chapter 11
23. a) constructive
b) constructive
c) partial
d) destructive
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
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
3
c) 2 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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
5.8959 10 m
sin 2.5 10 m
7
1
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
Solutions to End-of-chapter Problems
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
7
7
(6.5648 10 m) (6.5630 10 m)
avg 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°
Solutions to End-of-chapter Problems
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
W0. The gold will absorb all of the
E
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
W0, the photoelectric effect will
Since E
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
Solutions to End-of-chapter Problems
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
34
(6.626 10 J·s)(3.0 10 m/s)
Ekmax 2.19 107 m
8
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 Solutions to End-of-chapter Problems
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
2(6.626 1034 J·s)(3.0 108 m/s)
i (9.11 10 kg) 1(7.12 10 m/s)
(3.0 10 m/s)(7.12 10 m/s)
31
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
mvf cos (eq. 2)
cos 43° 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
5
2
8
5
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
Chapter 13
3 cm
1m
1a
28. a) 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
b) 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
c) 3.6 108
3.0 108 m/s
d) 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
e) 7.33 103
3 108 m/s
29. a) Snoopy must fly 50 km/h [N then E].
Let y resultant ground speed
y (130 km/h)2 (50 km/h)2
y 120 km/h
b) The Baron going west has a ground speed
of:
bvg bvw
wvg
50 km/h
bvg 130 km/h
bvg 180 km/h [W]
While going east,
bvg 130 km/h 50 km/h
bvg 80 km/h [E]
c) 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 w)(v 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
t
v2 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
Solutions to End-of-chapter Problems
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 0.7(3 108 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
L 7.58 1010 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
Solutions to End-of-chapter Problems
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
2
t t0 1
2c
The difference in the flying clocks compared
to the ones on Earth is:
t t t0
v2
2 t0
t t0 1
c
2r
t 1
v
v2
2 1
2c
(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
Solutions to End-of-chapter Problems
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:
t1a
(9.0 10 N·m C )(1.602 10 C) 1 (0.6)
r (9.11 1031 kg)[0.6(3.0 108 m/s)]2
9
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
2
2
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
2 1
v
2c
1 2
c
v2
2 m0
m m0 1
2c
v
1
m m 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
4
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 0.999 999 999 67)
9.11 1031 kg
m 2(3.3 1010)
m 3.55 1026 kg
186
19
2
Solutions to End-of-chapter Problems
2
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
31
(9.1 10 kg)(3.0 10 m/s)(0.999 999 986)
B 8
(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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
Since v 75 103 m/s, or v
c, use the
low-speed binomial approximation:
v2
1
1
2c2
v2
1 2
c
v2
2 1
p m0v 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
2 1
Ek m0c2 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
27
31
(3.0 10 m/s) [(1.67 10 kg) (9.1 10 kg)]
V 1.60 10 C
8
2
19
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
Solutions to End-of-chapter Problems
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
Solutions to End-of-chapter Problems
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
Ek,
51. Since 232
92U → 90Th
2He
2
Ek [mU mTh m]c
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
19
(8.99 10 J·m/C )(1.6 10 C) (2)(92)
r (5.3 106 eV)(1.6 1019 J/eV)
9
2
2
r 5.0 1014 m
0
231
v
53. 231
1e
90Th → 91Pa
235
231
4
92U → 90Th
2He
54. The mass difference is:
∆m mn (mp me)
∆m [939.57 938.27 0.511] MeV/c 2
∆m 0.789 MeV/c2
55. From problem 54, the energy equivalent of
0.789 MeV/c2 is 0.789 MeV.
2
Thus (0.789 MeV) 0.526 MeV.
3
56. 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
pC (2.64 1021)2 (4.76 1021)2
pC 5.44 1021 N·s
p2
57. 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
19
(8.99 10 J·m/C )(1.6 10 C) (2)(79)
r (449 106 eV)(1.6 1019 J/eV)
9
2
2
r 5.07 1016 m
Solutions to End-of-chapter Problems
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
R
1
2
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.
T 14 d
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 t
N N0 T
2
1 t
0.0218 0.0383 T
2
0.0218
t
1
log log 8
0.0383
7.1 10 a
2
1
2
1
2
1
2
7.0 min
198 720 min
7.0 min
3.1 min
1
2
0.0218
log (7.1 108 a)
0.0383
t 1
log 2
8
t 5.78 10 a
1
2
192
1
2
1
2
2000 a 650 a
5730 a
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
Solutions to End-of-chapter Problems
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
Ek. To determine the
66. 137
1 e v
55Cs → 56Ba
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
activity time energy percentage absorbed
67. Dose mass
(3700 Bq)(365 24 60 60 s)(1.0 10 eV)(1.6 10 J/eV)(5%)
Dose 70 kg
6
19
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 1.2AN
3
3
rs 1.24 1.214
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
19
(9.0 10 J·m/C )(1.6 10 C) (2)(7)
Ek (4.8 1015 m)(1.6 1013 J/MeV)
9
2
2
Ek 4.2 MeV
73. The half-life of hassium-269 is T 9.3 s.
The original amount of hassium is
1
2
mass of hassium Avogadro’s number
N mass per mole
Using the activity equation:
0.693N
Activity T
1
2
Activity 3
1
(1.0 10 g)(6.022 10 mol )
0.693 269 g/mol
23
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
3
1
(1.0 10 g)(6.022 10 mol )
Activity (7.18%) 23
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
Solutions to End-of-chapter Problems
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
310n,
80. For the reaction 235
ZY
92U
0n → 56Ba
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
Solutions to End-of-chapter Problems
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
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 (9000.5 MeV)2 (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
qB
The cyclotron frequency, f , yields:
2m
2mf
B q
2(2.3856 1027 kg)(20 106 Hz)
B 1.6 1019 C
B 1.87 T
2
1
1
87. 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
88. a) lambda (baryon)
b) pion or rho (mesons)
c) b-zero (meson)
d) eta-c (meson)
89. A neutron consists of udd, therefore an antineutron is u d d .
90. 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.
91. 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
92. t v
2.4 1015 m
t 3 108 m/s
t 8 1024 s
93. 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.
Solutions to End-of-chapter Problems
195
W → n
94. Antiproton decay: p → n
0
1
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
1
ttb 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 broken ends could be u and u or d and d , according to ss → su su or sd sd. These particles
are known as mesons.
196
Solutions to End-of-chapter Problems