ANSWERS AND SOLUTIONS
S
N
A
P
Student Notes and Problems
PHYSICS 12
British Columbia
ANSWERS AND SOLUTIONS
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CASTLE ROCK RESEARCH
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ANSWERS AND SOLUTIONS
Lesson 2—Special Theory of Relativity
MEASUREMENT OF
MOTION
Lesson 1—Frames of Reference
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
t
t0
1
1.
2.
2.5 m/s south  1.0 m/s south  3.5 m/s south

v2
c2
1.25 years
2
 2.40  108 m/s 
1 

8
 3.00  10 m/s 
 2.08 years
2.5 m/s north  1.0 m/s south  1.5 m/s north
3.
2.
t
t0
1
R  (v1 ) 2  (v2 ) 2
  2.5 m/s   1.0 m/s 
 2.7 m/s
opposite
tan 
adjacent
1.0 m/s

2.5 m/s
 1.0 m/s 
  tan 1 

 2.5 m/s 
 22
2

2
 7.50  107 m/s 
1 

8
 3.00  10 m/s 
 3.10 years
3.
5.
v2
c2
1.50 h
 44.4 m/s 
1 

8
 3.00  10 m/s 
 1.50 h
4.
t
2
t0

The spacecraft accelerates relative to the stars
because of the force acting on it. However, there is
no force acting on the floating object. By Newton’s
first law, the object remains in its state of motion
with the same velocity relative to the stars. To the
astronaut, who is fixed to the spacecraft, it seems
that the object is accelerating in the opposite
direction to the spacecraft, i.e., toward him.
2
t0
1
v2
c2

v2 
t0  t  1  2 

c 

2

 2.80  108 m/s  
  50.0 years   1  

8

 3.00  10 m/s  

 18.0 years
From the point of view (reference frame) of the
shark, it appears that the fish is approaching it at
(64–15) km/h.
Therefore, the motion of the fish appears to be
49 km/h west.
Not for Reproduction
t
1
R  2.7 m/s 22 S of W
4.
v2
c2
3.00 years
1
Physics 12 SNAP
ANSWERS AND SOLUTIONS
5.
t
8.
t0
1
2
v
c2

v2 
t0  t  1  2 

c 

2

 2.90  108 m/s  
  35.0 years   1  

8

 3.00  10 m/s  

 8.96 years
1
1
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
v2
1 2
c
10.0 years
18.0 years 
1
Lesson 3—Length Contraction
t0
t
6.
v2
8
v2
 3.00 10
v
8
m/s 
2
2
 3.00 10
8
m/s 
2
v2
 3.00 10
8
1.
 3.00 10
m/s 
2

m/s 
2
10.0 years
18.0 years
 10.0 


 18.0 
1
60.0 years 
1
8
m/s 
8
m/s 
2
2
 18.4 m
2
 10.0 
 1 

 18.0 
2.
2
L  L0 1 
L0 
 0.691
v2
c2
L
v2
c2
80.0 m
1

1   0.900 
 184 m
v2
c2
15.0 years
v2
 3.00 10
v2
 3.00 10
8
1
v2
c2
2
t0
t
1
L  L0 1 
 2.55 10
  35.0 m  1 
 3.00 10
v  (0.691)(3.00 108 m/s) 2
 2.49 108 m/s
7.
The teacher should give 2.5 h according to the
space bus clock. Not only would the space bus
clock slow down as the space bus travels, but the
biological clocks of all the passengers would slow
down at the same rate. If you travel at a high speed
through space, you are not aware that there is
anything different. Writing an exam for 2.5 h on
the space bus will seem the same to you as writing
the same exam in your class room.
v
m/s 
2
2
 3.00 10
8
m/s 
2
v2
 3.00 10
8
m/s 
2

8
3.
L  L0 1 
m/s 
2.85light years =  6.25light years  1 
2
 3.00  10
8
m/s 
2
2.85 light years
v2
 1
6.25 light years
(3.00  108 m/s) 2
v2
2
 0.456   1 
(3.00  108 m/s) 2
2
v
2
 1   0.456 
8
2
(3.00  10 m/s)
2
 15.0 years 
 1 

 60.0 years 
 0.938
2
v  (0.792)(3.00 108 m/s) 2
 2.67 108 m/s
v  (0.938)(3.00 108 m/s) 2
 2.90 108 m/s
CASTLE ROCK RESEARCH
v2
Note: You do not have to convert light years.
15.0 years
60.0 years
 15.0 years 


 60.0 years 
v2
c2
2
Copyright Protected
ANSWERS AND SOLUTIONS
4.
L  L0 1 
v2
c2
2.
m0
m
1
  90.0 m  1   0.800 
 54.0 m
NOTE: Height does not change. It is only the
dimension parallel to the direction of travel that
changes.
2
5.
L  L0 1 

v2 
m0  m  1  2 

c 

2

 2.50  108 m/s  
  7.20  1020 kg   1  

8

 3.00  10 m/s  

 3.98  1020 kg
v2
c2
 3.10 10
 1.20 km  1 
 3.00 10
8
m/s 
8
m/s 
2
3.
2
L  L0 1 
4.
2
1  (0.800) 2

kg
v
c
m0
1  x2
2

m0
2

 1.96 10 m/s
8
v2
c2
1.67  1027 kg
1   0.95 

 6.05  10 27 kg 
 1 

—27
kg 
 8.00  10
 0.654
v  xc
  0.654  3.00 108 m/s
1

10
m 
x  1  0 
 m
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
m
let x 
m
Lesson 4—Mass Increase
1.
v2
c2
v2
m0  1  2
c
  5.50  10-10 kg 
 3.30  10
v2
c2
 1.00 m  1   0.600 
 0.800 m
m0
m
1
You will soon discover that the mathematics is
impossible to do as one cannot find the square root
of a negative value. According to Einstein’s
special theory of relativity, nothing can travel
faster than light (3.00 × 108 m/s). The salesman
was not telling the truth.
6.
v2
c2
5.
2
 5.3  1027 kg
let x 
m
v
c
m0
1  x2
m 
x  1  0 
 m
2
2
1
 1  
3
 0.943
v  xc
  0.943 3.00 108 m/s


 2.83 10 m/s
8
Not for Reproduction
3
Physics 12 SNAP
ANSWERS AND SOLUTIONS
6.
let x 
m
v
c
m0
Lesson 5—Relativistic Addition of
Velocities
1  x2
m 
x  1  0 
 m
2
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
2
1
 1  
4
 0.968
v  xc
  0.968  3.00 108 m/s

u
2.
u

 2.90 10 m/s
8
7.
E  mc 2
 10.0 kg   3.00 108 m/s 
2
 9.00  1017 J
8.
9.
E  mc 2
E
m 2
c
7.25  1016 J

(3.00  108 m/s) 2
 0.806 kg
v  u
vu 
1 2
c
0.80c  0.75c

 0.80c  0.75c 
1
c2
 0.13c or 0.13c toward
NOTE: The negative sign indicates that the object
is moving toward the stationary observer.
E  mc 2
E
m 2
c
5.3  1010 J

(3.00  108 m/s) 2
 5.9  10 7 kg
NOTE: A grain of sand may have a mass
approximately 2.00 × 10–6 kg.  5.9 × 10–7 kg is
very small amount of gasoline, approximately onethird of the mass of a grain of sand.
NOTE: in order to obtain 5.3 × 1010 J of energy by
chemical means, you would burn approximately
1500 L of gasoline in your car.
16 000 km
 1500 L
10.6 km/L
CASTLE ROCK RESEARCH
v  u
vu 
1 2
c
0.80c  0.75c

 0.80c  0.75c 
1
c2
 0.97c away
1.
4
v  u
vu 
1 2
c
0.75c  0.60c

 0.75c  0.60c 
1
c2
 0.93c toward
3.
u
4.
u
v  u
vu 
1 2
c
0.80c   c 

 0.80c  c 
1
c2
 c or c toward the observer
Copyright Protected
ANSWERS AND SOLUTIONS
5.
v  u
vu 
1 2
c
0.80c  c

 0.80c  c 
1
c2
 c toward the observer
3.
u
b) Observer 2 will observe light from light 1 and
2 at precisely the same time because the light
from both sources travels precisely the same
distance.
NOTE: It does not matter if the vehicle is
travelling toward or away from the stationary
observer. The velocity of light is the same.
6.
u

4.
v  u
vu 
1 2
c
2.5  108 m/s  2.0  108 m/s
 2.5 10 m/s  2.0 10
 3.00 10 m/s 
8
1
8
a) Observer 1 will observe the illumination of
light 1 before light 2 because for observer 1,
light 1 travels less distance than light 2.
8
a) Proper time is measured by the observer who
is at rest with respect to the event. José is at
rest with respect to the event.
t
b)
m/s 
18.0 years 
2
v2
(3.00  108 ) 2
v  2.96  108 m/s
 2.9  10 m/s away
v  u
vu 
1 2
c
2.5 m/s  6.0 m/s

(2.5 m/s)(6.0 m/s)
1
(3.00 108 m/s) 2
 8.5 m/s away
u
5.
Proper time is the time measured by an observer
who is at rest with respect to the event. Proper
time is also less than the time measured by an
observer who is not at rest with respect to the
event. Note that for non-relativistic speeds,
the discrepancy from the proper time is
extremely small.
t0
t
2
v
1  
c
6.
Proper length is defined as the length as
determined by the observer who is at rest with
respect to the length determined. But, unlike
proper time, proper length is greater than the length
determined by the observer that is moving with
respect to the length determined.
NOTE: This is the same answer that you would
get if you just added the two velocity vectors
together as you added vectors earlier in this course.
Relativistic effects are only observable at very high
velocities.
Practice Test
ANSWERS AND SOLUTIONS
1.
2.
v2
c2
where L0 is the proper length.
Michelson and Morley designed an experiment to
measure the speed of light relative to ether. Their
result is known as the null result, because they did
not detect any evidence of the ether. This made
many scientists question its existence.
L  L0 1 
7.
An inertial frame of reference is any frame of
reference that is at rest or is moving at a constant
velocity.
Not for Reproduction
2
v
1  
c
3.00 years
1
8
7.
t0
5
Einstein’s special theory of relativity has two
postulates:
1. the laws of physics in all inertial frames of
reference are the same
2. the speed of light in space is the same for all
observers, regardless of the motion of the
source or observer.
Physics 12 SNAP
ANSWERS AND SOLUTIONS
8.
If Einstein’s postulates are correct, then space and
time are not absolute. It also follows that mass and
energy are equivalent.
9.
The length contraction will only be in the direction
of the motion. Therefore, side Y will remain at 50
m. There is length contraction of side Z so it will
be shorter.
Again Fg2 = 1.19Fg1
= (1.19)(7.74 N)
= 9.21 N = 9.2 N
2.
10. An observer will have the same experiences in any
inertial frame of reference. Pavel will not observe
any change to his watch other than the time
changing as he would normally notice.
FORCES CAUSE MOTION
 Fx  0
FT – Fgx  0
Fgx  FT  5000 N
Lesson 1—The First Condition
of Equilibrium
Fgx  Fg sin
Fgx
5000 N
Fg 

sin  sin 25.0
 1.18  10 4 N
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
3.
Fg1x  Fg1cos1
 Fg1cos40
 0.766Fg1
Fg1 y  Fg1sin1
 Fg1sin40
 0.643Fg1
Fg 2 x  Fg2 cos 2
 Fg2 cos50
 0.643Fg2
Fg2 y  Fg2 sin 2
 Fg2 sin50
 0.766Fg2
FTx  FT cos
 FT cos63.0
 0.454FT
 Fy  0
FTy  Fg  0
FTy  Fg  20.0 N
 Fx  0
Fg 2 x  Fglx  0
0.643Fg2  0.766Fg1  0
0.766 Fg1
Fg 
0.643
Fg 2  1.19Fg1
0
Fg1 y  Fg2 y – Fg3  0
0.643Fg1  0.766Fg2 –12 N  0
1.55Fg1  12 N
Fg1  7.74 N
y
Fapp
6
FTy
20.0 N

0.891 0.891
 22.4 N
F
 x0
Fapp  FTx  0
 0.454FT  0
Fapp   0.454  22.4 N 
 10.2 N
FT 
F
CASTLE ROCK RESEARCH
FTy  FT sin
 FT sin63.0
 0.891FT
Copyright Protected
ANSWERS AND SOLUTIONS
 Fx  0
FTx  F  0
FTx  39.7 N
FTx  0.866FT
F
39.7 N
FT  Tx 
0.866 0.866
 45.9 N
 Fy  0
FTy  Fg  0
0.500 FT  Fg
 Fg   0.500  45.9 N 
 22.9 N
4.
FT1x  FT1cos1
 FT1cos22.0
 0.927FT1
FT2x  FT2 cos 2
 FT2 cos80.0
 0.174FT2
FT1y  FT1sin1
 FT1sin22.0
 0.375FT1
FT2y  FT2 sin 2
 FT2 sin80.0
 0.985FT2
Fg  mg
Fg
22.9 N
m

g 9.81 m/s 2
 2.3 kg
 Fx  0
FT2x  FT1x  0
0.174FT2  0.927FT1  0
0.927 FT1
FT2 
0.174
 5.33FT1
 Fy
FT2y  FT1y – Fg
0.985FT2  0.375FT1
 0.985 5.33FT1  – 0.375FT1
4.88FT1
FT1
6.
0
0
 75.0 N
 75.0 N
 75N
 15.4 N
FTx  FT cos
 FT cos78
 0.208FT
 FT2 = 5.33 (15.4 N) = 82.1 N
5.
FTy  FT sin
 FT sin78
 0.978 FT
 Fx  0
FTx  F  0
F  FTx
 0.208FT
F
FTy  FT sin
FTx  FT cos
 FT sin30
 FT cos30
 0.500 FT
 0.866FT
F  Ffr  μmgcos
  0.27 15 kg  9.81 m/s  cos 0 
 39.7 N
Not for Reproduction
y
FTy  Fg
FTy
0.978FT
FT
7
0
0
 Fg
 735 N
735 N

0.978
 751 N
 7.5  102 N
Physics 12 SNAP
ANSWERS AND SOLUTIONS
FT1y  FT1sin1
FT1x  FT1cos1
7.
 FT1cos26.5
 FT1sin26.5
 0.895FT1
 0..446FT1
FT2x  FT2 cos 2
 FT2 cos63.5
 0..446FT2
FT1x  FT1cos1
  96 N  cos 60
  96 N  sin 60
 48 N
 83 N
FT2x  FT2 cos 2
 FT2 sin63.5
 0.895FT2
 Fx  0
FT1x  FT2x  0
0.895FT1  0.446 FT2  0
0.446 FT2
FT1 
0.895
 0.498FT2
0.498FT2  FT1
FT2y  FT2 sin 2
 FT2 cos30
 FT2 sin30
 0.866FT2
 0.500 FT2
 Fx  0
FT2x  FT1x  0
48 N  0.866 T2
48 N
FT2 
0.866
 55.43 N
FT2y  FT2sin 2
FT1y  FT1sin1
55 N
0.498
FT2  110.4 N or 1.1102 N
FT2 
55 N
F
y
 Fy  0
FT1y  FT2y – Fg  0
83 N  0.500 FT2  Fg
83 N   0.500  55 N   Fg
0
FT1y  FT2y – Fg  0
.446FT1  0.895 FT 2  Fg
0.446  55 N   0.895 110 N   Fg
 Fg =123 N or 1.2 102 N
8.
9.
For the whole system,
 Fy  0
FT1  Fg1  0
FT1  Fg1
 25 N  12 N
 37 N
For the lower block,
 Fy  0
FT2  Fg  0
FT2  Fg
 12 N
Find 1
op 3.0 m

hyp 6.7 m
1  26.5
sin1 
Find  2
op 6.0 m

hyp 6.7 m
 2  63.5
sin 2 
CASTLE ROCK RESEARCH
8
Copyright Protected
ANSWERS AND SOLUTIONS
2.
10.
F  0
F
y
x
|
FTx  F  0
a)
  0
1 –  2  0
r1 F1 sin1 – r2 F2 sin 2  0
0
FTy  Fg  0
FTx  F
FTx  410 N
 L  FT  sin 40 –  12 L  Fg  sin 90  0
0.643LFT –  200 N  L  0
FTy  Fg
 675 N
0.643FT  200 N
200 N
0.643
 311 N
 3.1102 N
FT 
b)
op
adj
410 N

675 N
  31.3
tan 
Lesson 2—The Second Condition of
Equilibrium
FTx  FT cos 
  311 N  cos 40 
 238 N
FTy  FT sin 
  311 N  sin 40 
 200 N
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
  rF sin 

F
r sin 
F
x
45 N  m
(0.35 m)(sin 90°)
 129 N or 1.3  102 N
0
FH – FTx  0

FH  FTx
FH  238 N
F
y
0
FTy  Fg  FV  0
200 N – 400 N  FV  0
Fv  238 N
Fv  200 N
Not for Reproduction
9
Physics 12 SNAP
ANSWERS AND SOLUTIONS
 Fx  0
FT2 – Fgb – Fgd  FT1  0
3744 N – 400 N – 735 N  FT1  0
3.
FT1  –3744 N  400 N  736 N
 FT1  –2605 N or – 2.61103 N
5.
  0
a)
1 –  2   3  0
r1 F1sin1 – r2 F2sin2  r3 F3sin3  0
 5.00 m  FTR  sin90
–  2.50 m  250 N  sin90
– 1.50 m  650 N  sin90  0
 5.00 m  FTR  – 625 N – 975 N  0
  0
1 –  2  0
r1 F1 sin1 – r2 F2 sin 2  0
 L  FT  sin 30 – 12 L  Fgb  sin 90  0
0.500 FT –100 N  0
0.500 FT  100 N
100 N
FT 
0.500
 200 N
=2.0 102 N
(625 N  m  975 N  m)
 FTR 
5.00 m
 FTR  320 N
b)
 
 Fx  0
FTR  FTL – Fgs – Fgb  0
320 N  FTL – 650 N – 250 N  0
FTL  900N – 320 N
 FTL  580 N
6.
4.
  0
1 –  2  0
r1 F1 sin1 – r2 F2 sin 2  0
point of rotation is arbitrary
  0
1 –  2   3  0
r1 F1sin1 – r2 F2 sin 2  r3 F3sin3  0
 L  FT  sin 50 –  12 L  Fgb  sin 40  0
0.766FT – 64.3 N  0
r1 FT2 – r2 Fgb – r3 Fgd  0
0.766FT  64.3 N
64.3 N
0.766
 83.9 N
 L  F   L  (400 N) 
1
4
T2
FT 
1
2
( L)(75.0 kg  9.81 m/s 2 )  0
1
FT2 – 200 N – 736 N  0
4
 FT2  4  936 N 
 3744 N or 3.74 103 N
CASTLE ROCK RESEARCH
=84 N
10
Copyright Protected
ANSWERS AND SOLUTIONS
Lesson 3—Uniform Circular Motion
6.
mv 2
r
(925 kg)(25 m/s) 2

72 m
 8.03  103 N
Fc 
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
2.
Here the frictional force is equal to the centripetal
force.
mv 2
r
(925 kg)(22 m/s) 2

75 m
 6.0  103 N
Fc 
Again
FN  Fg  mg
  925 kg   9.81 m/s 2 
 9.07  103 N
C  2 r
 (2 )  3282 m 
 2.06 104 m
v
Ffr   FN
F
  fr
FN
8.03  103 N

9.07  103 N
 0.89
d
t
2.06  104 m
(2.0 min)(60 s/m)
 172 m/s

2
v
r
(172 m/s) 2

3 282 m
 9.0 m/s 2
ac 
3.
4.
5.
mv 2
Fc 
r
(822 kg)(28.0 m/s) 2

105 m
 6.14  103 N
2 r
T
2 (2.8 m)

(0.98 s)
 18 m/s
mv 2
Fc 
r
2.00
 102 m  4.60  1014 N 

v2 
(9.11 1031 kg)
mv 2
r
(2.7  03 kg)(4.7  103 m/s) 2

1.8  107 m
3
 3.3  10 N
7.
Fc 
8.
a)
v
b)
Fc 
c)
Find time to reach the ground using the
vertical component
v0
v
a
d
t
1.2 m
×
v
(4.60  1014 N)(2.00  102 m)
9.11 1031 kg
 3.18  107 m/s
v
Not for Reproduction
2 r
T
2 (0.90 m)

0.30 s
 19 m/s
mv 2
r
(3.7 kg)(19 m/s) 2

0.90 m
 1.5  103 N
0
11
×
9.81 m/s
2
Physics 12 SNAP
ANSWERS AND SOLUTIONS
1
d y  v0 t  at 2
2
1
1.2 m  (9.81 m/s 2 )(t 2 )
2
2(1.2 m)
t
9.81 m/s 2
 0.495 s
mv 2
r
Fc r
v
m
(135 N)(1.10 m)

2.00 kg
 8.62 m/s
10. Fc 
Horizontally
d
vx  x
t
d x  vx t
 19 m/s  0.495 s 
 9.3 m
9.
11. a)
Ffr   FN
  0.95  9143 N 
 8.69 103 N
Speed
2r  Diameter
30.0 102 m
r
2
 15.0 102 m
T
FN  mg
  932 kg   9.81 m/s 2 
 9143 N
mv 2
r
(932 kg)(v 2 )
3
8.69  10 N 
82 m
(8.69  103 N)(82 m)
v
932 kg
 28 m/s
Fc 
1
f
1m
f  33 1 rev/minute 
3
60 s
1
33
 3 Hz
60
60
T
1
33 Hz
3
 1.80 s
2 r
v
T
2 (15.0  102 m)

1.80 s
 0.524 m/s
b)
Ffr   FN
  0.40  9143 N 
 3.66 103 N
mv 2
r
(932 kg)(v 2 )
3
3.66  10 N 
82 m
(3.66 103 N)(82 m)
v
932 kg
 18 m/s
Fc 
Acceleration
v2
ac 
r
(0.524 m/s) 2

15.0  10 2 m
 1.83 m/s 2
CASTLE ROCK RESEARCH
12
Copyright Protected
ANSWERS AND SOLUTIONS
FT  0
 Fc  Fg
mv 2
 mg
r
v  rg
4.
Lesson 4—Vertical Circular Motion
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
 (2.90 m)(9.81 m/s 2 )
 5.33 m/s
FT  0
 Fc  Fg
mv 2
 mg
r
v  rg
Fg  mg
5.
 1.50 kg   9.81 m/s 2 
 14.7 N
 (5.00 m)(9.81 m/s 2 )
 7.00 m/s
2.
Fc  FT – Fg
 186 N – 14.7 N
 171 N
mv 2
Fc 
r
Fc r
v
m
(171 N)(1.90 m)

1.50 kg
 14.7 m/s
FT  0
 Fc  Fg
mv 2
 mg
r
v  rg
 (0.95 m)(9.81 m/s 2 )
 3.1 m/s
3.
Fc  FN – Fg
 1.96  103 N – 655 N
 1.305 103 N
6.
Fg  mg
Fg
m
g
655 N

9.81 m/s 2
 66.8 kg
4 2 rm
T2
(4 2 )(1.0 m)(2.2 kg)

(0.97 s) 2
 92.3 N
Fc 
Fg  mg
  2.2 kg   9.81 m/s 2 
 21.6 N
Fc  FT  Fg
FT  Fc  Fg
 92.3 N – 21.6 N
 71 N
mv 2
r
Fc r
v
m
(1.305  103 N)(18.0 m)

66.8 kg
 18.7 m/s
Fc 
Not for Reproduction
a)
13
Physics 12 SNAP
ANSWERS AND SOLUTIONS
Fc  FT  Fg
FT  Fc  Fg
 92.3 N  21.6 N
 114 N
b)
b)
Fg  mg
  2.5 kg   9.81 m/s 2 
 24.5 N
mv 2
r
(2.5 kg)(12 m/s) 2

0.75 m
 4.8  10 2 N
Fc 
i) The tension of the string at its low point is
FT  Fc  Fg
 480 N  24.5 N
 5.0  102 N
FT  0
 Fc  Fg
mv 2
 mg
r
v  rg
7.
ii) The tension of the string at its high point is
FT  Fc – Fg
 480 N – 24.5 N
 4.6  102 N
 (48 m)(9.81 m/s 2 )
 22.7 m/s or 22 m/s
Practice Test
FT  0
 Fc  Fg
mv 2
 mg
r
v  rg
8.
ANSWERS AND SOLUTIONS
1.
 (43 m)(9.81 m/s 2 )
 20.5 m/s or 21 m/s
Force of friction depends on the surfaces
(coefficient of friction, ) and the normal force
(FN = mg cos).
 Ffr = mg cos
9.
a)
2.
1   2
m1 gL1  m2 gL2
 32 kg 1.5 m    42 kg  d  .
d  1.14 m
1.1 m
High
Fc  FT  Fg
FT  Fc – Fg
Low
Fc  FT  Fg
FT  Fc  Fg
3.
Rotational motion is related to torque. Therefore, if
a wooden beam has no rotational motion, then the
torques acting on the beam must be balanced.
It is clear from the above formulas that the
tension in the string greater at its low point
than at its high point
CASTLE ROCK RESEARCH
14
Copyright Protected
ANSWERS AND SOLUTIONS
4.
7.
Fnet  0
Ffr  Fg  x 
Fg  x   Fg sin
  2.55 N   9.81 m/s 2   sin 30 
 12.5 N
F  0
FT1 – Fg  0
FT1  Fg  mg
 18 kg   9.81 m/s 2 
5.
 177 N or 1.8  102 N
8.
  0
1   2  0
m1 gd – m2 g 1 m – d   0
 3.00 kg  d  –  5.00 kg 1 m – d   0
 3.00 kg  d – 5.00 kg  m   5.00 kg  d  0
8.00 kg  d  5.00 kg  m
d  0.625 m
or 62.5 cm
In circular motion, the velocity of a moving object
is tangent to the path.
9.
6.
In circular motion, the net force ( Fc ) is toward the
centre of the path, and the acceleration of an object
is always in the direction of the net force.
10. a)
Fnet
Fnet
Ffr
Fg  x 
0
 Fg  x  – Ffr
 Fg  x 
 Fg sin
The tension in the cord will be directed toward the
centre of the circle wherever the mass is in its path
in the vertical circle. The force due to gravity will
be directed straight down to Earth’s surface
wherever the mass is in its path in the vertical
circle. Therefore, when the mass is precisely at the
top of the vertical circle, both the force due to
gravity and the tension are directed down through
the centre of the circle.
  3.0 N   9.81 m/s 2   sin 37 
 17.7 N
Ffr   mgsin
Ffr

mg cos 
(17.7 N)

(3.0 kg)(9.81 m/s 2 )(cos 37)
 0.75
Not for Reproduction
15
Physics 12 SNAP
ANSWERS AND SOLUTIONS
10. b)
At the top of the vertical circle, both the tension
and the force of gravity act toward the centre of the
circle. Therefore, the magnitude of the centripetal
force at that moment is equal the sum of both
forces.
Fc  Fg  FT
11. The key to drawing this graph is recognizing that
the object is twirled at a constant speed. If the
speed remains constant, the tension in the cord will
remain constant. This is represented in a graph that
looks like this:
Fg 
3.
Fg 
Fc  Ffr
mv 2
  mgcos
r
v  r  g cos 
Fg 
5.
Fg 
.
FORCES CAUSE MOTION
m2 
Lesson 1—Newton’s Law of
Universal Gravitation
Gm1m2
r2
Fg r 2
Gm1
(3.2  106 N)(2.1 101 m) 2

2

11 N  m 
1
 6.67  10
 (5.5  10 kg)
kg 2 

 3.8 101 N
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
Gm1m2
r2
2

11 N  m 
(70.0 kg)(52.0 kg)
 6.67×10
2 
kg 


(1.50 m) 2
–7
 1.08  10 N
Fg 
CASTLE ROCK RESEARCH
Gm1m2
r2
2

11 N  m 
(5.98 1024 kg)
 6.67 10
2 
kg 

(6.50 102 kg)

(4.15 106 m  6.37 106 m) 2
 2.34 103 N
4.
 (40.0 m)(0.50)(9.81 m/s 2 )
 14 m/s
1.
Gm1m2
r2
As both the objects have equal mass,
Fg r 2
m2 
G
(2.30×10 8 N)(10.0 m) 2

N  m2
6.67×10 11
kg 2
4
2
 3.45  10 kg
m  1.86  10 2 kg
12. Centripetal force is the net force that causes an
object to travel in a circular path.
13.
Gm1m2
r2
2

11 N  m 
(5.98×1024 kg)
 6.67×10
2 
kg


(7.35×1022 kg)

(3.84×108 m) 2
 1.99  1020 N
2.
6.
Gm1m2
r2
Gm1m2
r
Fg
Fg 
2

11 N  m 
(2.0 10 2 kg)
 6.67  10
2 
kg


(2.0  102 kg)

3.7  106 N
 8.5 101 m
16
Copyright Protected
ANSWERS AND SOLUTIONS
7.
8.
Gm1m2
r2
2

11 N  m 
(5.98  10 24 kg)
 6.67  10
2 
kg


(70.0 kg)

(6.38  106 m) 2
2
 6.86 10 N
 Fg changes by a factor of
Fg 
Fg   6.86 102 N   0.444 
 3.05 102 N
11.
Gm1m2
r2
Fg  m
Fg 
Fg1  Fg2 , and since these forces are in opposite
directions, the net force will be zero.
Mass changes by a factor of:
To
35.0 kg

From 70.0 kg
 0.500
12.
Gm1m2
r2
2

11 N  m 
 6.67  10
 (10.0 kg)(10.0 kg)
kg 2 


(5.00  10 1 m) 2
 2.67  108 N
Fg1 
Mass changes by a factor of 0.500
 Fg will change by a factor of 0.500
Fg   6.86 102 N   0.500 
 3.43 102 N
9.
Gm1m2
r2
1
Fg  2
r
Fg 
Gm1m2
r2
2

11 N  m 
6.67

10

 (10.0 kg)(15.0 kg)
kg 2 


(5.00  101 m) 2
–8
 4.00  10 N
Fg2 
Distance changes by a factor of:
to
1.276 107 m

from 6.38 106 m
 2.00
Fnet  Fg2  Fg1
 4.00  10 –8 N – 2.67  10 –8 N
 1.33  10 –8 N
Distance changes by a factor of 2.00
 Fg changes by a factor of 0.250 =
1
= 0.444
(1.50) 2
1
(2.00) 2
Fg   6.86 102 N   0.250 
13. Fg 
 1.72 102 N
Fg 
Gm1m2
r2
m1m2
r
2
(2)(2)
(3) 2
 0.444

Gm1m2
r2
1
Fg  2
r
10. Fg 
Fg(new)   3.24 10–7 N   0.444 
 1.44 10–7 N
Distance changes by a factor of:
to
9.57 106 m

from 6.38 106 m
 1.50
Distance changes by a factor of 1.50
Not for Reproduction
17
Physics 12 SNAP
ANSWERS AND SOLUTIONS
Lesson 2—Field Explanation
GM
r2
GmY
 2
r
GmY
r
g
g
ANSWERS AND SOLUTIONS
PRACTICE EXERCISES
1.
2.
GM
r2
Gm
 2M
r
2

11 N  m 
23
 6.67 10
 (6.37 10 kg)
kg 2 


(3.43 106 m) 2
 3.61 N/kg
2

11 N  m 
(4.83 1024 kg)
 6.67  10
2 
kg

 
1.67 N/kg
 1.39  107 m
g
5.
GM
r2
Gm
 2E
r
GmE
r
g
g
g new   9.81 N/kg  0.25 
 2.45 N/kg
2

11 N  m 
24
 6.67  10
 (5.98 10 kg)
kg 2 


7.33 N/kg
6
 7.377  10 m
6.
From surface:
d  7.377 106 m – 6.38 106 m
 1.0 106 m
3.
g
GM
r2
1
g 2
r
1

(2) 2
 0.25
g
GM
r2
Gm
 2B
r
2

11 N  m 
22
6.67

10

 (4.00 10 kg)
kg 2 


(6.0 105 m) 2
 7.4 N/kg
g
Fg
m
63.5 N

22.5 kg
 2.82 N/kg
7.
gB
4.
g
Fg
gA
gB
m
50.0 N

30.0 kg
 1.67 N/kg
rA 2
rA
rA
CASTLE ROCK RESEARCH
GM
rA 2
GM
 2
rB
rB 2
 2
rA
g
 rB 2 B
gA
1
 rB
(
1.20
  0.913 rB
gA 
18
g A  1.20 g B )
Copyright Protected
ANSWERS AND SOLUTIONS
8.
GM
r2
Gm
 2m
r
2

11 N  m 
(7.35 10 22 kg)
 6.67 10
2 
kg 


(1.74 106 m) 2
 1.62 N/kg
g
g
4.
Fg
1 1
Ep  Gmmoon mmeteor   
 ri rf 


2

Nm 
  6.67  10 11
(7.35  10 22 kg)(1 750 kg)
2 
kg


1
1





6
6
15
000
m

1.74

10
m
1.74

10
m


7
 –4.2110 J
Ek  – Ep  4.21 107 J
Ek   Ek f –  Ek i
1
1
 mv 2  mvo 2
2
2
1
2
 m  v  vo 2 
2
m
Fg  mg
  20.0 kg 1.62 N/kg 
 32.4 N
Lesson 3—Gravitational
Potential Energy
4.21  10 J 
7

2
v 
2

24
 (5.98  10

9.90 106 m
Nm
kg
2


2 4.21  10
7
1750 kg
3

2 4.21  10 J
7
  1750 kg
J
 1.00  10 m/s 
2
3

2

2
2
kg)(5.00  10 kg)
3
5.
W  Ep
1 1
Ep  GmE ms   
 ri r f 


6.

N  m2 
 (5.98  1024 kg)(5.00  103 kg)
  6.67  1011
2 

kg




1
1
 


6
6
 6.38  10 m 9.90  10 m 
 1.11 1011 J
3.
3
2
 v  1.02  103 m/s
 –2.011011 J
2.
2

(1750 kg) v  1.00  10 m/s
v  1.00  10 m/s
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
GmE ms
1. Ep  
r

11
 6.67  10
 
1
GmE mo
r
2

11 N  m 
24
 6.67  10
 (5.98  10 kg)(10.0 kg)
kg 2 


6.38  106 m
 –6.25  108 J
Ep  
1 1
Ep  GmE ms   
 ri r f 


2

11 N  m 
24
3
  6.67  10
 (5.98  10 kg)(2.50  10 kg)
kg 2 

1
1





6
6
 6.38  10 m 6.90  10 m 
10
 1.18  10 J
W  Ep
The work done to lift the satellite into orbit was
1.11 1011 J , so the change in gravitational potential
energy was 1.11 1011 J .
Not for Reproduction
19
Physics 12 SNAP
ANSWERS AND SOLUTIONS
Lesson 4—Satellites: Natural &
Artificial
v
3.
v
4.
v
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
2.
2GM
r
2(6.67  1011 N  m 2 /kg 2 )(7.35  1022 kg)

(1.74  106 m)
3
 2.37  10 m/s
v
2GM
r
2(6.67  10-11 N  m 2 /kg 2 )(1.9  1027 kg)

(7.18 107 m)
 5.94 104 m/s
GmE
r
2

11 N  m 
24
 6.67 10
 (5.98 10 kg)
kg 2 


(4.4 105 m  6.38 106 m)
 7.65 103 m/s
2.
v
GmJ
r
2

11 N  m 
6.67

10
(1.90 10 27 kg)

2 
kg

 
(5.60  106 m  7.18 107 m)
 4.06  104 m/s
Gms
r
N  m2
)(1.98 1030 kg)
kg 2

1.50 1011 m
4
 2.97 10 m/s
(6.67 1011
3.
2GM
r
v2 r
M 
2G
(9.0  103 m/s) 2 (7.2  106 m)

2(6.67  10 11 N  m 2 /kg 2 )
 4.37  1024 kg
v
3
5.
T
2 r 2
Gm
3
(2 )(2.3 1011 m) 2

4.
2

11 N  m 
(1.98 1030 kg)
 6.67  10
2 
kg


 6.0 107 s or 1.9 Earth years
2GM
r
v2 r
M 
2G
(2.92  103 m/s) 2 (2.57  106 m)

2(6.67  1011 N  m/kg 2 )
 1.64 10 23 kg
v
Lesson 6—Electric Forces
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
Lesson 5—Satellites in Orbit
1.
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
GmE
r
2

11 N  m 
(5.98 10 24 kg)
 6.67 10
2 
kg

 
3.84 108 m
 1.02 103 m/s
v
CASTLE ROCK RESEARCH
20
Gm1m2
r2
2

11 N  m 
(70.0 kg)(52.0 kg)
 6.67  10
2 
kg 

(1.50 m) 2
 1.08  107 N
Fg 
Copyright Protected
ANSWERS AND SOLUTIONS
2.
3.
4.
5.
6.
Gm1m2
r2
2

11 N  m 
(5.98 1024 kg)
 6.67 10
2 
kg


 (7.35 10 22 kg)

(3.84 108 m) 2
 1.99 1020 N
Fg 
7.
kq1q2
r2
q1q2
Fe  2
r
(3)(3)

(2) 2
 2.25
Fe   4.5 10–3 N   2  25
Fe 
 1.0 10–2 N
kq1q2
r2
2

9 Nm 
6
 9.00  10
 (4.00 10 C)
C2 

 (3.00  106 C)

(2.00 102 m) 2
2
 2.70  10 N
Fe 
8.
kq1q2
r12
2

9 Nm 
6
6
 9.00  10
 (2.0 10 C)(2.0 10 C)
2
C 

(0.40 m) 2
 2.25 10 –1 N
kq q
F2  22 3
r2
2

9 Nm 
6
6
 9.00  10
 (2.0  10 C)(3.0  10 C)
C2 


(0.40 m) 2
 3.37  101 N
Fnet  F1  F2
 2.25 101 N  3.37  101 N
 1.110 –1 N
 1.110 –1 N left
F1 
kq1q2
r2
F
r2
q2  e
k
(3.40  102 N)(1.00 10 1 m) 2

N  m2
9.00  109
C2
14
2
 3.78  10 C
q  1.94  107 C
Fe 
kq1q2
r2
kq
q
r2  1 2
Fe
2

9 Nm 
6
9.00

10

 (2.0  10 C)
2
C


 (4.0  106 C)

(5.6  101 N) 2
–1
 1.28  10 m 2
r  3.6  10 –1 m
Fe 
Therefore the magnitude of the net electric force is
1.1 10 –1 N
9.
kq1q2
r2
1
Fe  2
r
1
 2
(3)
 0.111
Fe 
Fe   6.20 10–2 N   0.111
 6.89 10
Not for Reproduction
–3
kq1q2
r2
1
Fe 2
r
Distance changes by a factor of:
to
4.04 101 m

from 3.11101 m
 1.30
Distance increases by a factor of 1.30.
1
Fe 
(1.30) 2
 0.592
Fe   5.2 10–4 N   0.592 
Fe 
 3.110–4 N
N
21
Physics 12 SNAP
ANSWERS AND SOLUTIONS
10.
kq1q2
r2
12. Fe 
2

9 Nm 
6
6
 9.00  10
 (2.00  10 C)(3.00  10 C)
2
C 

(3.50  101 m) 2
–1
 4.40  10 N
F1 
kq1 q2
r12
Fnet  ma
F
a  net
m
4.40 101 N

2.00 105 kg
=2.20 104 m/s 2
2

9 Nm 
6
6
 9.00  10
 (3.0  10 C)(4.0  10 C)
2
C


(0.60 m) 2
 3.00  10 –1 N
kq q
F2  22 3
r2
2

9 Nm 
6
2
 9.00  10
 (4.0  10 C)
2
C 

(0.60 m) 2
 4.00  10 –1 N
q2
r12
q2
 2
r2
r2
 12
r2
13. FA on B 
FC on B
FC on B
FA on B
1.5 cm 

2
 4.5 cm 
2
 0.11
Lesson 7—Electric Fields
Fnet  ( F1 )2  ( F2 )2
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
 (3.00 101 N) 2  (4.00 101 N) 2
 5.0 10–1 N
11.
kq1q2
r2
2

9 Nm 
6
2
9.00

10

 (1.50  10 C)
2
C


(2.00  101 m) 2
 5.06 10 –1 N
F2 
CASTLE ROCK RESEARCH
22
kq1
r2
2

9 Nm 
6
 9.00  10
 (8.00 10 C)
C2 


(7.50  10 1 m) 2
 1.28  105 N/C
1.
E 
2.
g
GM
r2
2

11 N  m 
(6.37 1023 kg)
 6.67 10
2 
kg 


(3.43 106 m) 2
 3.61 N/kg
Copyright Protected
ANSWERS AND SOLUTIONS
3.
8.
kq1
r2
kq
r
|E|
E 
E1  E2 
2

9 Nm 
6
 9.00  10
 (4.60  10 C)
2
C

 
2.75  105 N/C
 3.88 10 –1 m
2

9 Nm 
6
9.00

10

 (4.50  10 C)
C2 


(2.50  10 1 m) 2
5
 6.48 10 N/C
Fg
4.
g
5.
Fe  q
Enet  E1  E2
 6.48  105 N/C  6.48 105 N/C
 1.296  106 N/C towards right
Enet  1.296  106 N/C
m
63.5 N

22.5 kg
 2.82 N/kg
9.
to
1.60 1019 C

from 3.20 1019 C
 0.500
Fe   0.500  0.250 N 
 0.125 N
6.
7.
kq
r2
kq1
r12
2

9 Nm 
6
 9.00  10
 (3.0  10 C)
2
C


(4.0  101 m) 2
 1.688  105 N/C
E1 
Fe
q
7.11103 N

5.20 106 C
 1.37 103 N/C
E 
kq2
r2 2
2

9 Nm 
6
 9.00  10
 (6.0  10 C)
2
C


(4.0  101 m) 2
 3.375 105 N/C
Enet  E1  E2
 1.688  105 N/C  3.375  105 N/C
 1.7  105 N/C towards left
Enet  1.7  105 N/C
E2 
Fe
q
  3.20 10 –19 C  7.60  104 N/C 
E 
 2.43  10 –14 N
Fe  ma
F
a e
m
2.43  1014 N

6.65  1027 kg
 3.65 1012 m/s 2
10.
Since both the point charges have equal value and
their distance are same on both side of the test
charge
E1  E2
 Enet  0
Not for Reproduction
23
Physics 12 SNAP
ANSWERS AND SOLUTIONS
11. Fe  ma
 9.1110–31 kg 7.50 1012 m/s 2


Lesson 8—Electric Potential due
to a Point

 6.83 10–18 N
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
F
E  e
q
6.83 1018 N

1.60 1019 C
 4.27 101 N/C
V
2.
V
3.
V1 
kq
r2
1
E 2
r
12. E 
Distance changes by a factor of:
To
4.50 101 m

From 3.00 101 m
 1.50
kq1
r
2

9 Nm 
6
 9.00  10
 (2.5  10 C)
2
C 


6.0  102 m
5
 3.8  10 V
1.
Distance changes by a factor of 1.50.
kq1
r
2

9 Nm 
6
 9.00  10
 (  2.5 10 C)
C2 


25 102 m
4
 –9.0 10 V
2
 1 
 E will change by a factor of 
  0.444
 1.50 
E    3.60 105 N/C   0.444 
 1.60 105 N/C
13.
kq
r2
1
E  2
r
1
2
r 
r
E
E 
kq1
r1
2

9 Nm 
6
 9.00 10
 (5.0 10 C)
C2 


45 102 m
4
 9.99 10 V
kq1
r2
2

9 Nm 
6
9.00

10

 (5.0 10 C)
C2 


15 102 m
 3.00 105 V
V2 
1
E
Field is changed by a factor of:
To
4.20 104 N/C

From 2.10 106 N/C
 2.00
V  V2 – V1
 3.00 105 V – 9.99 104 V
 2.00 105 V
W  E  qV
  0.030  10 –6 C  2.00  105 V 
E changes by a factor of 2.00
 r changes by a factor of
1
 0.707
2.00
 6.0  10 –3 J
r    7.50 101 m   0.707 
 5.30  01 m
CASTLE ROCK RESEARCH
24
Copyright Protected
ANSWERS AND SOLUTIONS
4.
VT  V1  V2  V3
 1.20 105 V   –1.54 105 V    –1.63 105 V 
kq1
r1
2

9 Nm 
18
9.00

10

 (6.4 10 C)
2
C


2.0 1011 m
 2.88  103 V
V1 
 –2.0 105 V
6.
kq1
r2
2

9 Nm 
18
 9.00 10
 (6.4 10 C)
2
C

=

0.50 m
 1.15  10–7 V
V2 
r2  r12  r32
 (8.0 cm) 2  (6.0 cm) 2
 1.0 101 cm
W  E  qV
 1.60  10 –19 C  2.88  103 J 
 4.60  10 –16 J
kq2
r2
2

9 Nm 
6
 9.00 10
 (  3.0 10 C)
2
C


1.0 101 m
 –2.70 105 V
V2 
E  E k
1
Ek  mv 2
2
2 Ek
v
m
2(4.60  10 16 J)

1.67  10 27 kg
=7.4  105 m/s
5.
kq3
r3
2

9 Nm 
6
 9.00 10
 (4.0 10 C)
C2 


6.0 102 m
5
 5.99 10 V
V3 
kq1
r1
2

9 Nm 
6
 9.00 10
 (1.0 10 C)
2
C


7.5  102 m
 1.20  105 V
V1 
VT  V1  V2  V3
 1.12 105 V   –2.70 105 V   5.99 105 V
 4.4 105 V
kq2
r2
2

9 Nm 
6
 9.00 10
 (  3.0 10 C)
C2 


17.5 102 m
5
 –1.54 10 V
7.
V2 
Ek  Ep  Ek   Ep 
0  Ep  Ek   Ep 
kq1q2 1
kq q
1
 m1 (v1 )2  m2 (v2 )2  1 2
r
2
2
r
Both masses m1 and m2 are alpha particles, so they
are equal in size. Therefore, the speeds v1 and v2
are also equal. Therefore, m1 (v1 ) 2 and m2 (v2 )2
are like terms.
kq
V3  3
r3
2

9 Nm 
6
9.00

10

 (  5.0 10 C)
2
C


27.5  102 m
 –1.63 105 V
Not for Reproduction
kq1
r1
2

9 Nm 
6
 9.00 10
 (1.0 10 C)
2
C


8.0  102 m
 1.2  105 V
V1 
m1  m2  m
 v1 
25
2
  v2     v 
2
2
Physics 12 SNAP
ANSWERS AND SOLUTIONS
kq1q2 1
kq q
1
 m(v) 2  m(v) 2  1 2
r
2
2
r
kq1q2
kq1q2
2
 m(v) 
r
r
kq1q2 kq1q2
2
m(v) 

r
r
kq1q2 kq1q2

r
r
v 
m
2.
3.
2

9 Nm 
19
2
 9.00 10
 (3.20 10 C)
2
C
kq1q2 


r
2.5 1012 m
 3.69  1016 N  m
2

9 Nm 
19
2
 9.00 10
 (3.20 10 C)
2
C 
kq1q2 

r
0.75 m
 1.23 1027 N  m
V
d
V Ed
E 
V   2.0  103 V/m  7.3  10 –2 m 
 1.5  102 V
4.
Inserting the values into equation 1:
3.69  1016 N  m  1.23  10 27 N  m
6.65  1027 kg
5
 2.4  10 m/s
v 
8.
V
d
V
d
|E|
2.0 102 V

5.0 103 V/m
 4.0 10 –2 m
E 
5.
W  E  qV
E
V
q
4.40  105 J

3.00  106 C
 14.7 V
Ek
q
1.50 1015 J

3.20 1014 C
 4.69 103 V
V
E k
q
Ek  qV
 1.60  10 –19 C  7.20  10 2 V 
V
 1.15  10 –16 J
6.
V
Ep(e)
q
Ee(p)  qV
  3.20  10 –19 C  7.50  103 V 
Lesson 9—Electric Potential in a
Uniform Electric Field
 2.40  10 –15 J
kinetic energy gained by the alpha particle
Ek  Ep(e)
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1 2
mv
2
2 Ek
v
m
2(2.40 1015 J)

6.65  1027 kg
 8.50 105 m/s
 Ek 
1.
E 
V
d
12.0 V
9.00 102 m
 1.33 102 V/m

7.
CASTLE ROCK RESEARCH
26
Because there is no change in the potential energy
of the proton, there is no work done.
Copyright Protected
ANSWERS AND SOLUTIONS
8.
 energy (kinetic) of the alpha particle when it hit
the plate is:
Ek  4.90 10–18 J  1.70 10–17 J
 2.19 10–17 J
1
2
Ek  m  v 
2
2 Ek
 v  
m
2(2.19  1017 J)

6.65  1027 kg
 8.11 104 m/s
V
d
7.5  101 V

6.0  102 m
 1.25  103 V/m
E 
Fe
q
Fe  q E
E 
 1.60  10 –14 C 1.25  103 V/m 
 2.00  10 –16 N
work  Fd
  2.00 10–16 N  3.0 10–2 m 
 6.0 10
9.
–18
11. Find initial energy:
1
Ek1  mv12
2
1
 (9.11 1031 kg)(5.0  105 m/s) 2
2
 1.14  10 –19 J
J
Ek
q
V  E
V
Find energy of electron at plate:
1
Ek2  mv2 2
2
1
 (9.11 1031 kg)(1.0  105 m/s) 2
2
 4.56  10 –21 J
Energy is changed by a factor of:
to
9.00 1017 J

from 3.00 1017 J
 3.00
The energy changes by a factor of 3.00. Therefore,
the potential difference changes by a factor of 3.00.
V    4.20 102 V   3.00 
E  1.14 10–19 J – 4.56 10–21 J
 1.09 10–19 J
 1.26 103 V
E
q
1.09 1019 J

1.60  1019 C
 6.8 10 –1 V
V
10. Find initial kinetic energy first:
1
Ek  mv 2
2
1
 (6.65  10 27 kg)(7.15 10 4 m/s) 2
2
 1.70  10 –17 J
12.
Now find the change in energy using following
steps:
V
E 
d
V Ed
 1.70  102 V/m  9.00  10 –2 m 
 1.53  101 V
E
V
q
 E  qV
  3.20  10 –19 C 1.53  101 V 
V
d
1
E 
d
To
5.0 cm

From 7.0 cm
 0.714
E 
Distance changes by a factor of 0.714.
 the electric field will change by a factor of
1
 1.40
0.714
E    9.3 102 V/m  1.40 
 4.90  10 –18 J
 1.3 103 V/m
Not for Reproduction
27
Physics 12 SNAP
ANSWERS AND SOLUTIONS
V
d
3.00  10 2 V

5.00  102 m
 6.00  103 V/m
Fe
q
5.30 1014 N

3.20 1019 C
 1.66 105 N/C
E 
13.
16. E 
V
d
V Ed
E 
1 2
mv
2
1
 (9.111031 kg)(6.00 106 m/s) 2
2
 1.64 10–17 J
Ek 
14.
 1.66 105 N/C  7.50 10 –2 m 
 1.24  104 V
E  1.64 10–17 J  0
 1.64 10–17 J
17. When a charged object is moved perpendicular to
an electric field, no work is done.
E
q
1.64  1017 J

1.60  1019 C
 1.02  102 V
V
18. a)
15. Uniform accelerated motion:
vi  0
t  2.50 10 –5 s
d  4.00 10 –2 m
b)
1
d  vi t  at 2
2
2d
a 2
t
2(4.00  10 2 m)

(2.50  10 5 s) 2
 1.28  108 m/s 2
Ek  Ep  0
1 2 1
mv  mv0 2  qV  0
2
2
1
(1.67  1027 kg)(v 2 )
2
(1.60 1019 C)(60.0 V)  0
 v  1.07 105 m/s
Ek  Ep  0
1 2 1
mv  mv0 2  qV  0
2
2
1
(9.111031 kg)(v 2 )
2
(1.60 1019 C)(60.0 V)  0
 v  4.59 106 m/s
1 2
mv
2
E  qV
1
qV  mv 2
2
v 2  V or V  v 2
19. Ek 
Fnet  Fe  ma
  6.65 10–27 kg 1.28 108 m/s 2 
 8.5110–19 N
Fe
q
8.51 1019 N

3.20  1019 C
 2.66 N/C
E 
speed changes by a factor of:
To
7.25 104 m/s

From 2.90 104 m/s
 2.50
 V changes by a factor of  2.50  6.25
2
V    6.25  2.50 105 V 
 1.56 106 V
CASTLE ROCK RESEARCH
28
Copyright Protected
ANSWERS AND SOLUTIONS
E  qV
 1.60  10 –19 C  5.00  103 V 
22. Draw a free-body diagram to identify the forces
involved.
 8.00  10 –16 J
20.
E  E k  0
Ek  E
1
Ek  mv 2
2
2 Ek
v
m
2(8.00  10 16 J)

9.11 10 31 kg
 4.19 107 m/s
E 
21. a)
Fg  mg
  0.500 kg   9.81 m/s 2 
 4.90 N
V
d
Considering equilibrium condition:
 Fy  0
FTy – Fg  0
FT sin  – Fg  0
FT sin 80.0 – 4.90 N  0
0.985FT  4.90 N
4.90 N
FT 
0.985
 4.975 N
 Fx  0
Fe – FTx  0
q E – FT cos   0
1 500 N/C  q  FT cos80.0
1 500 N/C  q  0.174FT
1 500 N/C  q   0.174  4.975 N 
q  5.77  10 –4 C
12.0 V
2.00 102 m
 6.00 102 V/m

Fe  q E
  3.20  0 –19 C  6.00  102 V/m 
 1.92 10 –16 N
b)
Fg  mg
  6.65 10 –27 kg   9.81 N/kg 
 6.52 10 –26 N
c)
Fnet  Fe – Fg
 1.92  10 –16 N – 6.52 10 –26 N
 1.92  10 –16 N
d)
Fnet  ma
F
a  net
m
1.92 1016 N

6.65 1027 kg
 2.89 1010 m/s 2
e)
23. Find how long the electron will be in the electric
field:
d
v
t
d
t
v
14.0  102 m

8.70  106 m/s
 1.61 10 –8 s
Fe  Fg
Fe  6.52 1026 N
Fe
q
6.52 1026 N

3.20 1019 C
 2.04 107 N/C
E 
Find the electric force on the electron:
Fe  q E
 1.60 10–19 C 1.32 103 N/C 
 2.1110–16 N
V E d
  2.04 107 N/C  2.00 102 m 
 4.07 109 V
Not for Reproduction
29
Physics 12 SNAP
ANSWERS AND SOLUTIONS
Fnet  ma
F
a  net
m
2.111016 N

9.111031 kg
 2.32 10 –14 m/s 2
3.
Now find the vertical component of displacement
of the electron:
vi  0
t  1.6110 –8 s
a  2.32 10 –14 m/s 2
4.
Lesson 11—Magnetic Forces on
Current-Carrying Conductors
Lesson 10—Magnetic Forces and
Fields
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
2.
B  0 In
 800 
 (4  10 7 N/A 2 )(2.0 A) 

 0.30 m 
 6.7  10 –3 T
Magnitude of displacement
1
d  vi t  at 2
2
1
 (2.32 1014 m/s 2 )(1.61108 s) 2
2
 3.00 10–2 m
1.
775 turns
n

30.0 cm 100 cm
n  2 583
B  0 In
B
I
0 n
0.100 T


7 T  m 
 4 10
 (2583)
A 

 31 A
1.
Fm  qvB
F
B  m
qv
9.50 1014 N
(1.60 1019 C)(2.10  105 m/s)
 2.83 T
B  0 In

7 T  m 
 4  10
 (1.25 A)(1800)
A 

25.0  102 m
2
 1.13  10 T

2.
B  0 In
B
n
0 I
Fm  B Il
  7.50 10–4 T   0.960 A  0.222 m 
 1.60 10–4 N
Fm  1.60 10–4 N east
2.1 103 T

7 T  m 
 4  10
 (0.72 A)
A 

 2.32 103 / m
3
2.32  10
x

100 cm
25 cm
x  580

3.
Fm  qvB
 1.60  10 –19 C  3.52  105 m/s 
  2.80  10–1 T 
 1.58  10 –14 N
Fm  1.58  10 –14 N west
4.
Fm  qvB
  3.20 10–19 C  7.50 104 m/s   5.50 T 
 1.30 10–13 N
Fm  1.30 10–13 N west
CASTLE ROCK RESEARCH
30
Copyright Protected
ANSWERS AND SOLUTIONS
5.
Fm  qvB
F
B  m
qv
10.
Fm  qvB
F
v m
qB
1.70  1014 N
(1.60  1019 C)(1.40 104 m/s)
 5.59 T
B  5.59 T up

6.
4.1 1013 N
(1.60  1019 C)(7.20  101 T)
 3.56  106 m/s
1
Ek  mv 2
2
1
 (9.11 10 31 kg)(3.56  106 m/s) 2
2
 5.77  1018 J
E  qV
E
V
q
5.77  1018 J

1.60  1019 C
 3.6  101 V

Fm  qvB
F
B  m
qv
7.11014 N
(1.60 1019 C)(2.7  105 m/s)
 1.6 T
B  1.6 T west

7.
The alpha particle is moving parallel to the
magnetic field.
11. Fm  qvB
 1.60 10 –19 C 6.20 105 m/s

 there is no force.
Fm  qvBsin
0
8.
E  qV
 1.60  10 –19 C 1.70  103 V 
 2.72  10 –16 J
12. Fm  Fg
1 2
mv
2
2 Ek
v
m
2(2.72  1016 )

9.11 1031 kg
 2.44  107 m/s
Fm  qvB
 1.60  10 –19 C  2.44  107 m/s 
Ek 
Fg  mg
 1.76 102 kg   9.81 N/kg 
 1.72 101 N
 Fm  1.72 101 N
Fm  B Il
F
B  m
Il
  2.50 10 –1 T 
 9.77 10
Not for Reproduction

T
 2.28 10 N
F  ma
F
a
m
F
a m
m
2.28  1014 N

9.111031 kg
 2.50 1016 m/s 2
Fm  B Il
  5.00 10–1 T   3.60 A   2.50 10–1 m 
–13
  2.30  10
–1
14
 4.50 10–1 N
9.

1.72 101 N
(6.00 A)(1.90 102 m)
 1.51 T

N
31
Physics 12 SNAP
ANSWERS AND SOLUTIONS
6.
Lesson 12—Cathode Rays
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
Fm  Fc
mv 2
r
qB r
v
m
(3.20 1019 C)(4.22 101 T)(1.50 103 m)

6.65 1027 kg
4
 3.05 10 m/s
qvB 
2.
Fm  Fc
mv 2
qvB 
r
qB r v 2

m
v
qB r
v
m
3.20
 1019 C  3.60 10 1 T 8.20 10 2 m 


6.65  1027 kg
 1.42  106 m/s
1 2
mv
2
2
1
  6.65 1027 kg 1.42 106 m/s 
2
 6.7110–15 J
Ek 
Fm  Fc
mv 2
qvB 
r
mv
B 
qr
(1.67  1027 kg)(5.40  105 m/s)

(1.60  1019 C)(7.20  103 m)
 7.83  101 T
7.
E
q
E  qV
 1.60  1019 C  2.50  103 V 
V
 4.00  1016 J
1 2
mv
2
2 Ek
v
m
2(4.00  1016 J)

9.11 1031 kg
 2.96  107 m/s
Ek 
3.
Fm  Fc
mv 2
qvB 
r
q
v

m B r
(3.60  105 m/s)
(6.10  101 T)(7.40  10 2 m)
 7.98  106 C/kg

4.
8.
Fe  Fm
q E  qvB
E  vB
  7.80 105 m/s  2.20 101 T 
 1.72 105 N/C
5.
Fe  Fm
q E  qvB
9.
|E|
B
2.10  105 N/C

6.50  101 T
 3.23  105 m/s
v
CASTLE ROCK RESEARCH
32
1 2
mv
2
1
 (9.111031 kg)(4.75 107 m/s) 2
2
 1.03 10–15 J
E
V
q
1.03 1015 J

1.60 1019 C
 6.42 103 V
Ek 
E
q
E  qV
 (1.00 1014 C)(1.40 103 V)
 2.24 1016 J
V
Copyright Protected
ANSWERS AND SOLUTIONS
12. When charged particles travel parallel to a
magnetic field there is no force. If there is no
force, there is no deflection.
1 2
mv
2
2 Ek
v
m
2(2.24  1016 J )

9.11 1031 kg
 2.23  107 m/s
Ek 
13. Fm  Fc
mv 2
r
qB r
v
m
(1.60 1019 C)(0.75 T)(0.30 m)
v
1.67 1027 kg
7
 2.16 10 m/s
qvB 
Fm  Fc
qvB 
mv 2
r
mv
qB
(9.11 1031 kg)(2.23 107 m/s)

(1.60  1019 C)(2.20  102 T)
 5.74  103 m
r
14. Fm  Fc
mv 2
r
qB r
v
m
(1.60 1019 C)(2.2 104 T)(0.77 m)

9.11 1031 kg
7
 2.98 10 m/s
1
E  mv 2
2
1
 (9.11 10 31 kg)(2.98  107 m/s) 2
2
 4.03  10 –16 J
E
V
q
4.03 1016 J

1.60 1019 C
 2.5 103 V
qvB 
10. Fe  Fm
q E  qvB
|E|
B
4.20  105 N/C

1.40  102 T
 3.00 107 m/s
1
Ek  mv 2
2
1
 (9.11 10 31 kg)(3.00  107 m/s) 2
2
 4.10  10 16 J
E
V
q
4.10 1016 J

1.6 1014 C
 2.56 103 V
v
15. Fm  Fc
mv 2
r
qB r
v
m
 (1.0  104 C/kg)(9.10  101 T)  0.240 m 
 2.4  103 m/s
d  C  2 r
 2  0.240 m 
 1.51 m
d
v
t
d
t
v
1.51 m

2.40  103 m/s
 6.28  104 s
qvB 
11. Fm  Fc
qvB 
mv 2
r
mv
rB
(8.4 1027 kg)(5.6 105 m/s)

(3.5 102 m)(2.8 101 T)
 4.8 10 –19 C
4.8 1019 C
# e 
1.6 1019 C
 3
The particle carries an extra 3 electrons.
q
Not for Reproduction
33
Physics 12 SNAP
ANSWERS AND SOLUTIONS
Lesson 13—Electromagnetic
Induction
  N
5.
  N
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
area   r 2
  (5.0 10–2 m) 2
 7.85 103 m 2
  BAcos
  3.2  103 T  7.85  103 m 2  cos 0

 2.5  105 Wb
2.
3.
area   r 2
  (6.0 10 –2 m) 2
 1.13 10 –2 m 2
 f  BAcos
 (3.6 104 T)(1.13 102 m 2 )  cos 90 
0
 0  BAcos
 (3.6  104 T)(1.13 102 m 2 )  cos 0 
 4.07 106 Wb
   f   0
 0 – 4.07 106 Wb
 – 4.07  106 Wb

  N
t
 4.07  106 Wb 
 (200) 

0.015 s


 5.4  10 –2 V

t
 –50 15 Wb/s 
 –7.5  102 V
4.

t
 (t )
  
N
(0.92 V)(0.12 s)

100
 –1.10  103 Wb
  BAcos

B
A cos 
1.10 103 Wb

4.0 10-3 m 2
 2.8  101 T
6.
area   r 2
  (12.5 102 m) 2
 4.91102 m 2
 0  BAcos
 (2.7  103 T)(4.91 102 m 2 )  cos 0 
 1.33  104 Wb
 f  BAcos
 (2.7 103 T)(4.9110 2 m 2 )  cos 180 
 –1.33  104 Wb
   f   0
 –1.33 104 Wb –1.33  104 Wb
 –2.66 10–4 Wb

  N
t
 2.66  10 4 Wb 
 (10) 

0.30 s


 8.8  103 V
 f  BAcos
 (2.2 10–2 T)  0  cos 90 
0
 0  BAcos
 (2.2  10 –2 T)(2.5  10 –3 m 2 )  cos 0 
 5.5  10 –5 Wb
   f   0
 0 – 5.5 10 –5 Wb
 –5.5 10 –5 Wb

  N
t
 5.5  105 Wb 
 

0.10 s


–4
 5.5 10 V
7.
 0  BAcos
  0.40 T  (5.0 103 m 2 )  cos 0 
 2.0 103 Wb
 f  BAcos
  0  (5.0 103 m 2 )  cos 0 
0
   f   0
 0 – 2.0 10 –3 Wb
 –2.0  103 Wb
Direction: clockwise
CASTLE ROCK RESEARCH
34
Copyright Protected
ANSWERS AND SOLUTIONS

t
 2.0  103 Wb 
 (25) 

1.0 s


 5.0  102 V
  N
8.
a)
  IR

I
R
1.6  103 V

2.0 
 8.0  104 A
 0  BAcos
 1.6 T  (7.2 103 m 2 )  cos 0 
 1.15 102 Wb
b) counter-clockwise
Lesson 14—Electric Generator
 f  BAcos
  0  (7.2 103 m 2 )  cos 0 
0
   f   0
 0 –1.15  102 Wb
 –1.15  102 Wb

  N
t
1.15 102 Wb 


0.050 s
 2.3 101 V
b)

PRACTICE QUESTIONS—
ANSWERS AND SOLUTIONS
1.
Vback   – Ir
 120 V – 12.0 A  (6.0 )
 48 V
2.
a) initial
Vback   – Ir
  Vback
r
I
120 V  0
r
40.0 A
120 V

40.0 A
 3.00 
 IR
I

R
2.3 101 V

12.0 
 1.9 10 –2 A
9.
c)
clockwise
a)
area  l 2
 (4.0 102 m) 2
 1.6 103 m 2
 0  BAcos
  0.20 T  (1.6  103 m 2 )  cos 0 
 3.2  104 Wb
 f  BAcos
  0.50 T  (1.6  103 m 2 )  cos 0 
 8.0  104 Wb
   f   0
 8.0  104 Wb – 3.2  104 Wb
 4.8 104 Wb

  N
t
4.8  104 Wb

0.30 s
 –1.6  103 V
Not for Reproduction
b) operating full
Vback   – Ir
 120 V – 15.0 A  (3.00 )
 75.0 V
3.
a)
Vback   – Ir
 120 V –  9.0 A  (5.0 )
 75 V
b) no back emf
35
Physics 12 SNAP
ANSWERS AND SOLUTIONS
c)
initial
Vback  0
Lesson 15—EMF Continued
Vback   – Ir
0  120 V – I (5.0 )
120 V
I
5.0 
 24 A
4.
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
  B lv
2.
  Blv
b) Back emf has decreased
c)
B 

3.

, which tells us that the
t
induced emf varies directly with the rate of change
in the magnetic flux. The back emf is the emf
induced by the motor. Therefore, if the rate of
rotation is halved, the back emf is halved.
  0.150 T  0.220 m 1.25 m/s 
 0.0413 V
  IR

R
0.0413 V

2.25 
 1.83  102 A
counter-clockwise
4.
a) initial Vback  0
Vback   – Ir
  Vback
r
I
120 V  0
r
10.0 A
120 V

10.0 A
 12.0 
a)
  IR
  5.6  10 –2 A  1.5  
 8.4  10 –2 V
Again
  B lv
B 

lv
8.4  10 2 V
(15  10 2 m)(0.95 m/s)
 0.59 T

b) operating full speed
Vback   – Ir
 120 V –  3.0 A  (12.0 )
 84 V
CASTLE ROCK RESEARCH
  B lv
I
45.0 V
We can use    N
6.

lv
0.075 V
(0.28 m)(0.80 m/s)
 0.33 T
i) Vback   – Ir
 120 V –  3.6 A  (6.0 )
 98 V
ii) Vback   – Ir
 120 V – 8.4 A  (6.0 )
 70 V
5.
  0.75 T  0.35 m 1.5 m/s 
 0.39 V
a) Greater current—more heat
(Remember: P  I 2 r )
b) counter-clockwise
36
Copyright Protected
ANSWERS AND SOLUTIONS
5.
  B lv
W
t
W  Pt
  0.181 W  1.0 102 s 
P
  0.950 T  0.300 m 1.50 m/s 
 0.428 V
  IR
I

 1.8 103 J
R
0.428 V

3.25 
 0.132 A
Fm  BIl
  0.950T  0.132 A  0.300 m 
 3.75 10 –2 N
6.
  B lv
  4.70 10
 2.79 10
7.
–6
3
10.   IR
  5.2 10 –2 A   2.9  
 0.151 V
  Blv
v
0.151 V
(0.65 T)(0.50 m)
 0.46 m/s
T   6.25 m  95.0 m/s 
V
11. a)
  0.600 T  0.300 m  3.00 m/s 
 0.540 V
2

b) clockwise
12. a)
  0.75 T 1.2 m  2.5 m/s 
 2.25 V
  IR
  ( B lv) N
 1.1 T  0.18 m  2.7 m/s 5 
 2.67 V
  IR
I

R
2.67 V

3.5 
 0.76 A

I
2.25 V

0.45 A
 5.0 
9.
 1.30 T  0.625 m 1.80 m/s 
 1.463 V
  IR
R
1.463 V

1.50 
 0.975 A
  B lv
R
  B lv
I
R
(0.540 V) 2

2.25 
 0.130 W
W
P
t
W  Pt
  0.130W 15.0 s 
 1.94 J
8.
B l

  B lv
P

b) clockwise
  B lv
  0.95 T 1.0 m  3.0 m/s 
 2.85 V
P
2
R
(2.85 V) 2
45.0 
 0.181 W
Not for Reproduction
37
Physics 12 SNAP
ANSWERS AND SOLUTIONS
I s Vp

I p Vs
0.267 A 1.20  102 V

Ip
20 V
I p  4.4  10 –2 A
Lesson 16—Transformers
(Induction Coil)
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
2.
3.
Is Np

I p Ns
0.95 A 1.0 103

0.25 A
Ns
Ns  2.6 102
a)
Pp  I pVp
Pp
Ip 
Vp
5.0  101 W

1.20  10 2 V
 0.42 A
7.
Pp  I pVp
 1.0 A  1.20 102 V 
The power in the secondary coil is also 1.2  10 2 W
(same in both cells)
8.
Pp  I pVp
Pp
Ip 
Vp
1.0 102 W

9.0 V
 11.1 A
Is Np

I p Ns
Np
0.65 A

11.1 A 1.5  103
N p  88
Np
Vs N s
1.20  102 V 1.50 102

Vs
25
Vs  20 V
Vs  I s Rs
V
Is  s
Rs
20 V

75 
 0.27 A
CASTLE ROCK RESEARCH
6.
 1.2  102 W
Ps  I sVs
P
Is  s
Vs
2.4  103 W

1.20  102 V
 2.0  101 A

  3.0A  1.20  102 V 
Power delivered to the secondary coil is also
3.6  10 2 W
(same in both cells)
Pp  I pVp
Pp
Ip 
Vp
2.4  103 W

3.6  103 V
 0.67 A
Vp
Pp  I pVp
 3.6  102 W
Is Np

I p Ns
5.00 A 1.20  103

Ip
1.50  02
I p  0.625 A
b)
4.
5.
38
Copyright Protected
ANSWERS AND SOLUTIONS
5.
Practice Test
ANSWERS AND SOLUTIONS
1.
2.
Gm1m2
1
or Fg  2 (r is doubled, because
2
r
r
the final distance from Earth’s centre is 2r)
1
 Fg 
4
2.0 N
Fg 
4
 0.50 N
Fg 
kq1q2
r2
2

9 Nm 
6
 9.00  10
 (2.00  10 C)
2
C 

(2.00 10 6 C)

(2.00  102 m) 2
 90.0 N
Fe 
GM

r2
1
g  2
r

This is an inverse square relationship. An inverse
square relationship is graphed as:
g
The force between the pith balls is 90.0 N.
6.
F2  F1 
3.
The equation Ep  mgh cannot be used in the case
of an object separated from Earth’s surface by a
large distance because the magnitude of g is not
constant—g decreases as the distance from the
earth increases.
4.
Ep  
kq1q2
r2
2

9 Nm 
6
2
9.00

10

 (3.00 10 C)
2
C


(0.333 m) 2
 0.730 N
Gm1m2
r
2

11 N  m 
(5.98  1024 kg)(1.25  103 kg)
 6.67  10
2 
kg


4.00  106 m + 6.38  106 m
 –4.80  1010 J
Fnet  F12  F2 2
 (0.730 N) 2  (0.730 N) 2
 1.03 N
The net electric force on sphere B is 1.03 N.
Not for Reproduction
39
Physics 12 SNAP
ANSWERS AND SOLUTIONS
7.
Fnet
m
2.02 102 N

5.00  104 kg
 4.05 105 m/s 2
kq1q2
1
or Fe  2
r2
r
1
r decreases by
3
2
r

changes by 0.111
Fe changes by 9 times
New force = 9.0F
a
Fe 
The initial acceleration of sphere A
was 4.05  105 m/s 2 .
The new force between the two point charges
is 9.0F.
8.
10.
Fnet  ma
a  Fnet
In this case the net force is the electric force.
Fnet  Fe 
kq1q2
r2
1
r2
This expression tells that when the distance
1
increases 3 times, the force becomes .
9
1
 if the force decreases to , then according to
9
Fnet 
the expression, a  Fnet
The negatively charged rod will induce the
electrons in the electroscope to move away from
the rod, i.e., move from the head to the leaves.
Therefore, the head of the electroscope will be
positive.
11.
1
the acceleration is also .
9
New acceleration is
1.00 1010 m/s2  19
 1.11109 m/s 2
1
2
(6.65  10
27
kg)v  (3.20  10
2
v 
2
The initial acceleration of sphere A
was 1.11  109 m/s 2 .
9.
Ek  Ep  0
1 2 1
mv  mv0 2  qV  0
2
2
1 2 1
mv  m(0) 2  qV
2
2
19
2(3.20  10
C)(  2.00  10 V)
19
3
C)(  2.00  10 V)
3
(6.65  10
27
kg)
v  4.39 10 m/s
5
The maximum speed of the alpha particle is
4.39  105 m/s .
Fnet  ma
a  Fnet
12.
In this case the net force is the electric force.
kq q
Fnet  Fe  12 2
r
2

9 Nm 
6
2
 9.00  10
 (3.00 10 C)
2
C


(2.00  102 m) 2
 2.03  102 N
The electron will accelerate in the direction of the
electric force which as away from the negative
plate (opposite direction to the electric field).
13. Electric potential is the potential electric energy per
unit of charge.
CASTLE ROCK RESEARCH
40
Copyright Protected
ANSWERS AND SOLUTIONS
kqq
r2
According to this equation, as the distance between
the charges decreases, the electric force increases.
If the electric force increases, the electric potential
energy increases.
14. Fe 
19.
Np
I
 s
Ns I p
I
1
 s
10 I p
I p  10 I s
This shows that the current in the primary coil is 10
times greater than the current in the secondary coil.
15. The direction of an electric field is defined in terms
of the direction of the electric force on a positive
test charge.
20. A “step-up” transformer is used to increase (hence
step up) the voltage. But, if the transformer
increases the voltage, the current must decrease.
Remember the law of conservation of energy. The
power in each coil must be the same. Therefore, a
step-up transformer is used to increase voltage and
decrease current.
16. The electric field is away from a positive object
toward a negative object. Therefore, the sketch
should show that the field is directed down toward
the negative plate.
21. A transformer makes use of the principle of
electromagnetic induction. This means that there
must be a change in the magnetic field inside the
secondary coil. In order to change this magnetic
field, a switch is placed in the primary circuit.
kq1
r
1
V
r
17. V 
22. Lenz’s law is an example of conservation of
energy.
The magnitude of the electric potential (V) as a
function of the distance (r) from a point charge is
an inverse relationship. This can be shown as a
curve approaching the x-axis as it moves away
from the y-axis.
23. Use the right-hand rule. The opposing magnetic
force must be in the opposite direction to its motion
(i.e., to the left). Place your right palm to face the
left along the conductor, and direct your fingers
into the page (in the direction of the magnetic
field). Your thumb will now point along the
conductor from end B toward end A. This is the
direction of the induced electric current.
  Blv
and
I

R
From these mathematical relationships, if the
velocity decreases, the current also decreases.
Therefore, the current within rod AB will be
directed toward end A, and it will decrease as the
rod passes through the magnetic field.
18. Lenz’s law states that the induced field will oppose
the original change of flux of the magnet. In other
words, the induced magnetic field inside the
solenoid must oppose the motion of the magnet.
That means that the right side of the solenoid is the
induced magnetic north.
Use the right-hand rule to determine the direction
of the current within the circuit. Point your thumb
of your right hand to the right and curl your fingers
in the direction of the electric current. For this to
happen, the current flows from A to B through G.
Not for Reproduction
24. Note that the alpha particle is moving parallel to
the magnetic field. Therefore, there is no magnetic
force acting upon the alpha particle.
Fm  qvBsin
41
Physics 12 SNAP
ANSWERS AND SOLUTIONS
33. Apply Lenz’s law here. Use the right-hand rule for
conventional current.
In this scenario, as the north pole of the magnet
moves down into the loop, the magnetic field
within the loop will point upward out of the loop.
Point your right thumb in this direction and curl
your fingers to identify the direction of the current.
Your fingers should be pointing in a
counterclockwise direction.
However, when the magnet is pulled out of the
loop, the motion of the magnet, the direction of the
magnetic field, and the direction of the current all
change direction. Therefore, the direction of the
current as the magnet is pulled out of the loop is in
the clockwise direction.
Therefore, as viewed from above the loop, the
direction of the induced current in the loop would
first be counterclockwise and then clockwise.
25. Use the right-hand rule. Point your thumb east.
Point your fingers north. Your palm should be
directed up. The magnetic force on the proton is
directed upward.
26. Lenz’s law states that the induced current flows in
such a way that the magnetic field inside the coil
opposes the magnetic field of the magnet. Using
the right-hand rule, curl the fingers on your right
hand in a counterclockwise direction. Your right
thumb will thus point upward. The magnetic field
inside of the loop is directed upward.
27. Use the left-hand rule (electrons). Point your
thumb north. When your left-hand fingers curl
above your thumb, they should be pointing west.
Therefore, the magnetic field above the conductor
will be pointed west.
34. J.J. Thomson determined the charge-to-mass ratio
of the electron by balancing the electric and
magnetic forces.
28.   Blv
If the period (time) of the rotation decreases, the
loop will be rotating at a higher speed. From the
above equation, as the speed increases, so does the
emf. Therefore, if the period of rotation was cut in
half, the induced voltage would increase as well.
35. A cathode ray is a beam of electrons.
36. Statements i) and ii) are true. Cathode rays are
electrons, and charged particles can be deflected by
electric and magnetic fields. Therefore, cathode
rays can be deflected by both electric and magnetic
fields.
29. Generators convert mechanical energy into electric
energy.
30. Motors convert electric energy into mechanical
energy.
Statement iii) is incorrect. Only electromagnetic
radiation travels at 3.00  108 m/s .
Electromagnetic radiation does not consist of
charged particles as cathode rays do.
31. Both transformers and generators operate on the
principle of electromagnetic induction.
32.
37. The greater the deflection, the smaller the radius.
Therefore, in order to increase the deflection,
determine which variables in the following
equation will result in a smaller radius.
Fm  Fc
mv 2
qvB 
R
mv
R
qB
The radius depends on the mass, speed, charge, and
magnetic field strength. In order to decrease the
radius, increase the charge of the particle, increase
the strength of the magnetic field, or decrease
either the mass or speed of the particle.
Fm  qvB
Fm  v
and Fm does not depend on the mass.
 if the speed is halved, you will halve the
magnetic force.
 Fm =
F
2
CASTLE ROCK RESEARCH
42
Copyright Protected
ANSWERS AND SOLUTIONS
38. If electrons pass through superimposed electric and
magnetic fields without deflection, that means the
electric and magnetic forces acting on the electrons
must be equal but in opposite directions.
Fe  Fm
q | E | qvB
|E|
v
B
Therefore, the speed of the electrons can be
determined from the strengths of the superimposed
fields.
5.
MOMENTUM
6.
p  mv
  0.673 kg  3.0 m/s north 
 2.0 kg  m/s north
Lesson 1—Momentum
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
1.
2.
3.
p  mv
  4.0 kg 12.0 m/s east 
 48 kg  m/s east
7.
p  mv
p
v
m
25.0 kg  m/s west

5.0 kg
 5.0 m/s west
v  v0
t
v 0
2
–9.81 m/s 
0.25 s
v  –2.45 m/s
a
p  mv
  5.0 kg  –2.45 m/s 
 –12 kg  m/s
=12 kg  m/s down
p  mv
p
m
v
36.0 kg  m/s south

8.0 m/s south
 4.5 kg
8.
Impulse  Fnet t
 17.0 N east   2.5  102 s 
 4.3  101 N  s east
p  mv
p
m
v
29 kg  m/s east

2.0 m/s east
 14.5 kg
Impulse  Fnet t
Impulse
t
Fnet
7.00 N  s west

11.2 N west
 0.625 s
10. Consider north direction as positive
Fnet t  mv
Fnet  2.60 s    26.3 kg  –21.0 m/s 
Fnet  –212 N or 212 N south
Fg  mg
 14.5 kg   9.81 m/s 2 
 1.4 102 N
Not for Reproduction
d
t
2.6 m west

1.1 s
 2.36 m/s west
v
p  mv
  7.0 kg  2.36 m/s west 
 17 kg  m/s west
9.
4.
Fg  mg
Fg
m
g
6.6 N

9.81 m/s 2
 0.673 kg
43
Physics 12 SNAP
ANSWERS AND SOLUTIONS
11. Magnitude of impulse
Fnet t  mv
 31.6 N  t  15.0 kg 10.0 m/s 
t  4.75 s
17. Consider up as positive,
Fnet  ma
F
a  net
m
1.5  105 N

9.5  103 kg
 15.8 m/s 2
v  v0
a
t
v 0
2
15.8 m/s 
15 s
v  2.4  102 m/s
12. Fnet t  mv
Therefore, mv   25.0 N   7.20 101 s 
=18.0 N  s north
13. Impulse  mv
  5.00 kg 15.0 m/s east 
 75.0 kg  m/s east
14.
or
Fnet t  mv
1.5 10 N  15 s    9.5 103 kg  v
5
d
t
26.3 m west

3.2 s
 8.22 m/s west
v  v0
vav 
2
v 0
8.22 m/s west 
2
v  16.4 m/s west
p  mv
 11.0 kg 16.4 m/s west 
 181 kg  m/s west
vav 
v  2.4 102 m/s
18.
19. The magnitude of impulse
Fnet t  mv
 225 N  t  1.0 103 kg 5.0 m/s  2.0 m/s 
t  13 s

15. The magnitude of momentum
p  mv
p
v
m
6.0 kg  m/s

3.0 kg
 2.0 m / s
v 2  v0 2  2ad
20.
16.
Fnet t  mv
 95 N north 1.65 s   15 kg  v 
Lesson 2—Collisions
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
2
1.
p1  mv1
 1.0 kg  –2.0 m/s 
 –2.0 kg  m/s
p2  mv2
 1.0 kg 1.6 m/s 
 1.6 m/s
p  p2 – p1
 1.6 kg  m/s   2.0 kg  m/s 
 3.6 kg  m/s upward
CASTLE ROCK RESEARCH

v  1.0 101 m/s north
 2  9.81 m/s 2  d
d  0.20 m
The object has fallen 0.20 m when its momentum
is 6.0 kg m/s down.
 2.0 m/s 
Fnet t  mv
Fnet  0.75 s    5.4 kg  3.0 m / s east 
Fnet  22 N east
Consider right as positive and left as negative
m1  30.0 kg
m2  20.0 kg
v1  1.00 m/s
v2  –5.00 m/s
p1  30.0 kg  m/s
p2  –100 kg  m/s
 psys(before)  p1  p2  –70.0 kg  m/s
44
Copyright Protected
ANSWERS AND SOLUTIONS
3.
m1  30.0 kg
v1  ?
p1  ?
m1  925 kg
v1  18.0 m/s
p1  1.67 104 kg  m/s
m2  20.0 kg
v2  –1.25 m/s
p2  25.0 kg  m/s
m2  ?
v2  0 m/s
p2  0 kg  m/s
 pbefore  p1  p2  1.67 104 kg  m/s
psys(after)  psys(before)  70.0 kg  m/s
after collision
psys(after)  p1  p2  70.0 kg  m/s
 p1  70.0 kg  m/s  (25.0 kg  m/s)
 p1  45.0 kg  m/s
m?
v  6.50 m/s
p  pafter
=1.67 104 kg  m/s
p1  m1v1
p
v1  1
m1
45.0 kg  m/s

30.0 kg
 1.50 m/s
p
v
1.67  104 kg  m/s

6.50 kg
 2.57  103 kg
m2  2.57  103 kg  925 kg
 1.65  103 kg
m
The velocity of the 30.0 kg ball after the collision
is 1.50 m/s left
2.
4.
Consider east as positive
m1  4.50 103 kg
v1  5.0 m/s
p1  2.25  104 kg  m/s
m2  6.50  103 kg
v2  0 m/s
p2  0 kg  m/s
m1  0.050 kg
v1  ?
p1  ?
 pbefore  p1  p2  2.25 104 kg  m/s
m  m1  m2  7.05 kg
v  5.00 m/s
p  pafter  35.25 kg  m/s
pbefore  pafter  35.25 kg  m/s
p1  pbefore  35.25 kg  m/s
p
v1  1
m1
35.25 kg  m/s

0.050 kg
 705 m/s
m  m1  m2  1.10 104 kg
v ?
p  pafter =2.25 104 kg  m/s
p  mv
p
v
m
2.25  104 kg  m/s

1.10  104 kg
 2.1 m/s east
Not for Reproduction
m2  7.00 kg
v2  0 m/s
p2  0 kg  m/s
45
Physics 12 SNAP
ANSWERS AND SOLUTIONS
pafter  po +ps =0
ps  4.4 kg  m/s
5.
ps
m
4.4 kg  m/s

76 kg
 0.058 m/s right
vs 
Consider right as positive and left as negative
m1  40.0 g
v1  9.00 m/s
p1  3.60  10 2 g  m/s
7.
m2  55.0 g
v2  6.00 m/s
p2  3.30 102 g  m/s
m  1.13 103 kg
v0
p0
 pbefore  p1  p2  3.00 101 g  m/s
pbefore  0
m  m1  m2  95.0 kg
v ?
p  pafter  pbefore =3.00 101 g  m/s
Consider east as positive
mP  25 kg
mL  1.1 103 kg
vP  325 m/s
vL  ?
pL  ?
pP  8.13  103 kg  m/s
p
m
3.00 101 g  m/s

95.0 g
 0.316 m / s right
v
pafter  pbefore  0
pafter  pL +pP  0
pL  8.13  103 kg  m/s
6.
pL
mL
8.13  103 kg  m/s

1.1 103 kg
 –7.4 m/s
 7.4 m/s west
vL 
m  76 kg
v0
p0
pbefore  0
8.
m?
v0
p0
Consider left as negative
mo  0.20 kg
ms  76 kg
vo  22 m/s
vs  ?
po  4.4 kg  m/s
ps  ?
pbefore  0
pafter  pbefore =0
CASTLE ROCK RESEARCH
46
Copyright Protected
ANSWERS AND SOLUTIONS
mC  1.02 103 kg
vC  25 m/s
pC  2.55  104 kg  m/s
Consider right as positive
mG  4.5 102 kg
mV  ?
vG  1.4 103 m/s
vV  45 m/s
pV  ?
pG  6.3 105 kg  m/s
 pbefore  pT  pC  1.28 105 kg  m/s
pafter  pbefore =0
pafter  pV +pG  0
pV  6.3  105 kg  m/s
pV
vV
6.3  105 kg  m/s

45 m/s
 1.4  104 kg
mV 
m  1.12  104 kg
v ?
p  pafter  1.28  105 kg  m/s
p
m
1.28  105 kg  m/s

1.12  104 kg
 11 m/s north
v
9.
m  7.0 kg
v0
p0
pbefore  0
11. a)
Consider right as positive
m1  225 g
v1  30.0 cm/s
p1  6.75  103 g  cm/s
Consider right as positive
mB  5.0 kg
mA  2.0 kg
vB  ?
vA  10.0 m/s
pB  ?
pA  20.0 kg  m/s
m2  125 g
v2  10.0 cm/s
p2  1.25  103 g  cm/s
pafter  pbefore  0
pafter  pA +pB  0
pB  20.0 kg  m/s
 pbefore  p1  p2  8.00 103 g  cm/s
pB
mB
20.0 kg  m/s

5.0 kg
 4.0 m/s left
vB 
m1  225 g
v1  ?
p1  ?
m2  125 g
v2  24 cm/s
p2  3.00 103 g  cm/s
psys(after)  psys(before)  8.00 103 g  cm/s
10.
p1  5.00 103 g  cm/s
p
v1  1
m
5.00 103 g  cm/s

225 g
 22.2 cm/s right
Consider north as positive
mT  1.02 104 kg
vT  15 m/s
pT  1.53 105 kg  m/s
Not for Reproduction
47
Physics 12 SNAP
ANSWERS AND SOLUTIONS
b) Kinetic energy calculations, unlike
momentum, require converting units to
standard units. Also, consider speed.
Mechanical (kinetic) energy lost in collision
 1.075 102 J – 9.1445 103 J
 1.61103 J
 collision is inelastic.
12. a)
m1 = 0.225 kg
v1 = 0.300 m/s
m2 = 0.125 kg
v2 = 0.100 m/s
Consider right as positive
m1  10.0 g
m2  30.0 g
v1  20.0 cm/s
v2  0
p1  200 g  cm/s p2  0
m1 = 0.225 kg
v1 = 0.222 m/s
m2 = 0.125 kg
v2 = 0.240 m/s
 pbefore  p1  p2  200 g  cm/s
Kinetic energy of object 1 before
1
Ek1  m1v12
2
1
 (0.225 kg)(0.300 m/s) 2
2
 1.0125  10 2 J
m1  10.0 g
v1  –6.0 cm/s
p1  –60 g  cm/s
psys(after)  psys(before)
Kinetic energy of object 2, before
1
Ek2  m2 v2 2
2
1
 (0.125 kg)(0.100 m/s) 2
2
 6.25  104 J
m2  30 g
v2  ?
p2  ?
 200 g  cm/s
p2  200 g  cm/s  (60 g  cm/s)
 260 g  cm/s
p2
m
260 g  cm/s

30.0 g
 8.67 cm/s right
v2 
Kinetic energy of object 1, after
1
2
  m1  v1 
Ek1
2
1
 (0.225 kg)(0.222 m/s) 2
2
 5.5445  103 J
b) Kinetic energy calculations, unlike
momentum, require converting units to
standard units.
Kinetic energy of object 2, after
1
2
  m2  v2 
Ek2
2
1
 (0.125 kg)(0.240 m/s) 2
2
 3.60  103 J
m1 = 0.0100 kg
v1 = 0.200 m/s
m2 = 0.0300 kg
v2 = 0
m1 = 0.0100 kg
= –0.060 m/s
m2 = 0.0300 kg
v2 = 0.0867 m/s
Total kinetic energy before collision
 1.0125 102 J  6.25 104 J
 1.075 102 J
Total kinetic energy after collision
 5.5445 103 J  3.60 103 J
 9.1445 103 J
CASTLE ROCK RESEARCH
48
Copyright Protected
ANSWERS AND SOLUTIONS
Kinetic energy of object 1 before
1
Ek1  m1v12
2
1
 (0.0100 kg)(0.200 m/s) 2
2
 2.00  104 J
Lesson 3—Two-Dimensional
Interactions
PRACTICE EXERCISES
ANSWERS AND SOLUTIONS
Kinetic energy of object 1 after
1
2
  m1  v1 
Ek1
2
1
 (0.0100 kg)( 0.060 m/s) 2
2
 1.8  105 J
1.
m1  2.0 103 kg
v1  35 km/h north
p1  7.00 104 kg  km/h north
Kinetic energy of object 2 before
1
Ek2  m2 v2 2
2
1
 (0.0300 kg)(0) 2
2
0
m2  1.4  103 kg
v2  37.0 m/s west
p2  28.0 kg  m/s west
Kinetic energy of object 2 after
1
2
  m2  v2 
Ek2
2
1
 (0.0300 kg)(0.0867 m/s) 2
2
 1.13  104 J
m  m1  m2  3.4  103 kg
v ?
p?
Total kinetic energy before collision
 2.00 104 J  0
 2.00 104 J
First find out the momentum of the system before
collision ( psys(before) )
Total kinetic energy after collision is
 1.8 105 J  1.13 104 J
 1.31104 J
Mechanical (kinetic) energy lost in collision
 2.00 104 J –1.31104 J
 6.9  05 J
 collision is inelastic.
c)
The magnitude of the resultant momentum is
pR 
Most of this energy was converted to thermal
energy—some to sound, etc.
p12  p2 2
 (7.00  104 kg  km/h) 2  (5.18  104 kg  km/h)2
 8.71 104 kg  km/h
Now, find the direction
p
tan   2
p1
5.18 104 kg  km/h

7.00 104 kg  km/h
 0.750
  36.5
Not for Reproduction
49
Physics 12 SNAP
ANSWERS AND SOLUTIONS
Therefore
psys(before)  pR
 8.71 104 kg  km/h 36.5 W of N
p
m
33.6 kg  m/s 56 W of N

14.2 kg
 24 m/s 65 W of N
v
Again p  psys(after)  psys(before)  pR
p
m
8.71 04 kg  km/h 36.5 W of N

3.4  103 kg
 26 km/h 37 W of N
v
3.
m1  4.0 104 kg
v1  8.0 m/s west
p1  3.2 105 kg  m/s west
2.
m2  3.0 104 kg
v2  5.0 m/s south
p2  1.5 105 kg  m/s south
mA  6.2 kg
vA  3.0 m/s north
pA  18.6 kg  m/s north
mB  8.0 kg
vB  3.5 k  m/s west
pB  28.0 kg  m/s west
after collision
1+2
m  m1  m2  7.0  10 4 kg
v ?
p?
m  mA  mB  14.2 kg
v ?
p?
pR 
p12  p2 2
 (3.2  105 kg  m/s) 2  (1.5  105 kg  m/s) 2
 3.53  105 kg  m/s
pR 
pA  pB
2
p2
p1
1.5  105 kg  m/s

3.2  105 kg  m/s
 0.469
  25
tan  
2
 (18.6 kg  m/s)2  (28.0 kg  m/s)2
 33.6 kg  m/s
pB
pA
28.0 kg  m/s

18.6 kg  m/s
 1.51
  56.4
56
tan  
p  pR  3.53 105 kg  m/s 26 S of W
p
m
3.53 105 kg  m/s 26 S of W

7.0 104 kg
1
 5.0 10 m/s 25 S of W
v
p  pR  33.6 kg·m/s 56 W of N
CASTLE ROCK RESEARCH
50
Copyright Protected
ANSWERS AND SOLUTIONS
4.
a)
m1  50.0 kg
v1  ?
p1  ?
m2  60.0 kg
v2  0
p2  0
p2 x  p2 cos 
  378 kg  m/s  cos 322 
 298 kg  m/s
p2 y  p2 sin 
  378 kg  m/s  sin 322 
 – 233 kg  m/s
m1  50.0 kg
v1  6.00 m/s 50.0 N of E
p1  300 kg  m/s
50.0 N of E
Do the vector addition for the components of
the momentum
 px  p1x  p2 x
=193 kg  m/s  298 kg  m/s
 491 kg  m/s
m2  60.0 kg
v2  6.30 m/s 38.0 S of E
p2  378 kg  m/s 38.0 S of E
 py  p1y  p2 y
=230 kg  m/s   –233 kg  km/s 
 –3.00 kg  m/s
Find horizontal and vertical components of
300 kg·m/s 50.0° N of E.
Now add
p
x
and
p
y
using Pythagoras
theorem, to find out the magnitude of the
resultant momentum.
pR  ( px ) 2  ( p y ) 2
 (491 kg  m/s) 2  (3.00 kg  m/s) 2
 491 kg  m/s
tan  
To solve this problem consider east and north
direction as positive.
p1x  p1x cos 
  300 kg  m/s  cos 50.0 
 193 kg  m/s
px
3.00 kg  m/s

491 kg  m/s
  0.35
0
p1y  p1 sin 
  300 kg  m/s  sin 50.0 
 230 kg  m/s
psys(after)  pR  491 kg  m/s east
psys(after)  psys(before)  p1
p1  m1v1
p
v1 
m
491 kg  m/s east

50.0 kg
 9.82 m/s east
Find horizontal and vertical components of
378 kg·m/s 38.0° S of E.
38.0° S of E = 322° heading counter clockwise
from positive x-axis
Not for Reproduction
py
51
Physics 12 SNAP
ANSWERS AND SOLUTIONS
Mechanical (kinetic) energy lost in collision is
2.41 × 103 J – 2.10 × 103 J = 3.10 × 102 J,
 collision is inelastic. Kinetic energy was
not conserved.
b)
m1  50.0 kg
v1  9.82 m/s
m2  60.0 kg
v2  0
c)
5.
m1  50.0 kg
v1  6.00 m/s
a)
m2  60.0 kg
v2  6.30 m/s
m1  15.0 kg
v1  7.0 m/s east
p1  105 kg  m/s east
NOTE: Do not include direction with the
magnitude of velocity because direction is not
necessary in calculating kinetic energy.
Kinetic energy of object 1 before
1
Ek1  m1v12
2
1
 (50.0 kg)(9.82 m/s) 2
2
 2.41 103 J
m2  10.0 kg
v2  0
p2  0
m1  15.0 kg
v1  4.2 m/s 20.0 S of E
p1  63 kg  m/s 20.0 S of E
Kinetic energy of object 1 after
1
2
  m1  v1 
Ek1
2
1
 (50.0 kg)(6.00 m/s) 2
2
 9.00  10 2 J
m2  10.0 kg
v2  ?
p2  ?
pafter  pbefore
pbefore  105 kg  m/s east
 pafter  105 kg  m/s east
Kinetic energy of object 2 before
1
Ek2  m2 v2 2
2
1
 (60.0 kg)(0)2
2
0
Find horizontal and vertical components of
63 kg·m/s 20.0° S of E.
20.0° S of E = 340° heading counter clockwise
from positive x-axis
Kinetic energy of object 2 after
1
2
  m2  v2 
Ek2
2
1
 (60.0 kg)(6.30 m/s) 2
2
 1.20  103 J
p1x  p1x cos 
  63 kg  m/s  cos 340 
 59.2 kg  m/s
Total kinetic energy before collision
= 2.41 × 103 J + 0
= 2.41 × 103 J
p1y  p1 sin 
  63 kg  m/s  sin 340 
 –21.5 kg  m/s
Total kinetic energy after collision
= 9.0 × 102 J + 1.20 × 103 J
= 2.10 × 103 J
CASTLE ROCK RESEARCH
Most of the loss in mechanical energy was
converted to thermal energy—some to
sound, etc.
52
Copyright Protected
ANSWERS AND SOLUTIONS
Kinetic energy of object 1, before
1
Ek1  m1v12
2
1
 (15.0 kg)(7.0 m/s) 2
2
 3.7  102 J
After collision, the 10.0 kg object must have a
horizontal component of
105 kg·m/s – 59.2 kg·m/s = 45.8 kg·m/s east
After collision, the 10.0 kg object must have a
vertical component of
Kinetic energy of object 1, after
2
1
  m1 v1
Ek1
2
1
 (15.0 kg)(4.2 m/s) 2
2
 1.3 102 J
0 – (–21.5 kg·m/s) = 21.5 kg·m/s north
 
Now, add 45.8 kg·m/s east and
21.5 kg·m/s north using Pythagoras theorem.
Kinetic energy of object 2, before
1
Ek2  m2 v2 2
2
1
 (10.0 kg)(0) 2
2
0
pR  ( px ) 2  ( p y ) 2
 (45.8 kg  m/s) 2  (21.5 kg  m/s) 2
 50.6 kg  m/s
Kinetic energy of object 2, after
1
2
  m2  v2 
Ek2
2
1
 (10.0 kg)(5.1 m/s) 2
2
 1.30  102 J
Magnitude of velocity
p
v2  2
m
50.6 kg  m/s

10.0 kg
 5.1 m/s
Total kinetic energy before collision
 3.7 102 J  0
 3.7 102 J
Now find out the direction using
py
tan  
px
21.5 kg  m/s

45.8 kg  m/s
  25 N of E
v2  5.1 m/s 25 N of E
Total kinetic energy after collision
 1.3 102 J  1.3 102 J
 2.6 102 J
Mechanical (kinetic) energy lost in collision is
3.7  102 J – 2.6  102 J  1.1 102 J ,
 collision is inelastic. Kinetic energy was
not conserved.
b)
c)
m1 = 15.0 kg
v1 = 7.0 m/s
m1 = 15.0 kg
v1 = 4.2 m/s
Not for Reproduction
m2 = 10.0 kg
v2 = 0
6.
Most of the loss in mechanical energy was
converted to thermal energy—some to
sound, etc.
before explosion
m = 3m
v =0
p =0
m2 = 10.0 kg
v2 = 5.1 m/s
53
Physics 12 SNAP
ANSWERS AND SOLUTIONS
after explosion
Now add the x- and y-components of p3 .
1
2
m1  m
v1  15 m/s east
p1  15m  m/s east
m2  m
v2  10 m/s
45 S of E
p2  10m  m/s
45 S of E
3
m3  m
v3  ?
p3  ?
p3(R)  ( p3 x ) 2  ( p3 y ) 2
psys(after)  psys(before)
p1  p2  p3  0
 ((22.1m) m/s) 2  ((7.07 m) m/s) 2
 (23.2m) m/s
Find the horizontal and vertical components of p1 .
p1x  15m  m/s east
p1 y  0
tan  
p3 y
p3 x
(7.07m) m/s

(22.1m) m/s
  17.8 N of W
p3  p3R  (23.2m) m/s 17.8° N of W
Find the horizontal and vertical components of p2 .
45.0° S of E = 315° heading counter clockwise
from positive x-axis
p3
m
(23.2m) m/s 17.8° N of W

m
 23.2 m/s 17.8 N of W
v3 
Practice Test
p2 x  p2 cos 
  10m  m/s   cos 315 
  7.07m  m/s
ANSWERS AND SOLUTIONS
p2 y  p2 sin 
1.
  10m  m/s   sin 315 
   7.07m  m/s
 p  15m  m/s   7.07m  m/s
x
p
y
p
x
Ep = mgh
Since the height (h) increases, the gravitational
potential energy increases.
  22.1m  m/s
 0    –7.07m  m/s 
  –7.07m  m/s
p = mv
Like kinetic energy, since both the mass and the
velocity remain constant, the momentum remains
constant.
should be zero,
x-component of p3 = (–22.1 m) m/s
p
y
1 2
mv
2
Since both the mass and the velocity remain
constant, the kinetic energy remains constant.
Ek 
2.
Momentum is conserved in all collisions and
explosions.
3.
Ft = mv
Units of Ft = N·s
Units of mv = kg·m/s
 units of impulse can be: N·s or kg·m/s.
should be zero,
 y-component of p3 = (7.07 m) m/s
CASTLE ROCK RESEARCH
54
Copyright Protected
ANSWERS AND SOLUTIONS
4.
If the object has momentum, it has mass and
velocity; and if it has mass and velocity, it has
kinetic energy also.
10. Find the velocity of the spacecraft.
Mechanical energy is the sum of the gravitational
potential energy and the kinetic energy. Therefore,
if it has kinetic energy, it must also have
mechanical energy.
m  1.30  102 kg  2.80  103 kg
=2.93  103 kg
v 0
p  mv
0
Note: The object does not need to have potential
energy.
5.
6.
An elastic collision is defined as a collision in
which both the momentum and kinetic energy
are conserved.
m1  1.30 102 kg
m2  2.80 103 kg
v1  9.00 m/s
v2  ?
p1  1.17 103 kg·m/s p2  –1.17 103 kg  m/s
1 2
mv
2
2 Ek
v 
m
Ek 
p2
m2
1.17  103 kg  m/s

2.80 103 kg
 0.418 m/s
v2 
Again magnitude of the momentum
p  mv
 2 Ek 
 p  m
 m 


m 2 2 Ek
2
p 
m
p  m 2 Ek or 2mEk
7.
8.
9.
 relative velocity of the astronaut with respect to
the spacecraft is
9.00 m/s – (– 0.418 m/s) = 9.42 m/s
d
v
t
d
t
v
22.0 m

9.42 m/s
 2.34 s
p = mv
If you triple the velocity, you triple the momentum.
p   m  3v 
p   3mv
p  3 p
Inertia is a measure of mass only.
It will take 2.34 s for the safety line to reach its full
extension.
If the object slides along a horizontal frictionless
surface, there will be no acceleration (i.e., constant
velocity). If the velocity remains constant, then the
object’s momentum, potential energy, and kinetic
energy all remain constant.
11. p  m  vf – v0 
Note that the final velocity of the subatomic
particle has a magnitude approximately equal to the
initial velocity; but the direction is opposite to its
original direction.
i.e., if v0 = v, then vf = –v, or if v0 = –v, then
vf = v.
p  m  v   v  
 m  2v 
 2mv
The change in momentum for the subatomic
particle is 2mv.
Impulse can be defined as the change in
momentum. Therefore, when an object experiences
an impulse, it is experiencing a change in its
momentum.
Not for Reproduction
55
Physics 12 SNAP
ANSWERS AND SOLUTIONS
12. A perfect elastic collision is defined as a collision
in which both kinetic energy and momentum are
conserved.
13. impulse  mv
  0.25 kg  vf – v0 
  0.25 kg  –2.5 m/s – 3.0 m/s 
 1.4 N  s
The impulse on the ball was 1.4 N∙s.
14.
Consider east as positive
m1  1.1103 kg
v1  25 km/h
p1  2.75 104 kg·km/h
m2  2.3 103 kg
v2  –15 km/h
p2  –3.45 104 kg  km/h
pbefore  p1  p2  –7.0 103 kg·km/h
m  3.4  103 kg
v ?
p  –7.0  103 kg·km/h
p
m
–7.0 103 kg  km/h

3.4 103 kg
 –2.1 km/h or 2.1 km/h west
v
The velocity of the locked cars after the collision
was 2.1 km/h west.
CASTLE ROCK RESEARCH
56
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