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Holt Physics
Problem 6A
MOMENTUM
PROBLEM
The world’s most massive train ran in South Africa in 1989. Over 7 km
long, the train traveled 861.0 km in 22.67 h. Imagine that the distance was
traveled in a straight line north. If the train’s average momentum was
7.32 108 kg•m/s to the north, what was its mass?
SOLUTION
Given:
Unknown:
∆x = 861.0 km to the north
∆t = 22.67 h
kg•m
pavg = 7.32 × 108  to the north
s
vavg = ? m = ?
Use the definition of average velocity to calculate vavg, and then substitute this
value for velocity in the definition of momentum to solve for mass.
(861.0 × 103 m)
∆x
m
vavg =  =  = 10.55  to the north
∆t (22.67 h)(3600 s/h)
s
pavg = mvavg
kg•m
7.32 × 108 
pavg
s
m =  =  = 6.94 × 107 kg
vavg
m
10.55 
s
1. In 1987, Marisa Canofoglia, of Italy, roller-skated at a record-setting
speed of 40.3 km/h. If the magnitude of Canofoglia’s momentum was
6.60 × 102 kg•m/s, what was her mass?
2. In 1976, a 53 kg helicopter was built in Denmark. Suppose this helicopter flew east with a speed of 60.0 m/s and the total momentum of
the helicopter and pilot was 7.20 × 103 kg•m/s to the east. What was the
mass of the pilot?
3. One of the smallest planes ever flown was the Bumble Bee II, which had
a mass of 1.80 × 102 kg. If the pilot’s mass was 7.0 × 101 kg, what was
the velocity of both plane and pilot if their momentum was 2.08 ×
104 kg•m/s to the west?
4. The first human-made satellite, Sputnik I, had a mass of 83.6 kg and a
momentum with a magnitude of 6.63 × 105 kg•m/s. What was the
satellite’s speed?
54
Holt Physics Problem Workbook
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ADDITIONAL PRACTICE
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DATE _______________ CLASS ____________________
5. Among the largest passenger ships currently in use, the Norway has
been in service the longest. The Norway is more than 300 m long, has a
mass of 6.9 × 107 kg, and can reach a top cruising speed of 33 km/h.
Calculate the magnitude of the ship’s momentum.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. In 1994, a tower 22.13 m tall was built of Lego® blocks. Suppose a block
with a mass of 2.00 g is dropped from the top of this tower. Neglecting
air resistance, calculate the block’s momentum at the instant the block
hits the ground.
Problem 6A
55
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Momentum and Collisions
Chapter
6
Additional Practice 6A
Givens
Solutions
1. v = 40.3 km/h
p = 6.60 × 102 kg • m/s
6.60 × 102 kg • m/s
p
m =  = 
= 59.0 kg
v (40.3 × 103 m/h)(1 h/3600 s)
ptot = mh v + mp v
2. mh = 53 kg
v = 60.0 m/s to the east
ptot = 7.20 × 103 kg • m/s to
the east
7.20 × 103 kg • m/s − (53 kg)(60.0 m/s)
ptot − mhv
 = 
mp = 
v
60.0 m/s
7.20 × 103 kg • m/s − 3.2 × 103 kg • m/s 4.0 × 103 kg • m/s
mp =  =  = 67 kg
60.0 m/s
60.0 m/s
II
2
3. m1 = 1.80 × 10 kg
1
m2 = 7.0 × 10 kg
ptot = 2.08 × 104 kg•m/s to
the west
4
4
−2.08 × 10 kg•m/s
−2.08 × 10 kg•m/s
p
 = 
v = tot
= 
2
1
2.50 × 102 kg
m1 + m2 1.80 × 10 kg + 7.0 × 10 kg
v = −83.2 m/s = 83.2 m/s to the west
= −2.08 × 104 kg•m/s
4. m = 83.6 kg
HRW material copyrighted under notice appearing earlier in this book.
p = 6.63 × 105 kg•m/s
5. m = 6.9 × 107 kg
p 6.63 × 105 kg•m/s
v =  =  = 7.93 × 103 m/s = 7.93 km/s
m
83.6 kg
p = mv = (6.9 × 107 kg)(33 × 103 m/h)(1 h/3600 s) = 6.3 × 108 kg • m/s
v = 33 km/h
6. h = 22.13 m
m = 2.00 g
g = 9.81 m/s2
1
mgh = 2mvf 2
vf = 2g
h
p = mvf = m 2g
m/s
m)
h = (2.00 × 10−3 kg) (2
)(
9.
81
2)(2
2.
13
p = 4.17 × 10−2 kg • m/s downward
Additional Practice 6B
1. m = 9.0 × 104 kg
vi = 0 m/s
vf = 12 cm/s upward
F = 6.0 × 103 N
mvf − mvi (9.0 × 104 kg)(0.12 m/s) − (9.0 × 104 kg)(0 m/s)
∆t =  = 
F
6.0 × 103 N
(9.0 × 104 kg)(0.12 m/s)
∆t = 
= 1.8 s
6.0 × 103 N
Section Two—Problem Workbook Solutions
II Ch. 6–1