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kg
2.21. List the known and unknown quantities: 3 _ _ 1 _ m _

= × ×
w = 13,720 N 6 1 kg s
v = 91 km/h m _
= 0.5
s
p = ?

The equation for momentum is p = mv, which is already solved Th e rifle recoils with a velocity of 0.5 m/s.

for momentum (p). The weight unit must be first converted to a 2.23. List the known and unknown quantities:

mass unit:
Astronaut → w = 2,156 N Wrench → m = 5.0 kg
w _
w = mg ∴ m = Astronaut → v = ? m/s Wrench → v = 5.0 m/s
g
kg·m
_ Note that the astronaut’s weight is given, but we need mass for the
13,720
2
__ conservation of momentum equation. Mass can be found because
s
= m _ 2

9.8 the weight on Earth was given, where we know g = 9.8 m/s .

s 2 Thus, the mass is
13,720 kg·m
_ _ s _ 2 w _

= × w = mg ∴ m =

9.8 s 2 m g
kg·m
= 1,400 kg _
2,156
2
s

The km/h unit should be converted to m/s next. Using the = _

m _
conversion factor from inside the front cover gives 9.8
s 2
m _
0.2778 _ _ _ _ 2

2,156 kg·m
_
s
s
km

91

× = 9.8 2 ×
m
_
km

h s
h
= 220 kg
m _ _ _
km
h

0.2778 × 91 × × So the converted known and unknown quantities are
s km h
m _ Astronaut → m =220 kg Wrench → m =5.0 kg
25
s Astronaut → v =? m/s Wrench → v =5.0 m/s
Now, list the converted known and unknown quantities This is a conservation of momentum question, where the

m = 1,400 kg astronaut and wrench can be considered as a system of
interacting objects:
v = 25 m/s
p = ? Wrench momentum = astronaut momentum
and solve for momentum (p). (mv) w = (mv) a
(mv) w – (mv) a = 0
p = mv
m _
m _ (5.0 kg)( 5.0 ) – (220 kg)v a = 0

= (1,400 kg)( 25 ) s
s
m _

kg·m
_ ( 25 kg · ) – (220 kg·v a ) = 0
s
= 35,000
s
m _

25 kg·
s

2.22. List the known and unknown quantities: v a = _

220 kg
Bullet → m = 0.015 kg Rifl e → m = 6 kg kg
25 _ _ _
1
m
Bullet → v = 200 m/s Rifl e → v = ? m/s = _ × ×

220 1 kg s

Note the mass of the bullet was converted to kg. This is a m _

conservation of momentum question, where the bullet and rifl e = 0.11
s
can be considered as a system of interacting objects:

The astronaut moves away with a velocity of 0.11 m/s.
Bullet momentum = rifle momentum
2.24. (a) Weight (w) is a downward force from the acceleration of
(mv) b = (mv) r
gravity (g) on the mass (m) of an object. This relationship is

(mv) b − (mv) r = 0 the same as Newton’s second law of motion, F = ma, and
m _
(0.015 kg)( 200 ) – (6 kg)v r = 0 m _

2)
(

s w = mg = (1.25 kg) 9.8
m _ s

( 3 kg · ) − (6 kg·v r ) = 0 m _

s

= (1.25)(9.8) kg ×
m _ s 2

3 kg · = 6 kg·v r
s kg·m

= 12.25 _
m _ s 2
3 kg ·

s

v r = _ = 12 N

6 kg
648 APPENDIX E Solutions for Group A Parallel Exercises E-6