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tiL12214_appd_633-642.indd Page 634 09/10/10 8:35 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

Example 2.13, p. 46 Example 2.17, p. 49
_ 2
mv
Astronaut: m = 0.25 kg F =

r
a = 0.500 m/s 2 r = 0.25 m
m _ 2
(0.25 kg)(2.0 )
t = 1.50 s v = 2.0 m/s __
s

=
v i = 0 m/s F = ? 0.25 m
(
v f = ? _ 2
4.0 m

(0.25 kg) 2 )
__

_ = s

v f − v i

a = ∴ v f = at + v i 0.25 m
t 2
(0.25)(4.0) kg . m
1 _
m _
m _

(
2)

×

= 0.500 (1.50 s) + 0 = _ _ m
2
0.25
s
s s
kg . m
m _ s _ _

= (0.500)(1.50) × = 4.0
2
s 1 s 2
m _ = 4.0 N

= 0.750
s
Example 2.20, p. 51
Spacecraft:
2
–11
a = 0.250 m/s 2 G = 6.67 × 10 N . m /kg 2
24
t = 1.50 s m e = 6.0 × 10 kg
6
v i = 0 m/s d = 12.8 × 10 m
v f = ? g = ?
_ _
Gm e

v f − v i

a = ∴ v f = at + v i g =
t 2
d
m _
m _
2 )
(
2)

= 0.250 (1.50 s) + 0 ( 6.67 × 10 _ 2 (6.0 × 10 kg)

–11 N . m

24
s

s
kg
____

m _ s _ = 6
2
= (0.250)(1.50) × (12.8 × 10 m)

s 2 1 _ 2
–11 N . m
24

m _ ( 6.67 × 10 2 ) (6.0 × 10 kg)

= 0.375 ____
kg
s =

14
1.64 × 10 m 2
24
Example 2.15, p. 47 ___ _ 2 _ 1 _
–11
kg
(6.67 × 10 )(6.0 × 10 ) N . m

= × ×

Student and boat m = 100.0 kg 14 2 1 2
1.64 × 10 kg m
Student and boat v = ?
kg . m
_

Rock m = 5.0 kg
2
_
s

Rock v = 5 m/s = 2.44
kg
rock momentum – student and boat momentum = 0
m _
(mv) r – (mv) s&b = 0 = 2.4

s 2
m _
(5.0 kg) (5.0 ) – (100.0 kg)(v s&b ) = 0
s
m _ CHAPTER 3
(25 kg . ) – (100.0 kg)(v s&b ) = 0
s
Example 3.2, p. 63
m _
(25 kg . ) = (100.0 kg)(v s&b ) w = 50 lb

s
d = 2 ft
m _
25 kg .

_ W = ?
s
v s&b =
100.0 kg W = Fd
= (50 lb) (2 ft)
25 _
_ kg 1 _ m _
= × × = (50)(2) ft . lb

100.0 1 kg s
= 100 ft . lb
m _

= 0.25
s
634 APPENDIX D Solutions for Follow-Up Example Exercises D-2