HRW Solution Chapter 09.pdf

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Halliday
♦
Resnick
♦Walker
FUNDAMENTALS OF PHYSICS
SIXTH EDITION
Selected Solutions
Chapter 9
9.19
9.37
19. There is no net horizontal force on the dog-boat system, so their center of mass does not move. Therefore
by Eq. 9-16,
M
∆x
com
= 0 =
m
b
∆x
b
+
m
d
∆x
d
which implies
|∆x
b
|
=
m
d
|∆x
d
|
.
m
b
Now we express the geometrical condition that
relative to the boat
the dog has moved a distance
d
= 2.4 m:
|∆x
b
|
+
|∆x
d
|
=
d
which accounts for the fact that the dog moves one way and the boat moves the other. We substitute
for
|∆x
b
|
from above:
m
d
|∆x
d
|
+
|∆x
d
|
=
d
m
b
which leads to
|∆x
d
|
=
d
2.4
= 1.92 m
.
m
d
=
1 +
m
b
1 +
4.5
18
The dog is therefore 1.9 m closer to the shore than initially (where it was 6.1 m from it). Thus, it is now
4.2 m from the shore.
37. Our notation is as follows: the mass of the original body is
M
= 20.0 kg; its initial velocity is
v
0
= 200ˆ
i
ˆ in SI units; the
in SI units (m/s); the mass of one fragment is
m
1
= 10.0 kg; ; its velocity is
v
1
= 100j
mass of the second fragment is
m
2
= 4.0 kg; ; its velocity is
v
2
=
−500
ˆ in SI units; and, the mass of the
i
third fragment is
m
3
= 6.00 kg.
(a) Conservation of linear momentum requires
M v
0
=
m
1
v
1
+
m
2
v
2
+
m
3
v
3
which (using the above information) leads to
v
3
= 1000 ˆ
−
167 ˆ
i
j
in SI units. The magnitude of
v
3
is
v
3
=
1000
2
+ (−167)
2
= 1.01
×
10
3
m/s. It points at
tan
−1
(−167/1000) =
−9.48
◦
(that is, at 9.5
◦
measured clockwise from the +x axis).
(b) We are asked to calculate ∆K or
1
1
1
2
2
2
m
1
v
1
+
m
2
v
2
+
m
3
v
3
2
2
2
1
2
−
M v
0
= 3.23
×
10
6
J
.
2

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