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Km
v
m
m
ii
==
=
1
2
1
2
475
113
22
(. )
.
and the final kinetic energy is
vm
v
m
m
fff
=+=
+
=
1
2
1
2
1
2
350
200
81
1
2
2
2
(. ) (. )
. .
c
h
Kinetic energy is not conserved.
104. (a) We use Fig. 922 of the text (which treats both angles as positivevalued, even
though one of them is in the fourth quadrant; this is why there is an explicit minus sign in
Eq. 980 as opposed to it being implicitly in the angle). We take the cue ball to be body 1
and the other ball to be body 2. Conservation of the
x
and the components of the total
momentum of the twoball system leads to:
mv
1
i
=
mv
1
f
cos
θ
1
+
mv
2
f
cos
2
0 = –
mv
1
f
sin
1
+
mv
2
f
sin
2
.
The masses are the same and cancel from the equations. We solve the second equation for
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 Spring '08
 Any
 Physics

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