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(
)
(
)
()
(
)
(
)
11
2 2
1
5 kg 3.0 m/s
10 kg
2.0 m/s
(5 kg)
10 kg 2.5 m/s
iif
f
f
mv
v
+=+
+=
+
GGG
G
G
which yields
1
2.0 m/s
f
v
=
G
. Thus, the speed of the 5.0 kg block immediately after the
collision is
20
.ms
.
(b) We find the reduction in total kinetic energy:
(
)
(
)
(
)
22
2
2
5 kg 3 m/s
10 kg 2 m/s
5 kg 2 m/s
10 kg 2.5 m/s
1.25 J
1.3 J.
if
KK
−=
+
−
−
=−
≈−
(c) In this new scenario where
G
v
f
2
40
=
. m s , momentum conservation leads to
G
v
f
1
10
. m s and we obtain
40 J
K
Δ=
+
.
(d) The creation of additional kinetic energy is possible if, say, some gunpowder were on
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 Spring '08
 Any
 Physics, Mass, Momentum

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