Question
Two 21.0 kg ice sleds are placed a short distance apart, one directly behind the other, as shown in the figure. A 3.42 kg cat, initially standing on sled 1, jumps across to sled 2 and then jumps back to sled 1. Both jumps are made at a horizontal speed of 2.84 m/s relative to the ice.
What is the final speed of sled 2? (Assume the ice is frictionless.)
What is the final speed of sled 2? (Assume the ice is frictionless.)
Answers
First find the velocity of sled 1 (V):
p1=p2
0=mv+MV(1)
V(1)=-mv/M
For sled 2:
0+mv=(m+M)V
V=mv/(m+M)
When the cat leaves again:
(m+M)V=MV(2)-mv
V(2)=((m+M)V+mv)/M
(V comes from the equation above)
For sled 1:
MV(1)+mv=(M+m)V
V(1) from the first equation
p1=p2
0=mv+MV(1)
V(1)=-mv/M
For sled 2:
0+mv=(m+M)V
V=mv/(m+M)
When the cat leaves again:
(m+M)V=MV(2)-mv
V(2)=((m+M)V+mv)/M
(V comes from the equation above)
For sled 1:
MV(1)+mv=(M+m)V
V(1) from the first equation
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