physics

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.)

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  1. 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

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