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
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.)
1 answer
1 answer