Question

Two 24.0 kg ice sleds are placed a short distance apart, one directly behind the other, as shown in the figure. A 3.25 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 2.93 m/s relative to the ice.

What is the final speed of sled 2?
What is the final speed of sled 1?

Answer 1

By the law of momentum change of conservation energy
u=initial speed v=final speed m=mass
m1u1 + m2u2 = m1v1 + m2v2
final speed of sled 2:
mcat x ucat + mice x uice = mcat x vcat + mice x vice
when a cat jump over an ice, the initial speed of cat given 2.93m/s,
the initial speed of ice given zero. After the cat jump onto the ice, the final speed of the cat and the ice will be the same.
3.25x2.93 + 0 = (3.25+24)v
v=0.35m/s
final speed of sled 1:
mcat x ucat + mice x uice = mcat x vcat + mice x vice
at begining, the cat jump away from the sled 1, it gives a momentum change when the cat jump away the sled. Given the speed when the cat jump away from sled 2 is 2.93m/s
when the cat jump away sled 1:
(3.25+24)x0 = 24v +3.25x2.93
v=-0.4m/s (negative is just a vector of the ice, we can ignore it except in velocity)
the cat jump away from sled 2 to sled 1
3.25x2.93 + 24x0.4 = (3.25+24)v
v=0.7m/s