An 85.0 kg fisherman jumps from a dock into a 135.0 kg rowboat at rest on the west side of the dock. If the velocity of the fisherman is 4.30 m/s to the west as he leaves the dock, what is the final velocity of the fisherman and the boat?

Answer 1

We can use the principle of conservation of momentum to solve this problem. The total momentum of the fisherman and the boat before the jump is equal to the total momentum of the fisherman and the boat after the jump.

Let's denote the velocity of the fisherman and boat after the jump as V. The total initial momentum is given by:

(85.0 kg)(4.30 m/s) + 0 = (85.0 kg + 135.0 kg)V

Solving for V:

(85.0 kg)(4.30 m/s) = (85.0 kg + 135.0 kg)V
V = (85.0 kg)(4.30 m/s) / (85.0 kg + 135.0 kg)
V = (365.5 kg*m/s) / (220.0 kg)
V = 1.66 m/s

Therefore, the final velocity of the fisherman and the boat after the jump is 1.66 m/s to the west.