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Question: Note: Take East as the positive direction. A(n) 87 kg fisherman jumps from a dock into a 121 kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 4.1 m/s to the West as he leaves the dock, what is the final velocity of the fisherman and the boat?
The situation described in the problem involves conservation of momentum. The total momentum before the jump (the initial momentum) must be equal to the total momentum after the jump (the final momentum). Let's denote: Mf - the mass of the fisherman, Mb - the mass of the boat, Vf - the velocity of the fisherman (before the jump), Vfb - the final velocity of the fisherman and the boat. The initial momentum is given by Mf * Vf, and because the boat is initially at rest, its momentum is zero. So, the initial total momentum is Mf * Vf. The final total momentum is given by the sum of the momentum of the fisherman and the boat moving together, which is (Mf + Mb) * Vfb. By the conservation of momentum, these two must be equal: Mf * Vf = (Mf + Mb) * Vfb Now, just plug in the given quantities into the formula: 87kg * 4.1m/s = (87kg + 121kg) * Vfb Solving for Vfb gives you the final velocity: Vfb = (87kg * 4.1m/s) / (87kg + 121kg) = 1.6 m/s. So, the final velocity of the fisherman and the boat is 1.6 m/s to the west. Since east was considered positive and our result is westward, in position-time graph, the velocity would be -1.6 m/s.
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