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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?
Assuming a closed system with no external forces, the principle of conservation of momentum can be applied here. Momentum before = Momentum after Fisherman's momentum before = his mass * his velocity = 87 kg * 4.1 m/s = 356.7 kg*m/s towards West Boat's momentum before = its mass * its velocity = 121 kg * 0 m/s = 0 kg*m/s Total initial momentum = Fisherman's momentum + Boat's momentum = 356.7 kg*m/s towards West After the fisherman jumps into the boat, they will move together, so their final momentum will be the sum of their masses multiplied by their shared final velocity (Vf). The total momentum afterwards = (Mass of fisherman + Mass of boat) * Vf = (87 kg + 121 kg) * Vf = 208 kg * Vf By the principle of conservation of momentum, total initial momentum = total final momentum. So, 356.7 kg*m/s = 208 kg * Vf Therefore, Vf = 356.7 kg*m/s ÷ 208 kg = 1.71 m/s towards West As we took east as the positive direction, the velocity is -1.71 m/s when direction is considered. So the final velocity of the fisherman and the boat is -1.71 m/s.
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