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A 183 lb man sits in the middle of a 94 lb, 14 ft long boat. The boat’s prow touches the pier, but the boat isn’t tied to it. The man stands up and walks towards the pier. Assume there is negligible resistance between the boat and the water, and that the boat’s center of mass is in the middle of the boat.

By the time he reaches the boat’s prow, what is the distance between the prow and the pier?

Answer :

Final answer:

When the man walks towards the boat's prow, the boat slides backwards to conserve the system's total momentum. Thus, when the man reaches the prow, the distance between the boat's prow and the pier would be 14 feet.

Explanation:

The scenario described in the question is a classic problem in physics dealing with the principles of conservation of momentum and the center of mass. When the man walks towards the pier, the boat will move in the opposite direction in order to conserve total momentum, as the system is isolated. In this case, the boat moves such that the center of mass of the system (man+boat) remains at the same position. Assuming that the system's initial center of mass is at the middle of the boat (i.e., 7 ft from the pier), even when the man reaches the prow after walking 7 ft, the system's center of mass remains unchanged. Hence, the boat would have moved backwards by the same distance, i.e., 7 feet, to maintain the center of mass. Therefore, the distance between the boat’s prow and the pier would become 7 (initial) + 7 (backward movement) = 14 feet.

Learn more about Conservation of Momentum here:

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