Two people, one twice as massive as the other, attempt a tug-of-war with 12 meters of rope on frictionless ice. After a brief time, they meet. The heavier person slides a distance of Group of answer choices4 m.0 m.6 m3 m
Question
Two people, one twice as massive as the other, attempt a tug-of-war with 12 meters of rope on frictionless ice. After a brief time, they meet. The heavier person slides a distance of Group of answer choices4 m.0 m.6 m3 m
Solution
The heavier person slides a distance of 4 meters.
Here's why:
In this scenario, the center of mass of the system (the two people and the rope) doesn't change because there are no external forces acting on it. The center of mass is closer to the heavier person because they have more mass.
Since the heavier person is twice as massive as the lighter person, the center of mass will be 1/3 of the way along the rope from the heavier person and 2/3 of the way along the rope from the lighter person.
This means that when they meet, they meet at the center of mass of the system. So, the heavier person will have moved 1/3 of the total length of the rope, and the lighter person will have moved 2/3 of the total length of the rope.
Since the total length of the rope is 12 meters, the heavier person moves 12 meters * 1/3 = 4 meters.
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