Two boxes that are tied together with a string and sitting at rest in the middle

Question

Two boxes that are tied together with a string and sitting at rest in the middle of a large air table ( no friction). Between the two boxes is a compressed spring. The string tying the two boxes together is going to be cut suddenly so the spring will expand pushing the boxes apart. The box on the left has four times the mass of the box on the right. After the string is cut and the boxes lose contact with the spring, will the magnitude of the momentum of the box on the left be greater than, less than, or equal to the magnitude of the momentum of the box on the right? Explain.

Explanation / Answer

The two mass blocks have compressed the spring, so when the string is cut, blocks will move in opposite directions.

Before string is cut:

momentum of both boxes = 0 ( As boxes are in rest)

As the restoring force of spring is internal force for the system of boxes and spring.

After string is cut:

m1v1 + m2v2 =0 (As no external force is acting, momentum is conserved)

m1v1 = - m2v2

So, magnitude of both the boxes is equal. but in opposite directions.

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