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Question

Two bodies A and B of mases m and 2m respectively are placed on a smooth floor. They are connected by a spring. A third body C of mass m moves with velocity v0 along the line joining A and B and collides elastically with A as shown in figure. At a certain instant of time t0 after collision, it is found that the instantaneous velocities of A and B are the same. Further at this instant the compression of the spring is found to be x0. Determine (a) the common velocity of A and B at time t0 and (b) the spring constant.

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Solution

The correct option is **A**

v=v03, k=23mv20x20

Upon collision, C transfers its entire momentum (mv0) and kinetic energy (12mv20) to block C and comes to rest. Let the common velocity of A and B at x0 compression be v.

Applying conservation of momentum, mv0=mv+2mv⇒v=v03

From energy conservation, 12mv20=12mv2+12(2m)v2+12kx20

Solving we get, k=23mv20x20

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