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Q.

A trolley of mass M rests on a frictionless floor with a man of mass M/2 standing at its edge. The man jumps off from the trolley with a velocity u relative to the trolley. Then what will be the velocity of the man relative to the ground just after he jumps off from the trolley?

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a

u6

b

3u4

c

2u3

d

u3

answer is C.

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Detailed Solution

If v be the velocity of the trolley, then conserving momentum, M2(u-v)+M(-v)=0⇒v=u3Velocity of the man relative to the ground =u-u3=2u3

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A trolley of mass M rests on a frictionless floor with a man of mass M/2 standing at its edge. The man jumps off from the trolley with a velocity u relative to the trolley. Then what will be the velocity of the man relative to the ground just after he jumps off from the trolley?