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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?

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$\frac{u}{6}$

$\frac{3 u}{4}$

$\frac{2 u}{3}$

$\frac{u}{3}$

answer is C.

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Detailed Solution

If v be the velocity of the trolley, then conserving momentum,

$\frac{M}{2} (u - v) + M (- v) = 0 \Rightarrow v = \frac{u}{3}$

Velocity of the man relative to the ground $= u - \frac{u}{3} = \frac{2 u}{3}$

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