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Q.

A boy of mass ‘m’ stands at one end of a long trolley of mass M moving with uniform speed V on a smooth horizontal floor. If the boy runs forward with a speed $\frac{V}{2}$ what is the speed of the centre of mass of the trolley-boy system

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a

$\frac{(2 M - m) V}{2 (M - m)}$

b

$\frac{(M + m) V}{(M - m)}$

c

$\frac{V}{2}$

d

$V$

answer is D.

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Detailed Solution

The motion of the boy is due to internal force which can not change the velocity of the center of mass, so it remains as V.

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A boy of mass ‘m’ stands at one end of a long trolley of mass M moving with uniform speed V on a smooth horizontal floor. If the boy runs forward with a speed V2 what is the speed of the centre of mass of the trolley-boy system