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A man of mass $M$ is carrying a ball of the mass $M / 2$ . The man is initially in the state of rest at a distance $D$ from fixed vertical wall. He throws the ball towards the wall with a velocity $V$ with respect to earth at $t = 0$ . As a result of throwing, the man also starts moving backwards. The ball rebounds elastically from the wall. The man finally collects the ball. Assuming friction to be absent.

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The velocity of the man + ball system after the man has collected the ball is $\frac{2 V}{3}$

Impulse by ball on man during the time he catches it is $\frac{MV}{3}$

Impulse by ball on man during the time he catches it is $\frac{M V}{6}$

He catches the ball again at $t = \frac{4 D}{V}$

answer is A, C, D.

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Detailed Solution

Initially

$MV = MV / 2 V_{1} = V / 2$

After collision

$MV / 2 + MV / 2 = 3 {MV}^{'} / 2$

$V^{'} = 2 V / 3$

$∴ Impulse = M (\frac{2 V}{3} - \frac{V}{2}) = \frac{MV}{6}$

Time

$= \frac{D}{V} + \frac{D + \frac{V}{2} (\frac{D}{V})}{V - (V / 2)} = \frac{D}{V} + \frac{3 D}{V} = \frac{4 D}{V}$

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