A man of mass m on an initially stationary boat gets off of the boat by leaping to the left in an exactly horizontal direction. Immediately after the leap, the boat of mass M, is observed to be moving to the right with speed v.

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Work done by man on boat is $\frac{1}{2} (m) v^{2}$

Increase in kinetic energy of man is $\frac{1}{2} \frac{M^{2}}{m} v^{2}$

Velocity of centre of mass of system is v.

Increase in the mechanical energy of system of man and boat is $\frac{1}{2} (\frac{M^{2}}{m} + M) v^{2}$

answer is B, D.

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Detailed Solution

$W_{m a n} = Δ K E$
From momentum conservation $m v_{m a n} = M v$
For (A) : Work done by man on boat $Δ K E_{b o a t} = \frac{1}{2} M v^{2}$
For (B) : $W_{m a n t o t a l} = Δ K E_{s y s t e m} = \frac{1}{2} m \frac{M^{2} v^{2}}{m^{2}} + \frac{1}{2} M v^{2}$
For (C) : $V_{C M} = 0$ {momentum conservation}
For (D): $W_{m a n o n h i m s e l f} = Δ K E_{m a n} = \frac{1}{2} \frac{M^{2}}{m} v^{2}$