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(B) Let $v_1$ be the velocity of the man with respect to the ground and $v_2$ be the velocity of the trolley with respect to the ground.Since the syst

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$\frac{m u_{rel}}{m + M}$

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(B) Let $v_1$ be the velocity of the man with respect to the ground and $v_2$ be the velocity of the trolley with respect to the ground.Since the system is on a smooth horizontal surface,there is no external horizontal force acting on the system.Therefore,the momentum of the system is conserved.Initially,the system is at rest,so the initial momentum is $0$.$m v_1 + M v_2 = 0$$v_1 = -\frac{M}{m} v_2$We are given the relative velocity of the man with respect to the trolley as $u_{rel} = v_1 - v_2$.Substituting $v_1$ in the relative velocity equation:$u_{rel} = -\frac{M}{m} v_2 - v_2$$u_{rel} = -v_2 \left( \frac{M}{m} + 1 \right)$$u_{rel} = -v_2 \left( \frac{M + m}{m} \right)$$v_2 = -\frac{m u_{rel}}{M + m}$The magnitude of the velocity of the trolley is $\frac{m u_{rel}}{m + M}$.

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