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(C) Since the system (man + trolley) is on a smooth horizontal surface,there is no external horizontal force acting on it. Therefore,the center of mas

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$\frac{L}{{{u_{rel}}}}$

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(C) Since the system (man + trolley) is on a smooth horizontal surface,there is no external horizontal force acting on it. Therefore,the center of mass of the system remains at rest.Let $v_m$ be the velocity of the man and $v_t$ be the velocity of the trolley with respect to the ground.According to the conservation of linear momentum,$m v_m + M v_t = 0$,which implies $v_t = -\frac{m}{M} v_m$.The velocity of the man with respect to the trolley is given as $u_{rel} = v_m - v_t$.Substituting $v_t$,we get $u_{rel} = v_m - (-\frac{m}{M} v_m) = v_m (1 + \frac{m}{M}) = v_m \left( \frac{M+m}{M} \right)$.Thus,the velocity of the man with respect to the ground is $v_m = u_{rel} \left( \frac{M}{M+m} \right)$.The time taken by the man to cover the length $L$ of the trolley is $t = \frac{L}{u_{rel}}$.Since the trolley also moves,the man covers the distance $L$ relative to the trolley at a relative speed $u_{rel}$.Therefore,the time taken is $t = \frac{L}{u_{rel}}$.

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