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(A) Let $v_1$ be the velocity of the man and $v_2$ be the velocity of the trolley with respect to the ground.Since the system is on a smooth horizonta

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $v_1$ be the velocity of the man 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 $0$.$m v_1 + M v_2 = 0 \implies v_2 = -\frac{m}{M} v_1$.The relative velocity of the man with respect to the trolley is given by $u_{rel} = v_1 - v_2$.Substituting $v_2$ in the equation: $u_{rel} = v_1 - (-\frac{m}{M} v_1) = v_1 (1 + \frac{m}{M}) = v_1 (\frac{M + m}{M})$.Solving for $v_1$,we get $v_1 = \frac{M u_{rel}}{M + m}$.

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