The figure shows a man of mass m standing at | vedclass

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(A) 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

Vedclass · Accepted Answer

$\frac{ML}{m + M}$

Vedclass · Answer

(A) 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 $x_m$ be the displacement of the man and $x_t$ be the displacement of the trolley with respect to the ground.Since the center of mass does not move,we have $m x_m + M x_t = 0$. Taking the magnitude,$m |x_m| = M |x_t|$,so $|x_t| = \frac{m}{M} |x_m|$.The relative displacement of the man with respect to the trolley is given as the length of the trolley,$L = |x_m| + |x_t|$.Substituting $|x_t|$,we get $L = |x_m| + \frac{m}{M} |x_m| = |x_m| (1 + \frac{m}{M}) = |x_m| (\frac{M + m}{M})$.Solving for $|x_m|$,we get $|x_m| = \frac{ML}{m + M}$.

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