On a smooth horizontal plane,a uniform string | vedclass

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(B) Let the mass of the man be $M_m = M$ and the mass of the string be $M_s = M$. The length of the string is $L$.Since the plane is smooth,there is n

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$L / 4$

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(B) Let the mass of the man be $M_m = M$ and the mass of the string be $M_s = M$. The length of the string is $L$.Since the plane is smooth,there is no external horizontal force acting on the system (man + string). Therefore,the position of the center of mass of the system remains unchanged.Let the initial position of the man be $x_m = 0$ and the center of mass of the string be at $x_s = L/2$.The initial position of the center of mass of the system is:$X_{cm} = \frac{M_m x_m + M_s x_s}{M_m + M_s} = \frac{M(0) + M(L/2)}{M + M} = \frac{ML/2}{2M} = L/4$.After the man collects the entire string,the man and the string become a single system of mass $2M$ located at the final position of the man,$x_f$.The final position of the center of mass of the system is $X'_{cm} = x_f$.Since the center of mass does not move,$X_{cm} = X'_{cm}$,so $x_f = L/4$.The displacement of the man is $\Delta x = |x_f - x_m| = |L/4 - 0| = L/4$.

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