A monkey of mass m is at rest on a light rope | vedclass

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(B) Initially,the system is at rest,so the velocity of the center of mass $\vec{v}_{CM} = 0$.Since there are no external forces acting on the system,t

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

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(B) Initially,the system is at rest,so the velocity of the center of mass $\vec{v}_{CM} = 0$.Since there are no external forces acting on the system,the center of mass remains stationary.Let $d_1$ be the distance the monkey moves relative to the ground and $d_2$ be the distance the balloon moves relative to the ground.From the conservation of the center of mass: $m d_1 = M d_2$.Since the monkey climbs a distance $L$ relative to the balloon,the total distance covered by the monkey relative to the ground is $d_1 = L - d_2$.Substituting this into the conservation equation: $m(L - d_2) = M d_2$.$mL - m d_2 = M d_2$.$mL = (m + M) d_2$.Therefore,the distance covered by the balloon is $d_2 = \frac{mL}{m + M}$.

$A$ monkey of mass $m$ is at rest on a light rope attached to a balloon of mass $M$ in mid-air. If the monkey climbs up the rope and reaches the top of the rope,what is the distance covered by the balloon? (The total length of the rope is $L$)

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