A hot air balloon of mass $M$ is stationary (with respect to the ground) in mid-air. A passenger of mass $m$ slides down a rope with constant velocity $v$ with respect to the balloon.

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The potential energy of the system decreases with time.

The balloon rises up with a velocity of $\frac{m}{m + M} v$ upward with respect to ground.

The man goes down with a velocity of $\frac{m}{m + M} v$ with respect to ground.

The potential energy of the system increases with time.

answer is A, D.

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Detailed Solution

$(M + m) g = B S_{cm} = 0 \Rightarrow m \times (v_{b} - v_{m}) + {Mv}_{b} = 0 v_{m} = \frac{{Mv}_{b}}{m} + v_{b} = v v_{b} = \frac{mv}{m + M} C o m b i n e d c e n t r e o f m a s s o f t h e s y s t e m r e m a i n s s t a t i o n a r y . H e n c e c h a n g e i n p o t e n t i a l e n e r g y i s z e r o . V e l o c i t y o f t h e m a n r e l a t i v e t o t h e g r o u n d = v - v_{b} = \frac{M v}{M + m} (d o w n w a r d)$