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.

Engineering

Physics

Center of Mass

Question

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 decreases with time.

The potential energy of the system increases with time.

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(M + m)g = B
S_cm = 0
⟹ m × –v_m + Mv_b = 0

$v_{m} = \frac{{Mv}_{b}}{m} + v_{b} = v$

$v_{b} = \frac{mv}{m + M}$

ΔU_max + ΔU_b

= –mgS_max + Mg × S_b

$= - mg \times \frac{M}{m} v_{b} t + {Mgv}_{b} t = 0$ .

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