Question

A man of mass $m$ climbs on a rope of length $L$ suspended below a balloon of mass $M$. The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed $V_{rel}$ (relative to rope). In what direction and with what speed (relative to ground) will the balloon move?

• A. downwards, $\frac{m{V}_{rel}}{m+M}$
• B. upwards, $\frac{M{V}_{rel}}{m+M}$
• C. downwards, $\frac{m{V}_{rel}}{M}$
• D. downwards, $\frac{(m+m)V_{rel}}{M}$

Answer 1

The correct option is A downwards, $\frac{m{V}_{rel}}{m+M}$

Let,$V_{man,ground},V_{balloon,ground},V_{man,balloon}$ is velocity of man respect to ground , velocity of balloon respect to ground , relative velocity between man and balloon$V_{rel}$ respectively.

Applying conservation of momentum

$m{V}_{man,ground}+M{V}_{ballon,ground}=0$

m($V_{man,balloon}+V_{balloon,ground})+M{V}_{balloon,ground}=0$

$(m+M)V_{balloon,ground}=-m.V_{man,balloon}$

$V_{balloon,ground}=-\frac{m}{M+m}{V}_{man,balloon}=-\frac{m}{M+m}{V}_{rel}$

Hence,option A is correct