Question

A man of mass $m$ climbs 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 $v_{\text{rel}}$ (relative to the rope), in what direction and with what speed (relative to the ground) will the balloon move?

• A. Downwards, $\frac{m{v}_{\text{rel}}}{m+M}$
• B. Upwards, $\frac{m{v}_{\text{rel}}}{m+M}$
• C. Downwards, $\frac{m{v}_{\text{rel}}}{M}$
• D. Downwards, $\frac{(M+m)v_{\text{rel}}}{M}$

Answer 1

The correct option is A Downwards, $\frac{m{v}_{\text{rel}}}{m+M}$


Initially the system of balloon and man is at rest and when the man starts climbing up on the rope, $F_{ext}=0$ on the system.

$m$ = mass of the man

$M$ = mass of the balloon

Thus applying law of momentum conservation in vertical direction on the system of man and balloon.

$p_f=p_i$

$\Rightarrow m{v}_m-M{v}_b=0.....\left(i\right)$

Taking upwards as $+\text{ve}$ direction.

Here $v_m$ is the velocity of man w.r.t ground and $v_{\text{rel}}$ is the man's velocity w.r.t the rope.

$v_b$ is velocity of balloon w.r.t ground

$\Rightarrow v_m=v_{\text{rel}}+(-v_b)$

or, $v_m=v_{\text{rel}}-v_b.....\left(ii\right)$

Since $p_i=0$, the balloon will tend to move downwards in response of the man climbing upwards such that the final momentum remains zero.

From Eq. $\left(i\right)$ & $\left(ii\right)$.

$m(v_{\text{rel}}-v_b)-M{v}_b=0$

or, $m{v}_{\text{rel}}=m{v}_b+M{v}_b$

$\Rightarrow v_b=\frac{m{v}_{\text{rel}}}{M+m}$


$\begin{array}{c}\text{Why this question}?\\ \text{Caution: While applying momentum conservation,}\\ \text{pay attention to keep all the velocities w.r.t ground}\\ \text{along with the proper signs as per the reference direction.}\end{array}$