Question

A balloon of mass M with a light rope and monkey of mass m are at rest in mid air. If the monkey climbs up the rope and reaches the top of the rope, the distance by which the balloon descends will be- (Total length of the rope is L)

• A. $\frac{mL}{(m+M{)}^2}$
• B. $\frac{mL}{m+M}$
• C. $\frac{(m+M)L}{m}$
• D. $\frac{(m+M)}{mL}$

Answer 1

Answer: (B)

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

Given that the initially the system is at rest i.e., $\overrightarrow{v_{CM}}=0$
Now as force of climbing is an internal force
So $\overrightarrow{v_{CM}}=constant=0$
or $\frac{m\overrightarrow{v}+M\overrightarrow{V}}{m+M}=0$
or $m\overrightarrow{v}+M\overrightarrow{V}=0$ ....(ii)
or $m\frac{\Delta \overrightarrow{r_1}}{\Delta t}+M\frac{\Delta \overrightarrow{r_2}}{\Delta t}=0$
$m\overrightarrow{d_1}+M\overrightarrow{d_1}=0$ [$\because \overrightarrow{d_2}$ is opposite to $\overrightarrow{d_1}$]
or $m{d}_1=M{d}_2$.....(ii)
Now as the monkey climbs up L towards the balloon (relative to balloon ), the balloon will descend a distance $d_2$ downwards relative to the ground so that upward displacement of man relative to ground will be
$d_1=L-d_2$ ....(iii) (i.e. $d_1+d_2=L$)
The equation (ii) and (iii)
$m(L-d_2)=M{d}_2$
$d_2=\frac{mL}{m+M}$