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 $\overrightarrow{v_{rel}}$ (relative to rope). Find velocity of balloon relative to the ground ?

• A. $\overrightarrow{V}=\frac{m}{M}{\overrightarrow{v}}_{rel}$
• B. $\overrightarrow{V}=-\frac{m}{M}{\overrightarrow{v}}_{rel}$
  
• C. $\overrightarrow{V}=-\frac{m}{m+M}{\overrightarrow{v}}_{rel}$
• D. $\overrightarrow{V}=+\frac{m}{m+M}{\overrightarrow{v}}_{rel}$

Answer 1

The correct option is C $\overrightarrow{V}=-\frac{m}{m+M}{\overrightarrow{v}}_{rel}$

Given that initially the system is at rest,
i.e., ${\overrightarrow{V}}_{CM}=0$
So ${\overrightarrow{V}}_{CM}=\text{constant}=0$
i.e., $\frac{m\overrightarrow{v}+M\overrightarrow{V}}{m+M}=0$
or $m\overrightarrow{v}+M\overrightarrow{V}=0\left[\text{as}\right(m+M)=\text{finite})]$
i.e., $M\overrightarrow{V}=-m\overrightarrow{v}\left(1\right)$
Furthermore, here it is given that;
${\overrightarrow{v}}_{rel}=\overrightarrow{v}-\overrightarrow{V}\left(2\right)$
Putting the value of $\overrightarrow{v}$ from Eq. (2) in Eq. (1), we get;
$M\overrightarrow{V}=-m({\overrightarrow{v}}_{rel}+\overrightarrow{V})$
or $\overrightarrow{V}=-\frac{m{\overrightarrow{v}}_{rel}}{(m+M)}\left(3\right)$