Question

The balloon, the light rope and the monkey shown in figure are at rest in the air. If the monkey reaches the top of the rope , by what distance does the ballon descent? Mass of the balloon = M , mass of the monkey = m and the lenght of the rope ascended by the monkey = L.

• A. $\frac{mL}{m+M}$ upwards
• B. $\frac{mL}{m+M}$ downwards
• C. $\frac{ML}{m+M}$ upwards
• D. $\frac{ML}{m+M}$ downards

Answer 1

The correct option is B

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

I think by now you may have gotten the basic ideo.
Yes ,
$△$ $X_{COM}=\frac{m_1△X_1+m_2△X_2}{X_1+X_2}$
Here , the motion is in vertical is in vertical direction,
$\therefore$ for simplicity ,
$△$ $y_{com}$ = $\frac{m_1△y_1+m_2△y_2}{m_1+m_2}$
But we know that COM does not moves $△$ $y_{com}$ = 0 = $\frac{m_1△y_1+m_2△y_2}{m_1+m_2}$
Assume the coventional coordinate system i.e upwards as +ve & downwards as -ve
M$△$$y_b$ + m$△$$y_m$ = 0 ..(i)Now as we know by relative motion that
$△$ $y_{\frac{m}{b}}$ = $△$$y_m-△$$y_b$ .. (ii)
m - Monkey
b - Ballon
given $△$ $y_{\frac{m}{b}}$ = L
from (ii) L = $△$$y_m$ - $△$$y_b$ $\Rightarrow$ $△$$y_m$ = L + $△$$y_b$
from (i)
M$△$$y_b$ + m(L + $△$$y_b$) = 0
$△$$y_b$ = $\frac{-mL}{M+m}$ where -ve sign indicated that ballon descends.