Question

In a gravity free room a man of mass $m_1$ is standing at a height $h$ above the floor. He throws a ball of mass $m_2$ vertically downwards with a speed $u$. Find the distance of the man from the floor when the ball reaches the ground.

• A. $\left(\frac{m_1+m_2}{m_1-m_2}\right)h$
• B. $\frac{h}{m_1}$
• C. $(1+\frac{m_2}{m_1})h$
• D. $\frac{m_2}{m_1}h$

Answer 1

The correct option is C $(1+\frac{m_2}{m_1})h$

In the gravity free room, $F_{ext}$ = 0
Time taken by ball to reach the floor is,
$t=\frac{h}{u}(\because u$ = constant and $a=0)$
On throwing ball downwards the man will move in upward direction.
Applying linear momentum conservation on (ball + man).
$m_{1}V=m_{2}u$
$\Rightarrow V=\frac{m_{2}u}{m_1}$
where $(m_1\to$ mass of man, $V\to$ velocity of man)
Distance moved along upward direction by man, till ball reaches floor is,
$h^{\prime }=Vt=\left(\frac{m_{2}u}{m_1}\right)\times \frac{h}{u}$
$\Rightarrow h^{\prime }=\frac{m_2}{m_1}h$
Hence total distance of man from floor,
$h_{\text{total}}=h+h^{\prime }$
$\Rightarrow$ Total distnace of the man from floor is = $h+\frac{m_2}{m_1}h=(1+\frac{m_2}{m_1})h$