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.
(Neglect the height of the man)

• 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 the ball to reach the floor is,

$t=\frac{h}{u}$

$(\because u$ = constant and $a=0)$

On throwing the ball downwards the man will move upwards.

Applying linear momentum conservation on the system of ball and man where,

$m_1$ = Mass of the man.
$m_2$ = Mass of the ball
$v$ = velocity of the man
$u$ = velocity of the ball

$m_{1}v=m_{2}u$

$\Rightarrow v=\frac{m_{2}u}{m_1}$

Distance moved along the upward direction by the man, till the 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 the man from floor,

$h_{\text{total}}=h+h^{\prime }$

$\Rightarrow$ Total distance of the man from floor is,

$h_{\text{total}}=h+\frac{m_2}{m_1}h=(1+\frac{m_2}{m_1})h$