Question

In a gravity free space, a man of mass $M$ standing at a height $h$ above the floor, throws a ball towards the ground. When the ball reaches the floor, the distance of the man above the floor will be:

• A. $h(1+\frac{m}{M})$
• B. $h\left(2-m/M\right)$
• C. $2h$
• D. a function of $m,M,h$ and $u$

Answer 1

Answer: (A)

$h(1+\frac{m}{M})$

In the gravity free space, lets consider man and ball be our system.
For a system, the center of mass remains same if there are no external forces on the system i.e. internal forces cannot change the center of mass of system.
Therefore, net change in center of mass $=0$
After the ball is thrown let the man moves a distance $x$ in opposite direction.
Let, $\uparrow$ direction be positive and viceversa
So, $Mx+m(-h)=0\Rightarrow x=\frac{hm}{M}$
Final distance from floor $=$ initial distance $+$ change after ball is thrown $=h+\frac{hm}{M}=h(1+\frac{m}{M})$