Question

In a gravity free space, a man of mass $M$ standing at a height $H$ above the floor throws a stone of mass $m$ downwards with a speed of $u$. When the stone reaches the floor, distance of the man above the floor will be:
A) $h$
B) $2h$
C) $h - \dfrac{{2mh}}{M}$
D) $\dfrac{{mh}}{M} + h$

Answer 1

Hint-The vector sum of linear momentum of all the bodies of the system is maintained in an isolated system according to the law of momentum conservation and is not affected due to their reciprocal action or reaction. A device with no external force operating on it means an independent system.

Complete step by step answer:
There are no particles present in the system in gravity-free space. There is, therefore, no force acting on the iron man. The center of mass remains the same for a system if there are no external forces on the system, that is internal forces do not change the system's center of mass.
Therefore, net change in center of mass =0.

Given, in a gravity free space, a man of mass $M$ standing at a height $h$ from the ground and releasing a stone of mass $m$ downwards with velocity $u$, after the release a man is moving little upward with velocity $V$, because of gravity free space.

From the law of conservation of momentum,
$ \Rightarrow MV - mu = 0$
$ \Rightarrow V = \dfrac{{mu}}{M}$ ………………(1)
Here, $M$ is the mass of the iron man
$m$ is the mass of the stone.
$u$ is the velocity of stone
$V$ is the velocity of man.
Now, time taken by the stone to reach floor is given by,
$ \Rightarrow {t_{stone}} = \dfrac{d}{v} = \dfrac{h}{u}$ ………………..(2)
In that time, height reached by the iron man of mass m is given by,${t_{man}} = \dfrac{{{h'}}}{V}$ ………………….(3)
From equation (1) and (2), time taken in each case is the same during the upward and downward motion.
$ \Rightarrow {t_{stone}} = {t_{man}}$
$ \Rightarrow \dfrac{h}{u} = \dfrac{{{h'}}}{V}$
From above equation we get,
The height reached by the man after release of stone from height his given by,
  $ \Rightarrow {h'} = \dfrac{h}{u}V$
From equation (1) we have,
 $ \Rightarrow V = \dfrac{{mu}}{M}$
Substitute in above equation we get,
$ \Rightarrow {h'} = \dfrac{h}{u}\left( {\dfrac{{mu}}{M}} \right)$
 $ \Rightarrow \dfrac{{mh}}{M}$
Total height from the floor is given by,
$ \Rightarrow H = h + {h'}$
$\therefore H = h + \dfrac{{mh}}{M}$
This is the distance of iron man above the floor.

therefore Correct option is (D).

Note:
-Momentum is a property of a moving body, which is a product of mass and velocity.
-The law of conservation of linear momentum is universal that as it applies to both, microscopic as well as macroscopic systems.