First slide

Motion of centre of mass

Question

In a gravity free space, a man of mass M standing at a height h ft above the floor, throws a ball of mass m straight down with a speed u. When the ball reaches the floor, the distance of the man above the floor will be

Moderate

A

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

$h (2 - \frac{m}{M})$
C

2h
D

a function of m, M, h and u

Solution

As in gravity free space, displacement of centre of mass of man and ball is zero. If displacement of the ball be h than the displacement of man in upward direction.
$mh = {Mh}^{'} \Rightarrow h^{'} = \frac{mh}{M} - - - - - - - (i)$
Hence the position of man from ground
$H = h + h^{'} = h + \frac{hm}{M} = h [1 + \frac{m}{M}]$

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