Rotational Motion Question 108
Question: In a gravity free space, a man of mass $M$ standing at a height $h$ 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
Options:
A)$h(1+m/M)$
B)$h(2-m/M)$
C)$2h$
D)a function of $m$ , $M$ , $h$ and $u$
Show Answer
Answer:
Correct Answer: A
Solution:
[a]As in gravity free space displacement of centre of mass of man and ball system should not move.
If displacement of the ball be h then 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=h 1+\frac{m}{M} $
