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:
As in gravity free space displacement of centre of mass of man and ball system should not change.
If displacement of the ball is h, then the displacement of the man is in the upward direction.
$mh=Mh’\Rightarrow h’=\frac{mh}{M}$
(i) Hence the position of man from the ground $H=h+h’=h=h\ 1+\frac{m}{M}$
BETA
