A man of mass m_1 is standing at a height h f | vedclass

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(C) In a gravity-free room,the center of mass of the system remains at rest because there are no external forces acting on the system.Let the man move

Vedclass · Accepted Answer

$\left( 1 + \frac{m_2}{m_1} \right) h$

Vedclass · Answer

(C) In a gravity-free room,the center of mass of the system remains at rest because there are no external forces acting on the system.Let the man move upwards with velocity $v$ when he throws the ball downwards with velocity $u$.By the law of conservation of linear momentum,the initial momentum is zero:$m_1 v - m_2 u = 0 \implies v = \frac{m_2 u}{m_1}$.The time taken by the ball to reach the floor is $t = \frac{h}{u}$.During this time,the man moves upwards by a distance $d = v 	imes t = \left( \frac{m_2 u}{m_1} ight) 	imes \left( \frac{h}{u} ight) = \frac{m_2}{m_1} h$.The initial height of the man was $h$. Since he moves upwards by $\frac{m_2}{m_1} h$,his new distance from the floor is $h + \frac{m_2}{m_1} h = \left( 1 + \frac{m_2}{m_1} ight) h$.

$A$ man of mass $m_1$ is standing at a height $h$ from the floor in a gravity-free room. He throws a ball of mass $m_2$ downwards with a speed $u$. Find the distance of the man from the floor when the ball reaches the floor.

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