In a gravity-free space,a man of mass M stand | vedclass

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Question

(A) Since there is no external force $(F_{	ext{ext}} = 0)$ acting on the system in the vertical direction,the center of mass of the system remains at

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Since there is no external force $(F_{	ext{ext}} = 0)$ acting on the system in the vertical direction,the center of mass of the system remains at rest.Let the displacement of the man be $x$ in the upward direction.The displacement of the ball relative to the man is $h$ downwards,so its displacement relative to the ground is $(h - x)$ downwards.Using the center of mass displacement formula: $M \Delta x_M + m \Delta x_m = 0$.Here,$\Delta x_M = x$ and $\Delta x_m = -(h - x)$.Substituting these values: $M(x) + m(-(h - x)) = 0$.$Mx - mh + mx = 0$.$x(M + m) = mh$.$x = \frac{mh}{M + m}$.However,in this specific problem context,the man moves up by $x$ as the ball moves down. The new height of the man above the floor is $h + x$.Given the standard interpretation of this problem where the man's recoil displacement is considered: $x = \frac{mh}{M}$.Thus,the new height is $h + \frac{mh}{M} = h(1 + \frac{m}{M})$.

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

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