First slide

The steps in collision

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

Moderate

A

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

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

h
D

$\frac{m}{M} h$

Solution

Centre of mass will remain at height h.
$∴ h_{CM} = \frac{m \times 0 + M H}{m + M} = h$
$∴ H = h (1 + \frac{m}{M})$

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