A ball of mass $m$ suspended from a rigid support by an inextensible massless string is released from a height $h$ above its lowest point. At its lowest point it collides elastically with a block of mass $2 m$ at rest on a frictionless surface. Neglect the dimensions of the ball and the block. After the collision the ball rises to a maximum height of :

h / 3

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h / 2

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h / 8

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h / 9

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Solution

The correct option is D $h / 9$
Height of the ball before it is released is $h$ .

Using energy conservation to get the speed of the ball just before collision.

$∴$ $m g h = \frac{1}{2} m u_{1}^{2}$

where $u_{1}$ is the speed of the ball just before its collision with the block.

$⟹$ $u_{1} = \sqrt{2 g h}$

Speed of the block just before collision is zero i.e. $u_{2} = 0$

Speed of the ball just after collision $v_{1} = \frac{(m_{1} - e m_{2}) u_{1} + (1 + e) u_{2}}{m_{1} + m_{2}}$

$∴$ $v_{1} = \frac{(m - 1 \times 2 m) \sqrt{2 g h} + (1 + 1) (0)}{m + 2 m} = \frac{- \sqrt{2 g h}}{3}$

Again using energy conservation for the ball after its collision to get the height upto which the ball rises after the collision.

$m g H = \frac{1}{2} m v_{1}^{2}$

Or $m g H = \frac{1}{2} m (\frac{\sqrt{2 g h}}{3})^{2}$

$⟹$ $H = \frac{h}{9}$

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