In the arrangement shown, the pendulum on the left is pulled aside. It is then released and allowed to collide with other pendulum which is at rest. A perfectly inelastic collision occurs and the system rises to a height $\frac{h}{4}$ . The ratio of the masses $(m_{1} / m_{2})$ of the pendulum is :

1

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2

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3

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4

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Solution

The correct option is D $1$
For mass $m_{1}$ :
Velocity of mass $m_{1}$ just before collision $v = \sqrt{2 g h}$
After collision, the combined mass $(m_{1} + m_{2})$ moves with a common velocity $V$ and reaches a height $\frac{h}{4}$ .
Using conservation of momentum just before and after collision : $P_{i} + P_{f}$
$∴$ $m_{1} v + 0 = (m_{1} + m_{2}) V$
$∴$ $m_{1} \sqrt{2 g h} = (m_{1} + m_{2}) V$ $⟹ V = \frac{m_{1} \sqrt{2 g h}}{m_{1} + m_{2}}$ .............(1)
Using the relation $V = \sqrt{2 g \times \frac{h}{4}} = \frac{\sqrt{2 g h}}{2}$ in (1)
$∴$ $\frac{m_{1} \sqrt{2 g h}}{m_{1} + m_{2}} = \frac{\sqrt{2 g h}}{2}$ $⟹ m_{1} = m_{2}$

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