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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 $\frac{1}{4} h$ . The ratio of the masses of the pendulums is

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answer is 1.

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Detailed Solution

(Loss in K.E.of combined system) = (Gain in P.E. of combined system)

$\begin{array}{r} \Rightarrow \frac{1}{2} (m_{1} + m_{2}) V^{2} = (m_{1} + m_{2}) \frac{gh}{4} \Rightarrow V = \sqrt{2 g (\frac{h}{4})} \end{array}$

$\begin{array}{r} m_{1} gh = \frac{1}{2} m_{1} v^{2} \Rightarrow v = \sqrt{2 gh} \end{array}$

Further by Law of Conservation of Momentum

$(m_{1} + m_{2}) V = m_{1} v$

$\Rightarrow (m_{1} + m_{2}) \sqrt{\frac{2 gh}{4}} = m_{1} \sqrt{2 gh}$

$\frac{1}{2} (m_{1} + m_{2}) = m_{1} \Rightarrow m_{1} = m_{2}$

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