In the arrangement shown, the pendulum $A$ is pulled aside. It is then released and allowed to collide with other pendulum $B$ 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 $\frac{m_{A}}{m_{B}}$ is:

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Solution

let $\frac{m_{a}}{m_{b}} = k m_{a} = k m_{b}$
Speed of $A$ at the bottom $= \sqrt{2 g h}$
after a perfectly inelastic collision, both will move with same velocities.
Conserving momentum.
$k m v + m v = k m \sqrt{2 g h} v = \frac{k \sqrt{2 g h}}{(1 + k)}$
Since their height was $\frac{h}{4},$ final vel. has to be $\sqrt{\frac{g h}{2}}$
$\sqrt{\frac{g h}{2}} = \frac{k \sqrt{2 g h}}{(1 + k)} \frac{k}{1 + k} = \frac{1}{2} k = 1$

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In the arrangement shown, the pendulum A is pulled aside. It is then released and allowed to collide with other pendulum B which is at rest. A perfectly inelastic collision occurs and the system rises to a height 14h. The ratio of the masses of the pendulums mAmB is:

let mamb=kma=kmbSpeed of A at the bottom =√2ghafter a perfectly inelastic collision, both will move with same velocities.Conserving momentum.kmv+mv=km√2ghv=k√2gh(1+k)Since their height was h4, final vel. has to be√gh2 √gh2=k√2gh(1+k)k1+k=12k=1