Q 11-05-020NEETEAMCET 2013 (Medical)Medium
A ball of mass $m$ moving with a horizontal velocity $v$ strikes the bob of mass $m$ of a pendulum at rest. During this collision, the ball sticks with the bob of the pendulum. The height to which the combined mass rises is ($g$ = acceleration due to gravity)
Answer: (B) $\dfrac{v^2}{8g}$
Perfectly inelastic collision: $mv = 2mV \Rightarrow V = \dfrac{v}{2}$.
Then energy is conserved while swinging up:
$$h = \frac{V^2}{2g} = \frac{v^2}{8g}$$
Solution by Sreeraj P, M.Sc Physics