Q 11-05-129JEE MainJEE Main 2022 (27 Jun, Shift 1)Medium
A pendulum of length $2$ m consists of a wooden bob of mass $50$ g. A bullet of mass $75$ g is fired towards the stationary bob with a speed $v$. The bullet emerges out of the bob with a speed $\dfrac{v}{3}$ and the bob just completes the vertical circle. The value of $v$ is ______ $\text{m s}^{-1}$ (if $g = 10\ \text{m s}^{-2}$).
Numerical value type. Enter your answer.
Answer: 10
For the bob (on a string) to just complete the vertical circle, its speed at the bottom must be
$$u = \sqrt{5gl} = \sqrt{5\times10\times2} = 10\ \text{m s}^{-1}$$
Momentum conservation during the impact:
$$0.075\,v = 0.075\cdot\frac v3 + 0.05\times10$$
$$0.05\,v = 0.5 \Rightarrow v = 10\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics