Q 11-05-066JEE MainJEE Main 2024 (1 Feb, Shift 1)Easy
A simple pendulum of length $1\ \text{m}$ has a wooden bob of mass $1\ \text{kg}$. It is struck by a bullet of mass $10^{-2}\ \text{kg}$ moving with a speed of $2\times10^{2}\ \text{m s}^{-1}$. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is (use $g = 10\ \text{m s}^{-2}$)
Answer: (B) $0.20\ \text{m}$
Momentum is conserved in the collision:
$$v = \frac{10^{-2}\times200}{1 + 0.01} = \frac{2}{1.01} \approx 1.98\ \text{m s}^{-1}$$
Then energy is conserved during the swing:
$$h = \frac{v^2}{2g} = \frac{(1.98)^2}{20} \approx 0.20\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics