A block of mass $1.9\ \text{kg}$ is at rest at the edge of a table, of height $1\ \text{m}$. A bullet of mass $0.1\ \text{kg}$ collides with the block and sticks to it. If the velocity of the bullet is $20\ \text{m s}^{-1}$ in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take $g = 10\ \text{m s}^{-2}$. Assume there is no rotational motion and loss of energy after the collision is negligible.]
Answer: (A) $21\ \text{J}$
Momentum conservation: $0.1\times20 = 2\,v \Rightarrow v = 1\ \text{m s}^{-1}$, so the KE after the collision is $\tfrac12\times2\times1^{2} = 1$ J.
Falling $1$ m adds $mgh = 2\times10\times1 = 20$ J. KE just before hitting the floor $= 21$ J.
Solution by Sreeraj P, M.Sc Physics