A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10 m/s² . Assume there is no rotational motion and loss of energy after the collision is negligable.]

Solution

Physics sir · 4 steps

Use momentum for the sticking collision, then energy for the fall. Do not mix the two.

1. Collision uses momentum. The bullet sticks to the block, so the collision is perfectly inelastic and kinetic energy is not conserved. Momentum is still conserved: 0.1 × 20 = (0.1 + 1.9) v, so v = 1 m s⁻¹.
2. Energy after collision. The combined 2 kg moves off horizontally with kinetic energy 1/2 × 2 × 1² = 1 J. Most of the bullet's 20 J is lost inside the collision.
3. Fall uses energy. After the collision no energy is lost, so gravity adds mgh = 2 × 10 × 1 = 20 J of kinetic energy during the 1 m drop.

   K_(floor) = 1/2(m+M)v² + (m+M)gh = 1 + 20 = 21 J

4. So. The kinetic energy just before the system strikes the floor is 21 J, option (b).

The idea to keep. When bodies stick together, find the common velocity from momentum, never from energy. Energy conservation only applies to the motion after the collision.

Answer from the NTA final answer key. Solution by PrepWiser.