A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms⁻¹ gets embedded in it, then loss of kinetic energy will be :

Solution

Physics sir · 3 steps

The bullet sticks, so only momentum is conserved. Find the common speed, then compare kinetic energies.

1. Momentum is conserved. The bullet (m = 0.2 kg) embeds in the bag (M = 9.8 kg), so they move together after impact. This is a perfectly inelastic collision.

   m u = (m + M) v ⇒ v = (0.2 × 10)/(0.2 + 9.8) = 0.2 m s⁻¹

2. Kinetic energies. Before: 1/2 × 0.2 × 10² = 10 J. After: 1/2 × 10 × 0.2² = 0.2 J.
3. So. Loss of kinetic energy = 10 - 0.2 = 9.8 J, which is option (b).

The idea to keep. When a moving body sticks to one at rest, conserve momentum, not energy. The fraction of KE lost is M/(m+M), so a light bullet in a heavy block loses almost all of it.

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