A bullet of mass 0.1 kg moving horizontally with speed 400 ms⁻¹ hits a wooden block of mass 3.9 kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves 20 m before coming to rest. The coefficient of friction between the block and the surface is __________.
(Given g=10 m / s² )

Solution

Physics sir · 3 steps

Momentum gives the speed right after impact, then friction must eat up that kinetic energy over 20 m.

1. Speed after impact. The collision is perfectly inelastic, so kinetic energy is lost, but momentum is conserved. 0.1 × 400 = (0.1 + 3.9) v, which gives v = 40/4 = 10 m s⁻¹.
2. Friction slows it. On the rough surface, friction is μ M g, so the deceleration is a = μ g, whatever the mass. The combined block slows from 10 m s⁻¹ to rest over s = 20 m.

   v² = 2μ g s ⇒ 10² = 2 × μ × 10 × 20

3. So. 100 = 400 μ, so μ = 0.25, which is option (b).

The idea to keep. Split a bullet-and-block problem into two stages. Use momentum for the collision itself, because energy is lost there. Use energy or kinematics only for the motion after it.

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