A lead bullet of mass 0.05 kg is fired with a speed of 200 m/s into a lead block of mass 0.95 kg at rest…

Analyze the question

A bullet embeds in a block at rest, so this is a perfectly inelastic collision. Momentum is conserved. Kinetic energy is not conserved because some is lost as heat and sound. We need the kinetic energy of the block (with the bullet embedded) just after impact.

Concept

Conservation of momentum: m₁u₁ + m₂u₂ = (m₁ + m₂)v

Kinetic energy: KE = ½mv²

Solution

1. Momentum before impact: (0.05 × 200) + (0.95 × 0) = 10 kg·m/s

2. Momentum after impact (bullet and block move together): (0.05 + 0.95)v = 1.0 × v

3. Equate them: v = 10 / 1.0 = 10 m/s

4. Kinetic energy after impact: KE = ½ × 1.0 × (10)² = ½ × 100 = 50 J

Check against the options

The kinetic energy of the combined bullet and block is 50 J, which matches option C. If you took only the block's mass (0.95 kg), you would get ½ × 0.95 × 100 = 47.5 J, which is approximately 50 J. So the answer is the same either way, and C is the closest option.

For comparison, the bullet's initial KE was ½ × 0.05 × 200² = 1000 J. Only 50 J remains as KE, and the other 950 J is lost as heat and sound, which is why the collision is inelastic.

Tip

For "bullet embeds in block" problems, find v from momentum first, then use KE = ½mv². A shortcut is KE_after = KE_before × (m_bullet / total mass) = 1000 × 0.05 = 50 J. Never apply conservation of kinetic energy in an inelastic collision.