JAMB 2018 Physics · Question 30

A lead bullet of mass 0.05 kg is fired with a velocity of 200 m/s into a lead block of mass 0.95 kg. Given…

Question

A lead bullet of mass 0.05 kg is fired with a velocity of 200 m/s into a lead block of mass 0.95 kg. Given that the lead block can move freely, the final kinetic energy after impact is

A.100J
B.150J
C.50J
D.200J

Answer: C. 50J

Explanation written with AI and checked against the answer key

Analyze the question: A bullet embeds in a free block (a perfectly inelastic collision). We need the kinetic energy of the bullet and block together after impact. Concept: Momentum is conserved in the collision, but kinetic energy is not. Some of it is lost as heat and sound. So we find the final velocity from momentum, then compute the kinetic energy. Solution: Conservation of momentum: m₁u₁ + m₂u₂ = (m₁ + m₂)v (0.05 × 200) + (0.95 × 0) = (0.05 + 0.95)v 10 = 1.0 × v v = 10 m/s Final kinetic energy: KE = ½(m₁ + m₂)v² KE = ½ × 1.0 × (10)² KE = ½ × 100 KE = 50 J Check: The initial KE was ½ × 0.05 × 200² = 1000 J. Only 50 J remains after impact (5%), and the other 950 J is lost as heat and sound. This is what we expect in an inelastic collision. Exam tip: For a bullet embedded in a block, the fraction of KE remaining is m₁/(m₁ + m₂) = 0.05/1.0 = 0.05. So 5% of 1000 J = 50 J. This shortcut saves time in the exam.

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