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JAMB Physics 1999 Question 2

Question 2

Newton's Laws and Momentum

A lead bullet of mass 0.05 kg is fired with a velocity of 200 m s−1 into a lead block of mass 0.95 kg. Given that the lead block can move freely, the final kinetic energy after impact is

  1. A.50 J
  2. B.100 J
  3. C.150 J
  4. D.200 J
Show the answer and explanation

Answer: A (50 J)

Explanation

Momentum is conserved: 0.05 × 200 = (0.05 + 0.95)v, so v = 10 m s−1. Final KE = ½ × 1 × 102 = 50 J.