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A lead bullet of mass 0.05kg is fired with a velocity of 200ms\(^{-1}\) into a...

Physics
JAMB 1999

A lead bullet of mass 0.05kg is fired with a velocity of 200ms\(^{-1}\) into a lead block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is

  • A. 50 J
  • B. 100 J
  • C. 150 J
  • D. 200 J
Correct Answer: Option A
Explanation

From principle of conservation of linear momentum,
(0.05 x 200) + (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic).

10 + 0 = V

Thus V = 10m/s.

Recall Kinetic Energy = \(\frac{1}{2} mv^2\)

\(\therefore\) K.E = 1/2 (0.05 + 0.95) x 10\(^2\)

K.E = 1/2 (1 x 100) = 50 J.


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