A lead bullet of mass 0.05kg is fired with a velocity of ...
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
50 J
100 J
150 J
200 J
Correct answer is A
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 = 12mv2
∴ K.E = 1/2 (0.05 + 0.95) x 10^2
K.E = 1/2 (1 x 100) = 50 J.