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 = ½mv²

∴
K.E = 1/2 (0.05 + 0.95) x 10²

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