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: A
Explanation
From principle of conservation of linear momentum, (0.05 x 200) +Â (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.
