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.