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
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Correct Answer: Option 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.
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.