A lead bullet of mass 0.05 kg is fired with a velocity of...
A lead bullet of mass 0.05 kg is fired with a velocity of 200 m/s into a lead block of mass 0.95 kg. Given that the lead block can move freely, the final kinetic energy after impact is
100J
150J
50J
200J
Correct answer is C
From the law of conservation of linear momentum,
m1u1+m2u2=m1v1+m2v2
Since the collision is inelastic, we have
(0.05×200)−(0.95×0)=(0.05+0.95)V
10=V
V=10ms−1
Hence the Kinetic energy = 12(0.05+0.95)×102
= 12×100
= 50J
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