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

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
A. 100J
A. 150J
C. 50J
D. 200J

Correct Option: 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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