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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

A.

100J

B.

150J

C.

50J

D.

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=10ms1

Hence the Kinetic energy = 12(0.05+0.95)×102

= 12×100

= 50J