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 Answer:

Option C – 50J

Explanation

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 = ½(0.05+0.95)×10²
= ½×100
= 50J