Question

A bullet of 10 gram moves with speed of 200m/s strikes and remain embedded in target which is originally at rest but free to move. At what speed dose the target moves off?

Answer 1

Answer:

Explanation:

Total momentum before collision

= (mass of the bullet x velocity) + (mass of the target x velocity)

= (0.01kg x 200m/s) + 2kg x 0)

= 2kg m/s

Total momentum after collision

= (mass of the bullet + mass of the target) x velocity

= (0.01kg + 2kg) x v

According to the law of conservation of momentum.

Total momentum before collision = Total momentum after collision

Total momentum before collision = Total momentum after collision

= 2kg /s = (2.01)v

∴v=22−01

Hence, the target will move off with the speed of 0.99 m/s