Question

A 10 g bullet travelling at 200m/s strikes and remain embaedded in a 2 kg target which is originaly at rest but free to move . At what speed does the target move off?

Answer 1

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∴v=22−01
Hence, the target will move off with the speed of 0.99 m/s