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

Question

Philoid · Accepted Answer

Mass of bullet, m = 10 g

= 0.01 kg

Speed of bullet, v = 200 m/s

Mass of target, M = 2 kg

Combine speed after hitting target = V

Before collision, momentum of the system is = mv = 2

After collision, the bullet and target move together. Their combined mass is (m+M) = 2.01 kg

So, the momentum of the system after collision is = 2.01V

Applying conservation of momentum,

Momentum of the system before collision = momentum of the system after collision

2 = 2.01 V

V = 0.99 m/s

This is the combined velocity of the target and bullet after the collision.