Question

A bullet (m = 0.05 kg) with velocity of 300 m/s collides with a stationary piece of wood (m = 5 kg). The bullet is embedded in the piece of wood. What is their combined velocity after the collision? Express your answer to the nearest hundredth

Answer 1

Apply momentum conservation: initial momentum (momentum of the bullet) becomes final momentum (momentum of the bullet and the wood):

$$mv=(m+M)u,\\\space\\ u=v\frac{m}{m+M}=2.97\text{ m/s}.$$