Question

A 10.0-g bullet is fired into a stationary block of wood having a mass of 5.00 kg. The bullet embeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet?

Answer 1

We can find the original speed of the bullet from the Law of Conservation of Momentum:

$$m_{bullet}v_{bullet}+m_{block}v_{block}=(m_{bullet}+m_{block})v,$$ $$v_{bullet}=\dfrac{(m_{bullet}+m_{block})v}{m_{bullet}},$$ $$v_{bullet}=\dfrac{(0.01\ kg+5\ kg)\cdot 0.6\ \dfrac{m}{s}}{0.01\ kg}=300.6\ \dfrac{m}{s}.$$

Answer:

$v_{bullet}=300.6\ \dfrac{m}{s}.$