Question

A 9.8-g bullet is fired into a stationary block of wood having mass m = 5.04 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.609 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.)

Answer 1

Answer:

313.8 m/s

Explanation:

Momentum p must be conserved.

The momentum before the collision:

$p=m_{bullet}v_{bullet} + m_{block}v_{block}$

The momentum after the collision:

$p=(m_{bullet}+m_{block})v_{bullet+block}$

Solving both equations:

$v_{bullet}=\frac{(m_{bullet}+m_{block})v_{bullet+block}}{m_{bullet}}$