Question

A 0.11 kg bullet traveling at speed hits a 18.3 kg block of wood and stays in the wood. The block with the bullet imbedded in it moves forward with a velocity of 8.8 m/s. What was the velocity (speed) of the bullet immediately before it hit the block (in m/s)?

Answer 1

Step-by-step explanation:

The energy of the system before the collision must equal the energy after the collision.

After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is

`(18.41 kg (8.8 m/s)^2)/(2) = 713 J`

Before the collision only the bullet has kinetic energy.

So we can now determine the speed of the bullet using

`(0.11kg (v^2))/(2) = 713 J\\v = 114 m/s`