A 0.18kg bullet traveling at 411m/s strikes a 2.4kg block of wood sitting at rest. The bullet is embedded in the block. Assuming no friction what is the resultant velocity of the block?

Question

A 0.18kg bullet traveling at 411m/s strikes a 2.4kg block of wood sitting at rest. The bullet is embedded in the block. Assuming no friction what is the resultant velocity of the block?

Answer 1

Hi Chloe,

For this problem you have to implement momentum conservation laws. Because the bullet becomes embedded in the block, this is considered an inelastic collision and kinetic energy is not conserved (some energy goes into embedding the bullet in the block).

In order to use momentum conservation, we have to define the system over which momentum is conserved. In this case, the system is the block and the bullet. Therefore, we need to use our linear momentum equations to solve the momentum before and after the collision for our system:

p_initial = p_final

p_{block (before)} +p_{bullet (before)} = p_{block + bullet (after)}

m_{block (before)}*v_{block (before)}+m_{bullet (before)}*v_{bullet (before)} = (m_block+m_bullet)*v_{block +bullet (after)}

We know from the problem that the block is at rest before the collision ( v_{block (before)} = 0) and the bullet and block become one object after the collision, so we can make the final object's mass m_block+m_bullet. We want to solve for the final velocity, so doing some algebra and substituting v_{block (before)} = 0:

v_{block +bullet (after) =} [m_{bullet (before)}*v_{bullet (before)} ]/(m_block+m_bullet)

Substituting in the numbers from the problem:

v_{block +bullet (after)}= [0.18 kg*411 m/s ]/(2.4 kg+0.18 kg)

v_{block +bullet (after)}=28.67 m/s