A 0.005kg bullet is fired at 400m/s straight into a stationary 1.0kg block of wood. The bullet exits the far side of the block and continues on the same straight line. The block also moves off in this same direction with a velocity of 1.2m/s. Calculate the velocity of the bullet after it exits the block.

To calculate the velocity of the bullet after it exits the block, we can use the principle of conservation of momentum. Assuming that no external forces act on the system (the bullet and the wood block), the total momentum before the collision is equal to the total momentum after the collision.

Before the collision:
The momentum of the bullet is given by: p_bullet = m_bullet * v_bullet
Where,
m_bullet = mass of the bullet = 0.005 kg (given)
v_bullet = velocity of the bullet = 400 m/s (given)

The momentum of the block is given by: p_block = m_block * v_block
Where,
m_block = mass of the block = 1.0 kg (given)
v_block = velocity of the block = 1.2 m/s (given)

The total momentum before the collision is then: p_initial = p_bullet + p_block

After the collision:
Let v_bullet' be the velocity of the bullet after it exits the block.

The momentum of the bullet after the collision is given by: p_bullet' = m_bullet * v_bullet'

The momentum of the block after the collision is given by: p_block' = m_block * v_block'

The total momentum after the collision is then: p_final = p_bullet' + p_block'

Since the total momentum before the collision is equal to the total momentum after the collision, we can set the two equations equal to each other:

p_initial = p_final
m_bullet * v_bullet + m_block * v_block = m_bullet * v_bullet' + m_block * v_block'

Substituting the given values:
(0.005 kg * 400 m/s) + (1.0 kg * 1.2 m/s) = (0.005 kg * v_bullet') + (1.0 kg * 1.2 m/s)

Now, we can solve for v_bullet'.

(0.005 kg * 400 m/s) + (1.0 kg * 1.2 m/s) - (1.0 kg * 1.2 m/s) = (0.005 kg * v_bullet')

20 kg m/s = (0.005 kg * v_bullet')

Dividing both sides by 0.005 kg:

v_bullet' = 20 kg m/s / 0.005 kg

v_bullet' = 4000 m/s

Therefore, the velocity of the bullet after it exits the block is 4000 m/s.