Answer
The bullet's speed after passing through the block is 395 m/s.
Work Step by Step
Let $A$ be the bullet, and let $B$ be the block.
The kinetic energy of the block just after the collision will be equal to the potential energy at maximum height.
$K = PE$
$\frac{1}{2}m_B~v_{B2}^2 = m_B~gh$
$v_{B2} = \sqrt{2gh} = \sqrt{(2)(9.80~m/s^2)(0.0038~m)}$
$v_{B2} = 0.273~m/s$
We can use conservation of momentum to find the final speed of the bullet.
$m_A~v_{A2}+m_B~v_{B2} = m_A~v_{A1}$
$m_A~v_{A2} = m_A~v_{A1}-m_B~v_{B2}$
$v_{A2} = \frac{m_A~v_{A1}-m_B~v_{B2}}{m_A}$
$v_{A2} = \frac{(0.00500~kg)(450~m/s)-(1.00~kg)(0.273~m/s)}{0.00500~kg}$
$v_{A2} = 395~m/s$
The bullet's speed after passing through the block is 395 m/s.
