Answer
After the collision, the final speed of the block and bullet is $5.0~m/s$
Work Step by Step
By conservation of momentum, the final momentum of the system is equal to the initial momentum. We can find the final speed $v_f$ of the block and bullet:
$m_f~v_f = m_0~v_0$
$v_f = \frac{m_0~v_0}{m_f}$
$v_f = \frac{(0.050~kg)(100.0~m/s)}{0.050~kg+0.95~kg}$
$v_f = 5.0~m/s$
After the collision, the final speed of the block and bullet is $5.0~m/s$.
