Answer
The final velocity of the block is $3.0~m/s$ to the east.
Work Step by Step
By conservation of momentum, the final momentum of the system is equal to the initial momentum.
Let $m_t$ be the mass of the bullet. Let $v_{t0}$ be the initial velocity of the bullet and let $v_{tf}$ be the final velocity of the bullet. We can find the final velocity $v_b$ of the block:
$m_b~v_b +m_t~v_{tf} = m_t~v_{t0}$
$m_b~v_b = m_t~(v_{t0}-v_{tf})$
$v_b = \frac{m_t~(v_{t0}-v_{tf})}{m_b}$
$v_b = \frac{(0.020~kg)~[200.0~m/s-(-100.0~m/s)]}{2.0~kg}$
$v_b = 3.0~m/s$
The final velocity of the block is $3.0~m/s$ to the east.
