Answer
Before the collision, the original speed of the bullet was $346~m/s$
Work Step by Step
Let $m_f$ be the total mass of the block and bullet. We can find the rate of deceleration along the horizontal surface:
$F_N~\mu_k = m_f~a$
$m_f~g~\mu_k = m_f~a$
$a = g~\mu_k$
$a = (9.80~m/s^2)(0.400)$
$a = 3.92~m/s^2$
We can find the velocity $v_0$ of the block and bullet when they start sliding along the horizontal surface:
$v_f^2 = v_0^2 + 2ax$
$v_0^2 = v_f^2 - 2ax$
$v_0 = \sqrt{v_f^2 - 2ax}$
$v_0 = \sqrt{0 - (2)(-3.92~m/s^2)(1.50~m)}$
$v_0 = 3.43~m/s$
By conservation of momentum, the final momentum of the system just after the collision is equal to the initial momentum of the bullet. We can find the original speed $v_b$ of the bullet:
$(0.020~kg)~v_b = (2.00~kg+0.020~kg)(3.43~m/s)$
$v_b = \frac{(2.00~kg+0.020~kg)(3.43~m/s)}{0.020~kg}$
$v_b = 346~m/s$
Before the collision, the original speed of the bullet was $346~m/s$.
