Answer
The initial speed of the bullet was 265 m/s.
Work Step by Step
We can find the magnitude of deceleration of the block after the bullet hits it.
$ma = F_f$
$ma = mg~\mu_k$
$a = g~\mu_k = (9.80~m/s^2)(0.20)$
$a = 1.96~m/s^2$
We can find the initial speed of the block when it starts to slide.
$v_0^2 = v^2 - 2ax = 0-2ax$
$v_0 = \sqrt{-2ax} = \sqrt{-(2)(-1.96~m/s^2)(0.310~m)}$
$v_0 = 1.10~m/s$
We can use conservation of momentum to find the initial speed of the bullet.
$m_1~v_1 = m_2~v_2$
$v_1 = \frac{m_2~v_2}{m_1}$
$v_1 = \frac{(1.205~kg)(1.10~m/s)}{0.00500~kg}$
$v_1 = 265~m/s$
The initial speed of the bullet was 265 m/s.
