Answer
The bullet's speed just before impact is $~~1800~m/s$
Work Step by Step
Let $m_B$ be the mass of the block.
Let $m_b$ be the mass of the bullet.
We can find the rotational inertia of the system:
$I = (0.060~kg~m^2)+m_B~R^2+m_b~R^2$
$I = (0.060~kg~m^2)+(0.50~kg)(0.60~m)^2+(0.0010~kg)(0.60~m)^2$
$I = 0.24~kg~m^2$
We can use conservation of angular momentum to find the initial speed of the bullet:
$L_i = L_f$
$Rm_bv = I~\omega$
$v = \frac{I~\omega}{R~m_b}$
$v = \frac{(0.24~kg~m^2)(4.5~rad/s)}{(0.0010~kg)(0.60~m)}$
$v = 1800~m/s$
The bullet's speed just before impact is $~~1800~m/s$
