Answer
The bullet's speed just before impact is $~~1290~m/s$
Work Step by Step
Let $m_b$ be the mass of the bullet.
We can find the rotational inertia of the rod:
$I = \frac{1}{12}ML^2$
$I = \frac{1}{12}(4.00~kg)(0.500~m)^2$
$I = 0.0833~kg~m^2$
We can find the rotational inertia of the rod-bullet system after impact:
$I_f = (0.0833~kg~m^2)+m_b~L^2$
$I_f = (0.0833~kg~m^2)+(0.00300~kg)(0.500~m)^2$
$I_f = 0.08408~kg~m^2$
We can use conservation of angular momentum to find the initial speed of the bullet:
$L_i = L_f$
$Rm_bv~sin~\theta = I_f~\omega_f$
$v = \frac{I_f~\omega_f}{R~m_b~sin~\theta}$
$v = \frac{(0.08408~kg~m^2)(10~rad/s)}{(0.250~m)(0.0030~kg)~sin~60.0^{\circ}}$
$v = 1290~m/s$
The bullet's speed just before impact is $~~1290~m/s$
