#### Answer

The turntable's angular velocity just after this event is 50 rpm

#### Work Step by Step

We can find the moment of inertia of the disk before the blocks land on it.
$I_0 = \frac{1}{2}MR^2$
$I_0 = \frac{1}{2}(2.0~kg)(0.10~m)^2$
$I_0 = 0.010~kg~m^2$
We can find the moment of inertia of the system after the blocks land on it. Let $M_b$ be the mass of the blocks.
$I_f = I_0 + M_b~R^2+M_b~R^2$
$I_f = 0.010~kg~m^2 + (0.50~kg)(0.10~m)^2+(0.50~kg)(0.10~m)^2$
$I_f = 0.020~kg~m^2$
We can use conservation of angular momentum to solve this question.
$L_f = L_0$
$I_f~\omega_f = I_0~\omega_0$
$\omega_f = \frac{I_0~\omega_0}{I_f}$
$\omega_f = \frac{(0.010~kg~m^2)(100~rpm)}{0.020~kg~m^2}$
$\omega_f = 50~rpm$
The turntable's angular velocity just after this event is 50 rpm.