#### Answer

The angular velocity is 8.0 rad/s

#### Work Step by Step

We can find the rod's moment of inertia.
$I = \frac{1}{3}ML^2$
$I = \frac{1}{3}(0.75~kg)(0.50~m)^2$
$I = 0.0625~kg~m^2$
The angular impulse is equal to the change in angular momentum. We can find the angular impulse exerted by the force.
$\Delta L = \tau ~t$
$\Delta L = (r\times F)~t$
$\Delta L = (0.25~m)(1000~N)(2.0\times 10^{-3}~s)$
$\Delta L = 0.50~N~s$
Since the rod started from rest, the angular momentum just after the blow from the force is $0.50~N~s$. We can find the angular velocity immediately after the blow.
$L = I~\omega$
$\omega = \frac{L}{I}$
$\omega = \frac{0.50~N~s}{0.0625~kg~m^2}$
$\omega = 8.0~rad/s$
The angular velocity is 8.0 rad/s