Confusion about conservation of angular momentum 
A uniform rod of length l is slightly disturbed from its vertical position. Find the angular velocity of the rod just after if hits the step. (Friction is sufficient everywhere to prevent slipping)

I have obtained the angular velocity of the rod just before collision with the step using conservation of energy. It comes out as $\sqrt{\frac{3g}{2l}}$ but I couldn't obtain final angular velocity "just after" the collision as the question asks.
I came across a solution that uses conservation of angular momentum of the rod about the contact point on the step "just before" and "just after" the collision. I was confused by it because only the force of reaction from the step passes through that point of collision. There are two torques acting on the rod (one is gravity through COM and the other is the reaction force from the hinge attached to the ground) about the point of contact.
a) So, in this situation is it appropriate to use conservation of angular momentum about the point of contact only "just before" and "just after" the collision?
b) Is there any other way to approach this?
Any help would be appreciated.
 A: If you consider the angular momentum about the contact point, the force of reaction passes through the point and  it won't torque the rod, only gravity will. But just before and just after the contact you can conserve angular momentum about the contact point as during the small $\Delta t$ (time interval that starts just before and and ends just after the contact is made), how much can $mg$ even torque the road? It will contribute some torque, but it's negligible. Same thing goes for the static friction acting on the foot of the road to keep it in place. It would've been a different story if you had tried to conserve angular momentum about the center of mass, as the impulse from the contact point is high enough such that you can't neglect the torque it produces in the aforementioned small $\Delta t$.
To give an easier example, let's say a projectile blasts into many pieces mid-air. There is gravity(external force) in the y-direction, but the internal forces of the blast are so high that you can neglect the work gravity does in the time $\Delta t$(time interval that starts just before and ends just after the blast)$\Rightarrow$ you can conserve momentum in the y-direction as well(despite the external force) just before and just after the blast. Once you get the momentum just after the blast, then just let gravity do it's work and you can easily calculate  the dynamics after the blast.
