Will the rods rotate about the hinge after collision? 
The problem is like this: two blocks(considered to be point masses) of masses $m$ are simultaneously moving with uniform velocities $v_0$ towards the rods and they hit the edges of rod and stick to it(i.e, perfectly inelastic collision). Each of the rods have a mass $m$(same as the blocks) and have length $l$. They are hinged together in the middle by a friction less hinge. Find the velocity of the hinge just after collision and the angular velocities of the rods about the hinge. 
My approach:
Let the vel. of the hinge after collision be $v$. Then by conservation of linear momentum, $$mv_0 + mv_0 = 4mv \Rightarrow v= v_0/2$$
Note: I believe that the centre of mass of the system(two rods and the blocks sticked  to them) is at the hinge just after collision.

Let the angular velocity of the rods about the hinge be $\dot\theta$ 
Now, let us consider the angular momentum of the system abot point $P$.
$$2lmv_0=0+2l \cdot m \cdot(v_0/2+\dot\theta l)+ \frac{3l}{2} \cdot m \cdot(v_0/2+\dot\theta \frac{l}{2}) + \frac{ml^2}{12} \cdot \dot{\theta} + \frac{l}{2} \cdot m \cdot(v_0/2+\dot\theta \frac{l}{2}) - \frac{ml^2}{12} \cdot \dot{\theta}$$
On solving, we get $\dot{\theta}=0$
So, according to my equations, the angular velocity is zero.
Are my calculations oK? Or am I missing something? 
 A: The total angular momentum about the hinge is $0$ by symmetry, but that does not mean the rods are not rotating about the hinge.  It just means the rods are rotating at the same speed in opposite directions.  You can figure the rotation rate by considering the angular momentum of one rod+block about the hinge.  There is no torque around the hinge, so you can consider one side in isolation for this.  Just before the collision it is $mv_0l$.  The moment of inertia of a rod+block is $\frac 43ml^2$.  If we define $\phi$ as the angle of either rod with respect to the horizontal, but measured clockwise on the right and counterclockwise on the left, the angular momentum of one side is $\frac 43ml^2\dot \phi$  This gives $\dot \phi = \frac {3v_0}{4l}$ 
So after the impact the center of mass is moving at $\frac 12v_0$ upward and both rod+blocks are rotating upward around the hinge at $\dot \phi = \frac {3v_0}{4l}$. The center of mass is on the center line at $\frac 34$ of the way along the rods.
A: Your linear momentum calculation is pre-supposing that the entire system moves as a rigid body post-impact.
But each of the two hinged pieces could be rotating about the hinge (so that the overall system is not rigid). The center of mass of each piece is distance $\frac{3}{4}l$ from the hinge, so that balance of linear momentum gives
$$2mv_0 = 4m\left(v + \frac{3}{4}l\dot\theta\right)$$
rather than just $4mv$ on the RHS.
