Free-body diagram for a collision

Question

Edit - Consider an homogeneous bar of length $L$ and mass $M$. This bar can rotate on a horizontal plane with no friction around a point $A$. The distance between $A$ and the center $O$ of the bar is $a$.
Since we are on a plane, the gravity is not working.

Suppose that this bar is rotating with constant angular velocity $omega_0$.

Moreover, suppose that this bar hits a mass $m$ in a point $B$. The distance between $B$ and the center $O$ is $b$. Regardless the nature of the collision (elastic or inelastic), I was told that linear momentum is not conserved while angular momentum is.

The explanation I received is the following: during the collisions, an impulsive force arises on the fulcrum in $A$; this force is external and hence the linear momentum is not conserved, while the angular one is conserved since this impulsive force does not produce torque in $A$.

This explanation does not convince me totally.

I post a picture.

I would like to figure out which are the forces that arises during the collisions. Moreover, I would like to know in which points they act.

Eli · Answer

I try to solve this problem exactly 
begin{align*}
  &text{I)  The EOM's bevor impact:} \
  M,L,ddot{varphi}&=cos(varphi)left(F_2+M,gright)
  m,ddot{y}&=F_1-m,g
end{align*}
begin{align*}
  &text{II)  The EOM's after impact:} \
  M,L,ddot{varphi}&=cos(varphi)left(F_2+M,g-F_cright),,&(1)
  m,ddot{y}&=F_1-m,g+F_c,,quad& (2)
  tan(varphi)&=frac{y}{b},,quad text{The impact condition}  &(3)
end{align*}
So we habe three equations for three unknowns

$ddot{varphi},,quad ddot{y},quad$  and the impact force   $F_c$

Results:

begin{align*}
    ddot{y}&=frac{2,dot{y}^2,y}{1+y^2},,quad y(0)=h_0,,quad D(y)(0)=0
    ddot{varphi}&=f(y,dot{y},,F_1,,F_2,,b)
 F_c&=M,g+frac{M}{1+y^2}left(2,dot{y}^2,yright)+F_1
  end{align*}
The EOM's don't depent on the  "geometry parameter $a$" !!

I hope it is helpful for you ?

Free-body diagram for a collision

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