Trying to find the Lagrangian between two non-interacting particles So, lets say that our system consists of two particles. Without the gravitational force or any other long-range force, the only possible way for these two particles to interact is to collide with each other directly, at least if the initial conditions are such that they lead to the collision.
This problem is easily solved with Newtonian mechanics, but trying to find the Potential of this system (for the sake of simplicity let's say its  2D, and same mass particles) should be something like $$ L=\dfrac{1}{2}m(\dot{x}_1^2+\dot{x}_2^2+\dot{y}_1^2 +\dot{y}_2^2) - V$$ where
$$V=V(|\vec{r_1} - \vec{r_2}|,|\vec{u_1} - \vec{u_2}|)$$ and my thought was $$V=\Theta (|\vec{r_1} - \vec{r_2}|)\cdot \delta(|\vec{u_1} - \vec{u_2}|)$$ but still can't get the results needed.
Note, that for same mass particles, the potential just changed velocities on $$\vec{r_1}=\vec{r_2}$$
Any ideas? I can still solve this with Newtonian mechanics, but Lagrangian approximation should work the same somehow
EDIT: Newtonian solution
With any given initial conditions (u1,u2,r1(0),r2(0)):
The x coordinate is given for both particles
$$x_1 = u_1 cos(\theta_1)\cdot t + x_{10}$$
$$x_2 = u_2 cos(\theta_2)\cdot t + x_{20}$$
so solving $x_1=x_2$ gives $$t_{x_1=x_2}=\dfrac{x_{20}-x_{10}}{u_1 cos(\theta_1)-u_2 cos(\theta_2)}$$ and same thing for $y_1=y_2$
$$t_{y_1=y_2}=\dfrac{y_{20}-y_{10}}{u_1 sin(\theta_1)-u_2 sin(\theta_2)}$$
and in both cases $\theta$ is the angle of velocity
Collision happens only if$$t_{y_1=y_2}=t_{x_1=x_2}$$  and in that case there is velocity exchange (same mass)
 A: The potential you are looking for is in the form $\delta^{(2)}(\vec{r}_1 - \vec{r}_2) = \delta (x_1 - x_2) \delta(y_1 - y_2)$ (the delta function is symmetric in its arguments, so we do not need to add the absolute value). The dynamical equations then read
$$m\ddot{x}_1 = \delta'(x_1 - x_2)\delta(y_1 - y_2)$$
$$m\ddot{x}_2 = -\delta'(x_1 - x_2)\delta(y_1 - y_2)$$
$$m\ddot{y}_1 = \delta(x_1 - x_2)\delta'(y_1 - y_2)$$
$$m\ddot{y}_2 = -\delta(x_1 - x_2)\delta'(y_1 - y_2)$$
where $\delta'$ is the delta-function derivative. These are of course singular expressions which are best understood in integral form (I will only mention one of the $x$ equations from now on, the $y$ set is analogous) around the collision time $t_{coll}$
$$\int_{t_{coll}-\epsilon}^{t_{coll} + \epsilon} m \ddot{x}_1 dt = \int_{t_{coll}-\epsilon}^{t_{coll} + \epsilon} \delta'(x_1 - x_2) \delta(y_1 - y_2) dt$$
$$\int_{t_{coll}-\epsilon}^{t_{coll} + \epsilon} m \ddot{x}_2 dt = -\int_{t_{coll}-\epsilon}^{t_{coll} + \epsilon} \delta'(x_1 - x_2) \delta(y_1 - y_2) dt$$
Now we take the limit $\epsilon \to 0$, eliminate the right-hand sides and see that we get the usual collision relations
$$m \dot{x}_1(t_{coll} - 0) - m \dot{x}_1(t_{coll} + 0) = - (m \dot{x}_2(t_{coll} - 0) - m \dot{x}_2(t_{coll} + 0))$$
where the $\pm 0$ mean just before/after the collision. The integral $\int_{t_{coll}-\epsilon}^{t_{coll} + \epsilon} \delta'(x_1 - x_2) \delta(y_1 - y_2) dt$ itself cannot be evaluated directly but from the properties of the delta function and its derivatives we know that it is generally non-zero. 
However, with more considerations it is easy to realize that the output of this collision is ill-determined. For instance, it is mathematically admissible that the particles fly through each other without a change of velocity. In other words, there is no well-defined model of point particles in contact interactions.
To eliminate this ambiguity, one can either compute the differential cross-section of the process and input it into every collision by hand, or introduce a different collision model. One of the simplest "contact" collision models is the hard sphere $V \sim \delta^{(d)} (|\vec{r}_1 -\vec{r}_2| - 2R)$ where $R$ is the diameter of the sphere-particles.
