energy conservation - Collision and impulsive forces: a formal approach

Consider two bodies $m$ and $M$. Suppose that $m$ is moving with constant velocity $v_0 > 0$ along a certain axis (e.g., it is moving on the right on the $x$-axis), and at a certain time, it collides with $M$ at $t=0$. Suppose that after the collision, $m$ has velocity $v_1 \in \mathbb{R}$ and $M$ has velocity $w_1>0$.

During the collision, the two bodies are subject to impulsive forces. Given $f>0$, these forces are $-f\delta(t)$ on body $m$ and $f\delta(t)$ of body $M$ due to the third Newton's law. Notice that $\delta(t)$ is the Dirac delta. Moreover, $f$ is $[N\cdot s]$ and $\delta(t)$ is $[s^{-1}]$.

In this setup, I can say that the velocities of two bodies over time are the followings:

$$\begin{cases} v(t) & = & v_0 + (v_1 - v_0)H(t)\\ w(t) & = & w_1 H(t) \end{cases}, $$

where $H(t)$ is the Heaviside step. According to these, we can write that:

$$\begin{cases} \dot{v} & = & (v_1 - v_0)\delta(t)\\ \dot{w} & = & w_1 \delta(t) \end{cases}, $$

since $H'(t) = \delta(t)$.

The laws of motion for the bodies are:

$$\begin{cases} m\dot{v} & = & -f\delta(t)\\ M\dot{w} & = & f \delta(t) \end{cases}, $$

and hence:

$$\begin{cases} m(v_1-v_0)\delta(t) & = & -f\delta(t)\\ Mw_1\delta(t) & = & f \delta(t) \end{cases} \Rightarrow \\ \begin{cases} m(v_1-v_0) & = & -f\\ Mw_1 & = & f \end{cases} \Rightarrow\\ f = Mw_1 = m(v_0-v_1). $$

As a side note, $f$ and $-f$ are the variations of the momenta of $M$ and $m$, respectively. Thus, I have obtained the equation of the conservation of the momentum of the whole system $m+M$.

Suppose that we don't know if the collision is elastic or not. How can we evaluate the variation of kinetic energy without using formulas like $\frac{1}{2}mv^2$, but just looking to the laws of motion and the velocity expression of the two bodies?

Answer

I would have written this as a comment, not an answer, but I don't have the reputation!

Usually, this type of problem is rewritten in the centre of mass frame of the colliding pair. This automatically takes care of conservation of overall momentum. The problem then just involves the relative motion of the two bodies, not their absolute motion. Then, an inelastic collision is specified as a relation of some kind between the pre-collisional and post-collisional relative velocities, perhaps involving a coefficient of restitution.

You can't avoid inputting some kind of equation like this, so as to specify the physics. The collision is still impulsive, but you deduce the value of $f$ from that relation. The consequences for the change in kinetic energy follow from those equations. If the collision is elastic, of course, energy is conserved.