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$$ \newcommand{\D}{\Delta} \newcommand{\d}{\delta} \newcommand{\O}{\Omega} \newcommand{\o}{\omega} \newcommand{\e}{\varepsilon} \newcommand{\dd}{\mathrm{d}} \newcommand{\pd}{\partial} \newcommand{\abs}[1]{\left\lvert#1\right\rvert} \newcommand{\tup}[1]{\left(#1\right)} \newcommand{\atup}[1]{\left\langle#1\right\rangle} \newcommand{\v}{\vec} \newcommand{\mc}{\mathcal} \newcommand{\L}{\mc{L}} $$

What is a very simple way to model a 1D inelastic collision as a closed system in the Lagrangian or Hamiltonian framework? By closed I mean total energy is conserved. I'd like to include some other component that absorbs the KE lost in the collision and I'm wondering how to do that.

Let's make the setup concrete. Consider two particles of unit mass with 1D positions $x_1, x_2$ and velocities $v_1, v_2$. A simple (and standard) way to model a collision between them is using a repulsive inverse-square potential. The Lagrangian for this system is $\L = T - U$, where the kinetic energy term is $T = \frac{1}{2}v_1^2 + \frac{1}{2}v_2^2$ and the potential energy term is

$$ U = \frac{k}{\abs{x_2-x_1}} $$

for some constant $k$.

However, to model the kind of elastic collision often presented where no force is felt on the particles except at the moment of collision (e.g. as given here), we can use a piecewise interaction energy so that the particles feel no force until they are within a particular distance $\e$ with each other, at which point the interaction kicks in. By making the interaction distance $\e$ arbitrarily small we approximate an instantaneous collision. Here is the potential energy term:

$$ U = \begin{cases} k\tup{\frac{1}{\abs{x_2-x_1}} - \frac{1}{\e}} & \abs{x_2 - x_1} < \e \\ 0 & \mathrm{otherwise} \end{cases} $$

(The inverse square potential is offset so that the PE is 0 when $\abs{x_2 - x_1} = \e$, making $U$ a continuous function of $x_1,x_2$. It doesn't really matter what the non-zero part of the potential function looks like, so long as it explodes to infinity as the particles approach each other.)

My question is, what would be a simple way to modify this setup so that we can approximate an inelastic collision with some arbitrary coefficient of restitution between 0 and 1, such that total energy is conserved? To be specific, let suppose there is another degree of freedom with position $y$ and velocity $w$. What additional potential energy term can we add to $U$ s.t. KE is sucked out of the two particles during their collision and added to this third DoF, either in the form of potential or kinetic energy?

edit: I replaced the potential $1/(x_2-x_1)^2$ with $1/\abs{x_2-x_1}$ everywhere, which is the correct inverse-square law potential.

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To model the inelastic collision, we need a velocity dependent potential. The one you proposed looks awkward. I propose a combine of elastic potential $-kr$ and the $1/r^2$ potential, for the relative distance $r$. $$ U(r) = \Big\{ \begin{matrix} \frac{k_{in} (r_c - r)}{r^2}, \,\,\,\text{ for } r < r_c.\\ 0, \,\,\, \text{ for } r>r_c. \end{matrix} $$ The $k_{in}$ indicates the force constant during two colliders are approaching each other.

For inelastic collision, we make the potential softer by given an smaller $k_{out}$ during the two colliders are separating from each other: $$ U(r) = \Big\{ \begin{matrix} \frac{k_{out} (r_c - r)}{r^2}, \,\,\,\text{ for } r < r_c.\\ 0, \,\,\, \text{ for } r>r_c. \end{matrix} $$

The area difference between these two potentials in between $r_c$ and the turning point is the energy lost during the collision. And the ratio between these two force constants is an index for the coefficient of restitution: $$ \alpha = \frac{k_{out}}{k_{in}}. $$

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  • $\begingroup$ Thanks for the reply. If I may generalize your trick, we can define the potentials $U(r) = k_{in}f(r)$ and $U_{out} = k_{out}f(r)$ s.t. a repulsive force is felt as $r\to 0$ (i.e. $f$ explodes as $r\to 0$), and define $U(r) = U_{in}(r)$ while the particles are moving towards each other and $U(r) = U_{out}(r)$ while the particles are moving away. Then the total KE extracted from the system (negative work) is $(k_{in} - k_{out})f(r_0)$ where $r_0$ is the closest the particles get to each other (i.e. the point where they turn around). $\endgroup$
    – danabo
    Commented Sep 6, 2021 at 18:31
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    $\begingroup$ FYI, the potential $\frac{(r - r_c)}{r^2}$ is inverted. It should be $\frac{(r_c - r)}{r^2}$ to produce a repulsive force. $\endgroup$
    – danabo
    Commented Sep 6, 2021 at 18:32
  • $\begingroup$ Your solution models an open system that loses energy to the environment. I'm wondering how to model a third component (in addition to the two particles) which receives the KE lost in the collision, so that their combined interaction conserves total energy. $\endgroup$
    – danabo
    Commented Sep 6, 2021 at 18:37
  • $\begingroup$ You have to introduce some internal degree of freedom for the two colliders. such that the internal energy of the collider increase by the amount of the lost kinetic energy. Such as two bi-atom molecules, they absorbed part of the kinetic energy converted to increase their internal vibration energies. $\endgroup$
    – ytlu
    Commented Sep 7, 2021 at 12:59
  • $\begingroup$ @danabo In your model, the total extracted kinetic energy is not $( k_{in} - k_{out} ) f(r_0)$, but $( k_{in} - k_{out} ) \int_{r_0}^\infty f(r) dr$. $\endgroup$
    – ytlu
    Commented Sep 7, 2021 at 13:23

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