# What is the formula of the force in an partially inelastic collision

How can i find the force at a specific moment in a partially inelastic collision as a function of the distance between both bodies centers (d), the some of their radius (r), the coefficient of restitution (e) and their relative velocities (v)?

Obs: I want to be able to calculate the force during the collision at every moment given information about both bodies only in the very previous moment, not necessarly the inicial moment. I would like to know the energy loss in heat too.

• You have to be more specific about the geometry of the bodies? In real life even spheres deform in very complex ways that equations cannot predict (youtube.com/watch?v=AkB81u5IM3I) Mar 10, 2018 at 15:08
• I’m trying to create a computer program that simulates collisions with energy loss by heat. I just need a way to recreate this simulation, no need for extreme precision, I just need it to colide and change the relative velocity to a number betwen the inicial relative velocity and the inicial relative velocity times (e). Mar 10, 2018 at 15:28
• Is there a way? Mar 10, 2018 at 15:28
• You mean beyond conservation of momentum? You have to create some kind of penalty method in the contact. Like a non-linear spring that provides restoring force as a function of penetration. Read this and this to learn how to do simulations with contacts. The formula you request depends on the geometry and the material properties. Mar 10, 2018 at 15:32
• I am trying to decide if this question is essentially duplicate or similar to physics.stackexchange.com/questions/216340/…. If you have a force deflection model, then you can simulate dynamic contacts. Mar 10, 2018 at 15:40

## Simple linear spring damper model

1. Problem Definition

Two point bodies (no rotation considered) in contact with mass $m_1$ and $m_2$ respectively. The motion is in one dimensional (the x-axis for example) and they come in contact at exactly $t=0$.

Each body center has location $x_1$ and $x_2$ and problem is parametrized using the combined center of mass location $x_c$ and overlap $e$

$$x_c = \frac{m_1 x_1 + m_2 x_2}{m_1+m_2} \\ e = (x_1+r_1) - (x_2-r_2)$$

Here $r_1$ and $r_2$ are the radii of the objects.

2. Contact Force Model

The contact is a simple spring/damper system where the force $F$ depends on the overlap $e$ and the overlap speed $\dot{e}$

$$\begin{cases} F = k\, e + d\, \dot{e} & e\ge 0 \\ F =0 & e<0 \end{cases}$$

1. Body Equations of Motion

During the contact the contact force is applied at equal and opposite measure on the two bodies

\begin{aligned} m_1 \ddot{x}_1 &= -F = -k\,e - d\,\dot{e} \\ m_2 \ddot{x}_2 & = F = k\,e + d\,\dot{e}\end{aligned}

The accelerations of the two bodies are expressed in terms of the center of mass acceleration $\ddot{x}_c$ and the overlap acceleration $\ddot{e}$.

\begin{aligned} \ddot{x}_1 & = \ddot{x}_c + \frac{m_2}{m_1+m_2} \ddot{e} \\ \ddot{x}_2 & = \ddot{x}_c - \frac{m_1}{m_1+m_2} \ddot{e} \end{aligned}

Use the accelerations in the equations of motion and solve for $\ddot{x}_c$ and $\ddot{e}$

\begin{aligned} \ddot{x}_c & = 0 \\ \ddot{e} & = - \frac{m_1+m_2}{m_1 m_2} (k\,e + d\,\dot{e}) \end{aligned}

2. Overlap Equations of Motion

So besides the obvious that the center of mass velocity remains constant $\dot{x}_c = \mbox{(const)}$ the overlap obeys the 1D damped vibration equation

$$\mu\, \ddot{e} = -k\, e - d\,\dot{e}$$

Where $\mu = \frac{m_1 m_2}{m_1+m_2}$ is the reduced mass of the system.

3. Damped Harmonic Solution

We need to specify the initial conditions at $t=0$ as $e=0$ and $\dot{e}=v$, where $v = \dot{x}_1 - \dot{x}_2$ is the relative speed at impact. The general solution of the equations of motion is

$$e(t) = \frac{v}{\omega} \exp(-\beta t) \sin(\omega t)$$

Using the substitution $k = \mu \omega_n^2$ and $d = 2 \zeta \mu \omega_n$ the above solves the equations of motion when $\omega = \omega_n \sqrt{1-\zeta^2}$ and $\beta = \zeta \omega_n$

$$e(t) = \frac{v}{\omega_n} \left( \frac{ \exp(-\zeta \omega_n t) \sin (\omega_n t \sqrt{1-\zeta^2})}{\sqrt{1-\zeta^2}} \right)$$

The individual object positions are thus

\begin{aligned} x_1 (t) & = \dot{x}_c\,t + \frac{m_2}{m_1+m_2} (e(t)-r_1-r_2) \\ x_2 (t) & = \dot{x}_c\,t - \frac{m_1}{m_1+m_2} (e(t)-r_1-r_2) \\ \end{aligned}

4. Parametrized Solution

$$\begin{cases} v & = \dot{x}_1 - \dot{x}_2 & & \mbox{relative speed at }t =0 \\ \mu & = \frac{m_1 m_2}{m_1+m_2} & & \mbox{reduced mass} \\ \omega_n & = \sqrt{ \frac{k}{\mu} } & & \mbox{natural frequency} \\ \zeta & = \frac{d}{2 \mu \omega_n} & & \mbox{damping ratio} \\ \omega & = \omega_n \sqrt{1-\zeta^2} & & \mbox{damped frequency} \\ \varphi &= \frac{t}{\omega_n} & & \mbox{phase angle} \end{cases}$$

\boxed{ \begin{aligned} e & = \frac{v}{\omega} \exp(-\zeta \varphi) \sin\left( \frac{\omega}{\omega_n} \varphi \right) \\ \dot{e} & = v\, \exp(-\zeta \varphi) \left( \cos\left( \frac{\omega}{\omega_n} \varphi \right) - \frac{\zeta \omega_n}{\omega} \, \sin\left( \frac{\omega}{\omega_n} \varphi \right) \right) \end{aligned} }

5. Force vs. Time

Using the solution from above the force is evaluated from $F = k e + d \dot{e} = (\mu \omega_n^2) e + (2 \zeta \mu \omega_n) \dot{e}$ as

$$F = \mu v \exp(-\zeta \varphi) \left( 2 \zeta \omega_n \cos \left( \frac{\omega}{\omega_n} \varphi\right) + \frac{\omega_n^2 (1-2\zeta^2)}{\omega} \sin \left( \frac{\omega}{\omega_n} \varphi\right) \right)$$

Note that at critical damping with $\zeta = 1$ the overlap and force equations are $$e = \frac{v \varphi \, {\rm e}^{-\varphi} }{\omega_n}$$ $$F = \mu v \omega_n (2-\varphi) \, {\rm e}^{-\varphi}$$

6. Energy Loss due to Damping

Since the center of mass moves with constant velocity the change in kinetic energy is purely because of losses during the impact.

$$\Delta K = \frac{1}{2} \mu v^2 - \frac{1}{2} \mu \dot{e}^2$$

The end of the contact is when $e=0$, or specifically that the phase angle $\varphi = \frac{\pi}{\sqrt{1-\zeta^2}}$

$$\Delta K = \frac{1}{2} \mu v^2 \left(1 - \exp\left( - \frac{2 \pi \zeta}{\sqrt{1-\zeta^2}} \right) \right)$$

So now if you want to target specific energy loss ratio $\eta = \frac{\Delta K}{K}$ you need to use damping ratio

$$\zeta = - \frac{ \ln(1-\eta) }{\sqrt{ 4\pi^2 + \ln(1-\eta)^2}}$$

1. Coefficient of Restitution

Look at the velocity at the end of the contact when $\varphi = \frac{\pi}{\sqrt{1-\zeta^2}}$ as a fraction of the initial velocity $v$ defines the coefficient of restitution $\epsilon$

$$\epsilon = -\frac{\dot{e}}{v} = \exp \left( \frac{\pi \zeta}{\sqrt{1-\zeta^2}} \right)$$

Or in reverse, if one wants to find the damping needed for a specific COR

$$\zeta = - \frac{\ln(\epsilon)}{\sqrt{\pi^2 + \ln(\epsilon)^2}}$$