# A question regarding the coefficient of restitution

I recently learned collisions. I was introduced to the concept of the coefficient of restitution. My understanding is that the kinetic energy that is lost in the collision is converted into potential energy. My question is , does the coefficient of restitution take into account , the friction between the surfaces. If I were given the coefficient of restitution , would I be able to accurately calculate the velocity of a ball after it collides with a very rough surface (ignoring air resistance).

Also how does the simple formula $\frac{v_1 - v_2}{u_2 - u_1}$ tell us how much kinetic energy was lost in a collision.

After an inelastic colision, the sum of the kinetic energies of the bodies is not conserved. The kinetic energy loss is converted into internal energy (related with temperature, elasticity, cshape/plasticity and other properties), and heat. In this sense the coefficent of restitution is a GIGO machine (garbage in, garbage out): it just take the initial energy, the final energy and compute the energy loss. Since the energy is conserved in the whole system, the coefficent of restitution must take into account the friction and other variables.

To leave it clear: the coefficent of restitution takes in account: the friction between the surfaces, the dragging of the air, the change in the shape of the interacting bodies, the change of temperature of the interacting bodies, definitely everything that leads to a loss in kinetic energy.

Now suppose you have a ball moving towards a wall (we consider a wall cause have in this context infinite mass/inertia) so we can discard its dynamics from the equation. We are given that the coefficent of restitution is $e$. Then: $$e=\sqrt{\frac{E_{Kin,2}}{E_{Kin,1}}}\Rightarrow E_{Kin,2}=e^2 E_{Kin,1}\Rightarrow |v_2|=e|v_1|$$

The repidities of the ball before and after the collision are related, but we cannot be sure about the velocities. Since part of the energy is lost, the change of the shape in the ball or in the wall, the increase on the temperature, and other factors, will determine how much momentum is absorbed by the wall. Because we cannot control this variables, the velocity (i.e. the angle after the colision) remains unknown.

Recall that the kinetic energy and the momentum are related:

$$E_kin = \frac{p^2}{2m}$$

Therefore: $$e=\sqrt{\frac{E_{kin,2}}{E_{kin,1}}}=\frac{p_2}{p_1}$$ Suppose now we dont have a wall but two balls (A, and B) colliding. Now the dynamics is ruled by the conservation of momentum and the conservation of energy (taking in account this coefficent of restitution. For the sake of simplicity we will consider a head on colision (so an effective 1-dimensional problem) with balls of the same mass. Applying conservation of momentum:

$$p_{1,A}+{p}_{1,B}=e({p}_{2,A}+{p}_{2,B})\Rightarrow v_{1,A}+{v}_{1,B}=e({v}_{2,A}+{v}_{2,B})\Rightarrow \frac{v_{1,A}-{v}_{2,A}}{{v}_{2,B}-{v}_{1,B}}=e$$

So with the expression you gave, we can calculate the coefficent of restitution and from there the final kinetic energy: $$\Delta E = E_{kin,2}-E_{kin,1}=eE_{kin,1}-E_{kin,1} =(e-1)E_{kin,1}$$

However realize that this expression is only valid when the masses are equal in a one dimensional problem. If the masses differ, or the collision does not happen in a line but with some angle, the previous argument does not hold and some coefficents related with the masses and the angles must be introduced in the formula.