Does the coefficient of restitution matter in a collision?

Suppose two particles are colliding. Let initial velocities be $u_1$ and $u_2$ .Their final velocities are $v_1$ and $v_2$. Now $u_1$ and $u_2$ are known. By intuition I can understand that $v_1$ and $v_2$ will depend on the coefficient of restituition and hence on their material.

Now I have an equation from conservation of linear momentum and another for the coefficient of restituition. I can now find $v_1$ and $v_2$. So it looks as if $v_1$ and $v_2$ depend upon the coefficient only.

But in my physics textbook I have a solved problem. There, a ball hits a wall and initial and final velocities and the mass of the ball were given. They have basically calculated the average force and impulse. They are saying that the impulse remains same and the maximum force along with the time of collision will depend on the softness or hardness of the ball.

But that means we have the same $v_1$ and $v_2$ for different coefficients of restituition. So where is my thinking wrong?

• The coefficient of restitution is one way of parameterizing the (in)elasticity of the collision. – dmckee Aug 30 '14 at 16:56
• Can you just explain a bit more? – soumyadeep Aug 30 '14 at 16:58
• CoR = 1 is an elastic collision. CoR = 0 is a maximally/completely/totally inelastic collision. Numbers in between represent partially inelastic collisions, with larger values representing larger fraction of the internal kinetic energy remaining after the collision. – dmckee Aug 30 '14 at 17:02
• -1 Not clear what you are asking. In one problem you calculate final velocities using COR. In different problem you are given final velocity and use it to calculate average force. You have not explained how two different coefficients of restitution arise in the same problem. You do not appear to have any contradiction to explain. – sammy gerbil Mar 2 '18 at 15:57

Coefficient of restitution and hardness are not the same thing. The COR basically tells us how much energy gets lost in the collision process. In the case of a soft/hard ball with identical v1/v2 the COR=1, i.e. there is no energy loss. However, the collision of a perfectly elastic ball with a perfect wall will take a different amount of time, depending on how hard or soft the ball is. You can model the process with a perfect spring attached to the wall (or the ball) and a simple mass hitting the spring. The spring constant will then model the hardness of the ball, without introducing an energy loss.

• OK i get it but then what will the COR depend on? – soumyadeep Aug 30 '14 at 14:02
• Look at the definition of COR. If the collision is elastic, the COR=1 (I had to correct my mistake above). If v1=v2, then the collision is perfectly elastic. But even in an elastic collision, the contact time depends on the elasticity of the system, in this case the ball's "hardness". That's described by a second parameter, which could be a spring constant for a process that is both elastic and linear. – CuriousOne Aug 30 '14 at 14:07
• Yes I have understood that.But my question now is for a pair of bodies What will determine its COR?The definition of COR is given to be negative of the ratio of difference in final velocities and that of initial velocities which hardly gives any insight or physical sense. – soumyadeep Aug 30 '14 at 14:11
• The COR is determined by the loss of energy (i.e. the loss of velocity) in the collision. In this case there is no loss of velocity. If you need a physical model for a system with COR other than one and elasticity, you have to add a friction force acting in addition to the spring recoil. – CuriousOne Aug 30 '14 at 14:15
• Yes i get it.:D – soumyadeep Aug 30 '14 at 14:22

For calculations of a collision, you look at (kinetic) energy and momentum equations.

Momentum is conserved; energy may be dissipated. There is a direct exchange of momentum in the form of $F\Delta t$ - the same impulse that slows one object down accelerates the other, by the same amount. But the force may result in a deformation of the ball such that internal friction causes heating, for example - in that case, the kinetic energy of the balls after the collision will be lower than before. The COR essentially talks about how much of the energy is available after the collision.

It seems to me that if your book didn't bring up the COR, it is because they assumed it to be =1 (in other words, that conservation of energy and momentum applied to the ball).