I'm currently running a simulation for a physics project, seeing how the product of coefficient of restitution affects the percent kinetic energy lost in a system of two identical balls when they collide into each other on a frictionless surface (momentum is conserved). Their initial velocities are same in magnitude but opposite in direction. But when both coefficients of restitution were equal to zero, meaning their product was also zero, they were not perfectly inelastic. They bounced off of each other. When I searched online, everyone is saying when the coefficient of restitution is zero, it should be perfectly inelastic.

I don’t know how this is possible, because if the coefficient of restitution is zero doesn’t it mean there is no relative speed between the two objects after collision? Is there an explanation for this?

  • $\begingroup$ Sounds like a bug in your simulation. You may want to check on one of the coding SE sites instead $\endgroup$
    – Dale
    Commented Jan 9, 2022 at 16:28

1 Answer 1


I assume that you a discussing a one-dimensional collision?

You have to be careful as to what you mean by the phrase perfectly inelastic

Whereas perfectly elastic means that no kinetic energy is lost whereas, in general, perfectly inelastic does not mean all the kinetic energy is lost.

Given that $e$ is a measure of the relative velocity of two objects after collision when $e=0$, after collision one would expect the two objects to be moving with identical velocities.

. . . . . when both coefficients of restitution were equal to zero, meaning their product was also zero . . . . .

I do not understand how there can be two coefficients of restitution when considering one collision.

So either you have made an error in the theory or the coding.

The Wikipedia article coefficient of restitution might be of some interest?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.