Usually, in a perfectly inelastic collision, maximum amount of KE is lost. I guess it depends on the rigidity of that object collision if any KE at all will be converted to work done to deform.

Assume: coefficient of restitution = 0, no surface friction or air resistance.

An example of this phenomenon, I can think of, is throwing a lump of clay onto a wall; at the end both become stationary, meaning KE = 0. Sometimes, this really confuses me, because the initial sum of momentum is not equal to the final sum.


1 Answer 1


Energy is dissipated in the form of "internal energy", which means that all of the objects kinectic energy is transfered to internal movement of atoms and mollecules of both the object and the surface. When there is a large deformation and no restitution you can argue that some of the energy is stored in some kind of ellastic energy of the mollecular bonds that constitute the material, the remaining part of the dissipated energy is transfered to the material via sound waves (i.e vibratios and oscillations of the material).

There is a undergraduate book called "Matter and Interactions" from Ruth W. Chabay, Bruce A. Sherwood that treats collisions by explaining this energy transfer/dissipation in perfect inellastic collisions. I recomend you to check it out.

  • $\begingroup$ If it's stored in bonds, it got to be released sooner or later, like a spring, right? $\endgroup$ Aug 4, 2015 at 14:13
  • $\begingroup$ Not exactly, because the spring in this case is just an analogy. The interactions are far more complex than Hooke's law. $\endgroup$ Aug 4, 2015 at 14:29
  • $\begingroup$ But if there was initial velocity, there must be some final velocity? Where is that bit of velocity if both are stationary? $\endgroup$ Aug 4, 2015 at 20:45
  • $\begingroup$ The velocity is distributed onto all atoms that compose both materials. They seem to be stationary at the macroscopic scale, but at the microscopic level they are moving quite a bit. $\endgroup$ Aug 4, 2015 at 21:10
  • 1
    $\begingroup$ Depending on the object's structure this velocity will be larger/smaller, but you will never be able to prevent this movement on the microscopic level. You have to understand that there is no such thing as direct contact, surfaces never really touch, they just come really close so that the electrostatic force is large enough to prevent further approximation. $\endgroup$ Aug 4, 2015 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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