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The more I think about it, inelastic collisions produce heat and sound which imply motion at some scale, right?

Are inelastic collisions macroscopic events that ignore motion at microscopic levels? Maybe I'm just misunderstanding the definition of an inelastic collision.

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4 Answers 4

up vote 6 down vote accepted

Collisions can be elastic or inelastic.

Elastic collisions are collisions where the incoming and outgoing kinetic energies are the same and only the angles change

The same holds true classically, example: billiard balls ; and in the elementary particle framework. example: An electron hitting an electron has a probability of scattering elastically.

Billiard balls will turn some of the kinetic energy from linear to rotational when scattering, so completely elastic scatters are rare. Inelastic scattering of billiard balls would involve deformation of the balls.

At the quantum level if a scatter is not elastic, the energy changes form, example: atoms in a gas scattering off the electric field of each other , will lose kinetic energy into radiation, which generates the black body radiation and the eventual cooling of the gas.

When elementary particles scatter inelastically new particles and radiation will be formed.

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If you collide two ideal billiard balls, then that would be what you would call a perfectly elastic collision.

If you have a large dense collection of billiard balls, and you slam a new one into the collection, then there are a whole lot of elastic collisions, transfering energy and momentum in many ways that you would be hard-pressed to calculate exactly. It you think about it statistically, that is thermal energy. If some of the balls in the dense collection are tied together, then you get spinning pairs (rotational kinetic energy). If some of the balls have non-zero angular moment of inertia, then some of the balls will spin (rotational kinetic energy). If some of the balls are in groups of three, connected by springs, then they could end up vibrating (kinetic plus potential energy).

This is the way atoms and molecules behave, giving rise to thermodynamics.

So if you collide two beanbags together, making an inelastic collision, I'm inclined to agree with you - that's really just a zillion elastic collisions we can't compute. So we just call it heat.

Of course, that's all classical. If you want to get into quantum stuff, @Anna's answer has it covered.

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In a typical introductory physics class, the terms elastic and inelastic refer to whether the macroscopic kinetic energy is the same before and after the collision. By this I mean the large-scale motion of the objects that you typically consider in collisions, like carts or balls, not the detailed particle motion/potential.

You are correct that the total energy is the same before and after any collision. But this does not mean that the kinetic energy is the same before and after. Some of the initial energy could go into vibrational energy though. This form of energy is microscopic and not discussed in-depth at the introductory level. It typically comes with the internal energy label in such courses.

I suspect there is more one could say when discussing collisions at the quantum level, but I think this addresses your question at a classical level.

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I think you are misunderstanding the definitions. Every collision is basically Inelastic, not Elastic. Elastic collisions are ideal collisions, which require an ideal system with fixed optimum pressure and temperature, point-sized particles etc. The only collisions that can be assumed to be PERFECTLY Elastic are those of Atoms.

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I don't think atom-atom collisions are perfectly elastic. –  RGrey Mar 27 '14 at 20:56
I don't understand your answer. First you state that an Elastic collision is idealized, then you add a modifier "PERFECTLY" that seems to change the meaning of "elastic" to be the ideal case. That implies that without the modifier elastic would apply to non-ideal collisions. Please clarify. –  Rick May 6 at 11:49

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