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For example, must the shock wave in each body be of a particular form which influences the shape and material properties of the bodies?

I suspect part of the the answer is that the objects must be spherical, and the round-trip of the shock wave in each body must be the same, but this is just a guess.

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I would guess it has to do more with internal structural damping and resistance to shear waves than with purely the elastic properties. –  ja72 Jul 31 '11 at 10:45

2 Answers 2

For a collision to be elastic, by the usual definition, no internal degrees of freedom of the colliding bodies can be excited/de-excited by the collision.

The internal degrees of freedom that might change in a collision include vibration, rotation (although some might argue about this), electron orbitals, electron spin, nuclear spin, etc. etc. etc.

For macroscopic objects, the density of states for all these parameters is so high that in practice it is astronomically unlikely that a TRULY inelastic collision will ever occur. This is true even if the bodies are spherical, and have the same shock wave round-trip times, etc. etc. The energy "lost" in a collision may be so small that it is extremely difficult to measure, but any collision of macroscopic objects is sure to be inelastic.

The situation is very different if the colliding particles are single atoms. There, the density of internal states is so low (and coupling to various degrees of freedom so weak in the scattering) that it can be thousands of times more likely to have an elastic collision than an inelastic one.

Diatomic molecules are appreciably worse than atoms; it is relatively easy to excite/de-excite rotational motion of the molecule in a collision. Inelastic collisions are about as common as elastic collisions.

Where's the tipping point? By the time you've got something as big as, say, a virus, I'm sure it's extremely improbable to have elastic collisions. But for something like a buckyball? I have no idea.

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You took care of it, but I had so much to write in response so I put it in a separate answer. –  Alan Rominger Jul 30 '11 at 2:40
    
Well, pretty good answer but again, when dealing with macroscopic objects one can't take the expressions like elastic literally. Suppose we instead require that at most 1% of the energy is lost in the collision. What can you then say about the properties of the macroscopic bodies for this to happen? Obviously OP is after this answer and not the trivial distinction between particles and macroscopic objects... –  Marek Jul 30 '11 at 6:25
    
The main important point you've made is that the collision of two macroscopic bodies can never be elastic. So how elastic can it be in the ideal case of identical spherical bodies colliding? The propagation of the shock wave and velocities of the colliding bodies must have some sort of influence which you haven't touched upon. –  John McVirgo Jul 30 '11 at 14:18
    
Perhaps I should have rewritten the question to something of the sort "How could one minimize inelasticity in the collision of two macroscopic objects?" But given the manner in which the original question was phrased, it was not clear to me that the original poster understood collisions even at the level I was explaining. –  Anonymous Coward Jul 30 '11 at 16:42

Extended reply to Anonymous Coward:

Perfect, I'll argue that gaining rotation in a collision doesn't make it inelastic. The kinetic energy is still there, it's just in the form of rotational kinetic energy. But yes, that is strictly semantics. You pretty much covered the question.

Generally I would claim that no macro objects can collide fully elastically but elementary particles can. Of course, elementary particles only do so in the absence of an interaction and the "collision" is though a force like Coulomb repulsion. Molecules, atoms, and other things can blur the line by having a meaningful chance of colliding fully elastically as well as a meaningful chance of some energy conversion taking place with kinetic energy. Like nuclear cross sections, it is possible to provide numeric quantification of the probabilities of different interaction types.

For fun consider an equilibrium gas. Collisions should be on average net-zero kinetic energy $\Delta$ but it's possible that collisions frequently (almost always even) convert binding energy or some other internal stored energy into kinetic energy. I imagine this would be the case for gases with a large molecule size, it could even be of relevance to a chemist.

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Second paragraph is quite confusing. First, elastic collision means that the collision is ellastic in some approximation or idealization (otherwise the term would be completely useless for macroscopical objects, which it's not). Also, you might want to reword the bit about particles: they can also collide inelastically (unless you are only talking about idealized classical point-masses; but then there are also idealized solid bodies...). Perhaps you want to say that only collisions that are potentially elastic are the particle collisions. –  Marek Jul 30 '11 at 6:12
    
As for the last paragraph: using term internal energy here is again confusing. We already have internal energy of the whole gas which comprises both kinetic and rotational/vibrational energies of the molecules. Now, even if the molecules of the gas wouldn't interact with each other, they would hit the walls and this is ultimately why we can treat all those degrees of freedom together (using equipartition theorem). –  Marek Jul 30 '11 at 6:18
    
@Marek Points taken. I tried to reword to address some of those things. –  Alan Rominger Aug 1 '11 at 1:11

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