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I understand that in inelastic collisions thermal energy is given out, but why does that happen? Why can't they simply rebound without giving off energy? Also, why in some collisions more heat is given out than in others (i.e. what determines that one collision is more inelastic than the other)? Thanks.

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It might help if you could specify what sort of collisions you are talking about - theoretical point masses in classical mechanics, extended objects, particle physics? –  Claudius Oct 12 '12 at 19:13
    
@Claudius I'm thinking of simple collisions like that of two blocks. So I guess I'm talking about point masses and/or extended objects (will the answer be different for extended objects?). –  Alraxite Oct 12 '12 at 19:20
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Given your comments, I shall attempt an answer. This obviously depends on how you model the interaction between the two particles/masses.

First, let us see what happens if we bounce two particles off each other in a simple theoretical model where the particles don’t have any internal structure. To simplify things, we assume one dimension and use the following Hamiltonian:

$$ H = \frac{1}{2 } m_1 v_1^2 + \frac{1}{2 } m_2 v_2^2 + U(x_1,x_2) $$

where we define $v_i$ ($m_i$) to be the speed (mass) of the $i$-th particle and $U(x_1,x_2)$ is the interaction potential which describes how the particles interact. Usually, we want something like

$$ U(x_1,x_2) = \{ \infty\textrm{ if }x_1 - x_2 < d \quad,\quad 0\textrm{ otherwise }\}$$

So the two particles can only get within distance $d$ of each other, because otherwise the potential energy $U$ is infinite (corresponding to a physically impossible state). If you try to solve the system given by the above Hamiltonian you will notice that it only describes elastic collisions. This can be linked to the fact that it is a microscopic view: We know where each particle is and even if we tried to introduce something like ‘heat’, it would only manifest itself in a different speed for a given particle.

So in order to get an inelastic collision (and in lieu of more complicated microscopic systems which sometimes, especially in quantum mechanics, define special operators to model interactions which can also result in inelastic collisions), we need a different model: A macroscopic one.

Again, imagine two extended masses speeding towards each other. However, now we shall assume these masses to have an internal structure, to be composited of many small atoms, placed in a relatively rigid grid. Furthermore assume that in the beginning, there is no temperature, that is $T = 0$. This means that the atoms in each of the masses all move in the same direction and don’t vibrate at all.

What happens is that these atoms (at least in our model) interact with each other very similarly to the particles in the first example.

Note that we now have many, many different particles (about $10^{24}$ rather than two) and not just one, but three dimensions. Therefore, if the two large masses collide, these small atoms will bounce off each other in many different directions (like balls bouncing off each other in different directions depending on the exact angle of collision). These ‘random’ movements of the atoms are caught by the other atoms around them, so in effect, they don’t run off everywhere but merely start vibrating in place. And this vibration is heat.

Of course, the model can be expanded further: We could think of some atoms gaining so much momentum from the impact that they flew out of the grid and into space (the masses being damaged) or that the grid is somehow changed and the atoms moved further towards each other (the masses being deformed).

The key point, however, is that inelastic collisions are usually macroscopic rather than microscopic effects relying on the internal structure of the masses involved.

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Thank you for such a detailed answer. However, I should have mentioned that I haven't done any advanced physics, but I do understand your answer for the most part. So you are saying that when two bodies collide, different atoms collide with each other at different angles and hence they start vibrating as they can't go in a single direction as a whole. That means, more irregular the surfaces of contact (especially at the microscopic level), the more inelastic it would be, right? Also, a collision can be elastic only if all the atoms collide at exactly the same angle, correct? –  Alraxite Oct 12 '12 at 20:21
    
A collision can only be perfectly elastic if all the atoms collide at exactly the same angle and the grid formed by these atoms is perfectly rigid. This, obviously, never occurs in the real world. Crystal balls probably form the best approximation to it, but even these tend to be deformed. –  Claudius Oct 12 '12 at 20:24
    
Can you tell me what does a grid mean? Also, when the objects collide, the atoms are in some sense 'blasted off' in different directions but due to the prevailing attraction that those atoms have with the other atoms residing in the body, those atoms come back but they get too close to the other atoms and are again sent off by the repulsion and a kind of harmonic motion begins (which is the vibration) only this time it is a 'perfect' harmonic motion with no energy loss, right? Or is the answer related to something deep in quantum mechanics? In which case I wouldn't want to know. –  Alraxite Oct 13 '12 at 8:42
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Inelastic collisions are where kinetic energy is not a constant - this means some of the initial kinetic energy must have been converted to some other form of energy.

A perfectly elastic collision, for example, would be perfectly silent.

This should be giving you an idea of how it is that you don't get perfectly elastic collisions.

Any collision involves deforming the surfaces of the objects colliding - this means squashing the particles the objects are made out of together for a short time.

On the scale of individual particles - all objects can be thought of as balls connected to each other by stiff springs - shoving a bunch of the balls means that the springs compress - pressing on the balls nearby and so one until the energy spreads through the whole object. The whole thing starts to get wobbly - like a stiff jelly. Some of the wobbles reach the "surface" where they vibrate the surrounding air (sound) and energy gets lost that way. Some of the wobbles bounce around inside the object ending up all randomized (heat) ... and so not all the energy from the initial shove is available in the same form after the collision. Thus - the collision was not elastic.

Sometimes the collision was hard enough to leave a dent ... in that case the non-elastic nature should be clear.

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