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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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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