Originally posted by @joe-shmo
"By specifying that the objects merge you constrained the final kinetic energy and implicitly made a claim about the material properties of the colliders"
So this "implicit claim" must be a very powerful one. It seems to describe "all" real physical interactions of the collision possible without the need for anything physical, other than mass and veloc ...[text shortened]...
Also, I have no experience with the center of momentum frame, or what its implications are.
I made an error in the above, what I was referring to as the coefficient of elasticity is actually called the coefficient of restitution. It was invented by Newton as an ad hoc
quantity to use while he worked out his framework, and isn't terribly useful as it depends on the colliding objects material properties, their shape, and the speed of the impact.
The centre of momentum frame is just the frame of reference where the total momentum is zero. For the earth moon system this is just the point the two objects orbit (which is inside the earth from memory). It is convenient for doing calculations, especially collisional ones. The reason for using it is that we automatically subtract the final kinetic energy (it is zero in that frame).
The maximum energy available for deforming the colliders is just the total kinetic energy, for which the conventional symbol is T, in the centre of momentum frame:
T = 1/2 m (u1)^2 + 1/2 M (u2)^2
Since we require that the net momentum is zero we have:
m*u1 = M*u2
So that u2 = (m/M)u1
T = 1/2 m u1^2 (1 + (m/M))
The relative velocity is just v = u1 + u2 = u1(1 + m/M). So we now have:
T = 1/2 m v^2 / (1 + m/M)
and our final result:
T = 1/2 μ v^2.
where μ = Mm/(M+m) is the reduced mass of the system.
Were our colliders to coalesce then the final kinetic energy would be zero and so the energy available for plastic deformation and heating of the colliders is equal to the initial kinetic energy. If they bounce off each other perfectly elastically then their outgoing velocities would be equal to the incoming ones.
What might be causing you some problems is a misapprehension of how much energy is involved. Consider a collision between two steel balls. Suppose they each weigh 1 kilogram and are moving at 1m/s relative to one another. The total kinetic energy is then 0.25 joules. The heat capacity of steel is 466J/kgK, so the upper bound on the temperature change of the steel balls in the collision is:
dT < 0.25 / 2*466 = 0.00027 kelvin.
A 5.56 mm calibre bullet weighs about 3.56 grammes and travels at 870m/s, it has a total energy of 1,758 joules (according to the Wikipedia page on the SA80 I got the figures from). and when absorbed by a target will heat the area around the bullet's path through the body by of the order of less than a half of a kelvin.
The meteorite that wiped out the dinosaurs hit the Earth at orbital velocities (say of the order of 10,000 m/s) and weighed of the order of 8 trillion tonnes, for a total energy of 800 ZJ. About equal to the power output of the sun for 2 milliseconds. For the energy released to be at all big the object has to be huge and going really really fast.