- 16 Mar '18 17:57 / 1 editI'm looking to fill some apparent gap in knowledge I have. I may have a bit of difficulty in properly conveying my issue, so I hope a discussion can clean it up. Looking at perfectly inelastic collisions, I am somewhat taken back by the fact that kinetic energy is not conserved? Looking at a simple model. Two bodies, one of mass "m"and the other "M" are bound for collision with velocities "v", and 0, respectively. after the collision they join together and continue with velocity " v' "

From conservation of momentum we have: m*v + 0 = (m + M)*v', which means that v' = m/( m + M )*v ...(1)

At this point lets look at the kinetic energy before and after the collision.

Before Collision: KE_before = 1/2*m*v^2

After Collision ( substituting 1): 1/2*m^2/(m+M)*v^2

Then taking the ratio of after to before: KE_after/KE_before = m/( m+M) < 1, meaning KE_before > KE_after

So is someone going to tell me the quantity of energy 1/2*m*v^2 - 1/2*m^2/(m+M)*v^2 has been lost as heat ( some seemingly divine intervention of the Second Law of Thermodynamics )? I wasn't aware that I was operating under the pretext of entropy in this idealized case? We know absolutely nothing of the internal collision mechaincs that would generate said heat and potential energies of various forms, yet some how this example dictates a specific quantity of transformed kinetic energy in this collision ( it doesn't say where it goes, but it says it goes)? For instance, In Newtons Laws ( and even conservation of energy - as far as the models are concerned ) you can simply "turn off" entropy, and the mechanics will yield a limiting case. But here, the limiting case of conserved kinetic energy seems to be impossible to show.

So what is going on here? What don't I understand?

Any help greatly appreciated. - 16 Mar '18 19:09

You're not thinking about the physics going on during the collision. For the collision to be inelastic the colliding bodies have to be deformed, which is where your missing energy goes. Because momentum has to be conserved what your calculation does is tell you the energy that was available to*Originally posted by @joe-shmo***I'm looking to fill some apparent gap in knowledge I have. I may have a bit of difficulty in properly conveying my issue, so I hope a discussion can clean it up. Looking at perfectly inelastic collisions, I am somewhat taken back by the fact that kinetic energy is not conserved? Looking at a simple model. Two bodies, one of mass "m"and the other "M" are bo ...[text shortened]... to show.**

So what is going on here? What don't I understand?

Any help greatly appreciated.*irreversibly*deform the colliding bodies.

Imagine the head on collision of two billiard balls. In the centre of momentum frame, the entire kinetic energy of motion is turned into strain which is then released pushing the balls apart again. For idealised billiard balls we can imagine that process being lossless. However, in a real collision there will be losses as the collision creates sound waves inside the material of the billiard balls which partially dissipate as heat - as well as some energy going into making the cracking sound. If the material does not deform elastically then the shape of the objects changes irreversibly and much less energy is available to push the objects away from one another. In the extreme case the objects merge and none of the initial energy is available as kinetic energy, ignoring effects like ejecta. - 17 Mar '18 03:22

I see what you are saying, but I must not fully get it.*Originally posted by @deepthought***You're not thinking about the physics going on during the collision. For the collision to be inelastic the colliding bodies have to be deformed, which is where your missing energy goes. Because momentum has to be conserved what your calculation does is tell you the energy that was available to***irreversibly*deform the colliding bodies.

Imagine ...[text shortened]... rge and none of the initial energy is available as kinetic energy, ignoring effects like ejecta.

"In the extreme case the objects merge and none of the initial energy is available as kinetic energy"

Why has my example not produced this case? Why specifically has that quantity of kinetic energy been transformed, when none of the material properties are apart of the model? What are the parameters, and given my example how can the solution possibly be different? - 17 Mar '18 05:45 / 1 edit

The material properties are contained in the coefficient of elasticity, which is implicitly in your equation. The relative velocity after collision v and relative velocity before, u, are related by v = -eu. For a perfectly elastic collision e=1 and the kinetic energy after is the same as for before the collision, for a completely inelastic collision e = 0. By specifying that the objects merge you constrained the final kinetic energy and implicitly made a claim about the material properties of the colliders.*Originally posted by @joe-shmo***I see what you are saying, but I must not fully get it.**

"In the extreme case the objects merge and none of the initial energy is available as kinetic energy"

Why has my example not produced this case? Why specifically has that quantity of kinetic energy been transformed, when none of the material properties are apart of the model? What are the parameters, and given my example how can the solution possibly be different?

Just a point, in your example you have one object initially moving and its target stationary, call this the lab frame. In my comments I looked at it in the centre of momentum frame. So I think your example has produced this case. - 17 Mar '18 16:26

"By specifying that the objects merge you constrained the final kinetic energy and implicitly made a claim about the material properties of the colliders"*Originally posted by @deepthought***The material properties are contained in the coefficient of elasticity, which is implicitly in your equation. The relative velocity after collision v and relative velocity before, u, are related by v = -eu. For a perfectly elastic collision e=1 and the kinetic energy after is the same as for before the collision, for a completely inelastic collision e ...[text shortened]... I looked at it in the centre of momentum frame. So I think your example has produced this case.**

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 velocity?

Obviously total energy must be conserved, so you appear to be saying are this just puts a limit on the**maximum**possible energy that can be used for deformation? It will do nothing to divide out the actual energy of fusing, heat, sound, potential, etc. from the collision. So all colliding inelastic material of mas "m", "M", and velocity "v", will just have varying portions of those common collision energies I just mentioned. How its divided is dependent on the properties of the colliding bodies, but the quantity itself is not?

Also, I have no experience with the center of momentum frame, or what its implications are. - 17 Mar '18 19:09 / 1 editif you throw a ball, air fraction from collisions with air molecules can measurably slow it down and yet after the throw there would result no easily measured increase in the average speed of the randomly moving air molecules (although they do increase in speed) so where has the kinetic energy from the ball gone?

The answer is it has been converted to heat energy from friction but that heat energy has been so dispersed that it would be extremely difficult to measure or detect or notice. - 18 Mar '18 22:15

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*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.*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

and

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. - 19 Mar '18 13:21

Just an aside, that Mm/M+m is the same as parallel resistors. Meaningless but interesting.*Originally posted by @deepthought***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 ...[text shortened]... For the energy released to be at all big the object has to be huge and going really really fast. - 21 Mar '18 17:28 / 2 edits

"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"*Originally posted by @deepthought***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 ...[text shortened]... For the energy released to be at all big the object has to be huge and going really really fast.

Ok, I thought you were telling me they had zero kinetic energy in all frames after the collision. Now I see you were referring to the center of momentum frame only.

"What might be causing you some problems is a misapprehension of how much energy is involved."

That had less to do with it, but certainly could have contributed ( if they were equal masses in my example half of the initial kinetic energy is transformed - I though that was a significant portion ). I believe most of my confusion arose by not recognizing the constraints ( on kinetic energy) I was imposing, and because of this it appeared ( as an example) as though I had made no explicit assumption of dissipative forces, they none the less arrived in the result. Then having realized that they were not going away ( whether I liked it or not ). I was further taken by surprise that this transformed energy was not ( apparently) dependent on the material properties of the colliders. Then my puny brain exploded...