I know that momentum and energy are always conserved in collisions, but if we have a perfectly inelastic collision in which an object sticks to another object $m_1 v_1 + m_2 v_2 = (m_1+m_2)v_{12}$, the kinetic energy is not conserved. I know that kinetic energy converts into thermal or sound energy, but I don't see how this would account for the whole of the lost kinetic energy.

Does the kinetic energy transform into some sort of potential energy between the bonding of the two objects? For example, if the two objects stick together with a small magnet (assuming the initial attraction between the objects is negligible) does the kinetic energy transfer to some sort of magnetic potential?


Its funny you should ask this as I recently ran several simulations on matlab regarding the same thing except with atoms.

Effectively, I had a diatomic molecule (H-H for example) and an atom (F lets say). The atom and diatomic both had some momentum relative to each other and the collision was setup to be perfectly collinear.

Now, what I noticed is that the initial energy of the reactant (that is the incoming F atom) was deposited into two modes... Translational and vibrational energy.

Depending on the choice of the atom and diatomic more of one form over the other would be required for a successful reaction (Polanyi rules but we wont go into that).

Essentially, if the reaction was elastic then you would have an unreactive collision. The atom and diatomic coalesced to form a three body transition state and then the atom would just break off and head back in the direction it came from.

In a reactive collision, which was always inelastic, there was always a change in vibrational energy between the reactants and the products i.e. upon form a new diatomic (H-F from our above example) there was reduced or increased vibrational energy.

I don't know how useful this tangential discussion has been but I thought it might help you :)

EDIT : I also want to clarify that an unreactive trajectory could also be generated from an inelastic collision simply because the incoming atom had far too much kinetic energy.

I've included the potential energy surface of a similar reaction below. The black line represents the trajectory, the 'wiggliness' you see initially is the vibrational energy.... There's an obvious change in the 'wigliness' as you go around the bend in the PES i.e. as you go from Cl+H2 to H-Cl + Cl.

If you would like I can send you these source files.

PES of a Cl + H2 reaction

PES of a Cl + H2 reaction

  • $\begingroup$ It took me a while to grasp, but are you saying that as the molecules collide, the kinetic energy of the molecule decreases and converts into vibrational energy, or kinetic energy within the molecule? (which would explain why we don't directly observe the movement of the object as a whole) So in effect, the kinetic energy that we no longer see in the moving object truly is converted into thermal energy? $\endgroup$ – Couchy311 Apr 4 '14 at 5:44
  • $\begingroup$ Firstly, this an atom+molecule collision which is different from your scenario of two billiard balls colliding. I dont know how to answer your query properly but here's what I can say... Depending on the initial conditions in my model I found that yes as the incoming atom collided, formed a new bond and forced one of the atoms on the initial molecule to leave (F+HH ---> HF + H) there could be a transfer of linear translational energy to vibrational energy which then decreased slowly over time. $\endgroup$ – Ari Ben Canaan Apr 4 '14 at 7:17

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