I am currently studying collisions from an online lecture and the teacher in it says that both perfectly elastic and perfectly inelastic collision are practically not possible.

I am able to find the reason why elastic collisions are practically impossible that is because there is always some loss of kinetic energy of system as heat (due to friction ), sound etc.

But I cannot figure out why perfectly inelastic collision are not possible.

I think perfectly inelastic collisions are possible for example if we take a vacuum chamber and inside it place two identical modelling clay balls tied to identical strings and held at same angle as shown in figure. enter image description here Then we release the clay ball simultaneously and they hit each other and stop at the middle then it will be a perfectly inelastic collision.

Is there any fallacy in my statement or the statement of the teacher ?

  • 1
    $\begingroup$ Actually, in the microscopic world, perfectly elastic collisions are possible - consider two neutral helium atoms colliding at slow speed. As long as the relative velocity is low enough that there's not enough energy to raise an electron to a higher orbital, there's no where else the energy to go but to stay as relative velocity... $\endgroup$
    – poncho
    Oct 27, 2021 at 16:00
  • $\begingroup$ There's no such thing as a perfect vacuum $\endgroup$
    – eps
    Oct 27, 2021 at 16:19
  • 1
    $\begingroup$ Your teacher needs to work on his definitions game. A literal perfectly elastic collision is, in practice, not possible. Meaning that there is no practical way to eliminate 100% of the margins of error, however small. In the same style the teacher is speaking, it is practically not possible for your teacher to make physical contact with his keyboard. $\endgroup$
    – David S
    Oct 27, 2021 at 16:37

4 Answers 4


I suspect this is semantics.

Two balls of clay thrown together in the ISS will stick together. Momentum is conserved, and kinetic energy is lost to sound, macroscopic deformation, and microscopic degrees of freedom. (We can do it in the airlock, so their no sound, if that is preferable).

And other example is 2 bowling balls with a spring contraption that locks at maximum compression. Momentum is conserved as the 2 balls move away as one, and kinetic energy is stored as (macroscopic) potential energy in the spring.

In both cases, the "missing" kinetic energy goes into internal energy of the final state object, and it can be macroscopic and/or microscopic.

I have ignored angular momentum. The collisions need to be perfectly aligned for the final object to have have zero rotation, and hence zero rotational energy. Since "perfect alignment" is impractical, one can never achieve this.

Now the semantics: If the rotational energy is defined as "kinetic energy of the final state", then one cannot achieve a perfectly inelastic collision. There were always be some kinetic energy above the absolute minimum allowed by conservation of momentum.

If the rotational energy is defined as "internal energy of the final state" (rotational excitation), then it is no different from a loaded spring, and a perfectly inelastic collision is achieved.

  • $\begingroup$ Why would the kinetic energy make any difference? A collision is perfectly plastic if the objects stick together after the collision. They can have any amount of rotational or translational kinetic energy. $\endgroup$
    – Orbit
    Oct 27, 2021 at 9:28
  • $\begingroup$ I'm imagining a scenario with two plates spinning at different rates, where one is dropped onto the other, they stick together, and continue to spin. Isn't that an inelastic collision? My interpretation of an inelastic collision is one where there is some reference frame in which all the initial mass has exactly zero velocity after the collision. In this plate-spinning case that frame is non-inertial, but there still is a reference frame in which the post-collision object has exactly zero kinetic energy. I think all an inelastic collision requires is no relative movement between mass. $\endgroup$ Oct 27, 2021 at 16:31
  • $\begingroup$ @Orbit it makes a difference because I say it makes a difference. Hence: semantics. $\endgroup$
    – JEB
    Oct 28, 2021 at 0:32
  • $\begingroup$ @JEB: "If the rotational energy is defined as "kinetic energy of the final state", then one cannot achieve a perfectly inelastic collision." That is nonsense. It does not matter if the kinetic energy is translational or rotational. $\endgroup$
    – Orbit
    Oct 28, 2021 at 18:13

There is no substance that is perfecly plastic and not at least partialy elastic. In your example clay has elasticity.

If you throw mud at a wall it will form a radial splash patern with traces of varied sized mud drops spreading out.

Those drops are surface vibration wavelets with kinetic energies larger then the cohesion of the mud.

They got ejected and became teardrop projectiles. typical behavior of an elastic substance.

  • 1
    $\begingroup$ This is completely incorrect. A perfectly plastic collision is a collision where the objects stick together after colliding. This happens all the time. A piece of chewing gum that sticks to the wall, a fly on your number plate etc. $\endgroup$
    – Orbit
    Oct 27, 2021 at 9:31
  • $\begingroup$ Consider two magnets, which attract each other when near. Not only they stick together, but they may also be considered to have a negative COR. $\endgroup$ Oct 27, 2021 at 11:37
  • $\begingroup$ I don't really see how this reasoning applies to rigid bodies that do not eject mass during the collision. $\endgroup$ Oct 27, 2021 at 16:09

The problem isn't that there aren't any plastically interacting objects, the problem is that there aren't any ideal rigid bodies.


a dropped ball of soft clay landing on a concrete block is very, very close to a perfectly inelastic collision.

  • 3
    $\begingroup$ Yet somehow dropping a bowl of salsa on the floor always ends up with some in somebody's eye... But that is a whole other issue... $\endgroup$
    – Jon Custer
    Oct 26, 2021 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.