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If two balls of same mass with speed $v$ and $-v$ undergo an elastic collision, the kinetic energy will be the same after the collision as before.

However, during the collision, does it also remain the same? Isn't there a moment where both balls have zero velocity and hence zero kinetic energy?

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    $\begingroup$ It's called an _______ collision because you need to imagine the material of the balls as _______ $\endgroup$ – Joshua Ronis Dec 14 '18 at 16:37
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    $\begingroup$ Note that for most elastic collisions, there will not be a moment where both the balls have zero velocity. For example, if a ball moving at $v$ hits another ball at rest, there is never an instant where both of them are at rest. $\endgroup$ – Michael Seifert Dec 14 '18 at 17:10
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    $\begingroup$ @MichaelSeifert but two balls of the same mass colliding with equal and opposite velocity... $\endgroup$ – leftaroundabout Dec 14 '18 at 17:25
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    $\begingroup$ Some things I wrote about the treatment of collisions in an earlier answer might be helpful in thinking about this kind of question. $\endgroup$ – dmckee Dec 14 '18 at 19:29
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    $\begingroup$ One could consider a ball bouncing off the ground to eliminate the push back about a collision not ever reaching zero velocity. It still retains the essence of the question of turning kinetic energy into potential energy and then back again. $\endgroup$ – CramerTV Dec 14 '18 at 22:17
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Good point. The comparison of initial and final energies is done before and after contact. During contact there must be some work done to bring them to rest and turn around. But for an elastic collision these internal forces are conservative, like the elastic force. If you watch slow motion photography of a collision you will see the balls deform slightly then come back to their original shape. This is due to the elasticity of the materials in each ball. In real life there is no such material that is perfectly conservative (at least as far as I know) but it's a good approximation for many materials. So in short, while they are at rest for a moment the kinetic energy is stored as potential in the balls.

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    $\begingroup$ The deformation can be significant, see these examples of a golf ball and a squash ball. $\endgroup$ – Bas Swinckels Dec 15 '18 at 8:41
  • $\begingroup$ True, what about billiard balls? $\endgroup$ – ggcg Dec 15 '18 at 11:29
  • $\begingroup$ Those are much stiffer than a golf ball. They should obviously deform too, but probably by too little to be observed by eye. You do however clearly see the deformation of the cushions. $\endgroup$ – Bas Swinckels Dec 15 '18 at 11:56
  • $\begingroup$ I agree. But for ball on ball collisions at typical speeds they're pretty rigid. Point is, as the post suggests, no material behaves like the idea model, but has some spring like behavior. Some more than others $\endgroup$ – ggcg Dec 15 '18 at 12:16
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You are correct. While there is a universal principle that momentum is conserved for ALL interactions, and momentum of isolated systems will remain constant, there is no universal conservation of kinetic energy.

In the perfectly elastic collision system, the interaction forces are modeled as conservative spring-like forces. The interactions of the objects result in kinetic energy being efficiently transformed into potential energy of "springy" surfaces, then 100% transformed back to kinetic energy.

For partially-elastic collisions (real-world collisions), the transformation to and from elastic potential energy is not 100%. Some KE goes into sound waves, deformation/stress of material, and internal ("thermal") energy

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During this collision process, kinetic energy is converted to internal energy. More specifically, elastic potential energy! While it may surprise you, each ball can actually be modeled as compressible, like a spring, under the study of Hertzian Contact Mechanics. This is due to the compressibility and deformation of the balls during collision.

In fact, length of compression between the 2 balls can be defined as $$d^3=\frac{9F^2}{16E*^22/R},$$ where Poisson's ratio and the elastic moduli of the ball can affect $E*$.

Of course however, we are assuming no friction or energy loss to the surroundings, a key basis for Hertzian Contact Mechanics.

Consequently, this elastic potential energy will be converting back to kinetic energy.

You can read up on 2 research articles in 1975 and 1981 by N. Maw, J. R. Barber and J. N. Fawcett titled "The Oblique Impact of Elastic Spheres" and "The Role of Elastic Tangential Compliance in Oblique Impact" respectively.

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  • $\begingroup$ Do those articles have a derivation of the equation you cited? If not, can you please provide a reference. $\endgroup$ – Chet Miller Dec 14 '18 at 21:38
  • $\begingroup$ Those articles support Hertz's derivation of the formula. However, I am unable to find Hertz's initial 1882 derivation of it, so I cited these two. $\endgroup$ – QuIcKmAtHs Dec 15 '18 at 1:31
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The question reminds me of a similar one, related to waves: what happens to the energy when 2 waves of equal amplitude and moving in opposite directions meet, being in phase, so that a crest meets another crest? What happens is that, yes, the waves stop for an instant (kinetic energy vanishes), but only because amplitudes add up and are twice as large (the energy is stored as potential energy).

The opposite case is when a crest meets a trough. In this case for a moment amplitudes cancel out (there is no potential energy), but velocities add up and are twice as large (all is kinetic energy).

(I took this information from the post by Pygmalion in this thread.)

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