Argthtjtr

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?

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.

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

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.

The question reminds of a similar one, in this case referred 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).