Thank you, Manlob, that makes a lot of sense, that way. I think that that’s the answer that all the rest of us physicists should have posted. 
Stick around, post some more, you’re helping the board.
Not only does this example result in conservation of momentum with no loss of kinetic energy (i.e. a perfectly elastic collision), this situation could be obtained in experiments (No, I have not tried it, but did do the math).
Suppose there are 3 white billard balls, and 3 red billard balls. All six balls have the same diameter and mass, and each ball is a solid sphere with no hollow cavities. When a white ball moving with velocity V hits the end of two stationary white ones, a single white one exits the other end with velocity V while the other balls are left stationary. Same thing happens when one red ball strikes two stationary red ones. But after a white ball strikes two stationary red balls, the white one is left moving with velocity -V/3 and the two red balls move together in the opposite direction with velocity 2V/3.
What is the difference (besides color) between the red and the white balls? In the world of freshmen physics, objects are treated as rigid or as point masses, and all the balls would seem to be the same. In the real world, objects are deformable, and as described in a previous post, the waves travelling through the balls determine what happens in the collision. The answer is that the difference in the balls is how they deform-- they have a different modulus of elasticity and therefore waves move at different speeds through the two balls.
To be left with the white ball moving at -V/3 and the two red ones at 2V/3, the speed of sound in the red balls must be precisely twice that of the white ones. With this ratio of wave speeds, the wave fronts initiated from the point of impact later meet again at the impact point, so the balls separate there. See my previous post for an explanation of how the waves interact to cause the balls to separate when all the balls have the same modulus of elasticity. If the wave fronts do not meet exactly at a contact point between balls, the balls will separate, but they will be left “ringing” with residual compression and tension waves bouncing around inside of them. When this happens the collision is not perfectly elastic, because some of the energy is used by those waves.
In practice, it is easy to get the waves fronts to meet between balls when they are all the same, since it does not matter what the speed of sound is. To get an elastic collision in the laboratory of the type described above is more tricky because you need to make balls with a precise ratio of material properties, but it should be possible to get very close by using composite materials, for example.
Manlob writes,
One nitpick to throw into an otherwise excellent discussion. In the Herrmann and Seitz paper cited above, they discover that the speed of the propagation of the disturbance is based on the time required for the Hertzian contact forces to develop after contact, rather than the speed of sound in the material. In fact, they measure propagation speeds of approx. an order of magnitude less than the speed of sound in the material.
What this means is that, to obtain different collision properties, you could also change the radius of the balls. This idea might work better on the “Newton’s cradle” rather than the pool balls, because the Newton’s cradle masses could be oblong, with different radii at the points of contact than at other points on the body.
Even if you couldn’t get a 2:1 ratio, wouldn’t any difference show up in some departure from the “only the last ball moves” case?
I wonder if some golfer who reads this could try the experiment using golf balls from different manufacturers as the “red” and “white” balls.