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Let us consider a case of two identical billiards/snooker balls that are on a pool table. A cue hits one of the identical balls (preferably the white striker ball), setting it into pure rolling motion on the table. Assuming that both the balls are perfectly spherical, and of mass M and radius R; and that there is no air resistance, what all exchanges take place when the first ball strikes the second?

By exchanges, I mean energy conservation, momentum conservation, and angular momentum conservation. I also want to know exactly what would happen at the moment the two balls touch; will there be a slight deformation of the two balls, followed by them getting re-formed due to internal forces, or due to rigid-body constraints?

Finally, I would like to know exactly what happens if the balls don't collide head-on, but off-axis. Does the second (non-striker) ball get projected in the direction of the line that joins the centres of the two striker balls?

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    $\begingroup$ This question requires a long and complicated answer and should probably be split into smaller pieces. For part of the answer, see the supplement on the mechanics of billiards in Sommerfeld's Mechanics, p. 158. $\endgroup$ – NickD Nov 10 '17 at 17:37
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    $\begingroup$ -1. No research effort. These kind of questions are dealt with in introductory mechanics textbooks and online tutorials. $\endgroup$ – sammy gerbil Nov 12 '17 at 1:15
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By exchanges, I mean energy conservation, momentum conservation, and angular momentum conservation.

All these conservation laws always take place. When the balls collide, they share energy, momentum and angular momentum.

Energy may be spent on permanent deformations (in a car crash, plenty of energy is spent on crumbling the car fronts), or heat losses (usually from friction), but none of those will be very significant in usual billiard balls.

will there be a slight deformation of the two balls, followed by them getting re-formed due to internal forces, or due to rigid-body constraints?

Yes. This is called an elastic collision. The same thing goes on on a trampoline. The intermolecular bonds are flexible but will not break or displace in an elastic material - and a billiard ball is in this category.

If any bonds change, displace or break, then permanent deformations have happened, and energy was spend on doing so. In elastic collisions all energy spent on the deformations are stored as elastic potential energy - this is returned as kinetic energy for the motion as soon as the flexible bonds have bounced back.

Finally, I would like to know exactly what happens if the balls don't collide head-on, but off-axis. Does the second (non-striker) ball get projected in the direction of the line that joins the centres of the two striker balls?

The answer follows the law of momentum conservation,

$$\sum \vec p_{before}=\sum \vec p_{after}\\ \vec p_{1,before}+\require{cancel}\cancel{\vec p_{2,before}}=\vec p_{1,after}+\vec p_{2,after}$$

This is a vector relationship. Thus only parallel components interact. If ball 1 impacts ball 2 off-axis as you say, then only the perpendicular momentum component (perpendicular to the contact point/surface) has an effect and pushes on the other ball. The component parallel to the contact point/surface stays unaltered. So, the motion gained by the ball at rest is perpendicular away from the contact point/surface.

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Read up on elastic collisions to understand the conservation of kinetic energy and momentum.

But in real life collisions are inelastic and there are more insignificant changes.

For general physics of billiards, read up this article

Just getting you started if you want to get a general idea. I hope someone will give a detailed answer.

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