# Collision of Discs and Snooker Kicks

I woke up this morning thinking about spinning discs. Could someone verify whether my reasoning below is correct?

Problem 1

Suppose have two identical uniform discs constrained to move in a plane. Set disc $A$ spinning about its centre of mass, clockwise viewed from above. Now collide it with disc $B$.

Assume that friction is the only force acting, and the collision is elastic. I believe that I should be able to work out qualitatively what happens from conservation of linear and angular momentum.

The friction will provide a torque causing $B$ to spin anticlockwise about its centre of mass. Let $L$ be the line joining the centre of $A$ and $B$ at the moment of collision. But then by conservation of angular momentum $B$ must deviate to the right of $L$ after collision. And then by conservation of linear momentum disc $A$ will move to the left of $L$.

The other option for conserving angular momentum would be for the disc A to start spinning faster, but this is ruled out by conservation of energy. Correct?

Note: this effect is observed in snooker when left hand side is imparted to the cue ball.

Problem 2

Now suppose that the discs are stood on end on a table and we do the same experiment, assuming no friction between table and disc. Then conservation of angular momentum must cause the first disc to jump in the air upon collision, by the same reasoning as before. Correct?

Note: I guess this is the origin of the "kick" in snooker.

So really understanding how to prevent snooker kicks is the subtle problem of understanding precisely how much angular momentum is transferred between the balls. You ideally want to minimise this (i.e. have almost frictionless ball collisions, compared with the friction between cloth and ball).

• For those who don't know what snooker is: en.wikipedia.org/wiki/Snooker – Kyle Kanos Apr 13 '14 at 15:51
• I do not see the direct link from pb 1 to pb 2: the kick is an out-of-plane movement, and you have considered movements in the plane only in pb 1. I believe rotation in a plane normal to the table is a minimal ingredient you have to add in order to hope to tackle pb 2. – Joce Apr 15 '14 at 8:51
• Hi Joce, thanks for your comment. I think I didn't make the setup clear enough. In problem two the balls are rotating about an axis parallel to the plane of the table. In other words they have "topspin" or "backspin". So this example is just like the previous one, only now we have gravity and the table surface to contend with. Does this make sense? Thanks again! – Edward Hughes Apr 15 '14 at 11:11