The Physics of Pool/Billiards

When I was growing up, I had a pool (billiards) table in my basement (I will refer to it at pool in this post). Although I was not a very dedicated player, I would always go down and occasionally play either by myself or with my brothers. Even at Colgate, I still sometimes go to the SRS house (100 Hamilton) and play pool there with some of my friends. When I think about how pool works, I realized that we can essentially boil it down to momentum and collision (mostly elastic). 

Because we are assuming that the collisions are mostly elastic, we have conservation of momentum and conservation of kinetic energy. When hitting the cue ball into another ball, we can use the equation mAvA = mAvA’ + mBvB’ with A being the cue ball and B being the ball it hit (at rest). Because the masses of the cue ball are the same as the other balls on the table, we can say that mA and mB are the same. Therefore, we have vA = vA’ + vB’.

Because we are assuming it is an elastic collision, kinetic energy is conserved, meaning that we have KEi = KEf . We can rewrite as ½ mA(vA)2 = ½ mA(vA’)2 + ½ mB(vB)2 . Again, because mA and mB are the same, we have (vA)2 = (vA’)2 + (vB’)2 .

Therefore, using the two equations above, we can draw the velocity vectors to form a triangle as shown below, except V1A is vA and V2A is vA’ and V2B is vB’.

As a result, ball A moves in a direction perpendicular to that of ball B after the collision, assuming that the collision is not head-on.