Pool balls colliding head-on

Two pool balls collide head-on, so that the collision is restricted to one dimension. Pool balls are constructed so as to lose as little kinetic energy as possible in a collision, so under the assumption that no kinetic energy is converted to any other form of energy, what can we predict about the results of such a collision?

Pool balls have identical masses, so we use the same symbol m for both. Conservation of energy and no loss of kinetic energy give us the two equations

The masses and the factors of 1/2 can be divided out, and we eliminate the cumbersome subscripts by replacing the symbols v_{1i ,...} with the symbols A,B,C, and D:

A + B = C + D

A² + B² = C² + D² .

A little experimentation with numbers shows that given values of A and B, it is impossible to find C and D that satisfy these equations unless C and D equal A and B, or C and D are the same as A and B but swapped around. A formal proof of this fact is given in the sidebar. In the special case where ball 2 is initially at rest, this tells us that ball 1 is stopped dead by the collision, and ball 2 heads off at the velocity originally possessed by ball 1. This behavior will be familiar to players of pool.

Gory Details of the Proof Above

The equation A + B = C + D says that the change in one ball's velocity is equal and opposite to the change in the other's. We invent a symbol x = C - A for the change in ball 1's velocity. The second equation can then be rewritten as A²+B² = (A+x)²+(B-x)². Squaring out the quantities in parentheses and then simplifying, we get 0 = Ax - Bx + x². The equation has the trivial solution x = 0, i.e., neither ball's velocity is changed, but this is physically impossible because the balls cannot travel through each other like ghosts. Assuming x 6≠0, we can divide by x and solve for x = B - A. This means that ball 1 has gained an amount of velocity exactly sufficient to match ball 2's initial velocity, and vice-versa. The balls must have swapped velocities.