Pool Hall Pondering: Why does a cue ball make a 90 degree angle with the ball it hits?

I'm teaching my 9th graders how to analyze billiard ball collisions. Specifically, we are looking at the case when the cue ball hits one other lone ball (say the eight ball). I teach the following principles:

1. momentum should be conserved
2. the cue ball should go off at 90 degrees relative the motion of the ball it hits. Call this the 90 degree rule.

As for #1 -this is well established. But where did #2 -the 90 degree rule- come from? I remember this fact from my college days when I played a lot of pool and it was crucial in assessing the likelihood of a scratch. It was important to make sure the cue ball didn't land in a pocket, so I needed to know the direction it headed after impact. Thus the 90 degree rule was very handy.

I thought about why this rule should be true and came up with the following vector theoretic proof. It's a little beyond your typical 9th grade physics class, but not terribly advanced:

GIVEN: Vectors p1,i , p2,i , p1,f , p2,f respectively denoting the cue ball's and eight ball's momenta before (i) and after (f) the impact. By assumption, the eight ball is initially at rest so that p2,i = 0. We assume conservation of momentum, as well as an elastic collision (i.e. conservation of kinetic energy). We also assume that all the balls have the same mass, denoted by m.

PROVE: The cue ball and eight ball shoot of perpendicularly to each other after impact. In other words, p1,f , p2,f are orthogonal.

PROOF: Since p2,i = 0, the initial total momentum is just p1,i . Conservation of momentum implies that the initial total momentum equals the final total momentum. Therefore

p1,f + p2,f = p1,i .

Furthermore, with a little vector algebra, one can express a particle's kinetic energy in terms of the magnitude of its momentum and its mass as follows:

E = || p || ² / ( 2 m )

Elasticity -i.e. conservation of energy- and the fact that p2,i = 0 therefore imply that

|| p1,i || ² / ( 2 m ) = || p1,f || ² / ( 2 m ) + || p2,f || ² / ( 2 m )

We can get rid of the common factor and re-write as:

|| p1,f || ² + || p2,f || ² = || p1,i || ²

Thinking about it a little we see that we have a case of two vectors whose norms add up to the norm of their sum. In other words, let a = p1,f and b = p2,f. We have a + b = p1,i and ||a|| ² + ||b|| ² = || p1,i || ². This can simply be restated as:

||a|| ² + ||b|| ² = || a + b || ².

Draw this as a vector addition diagram. If a makes one side and b makes the second side starting from the head of a, then the third side -the vector sum a + b - has length that satisfies the Pythagorean theorem! But the Pythagorean theorem has a converse, which I'll show next, and which will guarantee that a is orthogonal to b thus completing the proof.

The final part is:

VECTOR VERSION OF PYTHAGORAS'S THEOREM:

||a||² + ||b||² = || a + b ||² if and only if a is orthogonal to b.

Proof: By definition the norm of the sum of the two vectors is:

|| a + b ||² = ( a + b ) º (a + b ).

Simplify the right hand side by FOIL'ing to a º a + 2 a º b + b º b . Using the definition of norm in terms of the dot product again we get:

|| a + b ||² = || a ||² + 2 a º b + || b ||².

Setting this equal to ||a||² + ||b||² we get

||a||² + ||b||² = || a ||² + 2 a º b + || b ||².

Subtracting ||a||² + ||b||² we get

0 = 2 a º b .

In other words, the original equation holds exactly when a º b = 0, so exactly when a is orthogonal to b.

QED