# Billiard Balls and the 90-Degree Rule

If you play pool, you may have noticed that when the cue ball strikes a target ball off-center, that they will separate at a 90-degree angle after the collision. This is because the collision is very elastic, and the balls are the same mass. All kinetic energy along the line connecting the two balls is transferred to the target ball, leaving the cue ball with no energy along that direction. Any remaining energy in the cue ball is along the line perpendicular to the line connecting the two balls. This is why they separate at a 90-degree angle. (Also note: if the cue ball strikes the target ball dead-center, there will be no energy along a perpendicular line, all energy will be transferred to the target ball, unless there is substantial inelasticity.)

So let’s explore the simple case, where we are assuming that there is negligible spinning and friction, and that the collision is perfectly elastic. Initially, the cue ball is put into motion toward the target ball, and the target ball is stationary.
First, we use conservation of momentum to get our first equation, where $m_1$ is the mass of the cue ball, $m_2$ is the mass of the target ball, $\overrightarrow{v}$ represents their velocity vectors, with the subscripts $i$ meaning “initial”, and $f$ meaning “final”.
$m_1 \overrightarrow{v}_{1,i} + m_2 \overrightarrow{v}_{2,i} = m_1 \overrightarrow{v}_{1,f} + m_2 \overrightarrow{v}_{2,f}$   (1)
Now, since the masses of the balls are equal and the initial velocity of the second ball is 0, we can simplify the first equation to:
$\overrightarrow{v}_{1,i} = \overrightarrow{v}_{1,f} + \overrightarrow{v}_{2,f}$   (2)
Second, we use conservation of energy to get our second equation:
$\alpha( \frac{1}{2} m_1 v_{1,i}^2 + \frac{1}{2} m_2 v_{2,i}^2) = \frac{1}{2} m_1 v_{1,f}^2 + \frac{1}{2} m_2 v_{2,f}^2$   (3)
Again, we can reduce this, since the initial velocity of the second ball is 0, the masses are equal, and the collision is perfectly elastic $\alpha = 1$.
$v_{1,i}^2 = v_{1,f}^2 + v_{2,f}^2$   (4)
Now we can take equation 2 and square it.
$v_{1,i}^2 = v_{1,f}^2 + v_{2,f}^2 + 2 \overrightarrow{v}_{1,f} \cdot \overrightarrow{v}_{2,f}$
$= v_{1,f}^2 + v_{2,f}^2 + 2 v_{1,f} v_{2,f} cos (\theta)$   (5)
Where $\theta$ is the angle between the two balls after the collision.
Now, if you compare equations 4 and 5, you’ll see that they are equal. However, there’s this cross term $2 v_{1,f} v_{2,f} cos (\theta)$ that shows up in equation 5, but not in 4. This means it’s equal to 0.
So then,
$cos(\theta) = 0$
and this can only be true if $\theta = \frac{\pi}{2}$, which is equivalent to 90 degrees!
(If the target ball is struck dead-center, then the cross term goes to 0 from $v_{1,f} = 0$.)

In a more general case, where we include inelasticity and the possibility of the balls having different masses, we arrive at a more general formula for $\theta$. We’ll still assume negligable friction, spinning, and that the target ball is initially stationary.
So we still have conservation of momentum:
$m_1 \overrightarrow{v}_{1,i} = m_1 \overrightarrow{v}_{1,f} + m_2 \overrightarrow{v}_{2,f}$   (6)
And conservation of energy, but with some of the energy going to sound and heat. This is represented by an elasticity factor $\alpha$, that ranges from 1 (perfectly elastic) to 0 (perfectly inelastic).
$\alpha \frac{1}{2} m_1 v_{1,i}^2 = \frac{1}{2} m_1 v_{1,f}^2 + \frac{1}{2} m_2 v_{2,f}^2$   (7)
So we can take equation 6, and just like last time, we square it.
$m_1^2 v_{1,i}^2 = m_1^2 v_{1,f}^2 + m_2^2 v_{2,f}^2 + 2 m_1 m_2 v_{1,f} v_{2,f} cos(\theta)$   (8)
Now divide equation 8 by $m_1$ on both sides, and you will have an expression for $m_1 v_{1,i}^2$
$m_1 v_{1,i}^2 = m_1 v_{1,f}^2 + \frac{m_2^2}{m_1} v_{2,f}^2 + 2 m_2 v_{1,f} v_{2,f} cos(\theta)$,
which can be plugged into the left side of equation 7.
So now, you can solve for $\theta$ in terms of the final velocities, masses, and elasticity factor.
The general result is then:
$cos(\theta) = \frac{1-\alpha}{2 \alpha} \frac{m_1}{m_2} \frac{v_{1,f}}{v_{2,f}} + \frac{1}{2 \alpha} (1 - \alpha \frac{m_2}{m_1}) \frac{v_{2,f}}{v_{1,f}}$
We can get back the previous result $cos(\theta) = 0$,  by setting $m_1 = m_2$ and $\alpha = 1$.