## Pool Part 1: The Basic Shot

Prerequisites: Vectors

In this two-part post, we'll go through some of the basic geometry of pool/billiards.

In Part 2, we'll derive simple methods to make one-rail bank and kick shots (to be defined below). We'll also go briefly into an example of a two-rail kick, with which it's extremely difficult to actually pocket the target ball, but at least you can work your way out of some sticky situations and avoid giving your opponent ball-in-hand.

This analysis also works for mini-golf on flat surfaces, by the way.

#### How to pocket a ball

This section involves a bit of physics, which I'll explain for those who don't know it already, but you may need to quickly read up on vectors here before proceeding.

Suppose we have the following set-up:

The white ball is the cue ball, and we want to hit it into the 2 ball (the blue one, also known as the object ball) to pocket the latter in the side pocket on the top of the image.

In order to accomplish this, we need the cue ball, upon contact with the 2, to impart a force upon the latter which makes it move in the direction of the pocket. The way to make the force be in that direction is to hit such that the point of contact of the two balls lies along the line between the center of the pocket (where we want the object ball to go), and the center of the object ball as in the next diagram:

It's a subtle difference, but please note that you are aiming for the point of contact, and not the center of the cue ball, to lie along the yellow line upon contact with the 2. The translucent cue ball in the above diagram shows the desired cue ball position upon contact.

If you get this right, and hit the cue ball at center with a reasonable speed, then the force imparted on the 2 will be along the yellow line, and thus it will roll along the yellow line and into the pocket. This is Newton's second law at work, which states that if the mass (i.e. how many kilograms) of the 2 ball is $m$, and the force the cue ball imparts on the 2 is ${\bf F}$ (note that this is a vector quantity, which is why it has a direction, while the mass is a scalar), then ${\bf F} = m {\bf a}$, i.e. the 2 will gain an acceleration ${\bf a}$ due to the force ${\bf F}$. The units of the acceleration are meters per second squared, and thus the units of the force are killograms*meters per second squared, also called Newtons after the same Isaac Newton we were just talking about.

Acceleration is the change, both of magnitude and direction, in velocity per unit time (thus change over one second, in how many meters per second the ball is traveling at that moment). Velocity is just the vector quantity whose magnitude is the speed (in units of meters per second) and whose direction is the direction of motion of the ball. These quantities can all vary over time, as can the force. The ball's mass $m$ is a scalar quantity that is constant over time and is a measure of how much matter is contained in the ball. The heavier the ball, the more force it takes to accelerate the ball by an equivalent amount. That's what the magnitude part of the vector equation ${\bf F} = m {\bf a}$ tells us. The direction part tells us that the acceleration is in the same direction as the force.

Make sense? Ok good- if there are questions on that, they can go in the comments section or maybe I can do a separate post, but my point was that since the cue ball is round and thus contacts the (also round) 2 at exactly one point, the cue ball must impart a force on the 2 i the direction of the yellow line in the diagram above, and thus the 2 will accelerate along that line after the contact. Since no forces act on the ball that would cause it to deviate off of that line after the contact, it will continue along that line and thus into the side pocket.

#### Where does the cue ball go after the contact?

That's an important question, and good players need to take this into account when planning a series of shots.

As it turns out, the cue ball bounces off perpendicular to the yellow line as in the next diagram:

To see why this is the case, we need to use conservation of momentum. What does this mean? Well, momentum is the vector quantity $m{\bf v}$ where $m$ and ${\bf v}$ are mass and velocity as above. To say that momentum is conserved means that the momentum vector of a system of objects (for a system of multiple objects, this would be the vector sum of the individual momenta) remains constant in the absence of a net external force. In our system of two balls, the force between the balls would not qualify as external. Gravity would, but it is counteracted by the force of the table pushing back up on the balls, which causes them to not fall to the ground. Thus, ignoring friction, energy loss due to the sound of the balls' hitting, etc., the momentum of the system of the two balls is the same right before and right after the collision.

In the diagram above, we have labeled the velocities, and let's assume the cue and 2 have the same mass $m$. Momentum is conserved before and after the collision, which means: $$m{\bf v}_0 = m{\bf v}_1 + m{\bf v}_2$$
Note that the 2 ball had no velocity initially, so the left-hand side of the equation has only the cue ball's momentum. The $m$'s cancel out to give the vector equation $${\bf v}_0 = {\bf v}_1 + {\bf v}_2$$ which actually comprises 2 algebraic equations, one in the $x$-component and one in the $y$-component:
\begin{align} v_{0x} &= v_{1x} + v_{2x} \tag{1}\\[2mm] v_{0y} &= v_{1y} + v_{2y} \tag{2} \end{align}Now, we can use $(1)$ and $(2)$ to obtain: \begin{align} \| {\bf v}_0 \|^2 &= v_{0x}^2 + v_{0y}^2 \\[2mm] &= (v_{1x} + v_{2x})^2 + (v_{1y} + v_{2y})^2 \\[2mm] &= \| {\bf v}_1 \|^2 + \| {\bf v}_2 \|^2 + 2v_{1x}v_{2x} + 2v_{1y}v_{2y} \\[2mm] &= \| {\bf v}_1 \|^2 + \| {\bf v}_2 \|^2 + 2({\bf v}_1 \cdot {\bf v}_2) \tag{3} \end{align} We also know that the energy of the system is conserved. The energy of an object of mass $m$ and speed $v$ is $\frac{1}{2}mv^2$. Technically, this is only the kinetic energy (energy due to motion of a massive particle), but there is no potential energy in this system (e.g. an object high up about to fall and gain speed, and thus kinetic energy, would have potential energy).

Conservation of energy tells us that \begin{align} &\frac{1}{2} m \| {\bf v}_0 \|^2 = \frac{1}{2} m \| {\bf v}_1 \|^2 + \frac{1}{2} m \| {\bf v}_2 \|^2 \\[2mm] \Longrightarrow \ &\| {\bf v}_0 \|^2 = \| {\bf v}_1 \|^2 + \| {\bf v}_2 \|^2 \tag{4} \end{align} Subtracting equation $(4)$ from equation $(3)$ shows that ${\bf v}_1 \cdot {\bf v}_2 = 0$, i.e. ${\bf v}_1$ and ${\bf v}_2$ are perpendicular. This means that the cue ball indeed bounces off at a right angle to the direction of the 2 after contact.

In part 2 of this post, we'll explore bank and kick shots...