The final shot in the pool game is yours, but the cue and the eight ball aren’t nicely lined up with any of the pockets. Looking around, the closest pocket is 45 degrees off the line between the two balls. You take aim for a glancing blow, the cue ball strikes the eight ball…what happens next? Do you sink the eight ball? Which way does the cue ball end up going? And how can you make the eight ball go off in a different direction?

Pool is a great example of physics in action. After every collision, the **momentum **of all the balls—the product of their mass and velocity—has to be conserved. That is, the total momentum before the collision has to be the same as the total momentum after the collision. And, roughly speaking, the **energy** must be conserved as well; the balls can’t fling away from each other with more energy than you give them. These two laws—the **conservation of energy** and the **conservation of momentum**—work together to steer the balls around the table.

In this project, you’ll experiment with colliding masses, see how they collide, and maybe learn how to use physics to plan the perfect pool shot!

### Problem

At what angle will two equal-mass balls move away from one another after a glancing collision?

### Materials

- 2 low friction masses of equal weight (hover pucks orair hockey pucks would work best)
- Smooth, flat surface (if using a pool table, try placing a foam board overthe surface to reduce friction)
- Protractor
- Tape
- String

### Procedure

- Place one puck on the surface and mark its starting position with the tape.
- Place the second puck a foot or so away from the first puck.
- Gently push the second puck towards the first puck, aimed so that it hits the puck at a glancing angle rather than straight on (this may take a few practice runs).
- Mark a couple of points along the paths both pucks tookafter the collision.
- Using the marks as a guide, lay a lengthof string along each of the paths taken by the pucks.
- Use the protractor to measure the angle between the strings—the angle at which the pucks moved away from each other.
- Repeat steps 1-6 several times and calculate the average angle between the strings. If the puck doesn’t hit at a glancing angle, then just skip that attempt and try again.

### Results

The angle between the pucks’ paths will be close to ninety degrees—a right angle.

### Why?

In an **elastic collision**, both momentum and kinetic energy are conserved. Momentum is given by *mv*and kinetic energy by ½*mv ^{2}*, where

*m*is mass and

*v*is velocity. If

*â€‹â€‹v*represents the velocity of the moving puck before the collision,

_{c}*â€‹â€‹v*is the velocity of the moving puck

_{a}*after*the collision, and

*â€‹â€‹v*is the velocity of the stationary puck after the collision, then conservation of kinetic energy leads to:

_{b}½*mv _{c}^{2}*=â€‹½

*mv*+ â€‹½

_{a}^{2}*mv*

_{b}^{2}Because all the masses are equal, the m’s cancel and you end up with:

*v _{c}^{2}*=

*v*

_{a}^{2 +}v_{b}^{2}This equation has the exact same form as the **Pythagorean Theorem**, *c** ^{2}* =

*a*

*+*

^{2}*b*

*â€‹where*

^{2}*a*and

*b*are the sides of a right triangle and

*c*is the hypotenuse. This only works if

*v*and

_{a}*v*are at right angles to one another.

_{b}The conservation of momentum adds some depth (and complexity). Momentum is a little more complicated because it has to be broken down into **components**: the momentum along the original direction of motion (*x*) and momentum perpendicular to that direction (*y*). Momentum in both directions has to be conserved. Initially, all the momentum is in the *x* direction:

*mv _{cx} = mv_{ax} + mv _{bx}*

Canceling the masses, you end up with

*v _{cx} = v_{ax} + v _{bx}*

In the *y* direction, there is initially *no* momentum. To make everything balance, that means the y-direction momentums after the collision must perfectly cancel:

*mv _{cy} = 0 = mv_{ay} + mv _{by}*

*0 = v _{ay} + v _{by}*

*v _{ay} = -v _{by}*

Putting this together with the conservation of energy, you find that all the velocity components after the collision have the same magnitude, with the *y* components pointing in different directions. You end up with the final velocities pointing at right angles away from each other.

In an **inelastic collision**, kinetic energy is *not* conserved; some energy is lost to the surroundings. This means that, while the *y*components of the velocity still have to cancel, the *x*components can be different. The balls will no longer bounce away at right angles.

In reality, perfectly elastic collisions rarely happen; some energy is always lost. Collisions between subatomic particles (protons and electrons) are very nearly elastic; so are atoms in an ideal gas. Space probes that slingshot around a planet behave the same way as elastic collisions as well.

### Going Further

What happens if you use pucks (or balls) with different masses? For example, if you’re using air hockey pucks, try making the stationary puck be two pucks stacked on top of one another. What changes?

What happens if you use a surface that isn’t smooth (for example, a carpet)? How does that change the angle? Is it less than or more than a right angle? Can you explain what you’re seeing using the equations mentioned above?

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