# Physics of Pool: Elastic Collision of Equal Masses

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

1. Place one puck on the surface and mark its starting position with the tape.
2. Place the second puck a foot or so away from the first puck.
3. 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 practise runs).
4. Mark a couple of points along the paths both pucks tookafter the collision.
5. Using the marks as a guide, lay a lengthof string along each of the paths taken by the pucks.
6. Use the protractor to measure the angle between the strings—the angle at which the pucks moved away from each other.
7. 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 MvAnd kinetic energy by ½Mv2, where MIs mass and VIs velocity. If ​​vCRepresents the velocity of the moving puck before the collision, ​​vAIs the velocity of the moving puck After The collision, and ​​vBIs the velocity of the stationary puck after the collision, then conservation of kinetic energy leads to:

½MvC2=​½MvA2+ ​½MvB2

Because all the masses are equal, the m’s cancel and you end up with:

VC2=VA2 +VB2

This equation has the exact same form as the Pythagorean Theorem, C2= A2+ B2​where AAndBAre the sides of a right triangle and C Is the hypotenuse. This only works if VAAnd VBAre at right angles to one another.

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 XDirection:

MvCx= mvAx+ mv Bx

Canceling the masses, you end up with

VCx= vAx+ v Bx

In the YDirection, there is initially NoMomentum. To make everything balance, that means the y-direction momentums after the collision must perfectly cancel:

MvCy= 0 = mvAy+ mv By

0 = vAy+ v By

VAy= -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 YComponents 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 NotConserved; some energy is lost to the surroundings. This means that, while the YComponents of the velocity still have to cancel, the XComponents 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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