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 or air hockey pucks would work best)
• Smooth, flat surface (if using a pool table, try placing a foam board over the 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 practice runs).
4. Mark a couple of points along the paths both pucks took after the collision.
5. Using the marks as a guide, lay a length of 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 mv and kinetic energy by ½mv², where m is mass and v is velocity. If ​​v_c represents the velocity of the moving puck before the collision, ​​v_ais the velocity of the moving puck after the collision, and ​​v_b is the velocity of the stationary puck after the collision, then conservation of kinetic energy leads to:

½mv_c² =​ ½mv_a²  + ​ ½mv_b² 

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

v_c² = v_a^{2 +} v_b² 

This equation has the exact same form as the Pythagorean Theorem, c²  = a² + b² ​ where a and b are the sides of a right triangle and c is the hypotenuse. This only works if v_a and v_b are 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 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?