What if two pucks hit off-center?

Olly slides an orange puck at a blue puck. The blue puck is sitting still.

What if the orange puck hits the blue one on its side, not in the middle?

Picture a very smooth table where pucks glide on a cushion of air. An orange puck slides toward a blue puck that is sitting still. The two are the same size and weight.

If the orange puck hits the blue one dead center, what happens is simple. But what if it clips the blue puck off-center, on one side? Which way does each puck go, and how fast? This lab lets you aim the hit and see.

Collisions in two dimensions need momentum treated as a vector: OpenStax resolves it into perpendicular components that are each conserved. Billiard balls, OpenStax notes, are a good everyday example, and so are some subatomic particle collisions.

The question: for two equal pucks, one at rest, in a bouncy (elastic) collision, how does an off-center hit set the directions and speeds afterward, and what stays the same no matter how you aim?

First, the orange puck hits the blue one right in the middle.

Bump! The orange puck stops. The blue puck zooms away just as fast.

Start with a straight, head-on hit. The orange puck slides at 2 meters per second into the still blue puck. In the model, the orange puck stops dead and the blue puck slides off at 2.00 meters per second, in the same direction.

The two pucks have simply swapped speeds. The total momentum, the 'amount of motion' measured as mass times speed, is 2.00 before and 2.00 after.

Head-on (impact parameter b = 0), equal masses of 1 kg, v₁ = 2 m/s, v₂ = 0, e = 1: the model gives v₁′ = 0 and v₂′ = 2.00 m/s. This matches the one-dimensional elastic result v₁′ = (m₁ − m₂)v₁/(m₁ + m₂) = 0 and v₂′ = 2m₁v₁/(m₁ + m₂) = v₁.

Momentum is 2.00 kg·m/s before and after, and no kinetic energy is lost. Equal masses exchange velocities completely.

Now the orange puck clips the side of the blue puck.

This time the orange puck does not stop. Both pucks slide away, in two different directions!

Now aim the orange puck so it clips the blue one off-center: halfway out from a head-on hit. In the model, the blue puck slides away at 1.73 meters per second, and the orange one keeps going at 1.00 meter per second, but turned to the side.

Measure the angle between their two paths: 90°, a perfect right angle. Try other aims, a little off-center or a lot. The speeds change, but the angle stays 90°.

With b = 0.5 (half the sum of the radii), the line of centers at contact is 30° from the orange puck's path. The impulse acts only along that line, so the blue puck leaves at 30° with 2 cos 30° = 1.73 m/s, and the orange puck leaves at 60° on the other side with 2 sin 30° = 1.00 m/s.

The model's 'angle between paths' readout shows 90° for b = 0.25, 0.5 and 0.75. The momentum magnitude stays 2.00 kg·m/s and the energy lost is 0%.

The pucks share their push. Together they keep all the motion the orange puck had.

When two same-size pucks share it like that, they always slide apart at a square corner.

Two rules are at work. First, the total momentum stays the same, so the two pucks together must still carry all the 'forward motion' the orange puck had. Second, in a bouncy hit no motion energy is lost.

For two pucks of the same mass, the only way to obey both rules at once is for them to slide apart at a right angle. Change the masses or make the hit less bouncy and the angle changes. The model's pucks are perfectly smooth and don't spin, which real objects can't quite manage.

Momentum: m v₁ = m v₁′ + m v₂′, so v₁ = v₁′ + v₂′ as vectors. Energy (e = 1): v₁² = v₁′² + v₂′². Squaring the first and subtracting the second leaves 2 v₁′·v₂′ = 0, so the velocities are perpendicular, the result OpenStax derives for equal-mass elastic collisions.

Break either condition and the rule fails. In the model, a 2 kg blue puck gives 120° and a 0.5 kg one 60°; with e = 0 at b = 0.5, the angle drops to about 49° and over a third of the energy is lost. The model's pucks are frictionless with each other and do not spin, so the impulse acts purely along the line of centers.

Pool players hit balls on a table. When one ball clips another, they roll apart at almost a square corner.

Good players use this to plan their shots!

Billiard and pool balls are almost exactly the same mass, and their collisions are nearly bouncy. OpenStax's physics textbook points out that if you play enough pool, you'll notice the two balls go off at very close to 90° after a glancing hit.

It isn't always exact. Putting a lot of spin on the ball changes the angle, because spin carries extra energy into the collision. Our model's pucks never spin, which is why its angle is a perfect 90°.

OpenStax describes billiard collisions as nearly elastic between equal masses: the separation angle is close to 90°, departing from it when heavy spin adds energy and angular momentum to the system. The same derivation is exact for some subatomic particle collisions between equal masses.

Our air-table model idealizes further: perfectly smooth, non-spinning pucks with constant restitution. Real collisions involve surface friction, spin and restitution that changes with impact speed, which the model leaves out.