What if a bowling ball hit a ping-pong ball?

A bowling ball is big and heavy. A ping-pong ball is tiny and light.

What if the bowling ball rolled into the ping-pong ball? Which one would go faster afterward?

When two balls of the same size and weight crash head-on, something neat can happen: the moving one stops, and the still one moves off at the same speed. But what if one is far heavier than the other, like a bowling ball rolling into a ping-pong ball?

Would the bowling ball stop? Would the ping-pong ball move at the bowling ball's speed, or faster? In this lab, two pucks on a smooth track stand in for the balls: a heavy orange puck and a very light blue one, one hundred times lighter.

Two quantities constrain a one-dimensional collision: total momentum, always conserved, and, for an elastic collision, total kinetic energy. OpenStax derives the final velocities from these two equations; for a target at rest they depend only on the mass ratio and the striker's speed.

This lab compares two cases in the collisions model with perfectly elastic restitution (e = 1) and a striker speed of 2 m/s: equal 1 kg masses, then a 10 kg striker on a 0.1 kg target, a 100:1 ratio standing in for a bowling ball and a ping-pong ball. How do the two outcomes differ, and what limit does the light ball's speed approach?

First, two balls of the same size and weight. The orange one rolls into the blue one.

Bump! The orange ball stops. The blue ball rolls away just as fast.

In the model, start with two pucks of the same mass, 1 kilogram each. The orange puck slides at 2 meters per second into the blue puck, which is standing still. Bump: the orange puck stops dead, and the blue puck moves off at exactly 2 meters per second.

All the motion has passed from one to the other. Nothing is lost, because the pucks are perfectly bouncy in this model. Keep this result in mind: when the masses are equal, the speeds simply swap.

With m₁ = m₂ = 1 kg, v₁ = 2 m/s, v₂ = 0 and e = 1, the model gives v₁′ = 0 and v₂′ = 2 m/s: equal masses exchange velocities. Momentum (2 kg·m/s) and kinetic energy (2 J) are both conserved.

This is the special case m₁ = m₂ of the general elastic result. The factor (m₁ − m₂) in v₁′ vanishes, so the striker stops; the factor 2m₁/(m₁ + m₂) in v₂′ equals 1, so the target leaves at the striker's speed.

Now a heavy ball rolls into a tiny light ball.

The heavy ball keeps rolling, almost as fast as before. The tiny ball zooms off nearly twice as fast!

Now make the orange puck 10 kilograms and the blue puck only 0.1 kilograms, one hundred times lighter. The heavy puck slides in at 2 meters per second and hits the light one.

The heavy puck barely notices: it slows only from 2 to about 1.96 meters per second. The light puck shoots off at about 3.96 meters per second, nearly twice as fast as the heavy puck was going. Speed up the heavy puck to 5 meters per second and the light one leaves at about 9.9.

At m₁ = 10 kg, m₂ = 0.1 kg, v₁ = 2 m/s the model gives v₁′ = 1.96 m/s and v₂′ = 3.96 m/s, matching (9.9/10.1) × 2 and (20/10.1) × 2. Momentum stays 20 kg·m/s and kinetic energy 20 J. At v₁ = 5 m/s, v₂′ = 9.90 m/s.

Reverse the roles (0.1 kg striker, 10 kg target) and the light puck rebounds at −1.96 m/s while the heavy one creeps forward at 0.04 m/s. With e = 0 instead, the heavy-hits-light pair moves together at 1.98 m/s and loses only 0.99% of its kinetic energy.

The tiny ball cannot slow the heavy ball much. It is too light.

But the bump pushes the tiny ball hard. So it bounces away very fast.

When the two pucks hit, they push on each other just as hard. That push changes the light puck's speed a lot, because it is so light, but hardly changes the heavy puck's speed at all.

Why nearly twice as fast? Think of it from the heavy puck's point of view: the light puck rushes toward it at 2 meters per second and bounces off it like a ball off a wall, going back out at 2. Add the heavy puck's own 2 meters per second, and you get nearly 4. Our model's pucks are smooth and don't spin, and the bump is perfectly bouncy.

Equal and opposite impulses J act on both bodies, so Δv = J/m: the light body's velocity change is m₁/m₂ = 100 times the heavy one's. In the heavy body's frame, nearly unchanged by the collision, the light body approaches at v₁ and, for e = 1, recedes at v₁; transforming back adds v₁, giving v₂′ → 2v₁ as m₂/m₁ → 0.

Model limits, from its documentation: smooth, non-spinning pucks, an instantaneous impact with constant restitution, no walls, and puck sizes set by the cube root of mass. Real balls also spin, rub and lose some energy to heat and sound.

Real balls are never perfectly bouncy. They lose a little energy as heat and sound.

In the model, a less bouncy bump sends the tiny puck off more slowly.

OpenStax works through the opposite case: a small object hitting a larger one that is standing still. The small one bounces backward, and the large one is knocked forward slowly, like a small car bouncing backward off a much bigger one. Try it in the model by swapping the masses.

OpenStax also explains that everyday collisions are never perfectly bouncy: some movement energy always turns into heat and sound. One nearly bouncy collision it names is two steel blocks on ice. The model can show what a less bouncy hit does: set the bounciness to 0.5, and the light puck leaves at about 2.97 meters per second instead of 3.96.

OpenStax's worked example has a 0.500 kg object at 4.00 m/s striking a 3.50 kg object at rest; the small object bounces backward and the larger is knocked forward at low speed, with internal kinetic energy unchanged at 4.00 J. It compares this to a compact car bouncing backward off a full-size SUV at rest.

In real collisions OpenStax notes that some kinetic energy is always converted to heat and sound; truly elastic collisions occur only with subatomic particles. The model shows what a less-than-perfect bounce does: at e = 0.5 the light puck leaves at 2.97 m/s instead of 3.96, and the heavy one keeps 1.97 m/s.