Stun Shot — Object Ball Speed

Here is the part of a cut shot that catches every improving player out: the object ball does not get all your speed, and on a thin cut it barely gets any. The balls are hard, smooth spheres, so during the collision the only force either can exert on the other points along the line joining their centres. Whatever component of the cue ball's velocity lies along that line gets transferred; whatever component lies across it cannot be touched. The along-the-line component is $u\cos\varphi$, and since the masses are equal and the collision is nearly elastic, the object ball takes all of it.

The cosine is gentle at first and then falls off a cliff, and because kinetic energy goes as the square, the energy transfer falls off twice as fast:

• 15° cut: the object ball gets 97% of the speed, 93% of the energy.
• 30° cut: 87% of the speed, 75% of the energy.
• 45° cut: 71% of the speed, 50% of the energy.
• 60° cut: 50% of the speed, 25% of the energy.
• 75° cut: 26% of the speed, 7% of the energy.

That last line is the one worth internalizing. A 75° cut delivers a fourteenth of the energy you put into the cue ball. If you want to send that object ball the length of the table, you have to hit the cue ball roughly four times as hard as you would for the same distance on a straight shot — and hitting a cue ball four times as hard while maintaining a thin, precise contact is exactly the skill thin cuts demand. Beginners under-hit thin cuts almost without exception, then blame their aim.

Two corrections, and I would rather state them than let you discover them at the table.

The balls are not perfectly elastic. A phenolic resin ball striking another gives a coefficient of restitution of about 0.95, so roughly 5% of the approach speed is lost — to sound, mostly, which is why a good hit is audible from across the room, and to heat in the brief compression of the two surfaces. About 10% of the energy goes with it. This formula therefore runs a little high on both counts, and the error is a fixed percentage rather than something that grows with angle, so the shape of the cosine is untouched.

The balls are not perfectly smooth either. Friction during contact throws the object ball a degree or two off the line of centres, so it does not depart in quite the direction the geometry says. That is a problem for aiming, not for speed — the speed correction from throw is negligible — but it means the direction this formula assumes for $v_{OB}$ is itself a degree or two off.

Run the relation backwards and it does something genuinely useful: given how fast the object ball needs to arrive somewhere, it tells you how hard the cue ball has to be struck. That is the arithmetic of every position shot in the game, and the reason a good player's stroke speed varies so much more than a beginner's.

One last framing, because it connects this page to the rest of the site. What you are looking at is a two-dimensional equal-mass elastic collision, resolved into components along and across the line of centres. Along the line of centres, equal masses in an elastic collision simply SWAP velocities — the cue ball surrenders $u\cos\varphi$ and the object ball, previously at rest, takes it. Across the line, nothing happens at all. Two lines of first-year mechanics, and they run every shot on the table.