Consider the following Diagram in which a Cue Ball (A) of mass M is shot twice at another pool ball with identical mass M.

Cue Ball (A) of mass M is shot twice at another pool ball with identical mass M.

When the force with which the cue ball (A) is hit (v1) is increased (v2) it seems that the angle which the cue ball is reflected is decreased, while the angle that the other ball is deflected is increased.

Is this thinking accurate? Or is this an illusion, and does the value X remain constant no matter what the force V?

Is there a way of calculating the exact of angle of deflection of the cue ball (x) and the angle of the deflection of the other ball (y), if you know its mass and the angle of incidence at which it hits? The second ball is resting and they have identical mass.

**Note: This is a simplified example where the ball is not spinning (Or in pool terms, "full stun."). But if you would like to provide a general formula that also accounts for spin, please do!

  • 1
    $\begingroup$ the way to calculate this is to use conservation of linear momentum, conservation of energy and conservation of angular momentum (since the cue ball is an extended object)...for point particle only the first two are sufficient, but not for the extended objects like cue balls.. $\endgroup$ – Bruce Lee Mar 3 '16 at 13:50
  • $\begingroup$ Although the angular momentum is only needed if one or both of the balls are initially spinning. But @BruceLee is absolutely correct - this is a classical mechanics scattering problem. For hard spheres it is often couched in terms of the 'impact parameter' which leads directly to the scattering angle and energies of the scattered particles. $\endgroup$ – Jon Custer Mar 3 '16 at 14:58
  • $\begingroup$ Is the first ball spinning before the impact or is it hit at its center of mass? $\endgroup$ – TheQuantumMan Mar 7 '16 at 20:45
  • $\begingroup$ @QuanticMan This is a simplified example where the ball is not spinning. But If you would like to provide a general formula that also accounts for spin, please do! $\endgroup$ – Code Whisperer Mar 7 '16 at 23:51


For a stun shot, the angle y of the object ball (OB) is along the center of the ghost cue ball to the OB. A ghost CB would be where the CB is at the moment of collision. The CB deflection angle x = 90 - y. From your picture, Y = 180 - y and X = 180 - x. enter image description here

The reason for this is on a stun shot, there are no transfer of angular momentum. The force is transmitted along the line from the center of the CB to the OB, hence the reason why OB is deflected at angle y. Since all the force is transmitted along this line, there are no linear momentum of the CB along this line, so the CB will deflect at 90 degree - y.

Speed of the CB has not effect on the angle with the caveat that the speed is small enough such that the deformation of the CB & OB is negligible.

If you do see an angle difference with increased speed, that's due to throw. On the throw, the CB is rolling forward with no sliding (assuming the distance you strike with the cue is far enough for the CB to get into rolling with no sliding). At contact, the angle of deflection remains the same; eg y for OB where y is the angle between the path of motion to the line from the center of CB to OB, and x is 90 degree - y. Because the CB is rolling, after contact, there would now be rolling with sliding, which produces a contact force which causes the CB to curve along the path of a parabola until it reach rolling without sliding. This contact force is wrcosx/sqrt(v^2 + 2vwrsinx + (wr)^2), where w is the angular speed of the CB at contact, r is the radius of the CB, v is the speed of the CB, and x is the angle of deflection.

The final angle that the CB would travel -- I'm still working on that.

| cite | improve this answer | |

Think of it this way, if you take a sledge hammer and nicely tap a plaster wall at a forty-five degree angle, the sledge hammer would theoretically bounce off of the wall at a forty-five degree angle, granted you did not hit the wall hard enough to do any noticeable damage. However, if you take the sledge hammer in the same way and smack the wall at a forty-five degree angle as hard as you can, the sledge hammer is going to break through the wall and will not deflect at a forty-five degree angle. Likewise, if you use an air cannon to blast a cue ball at another ball, one of the balls would likely break, causing the angle of deflection to change. In a perfect world where the structure of an object remained the same no matter what kind of force is applied to it, the ball would always bounce off another ball at the same angle and does for the most part in a regular game of pool. The angle of deflection does not change. ***The mass of the cue ball is not identical to the mass of the other balls in a conventional game of pool.

| cite | improve this answer | |
  • $\begingroup$ The additional mass of the cue ball is needed for the return mechanism of a "bar style" table. Officially, all balls should have the same mass. More sophisticated return mechanisms use a magnet inside the cue ball, and all balls are closer to the same mass. $\endgroup$ – Paul T. Mar 10 '16 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.