Say I have two balls of same radius, in the 2-D Plane. So like a pool (billiard) game. I have the cue ball, moving with the velocity vector V, the magnitude is not important, so we only need an angle to define the velocity, which is theta.

Depending on theta and r, the balls may or may not collide. So let's say we know that they will collide, and we know the initial positions, radii, as well as theta. So how do we calculate the angle of movement of the second ball ?

I hope the picture helps if I failed to define the problem. enter image description here


Although I stated the problem with theta, I'm also ok with a vector-form solution.

  • $\begingroup$ It also depends on friction. I assume the balls don't just grind to a halt by air drag? $\endgroup$ – John Dvorak Sep 6 '14 at 15:28
  • $\begingroup$ You could have removed the text ad, btw. $\endgroup$ – John Dvorak Sep 6 '14 at 15:29
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    $\begingroup$ I think he is asking about the "perfect" scencario. No spin, no friction $\endgroup$ – Asimov Sep 6 '14 at 15:30
  • $\begingroup$ @Asimov exactly. Jan, well sorry for the text :) I don't think it's an ad though. $\endgroup$ – jeff Sep 6 '14 at 15:30
  • $\begingroup$ @Asimov well, if you don't have friction, your balls won't stop. If you have friction with the table but not between two balls, it's unrealistic. $\endgroup$ – John Dvorak Sep 6 '14 at 15:31

When they collide, the point of impact between the two is along a plane (you can imagine a straight line between the two at impact) The resulting direction of the target ball is perpendicular to that plane. AKA The direction of the target ball is along the line of the center of the cue ball, to the center of the target ball.

(This isnt from my math, this is from my experience in playing pool)

  • $\begingroup$ "The resulting direction of the target ball is perpendicular to that plane" - absolutely not. Did you mean the change in velocity? $\endgroup$ – John Dvorak Sep 6 '14 at 15:30
  • $\begingroup$ Thanks, +1. @JanDvorak, I think what Asimov means is that the velocity is in the same direction as the line connecting to the two ball centers at the moment of collision, and I think this is correct, however we still need an explicit formulation for the cue ball's center location :) $\endgroup$ – jeff Sep 6 '14 at 15:32
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    $\begingroup$ @CengizFrostclaw the velocity absolutely isn't. If nothing else, this would break the conversation of momentum - unless you choose a reference frame that has the balls moving along the same line initially. The change in velocity is. $\endgroup$ – John Dvorak Sep 6 '14 at 15:33
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    $\begingroup$ See, this is why i despise physics. Its all about reference frames and changes in velocity/direction. I just wanna play pool! Also, the balls are presumed to be at rest before the shot begins, thats sort of how pool is played. $\endgroup$ – Asimov Sep 6 '14 at 15:36


It depends on exactly where on the Target the Cue hits. Assuming the simplest case - no friction, etc - the Cue will deliver an impact along the line of centres at the moment they hit, so that the Target will move off along that line.

  • $\begingroup$ Exactly what I said, except with better pictures. $\endgroup$ – Asimov Sep 6 '14 at 15:44
  • $\begingroup$ @almagest \\ Your diagram shows the path of the cueball continuing along the same path after the collision with the objectball. Actually, the path of the cueball at, and following, the collision is the path which is exactly perpendicular to the line connecting the centers of the two colliding balls. Also, the velocities as well are altered by the collision. I can give the mathematics of both factors, buy you will have to contact me via my contact information; I won't go to the trouble unless someone out in cyberspace is still interested in the problem. $\endgroup$ – Senex Ægypti Parvi Oct 31 '14 at 21:15
  • $\begingroup$ You must account for the motion of the striking "ball" as well. Its path changes at the instant of contact. The striking "ball"'s new path, assuming that the target "ball" is stationary at the instant of contact, will be perpendicular to the target "ball"'s new path as given in the above answer. But what if both "ball"s were in motion at the instant of contact? Take my word for it: This particular problem is not a walk in the park. Anyone who wants to correspond on this can contact me via my contact info. $\endgroup$ – Senex Ægypti Parvi Jan 7 '15 at 3:24

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