I'm no knowledgeable pool player, but I've noticed that sometimes when the cue ball hits another pool ball, they roll together; and sometimes the cue ball bounces back. And I have a very, very rough sense that a hard, sharp, and even strike of the cue ball tends to make it bounce back more while a slower or more angled strike will make it roll forward after collision. Can anyone give a more rigorous analysis of the phenomena, or point me to a resource for this? I've tried googling but haven't see anything that really seems to address this as far as I can tell.

[Edit: Upon more contemplation, I suppose a more general question is: In a collision, what determines how much of the combined momentum of the system is distributed to the parts? So in cars colliding, or pool balls, or a skater on ice throwing a baseball--what features of the system determine the amount of momentum imparted to each component?]

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    $\begingroup$ Related: physics.stackexchange.com/q/113923 $\endgroup$ Commented Jul 18, 2014 at 7:34
  • $\begingroup$ @John Rennie, I thought about topspin, and I could be wrong, but that doesn't seem like the effect that's going on here because the ball isn't spinning in a strange way for the effect I'm describing here. To put it another way, whether the cue comes back or not, in either case if it didn't collide with another ball, it would just travel in a straight forward path. $\endgroup$
    – Addem
    Commented Jul 18, 2014 at 7:40
  • $\begingroup$ It is just the spin. To make the cue ball follow the ball it strikes use top spin, and to make the cue ball move backwards us bottom spin. $\endgroup$ Commented Jul 18, 2014 at 8:04
  • $\begingroup$ If you ask this question on sports, you'll probably get the actual answer (which properties of the collision dominate), as opposed to a consideration of which properties are able to dominate under certain circumstances ;-) But the short version of the answer is what John says, top/backspin is the practical means to do this. The mechanism is different from getting a ball to move off the surface alone, though. You need less backspin to rebound the cue ball from an object ball than you need to reverse it on an empty table. $\endgroup$ Commented Jul 18, 2014 at 11:00

3 Answers 3



Those shots in which the cue ball "draws" backwards after hitting the target ball involve backspin. Without backspin, the cue ball cannot reverse direction.

Consider what happens when the cue ball is not spinning at all when it hits the target ball. The cue ball will come to a dead stop if it hits the target ball straight on. Think of Newton's cradle. The cue ball will continue moving forward (but at an angle) if a non-spinning cue ball hits the target ball obliquely.

The cue ball always moves forward after striking the target ball if the cue ball is rolling without slipping whilst hitting the target ball. A rolling cue ball will initially stop if it hits the target ball straight on. The cue ball will still be spinning, however, and this spin will soon make the cue ball start moving forward again. When a rolling cue ball hits the target ball obliquely, the collision will change the cue ball's direction and the spin will accentuate the forward motion.

The only way to combat these effects is to have the cue ball spinning backwards when it strikes the target ball. A backspinning cue ball that hits a target ball straight on will initially stop, but now the backspin will make the cue ball reverse direction.

So how can one make the cue ball have backspin? The answer is simple: Strike the ball below center. How much below depends on the distance to the target ball. This is easy if the target ball is close to the cue ball: Strike the cue ball a bit below center. You'll need to strike the cue ball a bit further below center if the target ball is further away. When the target ball is very far away (across the length of the table), it's very hard to have the cue ball spinning backwards at the point of collision.

You need to take care in your shot and how far from off-center you hit the cue ball. Hit the cue ball too far off-center and you'll hear a nasty "clink" sound. You've just miscued; the cue ball won't move anything like you planned. And maybe you've even ripped the table, bad move!

  • $\begingroup$ If the object ball is far away, putting a lot of power into the shot also helps maintain the spin as it travels across the table. But this often leads to massive miscue's if attempted by amateur players. $\endgroup$
    – Cruncher
    Commented Jul 18, 2014 at 14:14
  • $\begingroup$ @Addem, If you wish to do more research on hitting the cue ball off-center and improve your pool game, the term you're looking for is "english" (AmE only). Occasionally, more specific terms are used: "follow" (forward spin), "draw" or "screw" (backspin), and "side" (side spin). $\endgroup$
    – Brian S
    Commented Jul 18, 2014 at 14:24

First of all, if the collision is elastic, the distribution of momentum in between the components is completely determined by momentum and energy conservation!

This statement is most obvious in the center-of-mass frame where the total momentum is zero and the two objects are moving in opposite directions. The momentum conservation (the total momentum is zero) implies that they're moving in opposite directions after the collision as well and the ratios of the momenta are the same as they were before collisions. Energy conservation than determines that the absolute normalization of these final momenta has to agree with the initial momenta, too.

So in the center-of-mass frame, only the angles change! In particular, a straight collision of two idealized balls of the same mass has a clear outcome: if we describe the situation from the viewpoint of the table, the cue ball stops and transmits the whole energy to the previously static ball that was hit. This ball we hit is moving with the same speed that previously belonged to the cue ball.

This also pretty much settles the collisions of non-rotating objects without friction and deformation. Cars may be close to this idealization or not, depending on the amount of inner damage.

The deviations of the balls' final speeds from the unique, calculable speeds depends on the additional effects of the angular momentum, friction between the two balls, and friction between the balls and the table. And yes, the friction between the balls and the air which is responsible for some fancier effects not discussed here. In particular, if the ball is rolling forward – which is usually the case when you hit it "slowly" or "gradually" – it will tend to preserve its angular momentum and continue rolling forward. The angular speed $\omega$ of the ball will obey $R\omega=v$ where $v$ is the linear speed of the ball and $R$ is its radius. That corresponds to the frictionless rolling motion of the bottom side of the ball on the table.

If you hit it roughly, it's likely that the ball will be moving at some speed but rotate at a lower speed than what is needed for it to roll on the table without friction. I am not saying that the ball is actually rotating backwards. Instead, it is not rotating at all, so it will never try to revert the direction of motion, as you observed.

But if the cue ball isn't sufficiently rotating forward, it will be more likely to bounce back. The reason is simple. As the cue ball A hits the previously static ball B, the ball B starts to move forward and it also starts to roll forward, because of the friction between B and the table. But by momentum conservation, B has to act on A that starts to roll backwards (the point at which the two balls are meeting is moving up which has the "opposite" implications for the two balls). This backward rotation of the hard cue ball A will tend to send A in the backward motion, too.

All these considerations become much more complicated when the collision isn't straight, when it is left-right-asymmetric.

  • $\begingroup$ I feel like there's something still missing here, so let me ask something related. Here are two Physics problems I could imagine, both with the same initial setup: ball A travels toward a ball B that is stationary w.r.t. a frictionless surface supporting them both. The mass of A is 1kg and mass of B is 2kg. Ball A has a speed 5m/s directed straight into the center of ball B. One problem could specify that ball B travels away from the collision with speed 1m/s and you are asked to solve for the resulting speed of ball A... $\endgroup$
    – Addem
    Commented Jul 18, 2014 at 9:16
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    $\begingroup$ Another problem with the same initial conditions could specify that ball B travels with speed 10m/s resulting from the collision and then you are asked to solve for the speed of ball A. Are both scenarios possible, or were the initial conditions sufficient to determine the speed of each object after the collision, and so presumably both of these were non-physical problems? $\endgroup$
    – Addem
    Commented Jul 18, 2014 at 9:18

The direction the ball will take depends on the angular momentum. The velocity with which the ball moves or bounces backwards but the chief determinant is the spinning effect of the incoming ball.


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