# Trajectory of rolling billiard ball after collision with another billiard ball with same mass

the trajectory of white (cue) ball is always perpendicular to the line connecting center of ball A and ball B (please see image)

here's where my confusion comes in, I watched this video (00:52 - 1:36), I see that trajectory of the white ball is not always 90 degrees, instead, it gets closer to 0 degree as 'impact line gets more parallel to x axis' (I don't know how to explain it, please ask if my statement is confusing)

So, how exactly do I get the vector V2a? I don't think it's always 90 degrees like the picture stated, where do I got this wrong? additional information: cue ball is ball A, I'm assuming perfectly elastic collision, if that matters

• additional information: cue ball is ball A, I'm assuming perfectly elastic collision. and do forgive me if I should have posted this in math instead, I'm not really sure :/ – Rei Jul 22 '15 at 17:31
• You should add that to your question by editing it. – Gonenc Jul 22 '15 at 17:33
• @gonenc I've edited my question. do take a look :D – Rei Jul 22 '15 at 17:37
• When the two balls have exactly the same mass and experience a perfectly elastic collision where one ball was initially at rest, the final trajectory of the other ball is perpendicular to the line AB (in the figure). This is because as A strikes B, all of the momentum/velocity of it along line AB is transferred to B, which means the only velocity that can remain in A must be perpendicular to AB, which is tangent to the point of contact and couldn't cause any collision. The video shows the two balls have 90 degree trajectories... – Jim Jul 22 '15 at 18:40
• what people seem to forget, is that the balls are rolling. this makes a huge difference. your example assume that the balls are sliding on ice... – PinkFloyd Jul 22 '15 at 19:17

This actually makes sense if you look at the vector addition. If you add $\vec{v}_{2a}$ and $\vec{v}_{2b}$, you'll see they add up to $\vec{v}_{1a}$ for all elastic collisions with pool balls of equal mass. (Your vectors in your graphic are not to scale, so you can't do it graphically there.)

This makes sense, because when the cue hits the other ball, it is transferring all of the velocity in the $v_{2b}$ direction to the second ball. Not all of the speed is transferred, though, so the speed in the $v_{2a}$ direction remains for the cue ball. In math terms:

$$m_{cue} \pmatrix{v_{2a} \\ v_{2b}} + m_{ball} \pmatrix{0 \\ 0} = m_{cue}\pmatrix{v_{2a} \\ 0} + m_{ball} \pmatrix{0 \\ v_{2b}}$$

You'll see this meets the conditions for elastic collisions; the cue ball has $v_{2a}$ after the collision, and the other ball has $v_{2b}$ after it. As $v_{2a}$ decreases, you'll get a more head-on collision, because the cue ball has the velocity of $\pmatrix{v_{2a} \\ v_{2b}}$ before the collision, but small $v_{2a}$ means the cue ball really has a velocity closer to $v_{2b}$. If the cue ball has any velocity in the $\vec{v}_{2b}$ direction after colliding, it's not a purely elastic collision.

Pure elastic collisions don't often happen in real life. Most pool ball collisions are usually close enough to be considered purely elastic. Also, this does not account for spin of these balls; a slipping ball spinning a way which it is not already traveling can go in odd directions.

• You can use the pmatrix environment for writing matrices in $\LaTeX$ as usual. – Gonenc Jul 22 '15 at 19:58
• \begin{pmatrix} x & y & z \end{pmatrix} => $\begin{pmatrix} x & y & z \end{pmatrix}$ See physics.stackexchange.com/help/notation – ja72 Jul 22 '15 at 21:28
• @ja72 Thanks! That makes a better answer! – PipperChip Jul 22 '15 at 21:50

I also doubted it to begin with.

Then I considered the case of A colliding with B directly on the A-B line.

Then A stops cold and all its momentum is transferred to B.

Now deviate the collision angle to the left by a very small amount, like 1 degree. What happens?

A will almost stop. It will be left with a small velocity to the right, at right angles to the collision line!

Now consider the whole thing from B's point of view. This guy A comes at you with a certain speed toward the point of contact, and a certain speed at right angles to that. He hits you, and you take up all his speed toward the point of contact, and he keeps the speed at right angles to the point of contact.

Now to look at it in terms of vectors, $V_{1A}$ can be split into two components, one parallel to the line of contact, and one at right angles. The component parallel to the line of contact is lost to B, and the other component is not.

When the balls collide, they can transfer linear momentum. But the smooth surface means they transfer almost zero angular momentum.

In the case where the cue ball takes a glancing shot, it transfers little momentum. There might be some spin on the ball, but the large amount of linear momentum means that it behaves almost exactly as you see in the equation.

But when the cue ball shot is closer to full, almost all of the linear momentum is transferred. This makes the final angle the cue ball takes to be hugely dependent on interactions that the spin is making with cloth.

In a "normal" shot, the ball is rolling at the time of the collision. After the collision, it retains the same (forward) spin it had previously. This spin accelerates the ball forward and changes the angle. So all the "full" shots on the video show the cue ball moving forward of your expected 90 degree angle.

If the shots had backspin, they would go behind the expected 90 degree angle. Because the cue ball trajectory depends on the spin, they don't indicate it on the video. Only the object ball trajectory is completely determined by the strike point.

## protected by Qmechanic♦Jul 22 '15 at 22:01

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