# Snooker ball dissagreement

1. Jan 20, 2010

### nalA

Hello,

I'm having a discussion with my work colleagues, we can't agree about a physics question:

There is a red and a white snooker ball.
They are the same size.
The white snooker ball has a mass many, many, many times that of the red snooker ball.
Initially the red snooker ball is stationary
Initially the white snooker ball has a velocity x
The white snooker ball collides head on with the red snooker ball
What velocity will the red snooker ball have following the collision?

These are two real snooker balls, it is an inelastic collision.
Some kinetic energy will be lost to become other forms of energy after the collision.
Will the inelasticity of the collision change the outcome of the velocity of the red or white snooker balls? or will this effect be negligible? – bearing in mind that the mass of the red ball is negligible compared to that of the white ball?

Both sides of the discussion are completely convinced of the truth of their point of view! It would bring me great joy to learn I'm right ;-)

2. Jan 20, 2010

### DocZaius

If the collision was perfectly inelastic, both balls' velocities would be equal to MWVWB/(MR+MW)

M = mass
V = velocity
R = red
W = white
B = before collision

By the way if the collision was perfectly elastic, the red ball's velocity would be 2MWVWB/(MW+MR)

Last edited: Jan 20, 2010
3. Jan 20, 2010

### nalA

You've given me the two extremes - which I agree with by the way - but I'm interested in the middle ground - not perfectly elastic, or inelastic.

And simplifying:

If MW >> MR

then 2MWVWB/(MW+MR) $$\approx$$ 2VWB for perfectly elastic

and MWVWB/(MR+MW) $$\approx$$ VWB for perfectly inelastic

But what about for everything in-between say a coef of restitution of 0.2, quite an inelastic collision? This is the centre of the 'discussion' I'm having with my colleagues.

4. Jan 20, 2010

### DocZaius

5. Jan 20, 2010

### nalA

Ah yes, that does help.

So would it be correct to say that the velocity of the red ball after the collision would be ...

MW.VWB + MR.VRB + MW.CoR.(VWB - VRB)
-------------------------------------------
MW + MR

where CoR = Coefficient of Restitution

But since MW >> MR and VRB = 0

This will simplify to approximately:

VWB + CoR.VWB

Is this true?

6. Jan 21, 2010

### GRDixon

Assuming (1) the balls are not spinning, and (2) the red ball is initially at rest, and (3) the collision is perfectly elastic, the white snooker ball's final velocity will be practically x, and the red one's final velocity will be 2x. The problem is discussed in Sect. 10_4, "Collisions in One Dimension," in "Physics, Part 1," Halliday and Resnick. Assuming the balls are rolling, the collision is inelastic, etc., complicates things.

7. Jan 21, 2010

### nalA

Thanks for that :)

Yes - you can assume they're not spinning - imagine a collision in free space, it could be assumed to be a one dimensional problem.

One of the things we're arguing about is the speed of the white ball afterwards (for all types of collision: perfectly elastic, perfectly inelastic collisions and everything in-between) (I know that snooker balls are designed to produce close to perfectly elastic collisions - but our argument went down the inelastic collision route as well!)

A formula I found on the http://en.wikipedia.org/wiki/Coefficient_of_restitution" [Broken] that DocZaius kindly pointed out that describes all types of collision for the speed of the white ball afterwards is:

M = mass
V = velocity
R = red
W = white
B = before collision
CoR = coefficient of restitution

MW.VWB + MR.VRB + MR.CoR.(VR - VW)
--------------------------------------
MW + MR

If the red ball has a mass many, many, many times less than that of the white ball, is it reasonable to simplify the formula for the velocity of the white ball after the collision to:

VWB ?

Which would mean that the velocity of the white ball is not really affected in perfectly elastic, perfectly inelastic and everything in between?

Or is there a hole in my reasoning here? I'm very eager to know!

I'd also like a conformation of my statement: for inelastic collisions the red ball will have a velocity of

VWB + CoR.VWB

after the collision.

Last edited by a moderator: May 4, 2017
8. Jan 21, 2010

### nalA

Could this same problem be viewed from the frame of reference of the white ball?

For example - a white ball of normal mass is fixed securely to the top of a short pole that is fixed securely to the ground.

The red ball is dropped from a short distance above the white ball, such that it hits the white ball head on and bounces back up.

Because the white ball is securely attached to the rest of the planet - it effectively has a mass many times that of the red ball.

Would I be correct in saying that this description is exactly the same problem - but looked at from a different frame of reference?

Thanks,

Alan

Last edited: Jan 21, 2010
9. Jan 21, 2010

### DocZaius

To anyone who saw the weird edits I apologize. It turns out kinetic energy is conserved with the wikipedia equations and coefficient at 1.

Here are the graphs again showing speed of both balls, a white ball to red ball mass ratio of 1000, white ball's initial velocity of 1 and red ball's initial velocity of 0:

Red ball's velocity:

http://www.wolframalpha.com/input/?i=Plot[(1000+++1000+x)/1001,+{x,+0,+1}]

White ball's velocity:

http://www.wolframalpha.com/input/?i=Plot[(1000+-+x)/1001,+{x,+0,+1}]

Keep in mind the y axis' minimum in each graph is not 0 (and the function's maximum for the red ball's graph is not 2).

nalA's next post: I had made a mistake in my original graphs. I corrected them. Sorry about that!

Last edited: Jan 21, 2010
10. Jan 21, 2010

### nalA

edit: no problem. yes - they do appear to prove me right now :)

edit: well actually - I've realised a mistake I made in my original argument with my colleagues - I didn't think that an inelastic collision would affect the speed of the red ball either, but I can see that it obviously does now - but I mostly get to have the superior high ground with them for the rest of it ;-)

thanks! :D

It looks like the red ball gets very close to 2 for a perfectly elastic collision.

What is the opinion on my alternative frame of reference above? Is it describing the same problem? I think it makes the whole thing much more intuitive to look at it that way ... if so, then all you'd need to do is switch back to the original frame of reference afterwards to get the answer to the original question.

Last edited: Jan 21, 2010
11. Jan 21, 2010

### DocZaius

Although I do think that the alternative frame of reference you proposed above works (with the white ball having v=0 and the red ball v=1 this time), I am not sure how it is more intuitive.

To put the white ball on a planet, could introduce some misconceptions about the planet being a fixed point in the frame of reference (although it is not). We know of course that upon collision, the planet's velocity will be affected. I think to leave the situation as two balls colliding properly leaves out those possible misconceptions.

Another introduction with your new way of looking at it is the complications from the effects of gravity. We would have to set our point of reference as the planet's velocity right before collision, (to make it v=0). If we were to look at velocities right before the collision and right after, the original collision situation still applies of course. But leaving out an effect which is unimportant to the situation at hand (effects on momentum, kinetic energies, from collision with variable coefficient of restitution) would be best, IMO.

Out of curiosity, what were your workmates arguing?

Last edited: Jan 21, 2010
12. Jan 22, 2010

### nalA

However, I like it because it means the white ball can be an ordinary snooker ball and doesn't need to be made of a mysterious super dense material. The earth has such an enormous mass compared to that of a red snooker ball that it might as well be a fixed point since the change in the planet earth's velocity will be truly negligible as proved by the formulae above and agrees with intuition.

The gravity thing just means that the ball will either bounce to the same height it was dropped from in a perfectly elastic collision or a fraction of that height equal to the coefficient of restitution in an inelastic collision, due to the speed of the red ball immediately after the collision.

We were arguing over that the speeds of the balls after the collision.

I thought the red ball's speed would be twice that of the white ball after the collision, and the white ball's speed would be unaffected. I was also wrong in thinking that neither ball's speed would be effected by the collision being inelastic - but I realise my mistake now.

One colleague thought the speed of the red ball could be anywhere between zero and infinity - but impossible to calculate.

Another agreed with the speed of the red ball, but thought that the white ball would have a significant change in velocity too.