Billards Physics: Estimating Kinetic Energy Loss

[azaharak]

I've been thinking of developing a program which predicts the motion of two colloding billard balls, I wanted to encorperate some energy loss in the collision since its not an elastic collision.

My question is which root would be the valid or real result of a collision.

Imagine two billard balls, one with velocity V, the other at rest.
For two billards of same mass without kinetic energy loss, all of the energy gets delivered to the 2nd ball at rest, which acquires the same speed of the initially moving ball.

The other root of the quadratic yields zero velocity for the 2nd billard ball (initially at rest), this corresponds to no collision at all.

I was thinking to estimate the collision as being 90% conserving of the initial kinetic energy.

In doing so, we obtain two roots with momentum and energy analysis.

For instance, if ball #1 travels at an initial speed of 1m/s, (with a 90% conservation of kinetic energy) yields final velocities of .947 and .0527 m/s.

which root is nonsensical?

I was thinking that for 100% conservation of energy yields all the energy transferred to the 2nd ball, then maybe the solution should be the .947 for the impacted ball.

since the masses are the same My equations are

1= V2f +V1F (momentum)

(.9)*1^2= (V2f)^2 +(V1f)^2 (momentum)

(the 1 is the initial speed of ball 1)

Thanks

 

[cepheid]

For instance, if ball #1 travels at an initial speed of 1m/s, (with a 90% conservation of kinetic energy) yields final velocities of .947 and .0527 m/s.

which root is nonsensical?

Neither root is nonsensical. I don't know if you noticed, but these two velocity solutions (in red) add up to 1 m/s. So, if you choose v1 = 0.947 m/s, you'll get v2 = 0.0527 m/s by conservation of momentum. But if you choose v2 = 0.0527 m/s, then you'll get v1 = 0.947 m/s. Either way, *one* of the balls will be traveling at one of these velocities, and the *other* ball will be traveling at the other velocity. The rest is just down to how you label the balls.

You might protest, however, that there is a distinction between the ball labelled #1 and the ball labelled #2, since the former is initially moving and the latter is not. So I guess the two outcomes are qualitatively different. In one case, the initially moving ball retains most of its speed and is the faster of the two (after collision). In the other case, the initially moving ball gives up most of its speed and is the slower of the two after collision. So, maybe a better way of thinking about this result is that your condition that "90% of the KE is retained" doesn't provide a strong enough *constraint* for a unique solution. Both outcomes are possible and are consistent with the constraints you provided. Which one occurs depends on how that 90% kinetic energy that remains gets distributed amongst the two balls.

 

[azaharak]

Yes, I realized that both solutions are valid,

Your explanation of not enough constraint makes sense, but I have one other comment/question.

how would you break the symmetry (or model the system) to discern the true outcome.

My thinking is that the outcome where the 2nd ball (initially at rest) acquires the larger final speed seems more realistic. When kinetic energy is completely conserver, the 2nd ball picks up all of the speed (if they collide), 90% conservation can't be that much different than 100% so the 2nd ball must be described by the larger value.

Additionally, the 2nd ball must be faster after the collision since it would be spatially in front of the 1st ball?
Thanks again

 

[soothsayer]

yes, the second ball would be faster, it just would be slower than it would be if KE was perfectly conserved. This is easily tested by, well, playing pool! Shoot the cue ball (ball 1) at a stationary billiard ball (ball 2) you'll see ball #2 clearly pick up more speed than the first one. Your assumption that v₂ must necessarily be greater than v₁ after the collision because of the spatial separation is exactly correct.

 

[rcgldr]

I'm assuming that the collision is in the same direction as the first ball, resulting in both balls going in the same direction (or the 1st ball not moving) as the first ball before the collision.

During the collision and compression of the balls, you have a Newton pair of equal and opposing forces, decelerating the first ball and accelerating the second ball at the same rate of magnitude. So if the first ball is decelerated to zero velocity, then the second ball would be accelerated to the same velocity as that the first ball had if there are no losses.

If there are losses, then the impulse isn't enough to decelerate the first ball to zero velocity, and both balls end up moving forwards, with the second ball moving faster. If the collision is inelastic, then both balls end up remaining in contact and moving at half the speed of the first ball.

There is another issue that if the first ball is rolling at the start of the collision, then friction force from the felt will convert some of the angular energy of the first ball into linear energy during the collision.

 

[DaveC426913]

I wanted to encorperate some energy loss in the collision since its not an elastic collision.

Why do you think this? Do you think that every billiard ball collision results in dented billiard balls? They'd have to be replaced after every game.

 

[soothsayer]

Why do you think this? Do you think that every billiard ball collision results in dented billiard balls? They'd have to be replaced after every game.

No collision is perfectly elastic in nature, billiard ball collisions are pretty close, but just because the balls don't get dented every game doesn't mean the collisions are perfect.

 

[rcgldr]

I wanted to incorporate some energy loss in the collision since its not an elastic collision.

Why do you think this? Do you think that every billiard ball collision results in dented billiard balls?

Every billard ball collision results in the generation of some heat, as opposed to permanent deformation of the billard balls. It's a type of hysteresis, the amount of force that occurs during "recovery" (expansion) is less than the amount of force during "deformation" (compression) due to the heat loss.

 

[azaharak]

Ok great...


Now what about conservation of angular momentum, does this have to be considered.


For instance if two gears of the same mass and dimensions are brought to contact, what would be the final speed of gear #2 assuming it was initially at rest and gear #1 had an initial angular speed of (omega).

Can this be approached through angular momentum conservation. I was having difficulty with this because the two gears rotate in different directions (their angular momenum vectors are anti parallel).

One would think that both gears would tend toward the same final angular velocity, but this would result in a net zero angular momentum because of opposite spin.

Is angular momentum not conserved because I'm picturing some kind of axles holding the gears near each other, these axles introduce external forces?

I understand that this is becoming off topic but i couldn't help think of the angular aspect of the collision.

There is another issue that if the first ball is rolling at the start of the collision, then friction force from the felt will convert some of the angular energy of the first ball into linear energy during the collision.

Let me know if my thinking is correct:

Since the balls can only touch at their centers, that means that this frictional force which comes from having to move the 2nd ball, acts at the center of ball 1, this can not change the angular dynamics directly since its acting at radially inward toward where ball rotates about, however there will be a counter force the friction of from ball #1 which acts at the base and can change the angular dynamics.

there is also a frictional force between the balls (along the vertical direction), somewhat like the touching gears that I spoke of before, I'm not sure how to model this exactly.

Another question of mine is how to estimate the impact time. Impulse is force * time.

Thanks to everyone

AZ

 

[rcgldr]

Now what about conservation of angular momentum.

Unless you consider the pool table and the Earth as part of the system, angular momentum isn't conserved because the friction force from the felt on sliding or spinning balls and rolling friction change the angular momentum of the balls.

Since the balls can only touch at their centers

Coefficient of friction between the balls is lower than friction force between ball and the felt surface of the table. However the normal force during collision is much higher than the weight of the balls, so I'm not sure which is dominant, friction between balls or friction between ball and felt.

Another question of mine is how to estimate the impact time. Impulse is force * time.

You'd need to know the equation for compression versus normal force on a billiard ball. It would be a small number as the billiard balls only compress a tiny amount.

 

[DaveC426913]

Every billard ball collision results in the generation of some heat, as opposed to permanent deformation of the billard balls. It's a type of hysteresis, the amount of force that occurs during "recovery" (expansion) is less than the amount of force during "deformation" (compression) due to the heat loss.

Granted.

Is this enough to factor into his computations? Why? How long would billiard balls bounce around on a frictionless surface?

Seems penny-wise-pound-foolish to factor such a small variable in when there are other variables much larger.

Now, the rails - definitely inelastic collisions there. A realistic simulator would definitely model some loss in energy via rubber rails.

 

[rcgldr]

How long would billiard balls bounce around on a frictionless surface?

I've witness billard balls bounced on very hard surfaces that I assume to be nearly elastic, like metal, marble, ... and it's clear that the coefficient of restitution is good but less than 1.

As you mentioned, the losses between ball collisions are small compared to felt and cushion effects. I was just trying to answer the OP's question about the inelatic issue without getting into the details of how small of an effect that is.

 

[azaharak]

Granted.

Is this enough to factor into his computations? Why? How long would billiard balls bounce around on a frictionless surface?

Seems penny-wise-pound-foolish to factor such a small variable in when there are other variables much larger.

Now, the rails - definitely inelastic collisions there. A realistic simulator would definitely model some loss in energy via rubber rails.

I realize the losses are small, but my question rests on modelling the system with these losses, the other effects that are much larger are simple enough that someone taking an intro physics course could understand them.