Elastic collison: 2 interpretations

[abskebabs]

Imagine we have 2 equal masses, one at rest and the other coming towards it at a velocity v. If we take Kinetic energy to be conserved during the collision, and therefore that the collision is elastic we can analyse it in 2 different ways and yield different conclusions. For example, if we look at what I will call the 1 dimensional case:

 

[math]mv=mv_1+mv_2[/math]

Where [math]v_1[/math] and [math]v_2[/math] represent the velocities of both masses after the collision.

 

As the masses are equal, we can say:

[math]v=v_1+v_2[/math]

 

Also as energy is conserved:

[math]\frac{1}{2}mv^2=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2[/math]

 

Therefore:

[math](v_1+v_2)^2=v_1^2+v_2^2[/math]

 

So: [math]v_1^2+2v_1v_2+v_2^2=v_1^2+v_2^2[/math]

As we can see, for this to be true we need either [math]v_1[/math] or [math]v_2[/math] to equal zero. Therefore, if the collision is elastic, and the ball that has been hit starts moving, it will have all of the other ball's original energy and momentum and the other ball will have stopped in its tracks. Notice motion is confined to 1 dimension in this case.

 

We can look at this problem slightly differently however, if we look at the vector properties of momentum.

 

If:[math]\vec{p}=\vec{p_1}+\vec{p_2}[/math]

 

and we take the dot product:

[math]\vec{p}\cdot\vec{p}=p^2=(\vec{p_1}+\vec{p_2})\cdot(\vec{p_1}+\vec{p_2})[/math]

 

Then we get:

[math]p^2=p_1^2+2\vec{p_1}\cdot\vec{p_2}+p_2^2[/math]

 

If we now divide both sides by 2m to get an expression of the total Kinetic energy:

[math]\frac{p^2}{2m}=\frac{p_1^2}{2m}+\frac{p_2^2}{2m}+\frac{\vec{p_1}\cdot\vec{p_2}}{2m}[/math]

 

From observing the equation, we can see that if the collision is to be elastic the last term shown above has to equal zero. Therefore the dot product of both velocities after the collision has to equal zero, so that both particles will recoil from the collision travelling at velocities orthogonal to each other, as opposed to one stopping and the other moving.

 

As can be seen both results are different to each other. My question is how can they be reconciled? Also is the 1st one just the result if the collision is constrained in a situation where only one dimenson, and which of these results would be valid in a "real" elastic collision?

[ajb]

What does orthogonal mean in 1-d? What is a vector in 1-d?

 

Maybe thinking about that might shed some light on your problem.

[abskebabs]

What does orthogonal mean in 1-d? What is a vector in 1-d?

 

Maybe thinking about that might shed some light on your problem.

I see, so does that mean that the 1st result only applies when motion is constrained to one dimension?

 

I realise that the word orthogonal has no meaning in 1 dimensional motion. Actually, as opposed to what I have just put above does this mean that the vector sum of momenta in the original direction of motion of the ball after the collision equate to the momentum of the moving ball in that direction before the collision?

 

I can see how such a result would could actually make the results of both equations consistent.

 

Actually that would depend on the angles of each of these from the original direction of motion, as well as the individual magnitudes of the resultant momenta. I guess there are a lot of uncertainties involved in examining even elastic collisions. Do these equate to probabillities, if we carry out a statistical assessment of an experiment based on this?

 

I guess it would be intriguing to find if the individual outcomes are equally likely if the above premise is correct. If not, things would become very interesting I think...

[swansont]

[math]\vec{p_1}\cdot\vec{p_2}[/math]

 

Since you are in one dimension, this is [math]\pm p_1 p_2[/math]

 

The only way for it to be zero is for one of the terms to be zero, which gets you back to either the incident object being at rest, or the target being at rest (i.e. no collision)

[abskebabs]

[math]\vec{p_1}\cdot\vec{p_2}[/math]

 

Since you are in one dimension, this is [math]\pm p_1 p_2[/math]

 

The only way for it to be zero is for one of the terms to be zero, which gets you back to either the incident object being at rest, or the target being at rest (i.e. no collision)

Right! I see, the one dimensional case is a perfectly valid special case of the general solution to the problem. Of course you would get a possible solution being that the vector sum of the velocities remains unchanged, and the magnitudes of one of the velocities equalling zero, thus satisfying the the fact the identity you have quoted is true.

 

Just out of curiosity though, I would like to reiterate the question I had stated in my last post. Is there a difference in the likelihood of these events occurring?

 

For example is it more likely for the particles to take a trajectory perpendicular to each other after a collision, than the moving one coming to rest and the one initially at rest moving off at the original velocity of the 1st one?I guess that both could be equally likely, but the 2nd case where velocities are perpendicular would become much more likely as there would be far many more possible outcomes for which it could be true. Is such a conjecture correct?

[the tree]

If we take Kinetic energy to be conserved during the collision, and therefore that the collision is elastic

What? Kinetic energy doesn't need to be preserved in an elastic collision.

[swansont]

What? Kinetic energy doesn't need to be preserved in an elastic collision.

 

 

Um, that's the definition of an elastic collision.

[swansont]

Just out of curiosity though, I would like to reiterate the question I had stated in my last post. Is there a difference in the likelihood of these events occurring?

 

For example is it more likely for the particles to take a trajectory perpendicular to each other after a collision, than the moving one coming to rest and the one initially at rest moving off at the original velocity of the 1st one?I guess that both could be equally likely, but the 2nd case where velocities are perpendicular would become much more likely as there would be far many more possible outcomes for which it could be true. Is such a conjecture correct?

 

I think it would depend on the forces present during the collision. If there are no forces in the y direction, you must have a 1-d case. If there are, then you will have the orthogonal momenta; but now you have a continuum of possible results. The problem is then underconstrained, I think (I'd have to sit down for a few more minutes and confirm this) which means you can't say without other information.

[abskebabs]

I think it would depend on the forces present during the collision. If there are no forces in the y direction, you must have a 1-d case. If there are, then you will have the orthogonal momenta; but now you have a continuum of possible results. The problem is then underconstrained, I think (I'd have to sit down for a few more minutes and confirm this) which means you can't say without other information.

Hmmm... why would this be? There are no external forces involved in any of the equations, so why would the absence of an external orthogonal force cause a 1-d trajectory to be followed by the particles? I can understand why with external forces orthogonal to the original particle's velocity, it would be imposible for there to be a 1-d trajectory, but I cannot your statement above.

[swansont]

Hmmm... why would this be? There are no external forces involved in any of the equations, so why would the absence of an external orthogonal force cause a 1-d trajectory to be followed by the particles? I can understand why with external forces orthogonal to the original particle's velocity, it would be imposible for there to be a 1-d trajectory, but I cannot your statement above.

 

I didn't say it was an external force, I said the forces present during the collision. e.g. in pool, if the cue ball hits another ball dead-on, the motion will be 1-D. But if the collision is off-line, then there will be 2-d motion. In that case, the direction depends on the point of contact and how that aligns with the center-of-mass. But that's a third bit of information that would properly constrain the problem. Without it (or something else), you have an infinite number of solutions.

[abskebabs]

Even so, by the equations above we shouldn't need to specify this. Each of the solutions the 2d and the 1d should be possible without further information required to gain solutions. If what you say is true, is it really just an exposition of the limited applicabillity of the basic equations of classical mechanics like the ones above which do not take into account the extra information you have mentioned. If a particle hits another particle at rest shouldn't both the 2d and 1d trajectories be possible after the collision? Indeed if we carried out an experiment, shouldn't each of these trajectories occur. If not, I guess you're right and the architecture of these equations need updating. Is there a version of classical mechanics that incorporates this kind of extra info, e.g. lagrangian or hamiltonian?

[swansont]

If there are an infinite number of 2-d trajectories possible, the problem is underconstrained, so there is not a unique solution. You have more unknowns than equations. Saying it's 1-d is a constraint, i.e. that's extra information, but that's a subset of the 2-d case.

[abskebabs]

If there are an infinite number of 2-d trajectories possible, the problem is underconstrained, so there is not a unique solution. You have more unknowns than equations. Saying it's 1-d is a constraint, i.e. that's extra information, but that's a subset of the 2-d case.

I'm sorry to be pedantic, but I acknowledge that if the problem is underconstrained there will be a continuum of possible solutions. I'm guessing therefore from what you have said that each of the infinitude of outcomes will be of equal likelihood, given unconstrained conditions? More mathematically this would mean that if I tried to plot the probabillity density function, I would get a flat line.

 

Is this true or not. If not, is the function that describes the probabillity distribution, whether discrete or not, known?

[swansont]

I'm sorry to be pedantic, but I acknowledge that if the problem is underconstrained there will be a continuum of possible solutions. I'm guessing therefore from what you have said that each of the infinitude of outcomes will be of equal likelihood, given unconstrained conditions? More mathematically this would mean that if I tried to plot the probabillity density function, I would get a flat line.

 

Is this true or not. If not, is the function that describes the probabillity distribution, whether discrete or not, known?

 

It depends on the details of the scattering. e.g. if you assume the impact point is random within the cross-section, hard-sphere (or hard-circle for 2-d) scattering will have a sin(theta) term in it, so half of the scatters will be below 30 degrees and the other half above 30 degrees.

[abskebabs]

It depends on the details of the scattering. e.g. if you assume the impact point is random within the cross-section, hard-sphere (or hard-circle for 2-d) scattering will have a sin(theta) term in it, so half of the scatters will be below 30 degrees and the other half above 30 degrees.

Interesting. I'm guessing the angles you are mentioning here are the angles from the line of the momentum vector of the original moving particle right? Also can you perhaps elaborate in any more detail how mathematically we relate a function describing the shape or cross section of the surfaces to the distribution of velocities resulting from elastic collisions that occur between objects?(In CM of course) I have to say its nice learning about the little details or "digging deeper" here:-) .

[swansont]

Interesting. I'm guessing the angles you are mentioning here are the angles from the line of the momentum vector of the original moving particle right? Also can you perhaps elaborate in any more detail how mathematically we relate a function describing the shape or cross section of the surfaces to the distribution of velocities resulting from elastic collisions that occur between objects?(In CM of course) I have to say its nice learning about the little details or "digging deeper" here:-) .

 

The angle would be the direction the target particle would go, measured from the inident particle's direction.

 

Draw a circle and look at the trig. An incident particle will strike at some impact distance from the centerline. That distance is r*sin(theta), so at r/2, you're at 30 degrees. (This is without worrying about the incident particle also being a circle)

[abskebabs]

It's ironic in a way. A problem originally involving particles cannot get a unique solution unless we consider the shape and relative size of these particles, which initially we neglect. It seems you cannot have a full mechanics for idealised particles, even though they help a lot in simplifying a problem. They do not seem to be self consistent concepts, at least in classical mechanics anyway.