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[I edited the question according to Mark's and Grisha's answers.]

Consider two point-like particles of equal mass colliding centrally in 2D. The final directions of the momenta of the two particles are not determined by conservation of energy and momentum: they can go anywhere.

Under which circumstances the direction of the two particles is determined?

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Your question is now unreadable ... –  Cedric H. Nov 12 '10 at 11:52
    
I hope it's better now. –  Hans Stricker Nov 12 '10 at 15:28

3 Answers 3

In a frame of reference where one of the particles is initially at rest, the final direction of its momentum is determined by conservation of energy and momentum.

This statement is incorrect. Both particles can have transverse components of momentum equal in absolute value but into opposite directions. Just consider slightly non-central collision of two billiard balls.

[update]

Energy-momentum conservation law determines only so-called kinematics of the collision. The directions of final particles is the question of dynamics.

To determine the directions of final particles you should know the equations of motion at each moment of time (more generally, the interaction law). For classical mechanics it is Newton's laws of motion. Solving these equations gives you particle's trajectories.

Summarizing, there is no general prescription to determine the angular distribution of colliding particles, it depends on details how they interact.

Moreover, if you don't know the interaction law, the measuring of angular distribution is a good method to obtain some information about it. This problem is usually called as the inverse scattering problem.

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@Grisha: I talked about point-like particles colliding centrally. –  Hans Stricker Nov 12 '10 at 7:37
    
Doesn't matter. If there is no transverse momentum for each of particles, it doesn't imply that individual transverse momentum can't appear after collision (so that total transverse momentum remains zero). By the way, there is logical mistake in your comment ;) –  Grisha Kirilin Nov 12 '10 at 7:50
    
Which logical mistake ;-? –  Hans Stricker Nov 12 '10 at 7:53
    
But they might come from different directions than their "connection line", that is what I wanted to say. –  Hans Stricker Nov 12 '10 at 8:15
    
"point-like particles colliding centrally" is an oxymoron. There is no center of a point. I just would like to notice that energy-momentum conservation law doesn't care about centrality or non-centrality - It is satisfied always. By the way, a nontrivial angular distribution, which (I should repeat) doesn't determined by conservation of energy and momentum, indicates about a nontrivial internal structure of the colliding objects. You can read a corresponding article in Wikipedia about Rutherford experiment, which reveals the nuclear nature of atoms. –  Grisha Kirilin Nov 12 '10 at 8:22

(Edited to reflect updated question).

The solution has a single parameter, which we can take to be the angle the impacting particle goes after the collision. A particular solution is determined only after this parameter is given. Usually we would talk about the collision of, for example, two spheres, in which case this parameter could be related to whether the spheres are colliding head on or glancing.

For example, in the frame where one particle is initially moving and the other is not, one solution is for the moving particle to come to a halt and for the stationary particle to take up all its motion. Another solution is for both particles to fly off from the impact site at 45 degree angles to the original motion, each with $\sqrt{2}/2$ the initial velocity. Both of these scenarios conserve momentum and kinetic energy.

There are many other solutions. In this frame, the conservation of momentum says

$\vec{p} = \vec{p_1} + \vec{p_2}$

with $\vec{p}$ the momentum before the collision and $\vec{p_1}$ and $\vec{p_2}$ the momenta after. Squaring,

$p^2 = p_1^2 + 2\vec{p_1}\cdot\vec{p_2} + p_2^2$.

Kinetic energy is proportional to $p^2$, so conservation of kinetic energy gives

$p^2 = p_1^2 + p_2^2$.

These last two combine to give

$\vec{p_1}\cdot\vec{p_2} = 0$,

so the momenta point at a 90-degree angle. However, any one particle can choose any angle within 90 degrees of the initial direction of motion to go after the collision. Once this direction is chosen, the direction of motion of the other particle, as well as their momenta, are fixed.

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Thank you very much! –  Hans Stricker Nov 12 '10 at 10:27

Look up the phrase "impact parameter". Perhaps in combination with "Rutherford scattering". Your basic geometric model is not exactly right. Once you see it, everything will work out quantitatively, both in classical and quantum mechanics, using different other ideas, of course.

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