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I have a project I'm supposed to do in matlab. But, from my understanding, it doesn't seem to be solvable with a unique solution.

Here's the prompt:

Consider a ball A of mass 1 kg moving at 1 m/s in the negative x-direction. Another ball B of mass m moving at speed v in direction θ with respect to the positive x-axis collides with the ball A at the origin. Determine the velocity (direction and magnitude) of the two balls after the collision for:

I know Mass A, velocity of A, I also know Mass B, Velocity of B, and the angle of indecent (theta).

My issue, is, there isn't enough information given to solve this with a unique solution. Usually, the angle of one point mass after collision is given. In this case, its not.

Is there something I have missed? I can calculate the initial momentum of each point particle before impact, but I don't know what to do from there, since after the collision, each ball will have its own angle relative to the x axis. Given the initial conditions ($m_A$, $m_B$, $v_{iA}$, $v_{iA}$, $\theta_i$ ), I have a system of three equation (momentum in x direction, momentum in y-direction, conservation of energy) and four unknowns: $\theta_{Af}$, $\theta_{Bf}$, $v_{f1}$, and $v_{f2}$

Two ideas I had: Parametrize one of the angles? Or define a new coordinate system after collision and fix an axis parallel to the direction of one of the balls?

Thoughts?

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    $\begingroup$ Have you used the fact that one of the balls will move off along a line joining the centres of the balls during collision? $\endgroup$ – Farcher May 14 '17 at 6:09
  • $\begingroup$ @Farcher Could you explain why? $\endgroup$ – Oussama Boussif May 14 '17 at 9:31
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    $\begingroup$ The forces due to each ball acting in the other must act at the point of contact and be normal to the surfaces. $\endgroup$ – Farcher May 14 '17 at 9:37
  • $\begingroup$ @Farcher Could you please go into more detail. The 'balls' are really point particles. I don't quite understand why one of the balls will travel in the direction of the normal vector from the point of collision. How would I go about determining which point particle takes this path? $\endgroup$ – Ion B. May 14 '17 at 9:54
  • $\begingroup$ sounds like you need to set up formulas for both x and y directions and use the fact that momentum is conserved in both x and y ...? $\endgroup$ – PhysicsDave Sep 3 '18 at 2:04
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You are right that there is not enough information. The situation is not unlike the original Rutherford scattering experiment, although in that case the scatterer (gold nucleus) was so much more massive than the alpha particle that its recoil energy could be ignored.

Nevertheless, if you consider the motion of the two particles in the center of mass frame, it's easy to solve for their motion; you then add the velocity of the center of mass back in for the final solution.

The tricky thing is that you would normally have to assume something about the exact position of the two particles at the point of closest approach - this has a certain angle to their velocity, and determines the scatter angle. The best you can do is assume "one other thing", and derive things as a function of that. Probably, angle of incident particle after scattering is a good choice - it should be straightforward to generate a family of curves (for different incident and scattered angle) then.

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