# Solving 2D Elastic Collisions, unique solution?

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?

• Have you used the fact that one of the balls will move off along a line joining the centres of the balls during collision? May 14 '17 at 6:09
• @Farcher Could you explain why? May 14 '17 at 9:31
• The forces due to each ball acting in the other must act at the point of contact and be normal to the surfaces. May 14 '17 at 9:37
• @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? May 14 '17 at 9:54
• 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 ...? Sep 3 '18 at 2:04