I need to calculate the resultant velocities of 3 circles/masses/particles if they was to collide at the exact same time. I understand that this is theoretically impossible (or incredibly unlikely) to happen, but that's beyond the point.
Here is an example of what I mean
The blue line represents the velocity of the blue ball, the blue ball's X coordinate is exactly between the other two balls. The collisions are elastic.
I have asked my mathematics teachers about this problem and all of them said that it was impossible, since when you use conservation of momentum formula's you have too many unknowns. That being said, is there something that would give an accurate approximation?
Preferably, the radius' and mass's and velocities are variable, but I would be happy if they was kept constant if it means there is a formula (With the exception of the blue circles velocity).
Thank you.
Best Answer
Let's assume for now all the balls have the same mass $m$ and radius $r$, and the red balls are at rest, and the blue ball has velocity $v$. The total momentum of the system is $mv$, and total energy $\frac{1}{2}mv^2$. We assume the collision is elastic.
After collision, the red balls will be moving symmetrically with respect the blue line you drew, and the blue ball will be bounced directly backwards (downwards); all of these should be clear by symmetry and intuition about elastic collisions. At the point of collision, the centers of the balls form an equilateral triangle.
Let the final (upward) velocity of the blue ball be $v_b$, the horizontal component of velocity of the right red ball be $v_x$ and its vertical (upward) velocity be $v_y$. We can then write down the conservation of energy and momentum: $$ 0 = mv_x - mv_x $$ $$ mv = mv_y + mv_y + mv_b $$ $$ \frac{1}{2}mv^2 = \frac{1}{2}mv_b^2 + 2\left(\frac{1}{2}m(v_x^2+v_y^2)\right) $$ Simplifying, $$ v = 2v_y+v_b $$ $$ v^2 = v_b^2 + 2v_x^2+2v_y^2 $$ This is two equations for three unknowns, which cannot give a unique answer. However, we are still not using the fact that at contact, the three balls form an equilateral triangle, which means that $v_y = \sqrt{3}v_x$. Using this, you can solve to get $$ v_x = \frac{\sqrt{3}}{5} v $$ $$ v_y = \frac{3}{5} v $$ $$ v_b = -\frac{1}{5}v $$
In the case when the masses are all different, you just have to keep track of the mass ratios through this calculation. If the radii are different, then you have to analyze the geometric configuration of the balls when they are in contact and determine how the forces are directed to determine the directions each ball will go in.