How to handle multiple simultaneous elastic collisions?

Question

I\'m computing the result by colliding pairs of 2D convex objects (without rotation), using the basic equations on wikipedia. However, when there are dependencies, like two obj

Answer 1

Depending on the order I compute the collisions (1-3 first or 2-3 first), I will get different results.

That is correct. This is due to how the physics of collisions work. Consider this simple example, using your first figure:

m_1 = m_2 = m_3
u_1 = u_2
u_3 = 0
x_1 = x_2 + d

The only difference between 1 and 2 is that 1 is closer to 3 by d. 1 hits 3 first, stops, and v_3 becomes u_1 (u is initial and v is final velocity). Because u_2 and the new v_3 are the same, both objects 2 and 3 will proceed to the right at constant velocity with a constant distance, d, between them; they will never touch. If 1 and 2 are swapped, that is, if x_2 = x_1 + d, then 2 hits 3 and stops, and 1 trails after 3 by d.

The order of the collisions matters, and treating simultaneous collisions like two successive instantaneous collisions will give conflicting results depending on the order in which the collisions are processed.

Collisions occurring truly simultaneously is often a pathological case (mathematically) and is probably not necessary to resolve correctly for a game, or even for many scientific models.

If it really is important to correctly resolve multiple elastic collisions, the math for that can be worked out, but you would need to add additional assumptions. The elastic collision of two bodies is given by conservation of momentum:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

and conservation of energy:

(1/2) * m_1 u_1^2 + (1/2) * m_2 u_2^2 = (1/2) * m_1 v_1^2 + (1/2) * m_2 v_2^2

Given the initial velocities of the objects, the velocities of the two objects post-collision could be found. If you wanted to modify these equations to account for a third object,

m_1 u_1 + m_2 u_2 + m_3 u_3 = m_1 v_1 + m_2 v_2 + m_3 v_3
(1/2) * m_1 u_1^2 + (1/2) * m_2 u_2^2 + (1/2) * m_3 u_3^2 = (1/2) * m_1 v_1^2 + (1/2) * m_2 v_2^2 + (1/2) * m_3 u_3^2

a third independent equation would have to be introduced. A simple constraint could be that the momentum transfers of objects 1 and 2 should be proportional to their masses:

m_1^2 (v_1 - u_1) = m_2^2 (v_2 - u_2)

This would be nice for the situation depicted in figure 1: intuitively, I would expect 1 and 2 to have the same final velocity, and this constraint would give you that. Be warned, this equation doesn't have a clear physical basis and may give strange results in other scenarios. Experiment and see what looks right.

The equations you mention on Wikipedia (the standard, quite useful, textbook equations) assume that there is an instantaneous transfer of momentum between the two objects. This is not really true for anything in real life. When one billiard ball hits another, the balls deform very slightly, and this deformation takes time; this time is on the scale of milliseconds or less and is typically negligible.