Derivation of collision rules
My question is about the selected answer https://physics.stackexchange.com/a/176116/335402
The author of the question writes:
I want to be able to calculate the resulting velocities of 4 particles (hexagons, specifically) after they simultaneously (elastically) collide, given their initial velocities. They all have the same size and mass, and in the specific case I'm trying to solve currently, one hexagon is fired at three stationary hexagons (as shown below)
The author of the selected answer provides a solution for the general case, but in the middle of the explanation he uses the following system of equations that he calls "collision rules":
The collision rules are expressed along the contact normals where the final relative speed is a negative fraction of the impact relative speed $$\begin{align} \frac{1}{2}(v_A+\Delta v_A)-(v_B+\Delta v_B) & = -\epsilon (\frac{1}{2}v_A-v_B) \end{align}$$ $$\begin{align} (v_A+\Delta v_A)-(v_C+\Delta v_C) & = -\epsilon (v_A-v_C) \end{align}$$ $$\begin{align} \frac{1}{2}(v_A+\Delta v_A)-(v_D+\Delta v_D) & = -\epsilon (\frac{1}{2}v_A-v_D) \end{align}$$
Since all velocities before the collision except $v_A$ are equal to $0$, the system can be simplified for $\epsilon = 1$: $$\begin{align} \frac{1}{2}(v_A+\Delta v_A)-\Delta v_B & = -\frac{1}{2}v_A \end{align}$$ $$\begin{align} (v_A+\Delta v_A)-\Delta v_C & = -v_A \end{align}$$ $$\begin{align} \frac{1}{2}(v_A+\Delta v_A)-\Delta v_D & = -\frac{1}{2}v_A \end{align}$$
I understand how to derive these rules for the collision of two bodies from the conservation of energy and conservation of momentum laws. 512
But I fail to understand how the derivation could be done for the case described in the question.
The law of conservation of energy in this case for $\epsilon = 1$ is: $$(v_A+\Delta v_A)^2+(\Delta v_B)^2+(\Delta v_C)^2+(\Delta v_D)^2=v_A^2$$ The law of conservation of momentum in vector form is: $$(\mathbf{v_A}+\mathbf{\Delta v_A)}+\mathbf{\Delta v_B}+\mathbf{\Delta v_C}+\mathbf{\Delta v_D}=\mathbf{v_A}$$
The law of conservation of momentum along the normal from A to B: $$\frac{1}{2}(v_A+\Delta v_A)+\Delta v_B+\frac{1}{2}\Delta v_C-\frac{1}{2}\Delta v_D=\frac{1}{2}v_A $$ , along the normal from A to C (x axis): $$(v_A+\Delta v_A)+\frac{1}{2}\Delta v_B+\Delta v_C+\frac{1}{2}\Delta v_D=v_A $$ , along the normal from A to D: $$\frac{1}{2}(v_A+\Delta v_A)-\frac{1}{2}\Delta v_B+\frac{1}{2}\Delta v_C+\Delta v_D=\frac{1}{2}v_A $$
I tried to manipulate the equations but wasn't able to get to the aforementioned three rules. So my questions are: * how does one derive them? * is it possible to derive rules for the general case of N bodies? * will it work for the case when bodies B, C, D, ... are not stationary? 512