In the link below, there is a vector computation method for calculating post collision speeds for equal masses,
https://exploratoria.github.io/exhibits/mechanics/elastic-collisions-in-3d/index.html
The normal between two balls and relative velocity is
$$ \mathbf{n} = \frac{\mathbf{x}_1-\mathbf{x}_2}{|\mathbf{x}_1-\mathbf{x}_2|} $$
$$ \mathbf{v}_{\text{rel}} = \mathbf{v}_1-\mathbf{v}_2 $$
The relative velocity along the normal direction is
$$ \mathbf{v}_{\text{normal}} = (\mathbf{v}_{\text{rel}} \boldsymbol{\cdot} \mathbf{n}) \mathbf{n} $$
The article says "For balls of equal mass, momentum and velocity are interchangeable". Then final velocities $\mathbf{v}_{1f}$ and $\mathbf{v}_{2f}$ are
$$ \mathbf{v}_{1f} = \mathbf{v}_1 - \mathbf{v}_{\text{norm}}, \quad \mathbf{v}_{2f} = \mathbf{v}_2 + \mathbf{v}_{\text{norm}} $$
I am looking for a derivation of these vector formulas for final velocities, starting from convervation of momentum and energy assumption. I've seen a lot of literature on them using seperate components along the normal direction, etc, but no book seems to have the vector approach the post above has.
Thanks,
Momentum conservation is
$$ m_1 \vec{v}_1 + m_2 \vec{v}_2 = m_1 \vec{v}_1' + m_2 \vec{v}_2' $$
are $\vec{v}_1',\vec{v}_2'$ speed vectors after collision.
If momentum is conserved, loss of momentum in one ball is gain for another, and vica versa
$$ m_1 \vec{v}_1 = m_1 \vec{v}_1' - \Delta \vec{p} $$
$$ m_2 \vec{v}_2 = m_2 \vec{v}_2' + \Delta \vec{p} $$
In the idealized env momentum transfer can only happen on the line connecting centers of both balls. Say $P$ is the magnitude of momentum vector difference $\Delta \vec{p}$,
$$ \vec{v}_1' = \vec{v}_1 - (P / m_1) \vec{n} \qquad (1) $$
$$ \vec{v}_2' = \vec{v}_2 + (P / m_2) \vec{n} \qquad (2) $$
Find $P$ we find new speeds after collision.
The pic above shows $v_1$ ve $v_2$ are addition of two components each, then $v_1,v_1',v_2,v_2'$ is,
$$ \vec{v}_1 = a_1 \vec{n} + b_1 \vec{q}, \qquad \vec{v}_2 = a_2 \vec{n} + b_2 \vec{q} \qquad (3) $$
$$ \vec{v}_1' = a_1' \vec{n} + b_1' \vec{q}, \qquad v_2' = a_2' \vec{n} + b_2' \vec{q} \qquad (4) $$
$a_1,a_2,b_1,b_2$ are scalars.
Insert (3a) into (1)
$$ v_1' = a_1 \vec{n} + b_1 \vec{q} - (P/m_1) \vec{n} $$
$$ = (a_1 - p/m_1) \vec{n} + b_1 \vec{q} $$
$$ v_2' = a_2 \vec{n} + b_2 \vec{q} + (P/m_2) \vec{n} $$
$$ = (a_2 + P/m_2) \vec{n} + b_2 \vec{q} $$
We know the form $\vec{v}_1' = a_1' \vec{n} + b_1' \vec{q}$, $\vec{v}_2' = a_2' \vec{n} + b_2' \vec{q}$, look at the corresponding parts,
$$ a_1' = a_1 - (P/m_1), \qquad b_1' = b_1 \qquad (5) $$
$$ a_2' = a_2 + (P/m_2), \qquad b_2' = b_2 \qquad (6) $$
For $P$ scalar, we can use energy conservation formula. In single dim $1/2 m v^2$, multidim $\frac{1}{2} m \cdot \vec{v}\cdot\vec{v}$, or $\frac{1}{2} m <\vec{v},\vec{v}>$, or $\frac{1}{2} m ||v||^2$. Then,
$$ \frac{m_1}{2} ||v_1||^2 + \frac{m_2}{2} ||v_2||^2 = \frac{m_1}{2} ||v_1'||^2 + \frac{m_2}{2} ||v_2'||^2 $$
$||v_1||^2$ and $||v_1'||^2$ are easy, we look at the pic, ex $||v_1||$ is the hypotenus of a triangle with $a_1$ and $b_1$ sides.
$$ \frac{m_1}{2} (a_1^2+b_1^2) + \frac{m_2}{2} (a_2^2+b_2^2) = \frac{m_1}{2} (a_1'^2+b_1'^2) + \frac{m_2}{2} (a_2'^2+b_2'^2) $$
Previous (5),(6) values are inserted into the formula above
$$ \frac{m_1}{2} (a_1^2+b_1^2) + \frac{m_2}{2} (a_2^2+b_2^2) = \frac{m_1}{2} \left( \left(a_1-\frac{P}{m_1} \right)^2 + b_1^2 \right) + \frac{m_2}{2} \left( \left(a_2-\frac{P}{m_1} \right)^2 + b_2^2 \right) $$
$b_1^2$ are $b_2^2$ canceled. Rearrange with $P$ in LHS
$$ P = \frac{2 m_1 m_2 (a_1-a_2)}{m_1+m_2} $$
Plug that into (1) and (2)
$$ \vec{v}_1' = \vec{v}_1 - \frac{2 m_2 (a_1-a_2)}{m_1+m_2} \vec{n} $$
$$ \vec{v}_2' = \vec{v}_2 + \frac{2 m_1 (a_1-a_2)}{m_1+m_2} \vec{n} $$
The formula above is equivalent to
$$ \vec{v}_1' = \vec{v}_1 - \frac{2m_2}{m_1+m_2} \frac{< \vec{v}_1-\vec{v}_2, \vec{x}_1-\vec{x}_2 >}{||\vec{x}_1-\vec{x}_2||^2} (\vec{x}_1-\vec{x}_2) $$
$$ \vec{v}_2' = \vec{v}_2 - \frac{2m_1}{m_1+m_2} \frac{< \vec{v}_2-\vec{v}_1, \vec{x}_2-\vec{x}_1 >}{||\vec{x}_2-\vec{x}_1||^2} (\vec{x}_1-\vec{x}_2) $$
Equal mass gives,
$$ v_1' = v_1 - \left( (v_1-v_2) \cdot \vec{n} \right) \vec{n} $$
$$ v_2' = v_2 - \left( (v_2-v_1) \cdot \vec{n} \right) \vec{n} $$
where $\vec{n} = \frac{x_1-x_2}{|x_1-x_2|}$
Reference
https://www.gamasutra.com/view/feature/131424/pool_hall_lessons_fast_accurate_.php?print=1