Editing Elastic collision (section)

===Derivation of two dimensional solution===
The [[Impulse_(physics)|impulse <math>\mathbf J</math>]] during the collision for each particle is:
{{NumBlk|:|<math display="block">\begin{align}
\mathbf{p'_1}-\mathbf{p_1} &= \mathbf{J_1},
\\
\mathbf{p'_2}-\mathbf{p_2} &= \mathbf{J_2}
\end{align}</math>|{{EquationRef|2}}}}
Conservation of Momentum implies <math>\mathbf{J}\equiv\mathbf{J_1}=-\mathbf{J_2} </math>
<!--:<math>\mathbf{p'_1}+\mathbf{p'_2}=\mathbf{p_1}+\mathbf{p}_2</math>-->.

Since the force during collision is perpendicular to both particles' surfaces at the contact point, the impulse is along the line parallel to <math>\mathbf{x}_1-\mathbf{x}_2 \equiv\Delta \mathbf x </math>, the relative vector between the particles' center at collision time:
: <math>\mathbf J =\lambda\, \mathbf \hat n,</math> for some <math>\lambda</math> to be determined and <math>\mathbf \hat n \equiv \frac{\Delta \mathbf x}{\|\Delta \mathbf x\|}</math>
Then from ({{EquationRef|2}}):
{{NumBlk|:|<math display="block">\begin{align}
\mathbf{v'_1} &= \mathbf {v_1} + \frac{\lambda}{m_1} \mathbf \hat n,
\\
\mathbf{v'_2} &= \mathbf {v_2} - \frac{\lambda}{m_2} \mathbf \hat n
\end{align}</math>|{{EquationRef|3}}}}
From above equations, conservation of kinetic energy now requires:
:<math>\lambda^2\frac{m_1+m_2}{m_1 m_2}+2\lambda\,\langle \mathbf \hat n, \Delta \mathbf v\rangle = 0 ,\quad</math>with <math>\quad\Delta \mathbf v\equiv \mathbf{v}_1-\mathbf{v}_2.</math>
The both solutions of this equation are <math>\lambda = 0 </math> and <math>\lambda = -2 \frac{m_1 m_2}{m_1+m_2}\langle \mathbf \hat n, \Delta \mathbf v\rangle</math>, where <math>\lambda = 0 </math> corresponds to the trivial case of no collision. Substituting the non trivial value of <math>\lambda</math> in ({{EquationRef|3}}) we get the desired result ({{EquationRef|1}}).

Since all equations are in vectorial form, this derivation is valid also for three dimensions with spheres.