Editing Elastic collision (section)

====Derivation of solution====
To derive the above equations for <math>v_{A2},v_{B2},</math> rearrange the kinetic energy and momentum equations:

<math display="block">\begin{align}
m_A(v_{A2}^2-v_{A1}^2) &= m_B(v_{B1}^2-v_{B2}^2) \\
m_A(v_{A2}-v_{A1}) &= m_B(v_{B1}-v_{B2})
\end{align}</math>

Dividing each side of the top equation by each side of the bottom equation, and using <math>\tfrac{a^2-b^2}{(a-b)} = a+b,</math> gives:

<math display=block> v_{A2}+v_{A1}=v_{B1}+v_{B2}  \quad\Rightarrow\quad v_{A2}-v_{B2} = v_{B1}-v_{A1}</math>

That is, the relative velocity of one particle with respect to the other is reversed by the collision. 

Now the above formulas follow from solving a system of linear equations for <math>v_{A2},v_{B2},</math> regarding <math>m_A,m_B,v_{A1},v_{B1}</math> as constants:
<math display="block">\left\{\begin{array}{rcrcc}
v_{A2} & - & v_{B2} &=& v_{B1}-v_{A1} \\
m_Av_{A1}&+&m_Bv_{B1} &=& m_Av_{A2}+m_Bv_{B2}.
\end{array}\right.</math>
Once <math>v_{A2}</math> is determined, <math>v_{B2}</math> can be found by symmetry.