# Relative Speed in an Elastic Collision Elastic or not, total momentum is always conserved in any collision. So \begin{aligned}{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}&={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\\{{m}_{1}}({{u}_{1}}-{{v}_{1}})&={{m}_{2}}({{v}_{2}}-{{u}_{2}})\quad \quad \cdots \cdots (1)\end{aligned}

What is special about an elastic collision is that the total KE before and after the collision is unchanged. So \displaystyle \begin{aligned}\frac{1}{2}{{m}_{1}}{{u}_{1}}^{2}+\frac{1}{2}{{m}_{2}}{{u}_{2}}^{2}&=\frac{1}{2}{{m}_{1}}{{v}_{1}}^{2}+\frac{1}{2}{{m}_{2}}{{v}_{2}}^{2}\\\frac{1}{2}{{m}_{1}}({{u}_{1}}^{2}-{{v}_{1}}^{2})&=\frac{1}{2}{{m}_{2}}({{v}_{2}}^{2}-{{u}_{2}}^{2})\\\frac{1}{2}{{m}_{1}}({{u}_{1}}-{{v}_{1}})({{u}_{1}}+{{v}_{1}})&=\frac{1}{2}{{m}_{2}}({{v}_{2}}-{{u}_{2}})({{v}_{2}}+{{u}_{2}})\quad \quad \cdots \cdots (2)\end{aligned}

Dividing (2) by (1): \begin{aligned}{{u}_{1}}+{{v}_{1}}&={{v}_{2}}+{{u}_{2}}\\{{u}_{1}}-{{u}_{2}}&={{v}_{2}}-{{v}_{1}}\quad \quad \cdots \cdots (3)\end{aligned}

Notice that u1u2 represents the relative speed at which m1 approaches m2 before the collision, while v2v1 represents the relative speed at which m2 separates from m1 after the collision.

Physically, it means that in any elastic collision, the two bodies approach and separate from each other at the same relative speed.

Mathematically, it means that an elastic collision can be calculated more easily using equations (1) and (3),  rather than equation (2).