Consider an elastic collision.

Since the total KE remains the same after an elastic collision, we can write

\displaystyle \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}\text{    }\cdots \cdots(1)

Rearranging, we get

\displaystyle \begin{aligned}\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}

Elastic or not, total momentum is always conserved in any collision. So we can write

 \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}})\text{    }\cdots \cdots (3)\end{aligned}

Dividing (2) by (3), we get

\begin{aligned}{{u}_{1}}+{{v}_{1}}&={{u}_{2}}+{{v}_{2}}\\{{u}_{1}}-{{u}_{2}}&={{v}_{2}}-{{v}_{1}}\text{    }\cdots \cdots (4)\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. So, in showing that equation (1) can be simplified into equation (4), we have stumbled upon a curious fact about elastic collisions: From the perspective of m2, m1 came at it at the same speed as it went away.

I call equation (4) the RSOA=RSOS equation. (Relative speed of approach = relative speed of separation). Obviously, it is much simpler than equation (1). For this reason, do yourself a favour. Use the RSOA=RSOS equation instead of equation (1) to solve elastic collisions.

Video Explanation


Concept Test


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