3.3.3 RSOA=RSOS

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 u₁ – u₂ represents the relative speed at which m₁ approaches m₂ before the collision, while v₂ – v₁ 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.

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Video Explanation

RSOA=RSOS

Concept Test

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