# Kinetic Energy of Two Particles System Relative to Centre of Mass Frame

## Kinetic Energy of Two Particles System Relative to Centre of Mass Frame

Consider two particles of masses m₁ and m₂ having velocities v₁ and v₂. Total kinetic energy of two particle system is the sum of kinetic energy of system with respect to the centre of mass and kinetic energy of the centre of mass.

Kinetic energy of m₁ with respect to the centre of mass is:

$${{\left( K.{{E}_{1}} \right)}_{cm}}=\frac{1}{2}{{m}_{1}}{{\left[ \left( \frac{{{m}_{2}}\overrightarrow{{{v}_{1,2}}}}{{{m}_{1}}+{{m}_{2}}} \right) \right]}^{2}}=\frac{1}{2}\frac{{{m}_{1}}m_{2}^{2}}{{{\left( {{m}_{1}}+{{m}_{2}} \right)}^{2}}}|\overrightarrow{{{v}_{1,2}}}{{|}^{2}}$$.

Similarly, kinetic energy of m₂ with respect to the centre of mass is:

$${{\left( K.{{E}_{2}} \right)}_{cm}}=\frac{1}{2}\frac{{{m}_{2}}m_{1}^{2}}{{{\left( {{m}_{1}}+{{m}_{2}} \right)}^{2}}}|\overrightarrow{{{v}_{2,1}}}{{|}^{2}}$$.

Total kinetic energy of system of particles with respect to centre of mass:

$${{\left( K.{{E}_{system}} \right)}_{cm}}={{\left( K.{{E}_{1}} \right)}_{cm}}+{{\left( K.{{E}_{2}} \right)}_{cm}}$$.

$${{\left( K.{{E}_{system}} \right)}_{cm}}=\frac{1}{2}\frac{{{m}_{1}}m_{2}^{2}}{{{\left( {{m}_{1}}+{{m}_{2}} \right)}^{2}}}|\overrightarrow{{{v}_{rel}}}{{|}^{2}}+\frac{1}{2}\frac{{{m}_{2}}m_{1}^{2}}{{{\left( {{m}_{1}}+{{m}_{2}} \right)}^{2}}}|\overrightarrow{{{v}_{rel}}}{{|}^{2}}$$.

$${{\left( K.{{E}_{system}} \right)}_{cm}}=\frac{1}{2}\frac{{{m}_{1}}{{m}_{2}}}{({{m}_{1}}+{{m}_{2}})}|\overrightarrow{{{v}_{rel}}}{{|}^{2}}=\frac{1}{2}\mu |\overrightarrow{{{v}_{rel}}}{{|}^{2}}$$.

∴ $${{\left( K.{{E}_{system}} \right)}_{cm}}=\frac{1}{2}\mu |\overrightarrow{{{v}_{rel}}}{{|}^{2}}$$.

Where,

vrel = Relative velocity of the particle.

Always remember that vrel is independent of reference frame.

The kinetic energy of a system is minimum relative to the frame attached with centre of mass which is equal to K.E’ is called internal kinetic energy which does not depend on any reference frame. Thus K.E’ is an internal property of system.

Hence, total kinetic energy of two particles system relative to ground.

$${{\left( K.E \right)}_{Total}}={{\left( K.E \right)}^{‘}}+\frac{1}{2}Mv_{cm}^{2}$$.

$${{\left( K.E \right)}_{Total}}=\frac{1}{2}\mu v_{rel}^{2}+\frac{1}{2}Mv_{cm}^{2}$$.

Where, $$\mu =\frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}$$ = reduced mass.

$${{v}_{rel}}=|\overrightarrow{{{v}_{1}}}-\overrightarrow{{{v}_{2}}}|$$ and $$M=({{m}_{1}}+{{m}_{2}})$$.