Question

Consider a collection of a large number of particles each with speed $v$. The direction of velocity is randomly distributed in the collection. Show that the magnitude of the relative velocity between a pair of particles averaged over all the pairs in the collection is greater than $v$.

Answer 1

Consider two particles P and Q having velocities v and u

inclined to each other at an angle A .

Consider two particles P and Q having velocities ${\overrightarrow{v}}_p$ and ${\overrightarrow{v}}_q$

inclined at an angle A .

$|{\overrightarrow{v}}_p|=|{\overrightarrow{v}}_q|$ = u

$|{\overrightarrow{v}}_{qp}|$ = $\sqrt{u^2+u^2+2u^{2}cos(180-A)}$

= $2usin\frac{A}{2}$

Since, the velocities of the particles are distributed randomly ,

A can take values form 0 to $2\pi$

$v_{qp}\left(average\right)$ = $\frac{{\int }_0^{2\pi }2usin\frac{A}{2}dA}{{\int }_0^{2\pi }1dA}$

= $\frac{-4u(cos\pi -cos0)}{2\pi }$

= $\frac{4u}{\pi }$

= 1.27 u > u

Hence , proved