Question

Consider a collection of a large number of particles each with speed $v$ . The direction of velocity is randomly distributed in a 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

Hint: Consider the two particles of the same velocity. Calculate the relative velocity between the particles, by substituting the velocity as the same. Integrate itself with respect to the direction of the movement of the particles to find the relative velocity. Compare the relative velocity with respect to that of the velocity of the individual particles.

Formula used:
The trigonometric formula is given by
$\cos \left( {180 - \theta } \right) = {\sin ^2}\dfrac{\theta }{2}$
Where the $\theta $ is the angle between the particles.

Complete step by step solution:
It is given that the system is with many particles that moves in any direction with the velocity $v$ . So let us consider that the $P\,$ and $Q$ are the two particles in the system that possess the relative velocity of ${\vec v_p}$ and ${\vec v_q}$ respectively. Since the velocity of the particles in the system are same,
${\vec v_p} = {\vec v_q} = v$ ……………………..(1)
The relative velocity between the two particles is calculated as follows.
${\vec v_{qp}} = {\vec v_q} + {\vec v_p}$
To remove the vector sign on both sides, the modulus of the equations are taken.
$\left| {{{\vec v}_{qp}}} \right| = \left| {{{\vec v}_q} + {{\vec v}_p}} \right|$
$
  {v_{qp}} = \sqrt {{{\left( {{v_q} + {v_p}} \right)}^2}} \\
\Rightarrow {v_{qp}} = v_p^2 + v_q^2 + 2{v_p}{v_q}\cos \left( {180 - A} \right) \\
 $
Substituting the equation (1) in the above step, we get
$\Rightarrow {v_{qp}} = {v^2} + {v^2} + 2{v^2}{\sin ^2}\dfrac{A}{2}$
By further simplification, we get
$\Rightarrow {v_{qp}} = 2u\sin \dfrac{A}{2}$ ……………………..(2)
Since the velocity of the particle in the system is distributed randomly, the direction of it may vary from $0$ to $2\pi $.
Integrating the equation (2),
$
\Rightarrow {v_{qp}} = \int\limits_0^{2\pi } {2v\sin \dfrac{A}{2}} \\
\Rightarrow \dfrac{{ - 4v\left( {\cos \pi - \cos 0} \right)}}{{2\pi }} \\
 $
By further simplification, we get
$\Rightarrow {v_{qp}} = \dfrac{{4v}}{\pi } = 1.27v$
$\Rightarrow 1.27v > v$

Hence the relative velocity between the particles in the system is greater than $v$.

Note: Totally the angles vary from ${0^ \circ }$ to ${360^ \circ }$ , that means $2\pi $ . Since the above question provides the details that the particle moves randomly in any direction, its angle of the movement varies from the ${0^ \circ }$ to $2\pi $ . The term relative velocity is used if both the particles under consideration move and possess the different velocity.