66This question, if I remember correctly, is from NCERT physics for class XI... at least it was there when I was in class XI
Consider a collection of a large no. 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.
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66Consider two particles with velocities \vec{v}_1 and \vec{v}_2. The relative velocity of these particles
\vec{v}_r=\vec{v}_1-\vec{v}_2
Hence
|\vec{v}_r|=|\vec{v}_1-\vec{v}_2|=\sqrt{v_1^2+v_2^2-2\vec{v}_1\cdot\vec{v}_2}
i.e.
|\vec{v}_r|=\sqrt{2v^2-2v^2\cos\theta}=2v\left|\sin\dfrac{\theta}{2}\right|
here θ is the angle between the two velocities. Now if we consider the collection of particles, we see that the angle θ between each pair of particles varies randomly between 0 (for particles moving along the same direction) to \pi (for the pair of particles moving in the opposite directions). So if we consider the ensemble of particles to take the average over them, we simply figure out the average of |sin(θ/2)|, which is
\dfrac{1}{\pi}\int_0^\pi \left|\sin\frac{\theta}{2}\right|\ \mathrm d\theta = \dfrac{2}{\pi}
Hence, the average value of the relative velocity over all the particles is \dfrac{4v}{\pi} which is obviously more than v.
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