consider a collection of large number of particles, each moving with a speed-v.direction of v is randomly....?

Question

distributed in the collection. Show that the magnitude of relative velocity between a pair of particles averaged over all the pairs in the collection is greater than v.
ans- 1.273v (> v ) i specially dont understand how to approach this problem, and the meaning of the line -"relative velocity.....averaged over...."

Answer 1

The average of the relative velocity vector is zero, but its average magnitude (which is always a nonnegative real number) is not. The magnitude of the relative velocity of two particles with velocity vectors v1 and v2 is:

|v1-v2| = sqrt(v1^2 + v2^2 - 2v1*v2*cos(theta))
= v * sqrt(2 - 2*cos(theta))

where theta is the angle between the two velocity directions. The value ranges between 0 (when theta is 0) and 2v (when theta is 180 degrees or pi radians).

To calculate the average means to sum over all values of theta on the unit sphere (that is if one particle's velocity is chosen as the z-axis, the other velocity can be any point selected at random from the unit sphere):

<|v1-v2|> = v(1/4pi)*2pi*\integral_0^pi sqrt(2 - 2*cos(theta)) sin(theta) d theta
= (v/sqrt(2))*\integral_0^pi sqrt(1 - cos(theta)) sin(theta) d theta
= (v/sqrt(2))*\integral_-1^1 sqrt(1 - u) du
= (4/3)v

I see your answer (1.273 = 4/pi) comes from treating all values of theta as equally likely (just integrating over theta):

(1/pi) \integral 0^pi v * sqrt(2 - 2*cos(theta)) d theta
= 4/pi

That's incorrect (unless you're talking about a collection of particles in only two (rather than 3) dimensions, in other words averaging over the unit circle in two dimensions rather than the unit sphere in 3 dimensions.

***************

In any case you can also prove that this average must be > v using only algebra as follows:

For each pair of velocity vectors with an acute angle theta between them, there's another pair equally likely with the (supplementary) obtuse angle 180-theta (just obtained by reversing the direction of one of the particles to get an equally likely pair of velocities). So we can combine these two. The average for two pairs of velocities one with an angle theta and the other with angle 180-theta is: (defining u = cos(theta) = -cos(180-theta))

<|v1-v2|> = (1/2)v[sqrt(2-2u))+sqrt(2+2u)]
= (v/sqrt(2)) [sqrt(1-u))+sqrt(1+u)]
= (v/sqrt(2)) sqrt[ [sqrt(1-u))+sqrt(1+u)]^2]
= (v/sqrt(2)) sqrt[2+2sqrt(1-u^2)]
= v*sqrt[1+sqrt(1-u^2)]

which is always > v since u < 1.

Answer 2

It seems like this question is poorly worded. Relative velocity is the magnitude of the velocity vector between two objects. For instance, if you have two cars, each moving at 60 mph in opposite directions, the relative velocity between the two cars is 120 mph. If the two cars are moving in the same direction, the relative velocity is 0. If the two cars are moving at some angle to one another, you have to calculate the velocity vector of each car, then combine them and find the magnitude of the combined velocity vector in order to find the relative velocity. So, if one car were moving away at 40 mph at 60 degrees with respect to an arbitrary line, and the other car were moving towards the line at 90 degrees and 20 mph, the combined velocity vector would be [40 sin 60 - 20, 40 cos 60] = [14.641, 20], the magnitude of which is 24.7863 (the relative velocity between the two).

Now, in the problem you have 'n' objects all moving at the same velocity v moving in random directions. What you need to do is show that the average of all the relative velocities (as calculated above) is 1.273v.

If you need further help, add additional details.

[edit]
note that the answer 0, below, is wrong. The only angle at which the relative velocity is 0 is 0. There is a continuum of angles between 0 and 180 where the relative velocity is greater than 0 and the question isn't calling for the average velocity - it's calling for the average relative velocity.

Answer 3

The answer ought to be zero. If the number of particles is sufficiently large and the velocities are randomly distributed then there are always two equal velocities which are oppositely directed. Hence the sum total of all velocities must be zero.

Answer 4

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