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.

Options

A. zero
B. greater than`v`
C. less than `v`
D. `v`

Answer

To solve the problem, we need to find the average magnitude of the relative velocity between pairs of particles in a collection where each particle has the same speed \( v \) but with randomly distributed directions. Let's go through the steps systematically.

### Step 1: Define the Relative Velocity
Let \( \vec{V}_A \) and \( \vec{V}_B \) be the velocity vectors of two particles A and B, respectively. Since both particles have the same speed \( v \), we can express their velocities as:
\[ |\vec{V}_A| = |\vec{V}_B| = v \]
The relative velocity \( \vec{V}_{AB} \) of particle A with respect to particle B is given by:
\[ \vec{V}_{AB} = \vec{V}_A - \vec{V}_B \]

### Step 2: Calculate the Magnitude of the Relative Velocity
The magnitude of the relative velocity can be expressed as:
\[ |\vec{V}_{AB}| = |\vec{V}_A - \vec{V}_B| \]
Using the formula for the magnitude of the difference of two vectors, we have:
\[ |\vec{V}_{AB}| = \sqrt{|\vec{V}_A|^2 + |\vec{V}_B|^2 - 2 |\vec{V}_A||\vec{V}_B| \cos \theta} \]
where \( \theta \) is the angle between the two velocity vectors \( \vec{V}_A \) and \( \vec{V}_B \).

### Step 3: Substitute the Magnitudes
Substituting \( |\vec{V}_A| = v \) and \( |\vec{V}_B| = v \):
\[ |\vec{V}_{AB}| = \sqrt{v^2 + v^2 - 2v^2 \cos \theta} = \sqrt{2v^2(1 - \cos \theta)} = v \sqrt{2(1 - \cos \theta)} \]
Using the identity \( 1 - \cos \theta = 2 \sin^2(\theta/2) \):
\[ |\vec{V}_{AB}| = v \sqrt{2 \cdot 2 \sin^2(\theta/2)} = 2v \sin(\theta/2) \]

### Step 4: Average the Magnitude of Relative Velocity
To find the average magnitude of the relative velocity over all pairs of particles, we need to integrate over all possible angles \( \theta \) from \( 0 \) to \( 2\pi \):
\[ \langle |\vec{V}_{AB}| \rangle = \frac{1}{2\pi} \int_0^{2\pi} 2v \sin\left(\frac{\theta}{2}\right) d\theta \]
This simplifies to:
\[ \langle |\vec{V}_{AB}| \rangle = \frac{v}{\pi} \int_0^{2\pi} \sin\left(\frac{\theta}{2}\right) d\theta \]

### Step 5: Solve the Integral
To solve the integral, we can use the substitution \( u = \frac{\theta}{2} \), which gives \( d\theta = 2 du \):
\[ \int_0^{2\pi} \sin\left(\frac{\theta}{2}\right) d\theta = 2 \int_0^{\pi} \sin(u) du = 2 \left[-\cos(u)\right]_0^{\pi} = 2[(-\cos(\pi)) - (-\cos(0))] = 2[1 + 1] = 4 \]
Thus, we have:
\[ \langle |\vec{V}_{AB}| \rangle = \frac{v}{\pi} \cdot 4 = \frac{4v}{\pi} \]

### Step 6: Conclusion
Since \( \frac{4}{\pi} \) is greater than 1 (approximately 1.273), we conclude that:
\[ \langle |\vec{V}_{AB}| \rangle > v \]
This shows that the average magnitude of the relative velocity between pairs of particles is indeed greater than \( v \).