Two points mass `m` and `2m` are kept at a distance `a`. Find the speed of particles and their relative velocity of approach when separation becomes `a//2`.

Options

A. `2sqrt((3a)/(2Gm))`
B. `sqrt((a)/(2Gm))`
C. `2sqrt((2Gm)/(3a))`
D. `sqrt((6Gm)/(a))`

Answer

To solve the problem of finding the speed of the particles and their relative velocity of approach when the separation becomes \( \frac{a}{2} \), we can follow these steps:

### Step 1: Understand the System
Initially, we have two masses \( m \) and \( 2m \) separated by a distance \( a \). They are at rest, and since there are no external forces acting on them, the momentum of the system will be conserved.

### Step 2: Set Up the Conservation of Momentum
Let \( v_1 \) be the velocity of mass \( m \) and \( v_2 \) be the velocity of mass \( 2m \) when the distance between them becomes \( \frac{a}{2} \). According to the conservation of momentum:

\[ mv_1 - 2mv_2 = 0 \]

From this equation, we can express \( v_1 \) in terms of \( v_2 \):

\[ v_1 = 2v_2 \]

### Step 3: Apply Conservation of Energy
The mechanical energy of the system is conserved. Initially, the kinetic energy is zero (since both masses are at rest), and the potential energy due to gravitational attraction is given by:

\[ U_i = -\frac{G \cdot m \cdot 2m}{a} = -\frac{2Gm^2}{a} \]

When the separation becomes \( \frac{a}{2} \), the potential energy is:

\[ U_f = -\frac{G \cdot m \cdot 2m}{\frac{a}{2}} = -\frac{4Gm^2}{a} \]

The total mechanical energy conservation can be expressed as:

\[ K_i + U_i = K_f + U_f \]

Since \( K_i = 0 \):

\[ -\frac{2Gm^2}{a} = \frac{1}{2}mv_1^2 + \frac{1}{2}(2m)v_2^2 - \frac{4Gm^2}{a} \]

### Step 4: Solve for Velocities
Rearranging the equation gives:

\[ \frac{1}{2}mv_1^2 + mv_2^2 = \frac{2Gm^2}{a} \]

Substituting \( v_1 = 2v_2 \):

\[ \frac{1}{2}m(2v_2)^2 + mv_2^2 = \frac{2Gm^2}{a} \]

This simplifies to:

\[ \frac{1}{2}m(4v_2^2) + mv_2^2 = \frac{2Gm^2}{a} \]

\[ 2mv_2^2 + mv_2^2 = \frac{2Gm^2}{a} \]

\[ 3mv_2^2 = \frac{2Gm^2}{a} \]

Dividing both sides by \( m \):

\[ 3v_2^2 = \frac{2Gm}{a} \]

Thus,

\[ v_2^2 = \frac{2Gm}{3a} \]

Taking the square root:

\[ v_2 = \sqrt{\frac{2Gm}{3a}} \]

Now substituting back to find \( v_1 \):

\[ v_1 = 2v_2 = 2\sqrt{\frac{2Gm}{3a}} = \sqrt{\frac{8Gm}{3a}} \]

### Step 5: Find the Relative Velocity of Approach
The relative velocity of approach \( v_{rel} \) is given by:

\[ v_{rel} = v_1 + v_2 = \sqrt{\frac{8Gm}{3a}} + \sqrt{\frac{2Gm}{3a}} \]

Finding a common term:

\[ v_{rel} = \sqrt{\frac{2Gm}{3a}} \left( 2 + 1 \right) = 3\sqrt{\frac{2Gm}{3a}} \]

### Final Result
Thus, the speed of the particles when the separation becomes \( \frac{a}{2} \) is:

- \( v_1 = \sqrt{\frac{8Gm}{3a}} \)
- \( v_2 = \sqrt{\frac{2Gm}{3a}} \)
- Relative velocity of approach \( v_{rel} = 3\sqrt{\frac{2Gm}{3a}} \)