A particle of mass `m`, moving with velocity `v` collides a stationary particle of mass `2m`. As a result of collision, the particle of mass `m` deviates by `45^@` and has final speed of `v/2`. For this situation mark out the correct statement (s).

Explanation

To solve the problem, we need to analyze the collision between two particles and determine the correct statements regarding the angle of divergence and the nature of the collision.

### Step 1: Understand the Initial Conditions
- We have a particle of mass \( m \) moving with velocity \( v \).
- A stationary particle of mass \( 2m \) is at rest.
- After the collision, the particle of mass \( m \) deviates by \( 45^\circ \) and has a final speed of \( \frac{v}{2} \).

### Step 2: Analyze the Momentum Conservation
Since no external forces act on the system, we can apply the law of conservation of momentum.

#### For the x-axis:
- Initial momentum:
\[ p_{initial} = mv + 0 = mv \]
- Final momentum:
\[ p_{final} = m \left(\frac{v}{2}\cos(45^\circ)\right) + 2m(v_2 \cos \alpha) \]
where \( v_2 \) is the speed of the mass \( 2m \) after the collision, and \( \alpha \) is the angle it makes with the horizontal.

Setting the initial and final momenta equal:
\[ mv = m \left(\frac{v}{2\sqrt{2}}\right) + 2m(v_2 \cos \alpha) \]

#### For the y-axis:
- Initial momentum:
\[ p_{initial} = 0 \]
- Final momentum:
\[ p_{final} = m \left(\frac{v}{2}\sin(45^\circ)\right) - 2m(v_2 \sin \alpha) \]
Setting the initial and final momenta equal:
\[ 0 = m \left(\frac{v}{2\sqrt{2}}\right) - 2m(v_2 \sin \alpha) \]

### Step 3: Solve the Equations
From the x-axis equation:
\[ v = \frac{v}{2\sqrt{2}} + 2v_2 \cos \alpha \]
Rearranging gives:
\[ v - \frac{v}{2\sqrt{2}} = 2v_2 \cos \alpha \]
\[ v_2 \cos \alpha = \frac{v}{4\sqrt{2} - 2} \]

From the y-axis equation:
\[ \frac{v}{2\sqrt{2}} = 2v_2 \sin \alpha \]
Rearranging gives:
\[ v_2 \sin \alpha = \frac{v}{4\sqrt{2}} \]

### Step 4: Calculate the Angle of Divergence
The angle of divergence \( \theta \) between the two particles after collision is given by:
\[ \theta = 45^\circ + \alpha \]

To find \( \alpha \), we can divide the equations derived from the x and y momentum conservation:
\[ \frac{v_2 \sin \alpha}{v_2 \cos \alpha} = \frac{\frac{v}{4\sqrt{2}}}{\frac{v}{4\sqrt{2} - 2}} \]
This simplifies to:
\[ \tan \alpha = \frac{1}{\sqrt{2} - 1} \]
Since \( \tan \alpha < 1 \), we find that \( \alpha < 45^\circ \). Thus, \( 45^\circ + \alpha < 90^\circ \).

### Step 5: Determine the Nature of Collision
To determine if the collision is elastic or inelastic, we need to check the coefficient of restitution. Since the final velocities do not satisfy the condition for an elastic collision (where the coefficient of restitution \( e = 1 \)), we conclude that the collision is inelastic.

### Conclusion
- The angle of divergence is less than \( 90^\circ \) (or \( \pi/2 \)).
- The collision is inelastic.

### Correct Statements
- **B**: The angle of divergence between particles after collision is less than \( \pi/2 \) (Correct).
- **D**: Collision is inelastic (Correct).