A particle of mass 'm' is moving with speed '2v' and collides with a mass '2m' moving with speed 'v' in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass 'm' which move at angle `45^(@)` with respect to the original direction. The speed of each of the moving particle will be

Options

A. `sqrt(2) v`
B. `2sqrt(2) v`
C. `v//(2sqrt(2))`
D. `v//sqrt(2)`

Answer

To solve the problem, we will use the principle of conservation of momentum. Let's break down the solution step by step.

### Step 1: Understand the Initial Conditions
We have two particles:
- Particle A (mass = m) moving with speed = 2v.
- Particle B (mass = 2m) moving with speed = v in the same direction.

### Step 2: Calculate Initial Momentum
The initial momentum of the system can be calculated as follows:
- Momentum of Particle A = mass × velocity = m × 2v = 2mv.
- Momentum of Particle B = mass × velocity = 2m × v = 2mv.

Total initial momentum (P_initial) = Momentum of A + Momentum of B
\[ P_{\text{initial}} = 2mv + 2mv = 4mv. \]

### Step 3: Analyze the Final Conditions
After the collision:
- Particle A comes to a complete stop (final velocity = 0).
- Particle B splits into two particles, each of mass m, moving at an angle of 45 degrees with respect to the original direction.

Let the speed of each of the two new particles be \( v' \).

### Step 4: Calculate Final Momentum
The final momentum can be analyzed in two components (x-direction and y-direction).

#### X-Direction:
- The momentum of each of the two particles in the x-direction is \( m \cdot v' \cdot \cos(45^\circ) = m \cdot v' \cdot \frac{1}{\sqrt{2}} \).
- Since there are two particles, the total momentum in the x-direction after the collision is:
\[ P_{\text{final, x}} = 2 \left( m \cdot v' \cdot \frac{1}{\sqrt{2}} \right) = \frac{2mv'}{\sqrt{2}} = \sqrt{2} mv'. \]

#### Y-Direction:
- The momentum of one particle in the y-direction is \( m \cdot v' \cdot \sin(45^\circ) = m \cdot v' \cdot \frac{1}{\sqrt{2}} \).
- One particle moves in the positive y-direction and the other in the negative y-direction, so their momenta cancel each other out:
\[ P_{\text{final, y}} = m \cdot v' \cdot \frac{1}{\sqrt{2}} - m \cdot v' \cdot \frac{1}{\sqrt{2}} = 0. \]

### Step 5: Apply Conservation of Momentum
Since there is no external force acting on the system, the total momentum before the collision must equal the total momentum after the collision.

In the x-direction:
\[ P_{\text{initial}} = P_{\text{final, x}}. \]
\[ 4mv = \sqrt{2} mv'. \]

### Step 6: Solve for \( v' \)
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ 4v = \sqrt{2} v'. \]
Now, solve for \( v' \):
\[ v' = \frac{4v}{\sqrt{2}} = 2\sqrt{2}v. \]

### Final Answer
The speed of each of the moving particles after the collision is:
\[ v' = 2\sqrt{2}v. \]