Two bodies of masses `m` and `2m` are moving in same direction with speed `2v` and `v` respectively just after collision body of mass `m` come to rest and body of mass `2m` splits in two equal parts and move at `45^(@)` from initial direction of body of mass `m`. Find out speed of one part after collision

Options

A. `2sqrt(2)v`
B. `sqrt(2)v`
C. `2v`
D. None of these

Answer

To solve the problem step-by-step, we will use the principle of conservation of momentum.

### Step 1: Understand the initial conditions
We have two bodies:
- Body 1 (mass = m) moving with speed = 2v
- Body 2 (mass = 2m) moving with speed = v

### Step 2: Write down the initial momentum
The total initial momentum (before collision) can be calculated as:
\[ \text{Initial Momentum} = (m \cdot 2v) + (2m \cdot v) = 2mv + 2mv = 4mv \]

### Step 3: Analyze the situation after the collision
After the collision:
- Body 1 (mass = m) comes to rest, so its final velocity = 0.
- Body 2 (mass = 2m) splits into two equal parts, each of mass = m, moving at an angle of 45° from the initial direction of Body 1.

Let the speed of each part after the collision be \( V' \).

### Step 4: Write down the final momentum
The final momentum consists of:
- The momentum of Body 1: \( m \cdot 0 = 0 \)
- The momentum of the two parts of Body 2:
- For the first part: \( m \cdot V' \cos(45^\circ) \)
- For the second part: \( m \cdot V' \cos(45^\circ) \)

Since both parts move at 45°, their vertical components will also contribute to the momentum. The horizontal components of both parts will be:
\[ \text{Final Momentum} = 2 \cdot \left(m \cdot V' \cdot \cos(45^\circ)\right) = 2m \cdot V' \cdot \frac{1}{\sqrt{2}} = \frac{2mV'}{\sqrt{2}} \]

### Step 5: Set up the conservation of momentum equation
According to the conservation of momentum:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
\[ 4mv = \frac{2mV'}{\sqrt{2}} \]

### Step 6: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 4v = \frac{2V'}{\sqrt{2}} \]

### Step 7: Solve for \( V' \)
To isolate \( V' \), we multiply both sides by \( \sqrt{2} \):
\[ 4v \sqrt{2} = 2V' \]
Now, divide both sides by 2:
\[ V' = 2v \sqrt{2} \]

### Conclusion
The speed of one part after the collision is:
\[ V' = 2\sqrt{2}v \]