A particle of mass m moving towards west with speed v collides with another particle of mass m movies towards south. If two particles st ich t o each other the speed of the new particle of mass 2 m will be

Options

A. `vsqrt2`
B. `(v)/(sqrt2)`
C. v/2
D. v

Answer

To solve the problem, we need to analyze the collision of two particles and apply the principles of conservation of momentum. Let's go through the solution step by step.

### Step 1: Understand the scenario
We have two particles, each of mass \( m \):
- Particle 1 is moving towards the west with speed \( v \).
- Particle 2 is moving towards the south with speed \( v \).

When they collide, they stick together, forming a new particle with mass \( 2m \).

### Step 2: Set up the coordinate system
Let's establish a coordinate system:
- The west direction is negative x-axis.
- The south direction is negative y-axis.

### Step 3: Write down the initial momentum
The momentum of each particle before the collision can be expressed as:
- For Particle 1 (moving west):
\[ p_{1x} = -mv \quad (x \text{ component}) \]
\[ p_{1y} = 0 \quad (y \text{ component}) \]

- For Particle 2 (moving south):
\[ p_{2x} = 0 \quad (x \text{ component}) \]
\[ p_{2y} = -mv \quad (y \text{ component}) \]

### Step 4: Calculate total initial momentum
The total initial momentum in the x-direction and y-direction is:
- Total momentum in x-direction:
\[ p_{initial,x} = p_{1x} + p_{2x} = -mv + 0 = -mv \]

- Total momentum in y-direction:
\[ p_{initial,y} = p_{1y} + p_{2y} = 0 - mv = -mv \]

### Step 5: Write down the final momentum after collision
After the collision, the two particles stick together and move with a speed \( V \) at an angle \( \theta \) with respect to the negative y-axis. The momentum components of the combined mass \( 2m \) are:
- In the x-direction:
\[ p_{final,x} = -2mV \cos(\theta) \]

- In the y-direction:
\[ p_{final,y} = -2mV \sin(\theta) \]

### Step 6: Apply conservation of momentum
According to the conservation of momentum:
- In the x-direction:
\[ -mv = -2mV \cos(\theta) \]
Simplifying gives:
\[ v = 2V \cos(\theta) \quad (1) \]

- In the y-direction:
\[ -mv = -2mV \sin(\theta) \]
Simplifying gives:
\[ v = 2V \sin(\theta) \quad (2) \]

### Step 7: Divide the equations to find \( \theta \)
Dividing equation (1) by equation (2):
\[ \frac{v}{v} = \frac{2V \cos(\theta)}{2V \sin(\theta)} \]
This simplifies to:
\[ 1 = \frac{\cos(\theta)}{\sin(\theta)} \quad \Rightarrow \quad \tan(\theta) = 1 \]
Thus,
\[ \theta = 45^\circ \]

### Step 8: Substitute \( \theta \) back to find \( V \)
Substituting \( \theta = 45^\circ \) into equation (1):
\[ v = 2V \cos(45^\circ) \]
Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ v = 2V \cdot \frac{1}{\sqrt{2}} \quad \Rightarrow \quad v = \frac{2V}{\sqrt{2}} \quad \Rightarrow \quad v = \sqrt{2}V \]
Thus,
\[ V = \frac{v}{\sqrt{2}} \]

### Final Answer
The speed of the new particle of mass \( 2m \) after the collision is:
\[ V = \frac{v}{\sqrt{2}} \]