A particle of mass `m` moving eastward with a speed `v` collides with another particle of the same mass moving coalesce on collision. The new particle of mass `2 m` will move in the north - easterly direction with a velocity

Options

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

Explanation

To solve the problem, we need to analyze the collision of two particles and apply the principle of conservation of momentum. Here’s a step-by-step solution:

### Step 1: Understand the Situation
We have two particles, each of mass `m`. One particle is moving eastward with a speed `v`, and the other is moving northward with the same speed `v`. After the collision, they coalesce to form a new particle of mass `2m`, which moves in a north-easterly direction.

### Step 2: Define the Directions
- Let the eastward direction be the positive x-direction.
- Let the northward direction be the positive y-direction.
- The new particle moves at an angle θ (north-easterly direction) with respect to the east.

### Step 3: Write the Initial Momenta
- The initial momentum of the eastward-moving particle (mass `m`) is:
\[ p_{initial, x} = mv \quad \text{(in the x-direction)} \]
- The initial momentum of the northward-moving particle (mass `m`) is:
\[ p_{initial, y} = mv \quad \text{(in the y-direction)} \]
- The total initial momentum in the x-direction is:
\[ p_{initial, x} = mv + 0 = mv \]
- The total initial momentum in the y-direction is:
\[ p_{initial, y} = 0 + mv = mv \]

### Step 4: Write the Final Momentum
After the collision, the new particle has a mass of `2m` and moves with a velocity `v0` at an angle θ. The components of the final momentum are:
- In the x-direction:
\[ p_{final, x} = 2m \cdot v_0 \cos \theta \]
- In the y-direction:
\[ p_{final, y} = 2m \cdot v_0 \sin \theta \]

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

#### For the x-direction:
\[ mv = 2m v_0 \cos \theta \]
Dividing both sides by `m`:
\[ v = 2 v_0 \cos \theta \quad \text{(1)} \]

#### For the y-direction:
\[ mv = 2m v_0 \sin \theta \]
Dividing both sides by `m`:
\[ v = 2 v_0 \sin \theta \quad \text{(2)} \]

### Step 6: Equate the Two Equations
From equations (1) and (2):
\[ 2 v_0 \cos \theta = 2 v_0 \sin \theta \]
Dividing both sides by `2 v_0` (assuming `v_0` is not zero):
\[ \cos \theta = \sin \theta \]
This implies:
\[ \tan \theta = 1 \quad \Rightarrow \quad \theta = 45^\circ \]

### Step 7: Substitute θ Back to Find v0
Now substituting θ back into either equation (1) or (2). Using equation (1):
\[ v = 2 v_0 \cos(45^\circ) \]
Since \(\cos(45^\circ) = \frac{1}{\sqrt{2}}\):
\[ v = 2 v_0 \cdot \frac{1}{\sqrt{2}} \quad \Rightarrow \quad v = \frac{2 v_0}{\sqrt{2}} \quad \Rightarrow \quad v_0 = \frac{v \sqrt{2}}{2} \]

### Final Result
Thus, the velocity of the combined mass after the collision is:
\[ v_0 = \frac{v}{\sqrt{2}} \text{ m/s} \]