A particle of mass `m` moving eastward with a velocity `V` collides with another particle of same mass moving northwards with the same speed `V`. The two particles coalesce and the new particle moves in NE direction. Calculate magnitude and direction of velocity of new particle.

Options

A. `(v)/(2)` North - East
B. `(v)/(sqrt2)` South - West
C. `(v)/(2)` North - West
D. `(v)/(sqrt2)` North - East

Answer

To solve the problem, we will apply 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, both of mass \( m \):
- Particle 1 is moving eastward with velocity \( V \).
- Particle 2 is moving northward with velocity \( V \).

### Step 2: Set up the momentum equations
The momentum of each particle can be expressed as:
- Momentum of Particle 1 (eastward): \( \vec{p_1} = mV \hat{i} \)
- Momentum of Particle 2 (northward): \( \vec{p_2} = mV \hat{j} \)

### Step 3: Calculate the total initial momentum
The total initial momentum \( \vec{P_{initial}} \) before the collision is the vector sum of the momenta of both particles:
\[ \vec{P_{initial}} = \vec{p_1} + \vec{p_2} = mV \hat{i} + mV \hat{j} \]

### Step 4: Apply conservation of momentum
Since the two particles coalesce after the collision, the total momentum before the collision equals the total momentum after the collision:
\[ \vec{P_{initial}} = \vec{P_{final}} \]
Let \( \vec{V_{final}} \) be the velocity of the new particle after the collision. The total momentum after the collision can be expressed as:
\[ \vec{P_{final}} = (2m) \vec{V_{final}} \]
Thus, we have:
\[ mV \hat{i} + mV \hat{j} = 2m \vec{V_{final}} \]

### Step 5: Simplify the equation
Dividing through by \( m \):
\[ V \hat{i} + V \hat{j} = 2 \vec{V_{final}} \]
Now, divide both sides by 2:
\[ \vec{V_{final}} = \frac{V}{2} \hat{i} + \frac{V}{2} \hat{j} \]

### Step 6: Calculate the magnitude of the final velocity
To find the magnitude of \( \vec{V_{final}} \):
\[ |\vec{V_{final}}| = \sqrt{\left(\frac{V}{2}\right)^2 + \left(\frac{V}{2}\right)^2} \]
\[ |\vec{V_{final}}| = \sqrt{\frac{V^2}{4} + \frac{V^2}{4}} = \sqrt{\frac{2V^2}{4}} = \sqrt{\frac{V^2}{2}} = \frac{V}{\sqrt{2}} \]

### Step 7: Determine the direction of the final velocity
The direction of \( \vec{V_{final}} \) can be found from the components:
- The \( x \)-component is \( \frac{V}{2} \) (eastward).
- The \( y \)-component is \( \frac{V}{2} \) (northward).

This means the final velocity vector points in the northeast direction (45 degrees from both axes).

### Final Answer
The magnitude of the velocity of the new particle is \( \frac{V}{\sqrt{2}} \) and it moves in the northeast direction.

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