A particle at rest is constrained to move on a smooth horizontal surface. Another identical particle hits the stationary particle with a velocity `v` at an angle `theta=60^(@)` with horizontal. If the particles move together, the velocity of the combination just after the impact is equal to

Options

A. `v`
B. `v//2`
C. `sqrt(3)v//4`
D. `v//4`

Explanation

To solve the problem, we will use the principle of conservation of momentum. The steps are as follows:

### Step 1: Identify the given information
- Mass of each particle, \( m \)
- Velocity of the moving particle, \( v \)
- Angle of impact, \( \theta = 60^\circ \)
- The stationary particle is initially at rest.

### Step 2: Resolve the velocity of the moving particle into components
The velocity \( v \) can be resolved into horizontal (x-direction) and vertical (y-direction) components:
- Horizontal component: \( v_x = v \cos(60^\circ) = v \cdot \frac{1}{2} = \frac{v}{2} \)
- Vertical component: \( v_y = v \sin(60^\circ) = v \cdot \frac{\sqrt{3}}{2} \)

### Step 3: Write the initial momentum in the x-direction
Since the stationary particle is at rest, the initial momentum in the x-direction is only due to the moving particle:
- Initial momentum \( P_{initial} = m \cdot v_x = m \cdot \frac{v}{2} \)

### Step 4: Write the final momentum in the x-direction
After the collision, both particles stick together and move with a common velocity \( V \). The total mass after the collision is \( 2m \):
- Final momentum \( P_{final} = (2m) \cdot V \)

### Step 5: Apply the conservation of momentum
According to the conservation of momentum, the initial momentum must equal the final momentum:
\[ P_{initial} = P_{final} \]
Substituting the values we have:
\[ m \cdot \frac{v}{2} = (2m) \cdot V \]

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

### Conclusion
The velocity of the combination just after the impact is:
\[ V = \frac{v}{4} \]