A particle at rest is constrained to move on a smooth horizontal surface. Another identical particle hits the fractional particle with a velocity `v` at an angle `theta=60^(@)` with horizontal. If the particles move together, the velocity of the combination just after 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 principles of momentum conservation and break down the velocity of the incoming particle into its horizontal and vertical components.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Mass of both particles (m) is identical.
- Initial velocity of the first particle (at rest) = 0.
- Initial velocity of the second particle (incoming) = v at an angle θ = 60° with the horizontal.

2. **Break Down the Velocity into Components:**
- The horizontal component of the velocity (v_x) of the second particle is given by:
\[ v_x = v \cos(60°) = v \cdot \frac{1}{2} = \frac{v}{2} \]
- The vertical component of the velocity (v_y) of the second particle is given by:
\[ v_y = v \sin(60°) = v \cdot \frac{\sqrt{3}}{2} \]

3. **Apply Conservation of Momentum:**
- Before the collision, the total momentum in the horizontal direction is:
\[ p_{initial} = m \cdot 0 + m \cdot \frac{v}{2} = m \cdot \frac{v}{2} \]
- After the collision, both particles move together with a common velocity \( v' \). The total momentum after the collision is:
\[ p_{final} = (m + m) \cdot v' = 2m \cdot v' \]

4. **Set Up the Momentum Conservation Equation:**
- According to the conservation of momentum:
\[ p_{initial} = p_{final} \]
\[ m \cdot \frac{v}{2} = 2m \cdot v' \]

5. **Solve for the Final Velocity (v'):**
- Cancel the mass (m) from both sides:
\[ \frac{v}{2} = 2v' \]
- Rearranging gives:
\[ v' = \frac{v}{4} \]

6. **Conclusion:**
- The velocity of the combination just after the impact is:
\[ v' = \frac{v}{4} \]