A particle is projected up a `45^(@)` rough incline with a velocity v. The coeffciient of friciton is 0.5. The speed with which it returns back to the starting point is v'. Then v' `//`v is `:`

Options

A. `sqrt( 2) //sqrt( 3)`
B. `1//sqrt( 3)`
C. `1//2`
D. `1//sqrt(2)`

Answer

To solve the problem, we will analyze the motion of the particle projected up a rough incline and use the work-energy theorem to find the relationship between the initial velocity \( v \) and the final velocity \( v' \) when it returns to the starting point.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle:**
- When the particle is projected up the incline, it experiences two main forces:
- The gravitational force acting downwards, which can be resolved into two components: one parallel to the incline \( mg \sin(45^\circ) \) and one perpendicular to the incline \( mg \cos(45^\circ) \).
- The frictional force acting down the incline, which is given by \( f = \mu N = \mu mg \cos(45^\circ) \), where \( \mu = 0.5 \).

2. **Calculate the Work Done Against Gravity and Friction:**
- The work done against gravity when the particle moves a distance \( S \) up the incline is:
\[ W_g = -mg \sin(45^\circ) \cdot S = -mg \cdot \frac{1}{\sqrt{2}} \cdot S \]
- The work done against friction is:
\[ W_f = -f \cdot S = -\mu mg \cos(45^\circ) \cdot S = -0.5 \cdot mg \cdot \frac{1}{\sqrt{2}} \cdot S \]

3. **Apply the Work-Energy Theorem:**
- The work-energy theorem states that the change in kinetic energy is equal to the work done by all forces:
\[ \Delta KE = W_g + W_f \]
- The initial kinetic energy is \( \frac{1}{2} mv^2 \) and the final kinetic energy at the highest point (where the particle momentarily stops) is 0. Thus:
\[ 0 - \frac{1}{2} mv^2 = -mg \cdot \frac{1}{\sqrt{2}} \cdot S - 0.5 \cdot mg \cdot \frac{1}{\sqrt{2}} \cdot S \]
- Simplifying gives:
\[ -\frac{1}{2} mv^2 = -mg \cdot \frac{3}{2\sqrt{2}} \cdot S \]
- Canceling \( m \) and rearranging gives:
\[ S = \frac{v^2 \sqrt{2}}{3g} \]

4. **Calculate the Work Done When Returning:**
- When the particle returns, the work done by gravity is zero (as it moves back to the same height), but the work done by friction will act against the motion:
\[ W_f' = -2 \cdot \left(0.5 \cdot mg \cdot \frac{1}{\sqrt{2}} \cdot S\right) = -mg \cdot \frac{1}{\sqrt{2}} \cdot S \]
- The change in kinetic energy when returning is:
\[ \frac{1}{2} mv'^2 - 0 = -mg \cdot \frac{1}{\sqrt{2}} \cdot S \]

5. **Substituting the Value of \( S \):**
- Substitute \( S \) from step 3 into the work done equation:
\[ \frac{1}{2} mv'^2 = mg \cdot \frac{1}{\sqrt{2}} \cdot \frac{v^2 \sqrt{2}}{3g} \]
- Simplifying gives:
\[ \frac{1}{2} mv'^2 = \frac{1}{3} mv^2 \]
- Canceling \( m \) from both sides leads to:
\[ v'^2 = \frac{2}{3} v^2 \]

6. **Finding the Ratio \( \frac{v'}{v} \):**
- Taking the square root gives:
\[ \frac{v'}{v} = \sqrt{\frac{2}{3}} = \frac{v'}{v} = \frac{1}{\sqrt{3}} \]

### Final Answer:
The ratio \( \frac{v'}{v} \) is \( \frac{1}{\sqrt{3}} \).