A particle is projected directly along a rough plane of inclination `theta` with velocity u. If after coming to the rest the particle returns to the starting point with velocity v, the coefficient of friction between the partice and the plane is

Options

A. `(u^(2))/(v^(2))tan theta`
B. `(u^(2)-v^(2))/(u^(2)+v^(2))tan theta`
C. `(v^(2))/(u^(2)) tan theta`
D. `(u^(2)+v^(2))/(u^(2)-v^(2)) tan theta`

Answer

To solve the problem of finding the coefficient of friction between a particle and a rough inclined plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Particle:**
- When the particle is projected up the incline, the forces acting on it are:
- Gravitational force component down the incline: \( mg \sin \theta \)
- Normal force: \( N = mg \cos \theta \)
- Frictional force acting down the incline: \( F_f = \mu N = \mu mg \cos \theta \)

2. **Write the Equation of Motion for Upward Motion:**
- The net force acting on the particle when it moves up the incline is:
\[ F_{\text{net}} = -mg \sin \theta - \mu mg \cos \theta \]
- According to Newton's second law, this can be expressed as:
\[ ma = -mg \sin \theta - \mu mg \cos \theta \]
- Dividing by \( m \):
\[ a = -g \sin \theta - \mu g \cos \theta \]

3. **Use the Kinematic Equation:**
- The kinematic equation for the upward motion is:
\[ 0 = u^2 + 2as \]
- Substituting for \( a \):
\[ 0 = u^2 + 2(-g \sin \theta - \mu g \cos \theta)s \]
- Rearranging gives:
\[ u^2 = 2g s (\sin \theta + \mu \cos \theta) \quad \text{(Equation 1)} \]

4. **Analyze the Downward Motion:**
- When the particle returns down the incline, the forces acting on it are:
- Gravitational force component down the incline: \( mg \sin \theta \)
- Frictional force acting up the incline: \( F_f = \mu mg \cos \theta \)
- The net force while moving down is:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
- Thus, the acceleration is:
\[ a = g \sin \theta - \mu g \cos \theta \]

5. **Use the Kinematic Equation for Downward Motion:**
- The kinematic equation for the downward motion is:
\[ v^2 = 0 + 2as \]
- Substituting for \( a \):
\[ v^2 = 2(g \sin \theta - \mu g \cos \theta)s \]
- Rearranging gives:
\[ v^2 = 2g s (\sin \theta - \mu \cos \theta) \quad \text{(Equation 2)} \]

6. **Divide Equation 1 by Equation 2:**
- Dividing the two equations:
\[ \frac{u^2}{v^2} = \frac{\sin \theta + \mu \cos \theta}{\sin \theta - \mu \cos \theta} \]
- Cross-multiplying gives:
\[ u^2 (\sin \theta - \mu \cos \theta) = v^2 (\sin \theta + \mu \cos \theta) \]

7. **Rearranging to Solve for \( \mu \):**
- Rearranging terms:
\[ u^2 \sin \theta - u^2 \mu \cos \theta = v^2 \sin \theta + v^2 \mu \cos \theta \]
- Collecting \( \mu \) terms:
\[ u^2 \sin \theta - v^2 \sin \theta = u^2 \mu \cos \theta + v^2 \mu \cos \theta \]
- Factoring out \( \mu \):
\[ (u^2 - v^2) \sin \theta = \mu (u^2 + v^2) \cos \theta \]
- Solving for \( \mu \):
\[ \mu = \frac{(u^2 - v^2) \sin \theta}{(u^2 + v^2) \cos \theta} \]

### Final Answer:
The coefficient of friction \( \mu \) between the particle and the plane is given by:
\[ \mu = \frac{(u^2 - v^2) \sin \theta}{(u^2 + v^2) \cos \theta} \]