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, we will analyze the motion of the particle as it moves up and down the inclined plane and derive the expression for the coefficient of friction.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A particle is projected up a rough inclined plane with an initial velocity \( u \).
- It comes to rest and then returns to the starting point with a velocity \( v \).
- We need to find the coefficient of friction \( \mu \) between the particle and the plane.

2. **Forces Acting on the Particle (Moving Up)**:
- When the particle moves up, the forces acting on it are:
- Gravitational force component down the incline: \( mg \sin \theta \)
- Normal force: \( N = mg \cos \theta \)
- Frictional force (acting downwards): \( f = \mu N = \mu mg \cos \theta \)
- The net force acting on the particle while moving up is:
\[ F_{\text{net}} = mg \sin \theta + \mu mg \cos \theta \]
- According to Newton's second law:
\[ ma = mg \sin \theta + \mu mg \cos \theta \]
- Dividing through by \( m \):
\[ a = g \sin \theta + \mu g \cos \theta \]
- Since the particle is decelerating, we take \( a \) as negative:
\[ a = - (g \sin \theta + \mu g \cos \theta) \]

3. **Using Kinematics to Find Displacement (Moving Up)**:
- Using the kinematic equation \( v^2 = u^2 + 2as \) where \( v = 0 \) (final velocity when it comes to rest):
\[ 0 = u^2 - 2(g \sin \theta + \mu g \cos \theta)s \]
- Rearranging gives:
\[ s = \frac{u^2}{2(g \sin \theta + \mu g \cos \theta)} \]

4. **Forces Acting on the Particle (Moving Down)**:
- When the particle returns down the incline, the forces are:
- Gravitational force component down the incline: \( mg \sin \theta \)
- Normal force: \( N = mg \cos \theta \)
- Frictional force (acting upwards): \( f = \mu mg \cos \theta \)
- The net force acting on the particle while moving down is:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
- According to Newton's second law:
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
- Dividing through by \( m \):
\[ a = g \sin \theta - \mu g \cos \theta \]

5. **Using Kinematics to Find Displacement (Moving Down)**:
- Using the kinematic equation \( v^2 = u^2 + 2as \) where \( u = 0 \) (initial velocity when it starts moving down):
\[ v^2 = 0 + 2(g \sin \theta - \mu g \cos \theta)s \]
- Rearranging gives:
\[ s = \frac{v^2}{2(g \sin \theta - \mu g \cos \theta)} \]

6. **Equating the Two Displacements**:
- Since the displacement \( s \) is the same in both cases, we can set the two expressions for \( s \) equal:
\[ \frac{u^2}{2(g \sin \theta + \mu g \cos \theta)} = \frac{v^2}{2(g \sin \theta - \mu g \cos \theta)} \]
- Cross-multiplying gives:
\[ u^2 (g \sin \theta - \mu g \cos \theta) = v^2 (g \sin \theta + \mu g \cos \theta) \]

7. **Rearranging to Solve for \( \mu \)**:
- Expanding and rearranging:
\[ u^2 g \sin \theta - u^2 \mu g \cos \theta = v^2 g \sin \theta + v^2 \mu g \cos \theta \]
- Collecting \( \mu \) terms:
\[ u^2 g \sin \theta - v^2 g \sin \theta = u^2 \mu g \cos \theta + v^2 \mu g \cos \theta \]
- Factoring out \( \mu \):
\[ (u^2 - v^2) g \sin \theta = \mu g \cos \theta (u^2 + v^2) \]
- Finally, solving for \( \mu \):
\[ \mu = \frac{(u^2 - v^2) \sin \theta}{(u^2 + v^2) \cos \theta} \]

### Final Answer:
The coefficient of friction \( \mu \) is given by:
\[ \mu = \frac{(u^2 - v^2) \tan \theta}{(u^2 + v^2)} \]