A particle is projected up a `37^(@)` rough incline with velocity `v_(0)`. If `mu = 1//2`, the speed with which it returns back to the starting point is `v`. Then `v//v_(0)` is

Options

A. `(1)/(2)`
B. `(1)/(sqrt(3))`
C. `(1)/(sqrt(5))`
D. `(1)/(sqrt(2))`

Explanation

To solve the problem, we need to analyze the motion of a particle projected up a rough incline and determine the ratio of its return speed to its initial speed. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Forces Acting on the Particle
When the particle is projected up the incline:
- The gravitational force acting on it can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)
- The frictional force acting against the motion is given by \( F_f = \mu N = \mu mg \cos \theta \), where \( \mu \) is the coefficient of friction.

### Step 2: Calculate the Retardation While Going Up
The total retardation \( A_1 \) while the particle is moving up can be expressed as:
\[ A_1 = g \sin \theta + \mu g \cos \theta \]
Substituting \( \mu = \frac{1}{2} \):
\[ A_1 = g \sin \theta + \frac{1}{2} g \cos \theta \]

### Step 3: Use the Equation of Motion for Ascent
Using the equation of motion:
\[ v^2 = u^2 - 2 A_1 s \]
Where:
- \( v = 0 \) (final velocity at the highest point)
- \( u = v_0 \) (initial velocity)
- \( s \) is the distance traveled up the incline.

Thus, we have:
\[ 0 = v_0^2 - 2 A_1 s \implies v_0^2 = 2 A_1 s \]

### Step 4: Calculate the Retardation While Coming Down
When the particle comes down, the forces acting on it are:
- Gravitational force down the incline: \( mg \sin \theta \)
- Frictional force acting up the incline: \( F_f = \mu mg \cos \theta \)

The net acceleration \( A_2 \) while coming down is:
\[ A_2 = g \sin \theta - \mu g \cos \theta \]

### Step 5: Use the Equation of Motion for Descent
Using the equation of motion again:
\[ v^2 = u^2 + 2 A_2 s \]
Where:
- \( u = 0 \) (initial velocity at the top)
- \( v \) is the final velocity when it returns to the starting point.

Thus, we have:
\[ v^2 = 2 A_2 s \]

### Step 6: Relate the Two Equations
From the ascent:
\[ v_0^2 = 2 A_1 s \]
From the descent:
\[ v^2 = 2 A_2 s \]

Now, we can find the ratio \( \frac{v^2}{v_0^2} \):
\[ \frac{v^2}{v_0^2} = \frac{A_2}{A_1} \]

### Step 7: Substitute the Values of \( A_1 \) and \( A_2 \)
Substituting the expressions for \( A_1 \) and \( A_2 \):
\[ A_1 = g \sin \theta + \frac{1}{2} g \cos \theta \]
\[ A_2 = g \sin \theta - \frac{1}{2} g \cos \theta \]

### Step 8: Calculate the Ratio
Now substituting these into the ratio:
\[ \frac{A_2}{A_1} = \frac{g \sin \theta - \frac{1}{2} g \cos \theta}{g \sin \theta + \frac{1}{2} g \cos \theta} \]
This simplifies to:
\[ \frac{A_2}{A_1} = \frac{\sin \theta - \frac{1}{2} \cos \theta}{\sin \theta + \frac{1}{2} \cos \theta} \]

### Step 9: Substitute \( \theta = 37^\circ \) and Calculate
Using \( \sin 37^\circ = \frac{3}{5} \) and \( \cos 37^\circ = \frac{4}{5} \):
\[ \frac{A_2}{A_1} = \frac{\frac{3}{5} - \frac{1}{2} \cdot \frac{4}{5}}{\frac{3}{5} + \frac{1}{2} \cdot \frac{4}{5}} = \frac{\frac{3}{5} - \frac{2}{5}}{\frac{3}{5} + \frac{2}{5}} = \frac{\frac{1}{5}}{1} = \frac{1}{5} \]

### Step 10: Find the Final Ratio
Thus, we have:
\[ \frac{v^2}{v_0^2} = \frac{1}{5} \implies \frac{v}{v_0} = \frac{1}{\sqrt{5}} \]

### Final Answer
The ratio \( \frac{v}{v_0} \) is \( \frac{1}{\sqrt{5}} \).

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