Two particles at a distance `5 m` apart, are thrown towards each other on an inclined smooth plane with equal speed 'v'. Inclined plane is inclined at an angle of `30^@` with the horizontal. It is known that both particle move along the same straight line. The particle colllide at the point from where the lower is thrown. Find the value of velocity`["take " g = 10 m//s^2]`.

Options

A. `2.5m//sec`
B. `5m//sec`
C. `7.5m//sec`
D. `10m//sec`

Explanation

To solve the problem, we need to analyze the motion of two particles thrown towards each other on an inclined plane. Let's break down the solution step by step.

### Step 1: Understand the setup
We have two particles, A and B, starting 5 meters apart on a smooth inclined plane that is inclined at an angle of \(30^\circ\) with the horizontal. Both particles are thrown towards each other with the same initial speed \(v\).

### Step 2: Identify the forces acting on the particles
The only force acting on the particles is gravity. The component of gravitational acceleration acting along the incline is given by:
\[ g_{\text{incline}} = g \sin(\theta) \]
where \(g = 10 \, \text{m/s}^2\) and \(\theta = 30^\circ\). Thus,
\[ g_{\text{incline}} = 10 \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s}^2 \]

### Step 3: Determine the time taken for each particle to reach the point of collision
Let’s denote the time taken for particle A to reach the collision point as \(t_A\) and for particle B as \(t_B\). Since both particles are thrown with the same speed and collide at the midpoint, we have:
\[ t_A = t_B = t \]

### Step 4: Use the equations of motion
For particle A moving up the incline:
\[ s_A = v t - \frac{1}{2} g_{\text{incline}} t^2 \]
For particle B moving down the incline:
\[ s_B = v t - \frac{1}{2} g_{\text{incline}} t^2 \]
Since the total distance between them is 5 m, we have:
\[ s_A + s_B = 5 \]
Substituting the equations for \(s_A\) and \(s_B\):
\[ (v t - \frac{1}{2} g_{\text{incline}} t^2) + (v t - \frac{1}{2} g_{\text{incline}} t^2) = 5 \]
\[ 2(v t - \frac{1}{2} g_{\text{incline}} t^2) = 5 \]
\[ 2vt - g_{\text{incline}} t^2 = 5 \]

### Step 5: Substitute \(g_{\text{incline}}\)
Substituting \(g_{\text{incline}} = 5 \, \text{m/s}^2\):
\[ 2vt - 5t^2 = 5 \]

### Step 6: Rearrange the equation
Rearranging gives us:
\[ 5t^2 - 2vt + 5 = 0 \]

### Step 7: Solve for \(t\)
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 5\), \(b = -2v\), and \(c = 5\):
\[ t = \frac{2v \pm \sqrt{(-2v)^2 - 4 \cdot 5 \cdot 5}}{2 \cdot 5} \]
\[ t = \frac{2v \pm \sqrt{4v^2 - 100}}{10} \]

### Step 8: Set the discriminant to zero for collision
For the particles to collide, the discriminant must be zero:
\[ 4v^2 - 100 = 0 \implies 4v^2 = 100 \implies v^2 = 25 \implies v = 5 \, \text{m/s} \]

### Conclusion
The value of the velocity \(v\) is \(5 \, \text{m/s}\).