Two particles at a distance 5 m apart, are thrown towards each other on inclined smooth plane with equal speeds 'v'. It is known that both particles move along the same straight line. Find the value of v if they collide at the point from where the lower particle is thrown. Inclined plane is inclined at an angle of 30° with the horizontal. [take g = 10 m/s2]

To solve this problem, we need to analyze the motion of both particles on the inclined plane and determine the conditions under which they collide at the initial position of the lower particle.

1. **Understanding the Motion on the Inclined Plane:**
   The inclined plane is at an angle of \( 30^\circ \) with the horizontal. The acceleration due to gravity along the plane, \( g_{\text{plane}} \), is given by:
   \[
   g_{\text{plane}} = g \sin 30^\circ = 10 \times \frac{1}{2} = 5 \, \text{m/s}^2
   \]

2. **Coordinates and Directions:**
   Let the lower particle be at the origin (\( x = 0 \)) and the upper particle be at \( x = 5 \, \text{m} \). Both particles are thrown towards each other along the inclined plane.

3. **Equations of Motion:**
   - For the lower particle, assuming it is thrown up the incline with initial speed \( v \):
     \[
     x_1(t) = v t - \frac{1}{2} g_{\text{plane}} t^2
     \]
   - For the upper particle, assuming it is thrown down the incline with initial speed \( v \):
     \[
     x_2(t) = 5 - (v t + \frac{1}{2} g_{\text{plane}} t^2)
     \]

4. **Collision Condition:**
   For the particles to collide at the position where the lower particle was initially (i.e., \( x = 0 \)):
   \[
   x_1(t) = 0
   \]
   Solving for \( t \) in the equation of motion of the lower particle:
   \[
   0 = v t - \frac{1}{2} g_{\text{plane}} t^2 \implies t (v - \frac{1}{2} g_{\text{plane}} t) = 0
   \]
   This gives two solutions: \( t = 0 \) or \( v = \frac{1}{2} g_{\text{plane}} t \). Since \( t = 0 \) is not meaningful for collision, we take:
   \[
   t = \frac{2v}{g_{\text{plane}}}
   \]

5. **Collision Position for the Upper Particle:**
   Substitute \( t \) in the equation of motion of the upper particle and set it to 0:
   \[
   5 - \left( v \cdot \frac{2v}{g_{\text{plane}}} + \frac{1}{2} g_{\text{plane}} \left( \frac{2v}{g_{\text{plane}}} \right)^2 \right) = 0
   \]
   Simplifying, we get:
   \[
   5 - \left( \frac{2v^2}{5} + \frac{1}{2} \cdot 5 \cdot \frac{4v^2}{25} \right) = 0
   \]
   \[
   5 - \left( \frac{2v^2}{5} + \frac{2v^2}{5} \right) = 0
   \]
   \[
   5 - \frac{4v^2}{5} = 0
   \]
   \[
   5 = \frac{4v^2}{5}
   \]
   \[
   v^2 = \frac{25}{4}
   \]
   \[
   v = \frac{5}{2} \text{ m/s} = 2.5 \text{ m/s}
   \]

Therefore, the speed \( v \) with which the particles must be thrown towards each other to collide at the lower particle's initial position is \( 2.5 \text{ m/s} \).

Do you want more details or have any questions?

### Relative Questions
1. What is the effect of gravity on the motion of particles on an inclined plane?
2. How does the angle of inclination affect the acceleration of particles on the plane?
3. Can you derive the formula for acceleration along an inclined plane?
4. What would happen if the angle of inclination were increased?
5. How would the initial speeds of the particles change if the distance between them were doubled?
6. What are the energy considerations in this problem?
7. How does the presence of friction affect the motion of the particles on the inclined plane?
8. Can you solve the problem using energy conservation principles instead of kinematics?

### Tip
Always break down the components of motion along and perpendicular to inclined planes to simplify the problem-solving process.

Learn how to solve a physics problem involving the collision of particles on an inclined plane. Calculate the initial speed 'v' required for two particles to collide at the initial position of the lower particle. Includes step-by-step solutions, equations of motion, and calculations for acceleration on a 30-degree inclined plane. Suitable for high school students.

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