Two particles A and B are initially separated by a distance of 1.0 m and are at rest. The mass of A is 100 gm and that of B is 30.0 gm and they move under a constant mutual attractive force of 0.02 N .At what distance from the A's original position do the particles collide.

Explanation

To solve the problem, we need to find the distance from particle A's original position to the point where particles A and B collide. Here’s a step-by-step solution:

### Step 1: Identify the Given Information
- Mass of particle A, \( m_A = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Mass of particle B, \( m_B = 30 \, \text{g} = 0.03 \, \text{kg} \)
- Initial separation distance, \( d = 1.0 \, \text{m} \)
- Mutual attractive force, \( F = 0.02 \, \text{N} \)

### Step 2: Calculate the Acceleration of Each Particle
Using Newton's second law, we can find the acceleration of each particle.

**For particle A:**
\[ a_A = \frac{F}{m_A} = \frac{0.02 \, \text{N}}{0.1 \, \text{kg}} = 0.2 \, \text{m/s}^2 \]

**For particle B:**
\[ a_B = \frac{F}{m_B} = \frac{0.02 \, \text{N}}{0.03 \, \text{kg}} \approx 0.6667 \, \text{m/s}^2 \]

### Step 3: Set Up the Equations of Motion
Since both particles start from rest, we can use the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Where \( u = 0 \) (initial velocity).

**For particle A:**
\[ s_A = \frac{1}{2} a_A t^2 = \frac{1}{2} \cdot 0.2 \cdot t^2 = 0.1 t^2 \]

**For particle B:**
\[ s_B = \frac{1}{2} a_B t^2 = \frac{1}{2} \cdot 0.6667 \cdot t^2 \approx 0.3333 t^2 \]

### Step 4: Relate the Distances
Since the total distance between A and B is 1 meter, we have:
\[ s_A + s_B = 1 \]
Substituting the expressions for \( s_A \) and \( s_B \):
\[ 0.1 t^2 + 0.3333 t^2 = 1 \]
\[ 0.4333 t^2 = 1 \]
\[ t^2 = \frac{1}{0.4333} \approx 2.3077 \]

### Step 5: Calculate \( s_A \)
Now we can find the distance \( s_A \):
\[ s_A = 0.1 t^2 = 0.1 \cdot 2.3077 \approx 0.2308 \, \text{m} \]

### Step 6: Find the Distance from A's Original Position
The distance from A's original position to the point of collision is approximately:
\[ \text{Distance from A's position} = s_A \approx 0.2308 \, \text{m} \]

### Final Answer
The particles collide at a distance of approximately **0.2308 m** from A's original position.

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