A particle P of charge q and m , is placed at a point in gravity free to move. Another particle Q, of same charge and mass , is projected from a distance r from P with an initial speed `v_(0)` towards P , initial the distance between P and Q decreases and then increases.
What are the speeds of particles P and Q, when their separation is minimum ?

Options

A. `(v_(0))/(2), (v_(0))/(2)`
B. `0,v`
C. `(v_(0))/(3),(2v_(0))/(3)`
D. `0,0`

Answer

To solve the problem, we need to analyze the motion of the two charged particles, P and Q, and apply the principle of conservation of momentum. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
- Particle P is stationary and has a charge \( q \) and mass \( m \).
- Particle Q is projected towards P from a distance \( r \) with an initial speed \( v_0 \) and also has charge \( q \) and mass \( m \).

### Step 2: Identify the Minimum Separation
- When the particles are at their minimum separation, they will have the same speed due to the conservation of momentum and the symmetry of the problem.

### Step 3: Apply Conservation of Momentum
- The initial momentum of the system can be expressed as:
\[ \text{Initial Momentum} = m v_0 + m \cdot 0 = m v_0 \]
- At the point of minimum separation, let the speed of both particles be \( v_1 \). The total momentum at this point is:
\[ \text{Final Momentum} = m v_1 + m v_1 = 2m v_1 \]
- By the conservation of momentum, we have:
\[ m v_0 = 2m v_1 \]

### Step 4: Solve for \( v_1 \)
- We can simplify the equation by dividing both sides by \( m \):
\[ v_0 = 2 v_1 \]
- Rearranging gives:
\[ v_1 = \frac{v_0}{2} \]

### Step 5: Conclusion
- Therefore, when the separation between particles P and Q is at a minimum, both particles will have the same speed:
\[ v_P = v_Q = \frac{v_0}{2} \]

### Summary of Results
- The speeds of particles P and Q when their separation is minimum are:
\[ v_P = v_Q = \frac{v_0}{2} \]