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.
The potential energy of the system of particles P and Q, at closest separation is

Options

A. `(1)/(4pi epsilon_(0))(q^(2))/( r)+ (1)/(2) mv_(0)^(2)`
B. ` - (1)/(4pi epsilon_(0))(q^(2))/( r)+ (1)/(2) mv_(0)^(2)`
C. `(1)/(4piepsilon_(0))(q^(2)) /( r)+ ( mv_(0)^(2))/( 4)`
D. ` - (1)/(4piepsilon_(0))(q^(2)) /( r)+ ( mv_(0)^(2))/( 4)`

Answer

To find the potential energy of the system of particles P and Q at their closest separation, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
- We have two particles, P and Q, both with charge \( q \) and mass \( m \).
- Particle Q is projected towards P with an initial speed \( v_0 \) from a distance \( r \).

2. **Identify Closest Approach**:
- At the closest approach, the distance between P and Q is minimized. At this point, both particles will have the same velocity because the rate of change of distance between them is zero.

3. **Conservation of Momentum**:
- The initial momentum of the system is given by the momentum of Q since P is stationary:
\[ p_{\text{initial}} = mv_0 \]
- At the closest approach, let the common velocity of both particles be \( v' \). The total momentum at this point is:
\[ p_{\text{final}} = 2mv' \]
- By conservation of momentum:
\[ mv_0 = 2mv' \]
- Solving for \( v' \):
\[ v' = \frac{v_0}{2} \]

4. **Conservation of Mechanical Energy**:
- The initial mechanical energy consists of the kinetic energy of Q and the potential energy due to the separation \( r \):
\[ E_{\text{initial}} = \frac{1}{2}mv_0^2 + \frac{kq^2}{r} \]
- At the closest approach, the kinetic energy of both particles and the potential energy at the closest distance \( d \) (which we need to find) is:
\[ E_{\text{final}} = 2 \cdot \frac{1}{2}mv'^2 + U_{\text{final}} = \frac{1}{2}m\left(\frac{v_0}{2}\right)^2 + U_{\text{final}} \]
- This simplifies to:
\[ E_{\text{final}} = \frac{1}{2}m\frac{v_0^2}{4} + U_{\text{final}} = \frac{mv_0^2}{8} + U_{\text{final}} \]

5. **Equating Initial and Final Energies**:
- Setting the initial energy equal to the final energy:
\[ \frac{1}{2}mv_0^2 + \frac{kq^2}{r} = \frac{mv_0^2}{8} + U_{\text{final}} \]
- Rearranging gives:
\[ U_{\text{final}} = \frac{1}{2}mv_0^2 + \frac{kq^2}{r} - \frac{mv_0^2}{8} \]
- Simplifying further:
\[ U_{\text{final}} = \frac{4mv_0^2}{8} - \frac{mv_0^2}{8} + \frac{kq^2}{r} = \frac{3mv_0^2}{8} + \frac{kq^2}{r} \]

6. **Final Expression**:
- Therefore, the potential energy of the system at the closest separation is:
\[ U_{\text{final}} = \frac{3mv_0^2}{8} + \frac{kq^2}{d} \]
- Here, \( d \) is the closest distance between the two charges.