Two particles of masses `m_(1) and m_(2)` are intially at rest at an infinite distance apart. If they approach each other under their mutual interaction given by `F=-(K)/(r^(2))`. Their speed of approach at the instant when they are at a distance d apart is

Options

A. `sqrt((2K)/(d)[(1)/(m_(1))+(1)/(m_(2))])`
B. `sqrt((2K)/(d)[(1)/(m_(1))-(1)/(m_(2))])`
C. `sqrt((2K)/(d)[(m_(1)m_(2))/(m_(1)+m_(2))])`
D. `sqrt((2K)/(d)[(m_(1)m_(2))/(m_(1)-m_(2))])`

Answer

To find the speed of approach of two particles of masses \( m_1 \) and \( m_2 \) when they are at a distance \( d \) apart, we can use the principles of conservation of momentum and energy.

### Step-by-Step Solution:

1. **Understand the Forces**: The force between the two particles is given by \( F = -\frac{K}{r^2} \). This is an attractive force, meaning that as the particles move closer, they will accelerate towards each other.

2. **Conservation of Momentum**: Since the particles are initially at rest and there are no external forces acting on them, the total momentum of the system remains constant. Initially, the momentum \( P_i = 0 \). Thus, at any point, the total momentum \( P_f \) must also equal zero:
\[ m_1 v_1 + m_2 v_2 = 0 \]
This implies:
\[ v_2 = -\frac{m_1}{m_2} v_1 \]

3. **Conservation of Energy**: The total mechanical energy of the system must also be conserved. The initial potential energy when the particles are infinitely apart is zero, and as they approach each other, the potential energy at a distance \( r \) is given by:
\[ U = -\frac{K}{r} \]
The kinetic energy of the particles when they are at distance \( d \) apart is:
\[ KE = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Therefore, the conservation of energy gives us:
\[ 0 + 0 = -\frac{K}{d} + \left(\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2\right) \]

4. **Substituting for \( v_2 \)**: Substitute \( v_2 \) into the kinetic energy equation:
\[ -\frac{K}{d} + \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 \left(-\frac{m_1}{m_2} v_1\right)^2 = 0 \]
Simplifying this gives:
\[ -\frac{K}{d} + \frac{1}{2} m_1 v_1^2 + \frac{1}{2} \frac{m_1^2}{m_2} v_1^2 = 0 \]
\[ -\frac{K}{d} + \frac{1}{2} v_1^2 \left(m_1 + \frac{m_1^2}{m_2}\right) = 0 \]

5. **Solving for \( v_1^2 \)**: Rearranging gives:
\[ \frac{1}{2} v_1^2 \left(m_1 + \frac{m_1^2}{m_2}\right) = \frac{K}{d} \]
\[ v_1^2 = \frac{2K/d}{m_1 + \frac{m_1^2}{m_2}} \]

6. **Finding the Speed of Approach**: The speed of approach \( v \) is given by:
\[ v = v_1 + |v_2| = v_1 + \frac{m_1}{m_2} v_1 = v_1 \left(1 + \frac{m_1}{m_2}\right) \]

7. **Final Expression**: Substitute \( v_1 \) into the expression for \( v \):
\[ v = \sqrt{\frac{2K}{d \left(m_1 + \frac{m_1^2}{m_2}\right)}} \left(1 + \frac{m_1}{m_2}\right) \]