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 solve the problem of finding the speed of approach of two particles with masses \( m_1 \) and \( m_2 \) that are initially at rest and are attracted towards each other by a force given by \( F = -\frac{K}{r^2} \), we can follow these steps:

### Step 1: Understand the System
Initially, both particles are at rest and at an infinite distance apart. As they move towards each other, they will accelerate due to the mutual force acting on them.

### Step 2: Apply Conservation of Momentum
Since there are no external forces acting on the system, the momentum of the system is conserved. Initially, the momentum is zero (both particles are at rest). At a distance \( d \), let \( v_1 \) be the velocity of mass \( m_1 \) and \( v_2 \) be the velocity of mass \( m_2 \).

The conservation of momentum gives us:
\[ m_1 v_1 + m_2 v_2 = 0 \]
From this, we can express \( v_1 \) in terms of \( v_2 \):
\[ m_1 v_1 = -m_2 v_2 \implies v_1 = -\frac{m_2}{m_1} v_2 \]

### Step 3: Use the Work-Energy Theorem
The work done by the force as the particles move from infinity to a distance \( d \) is equal to the change in kinetic energy of the system. The work done by the force \( F = -\frac{K}{r^2} \) can be calculated as:
\[ W = \int_{\infty}^{d} -\frac{K}{r^2} dr \]
Calculating this integral:
\[ W = K \left[ \frac{1}{r} \right]_{\infty}^{d} = K \left( 0 - \frac{1}{d} \right) = -\frac{K}{d} \]

### Step 4: Relate Work Done to Kinetic Energy
The initial kinetic energy is zero (since both particles are at rest). The final kinetic energy \( KE \) when they are at distance \( d \) is given by:
\[ KE = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting \( v_1 = -\frac{m_2}{m_1} v_2 \):
\[ KE = \frac{1}{2} m_1 \left(-\frac{m_2}{m_1} v_2\right)^2 + \frac{1}{2} m_2 v_2^2 \]
\[ = \frac{1}{2} \frac{m_2^2}{m_1} v_2^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} v_2^2 \left(\frac{m_2^2}{m_1} + m_2\right) \]

### Step 5: Set Work Done Equal to Kinetic Energy
Setting the work done equal to the final kinetic energy:
\[ -\frac{K}{d} = \frac{1}{2} v_2^2 \left(\frac{m_2^2}{m_1} + m_2\right) \]

### Step 6: Solve for \( v_2 \)
Rearranging gives:
\[ v_2^2 = -\frac{2K}{d} \cdot \frac{1}{\left(\frac{m_2^2}{m_1} + m_2\right)} \]
Taking the square root to find \( v_2 \):
\[ v_2 = \sqrt{-\frac{2K}{d} \cdot \frac{1}{\left(\frac{m_2^2}{m_1} + m_2\right)}} \]

### Step 7: Find the Speed of Approach
The speed of approach \( V \) is given by:
\[ V = v_1 + v_2 = \left(-\frac{m_2}{m_1} v_2\right) + v_2 = v_2 \left(1 - \frac{m_2}{m_1}\right) \]

### Final Expression
After substituting back and simplifying, we arrive at the final expression for the speed of approach.