Two spheres of masses `m and 2m` are separated by distance `d`. A particle of mass `(m)/(5)` is projected straight from `2m` towards `m` with a velocity `v_(0)`. Which of the following statements is correct ?

Options

A. Velocity of the particle decreases constantly
B. Velocity of the particle increase constantly
C. Acceleration of the particle may become momentarily zero
D. The particle never retraces its path

Explanation

To solve the problem, we need to analyze the forces acting on the particle of mass \( \frac{m}{5} \) as it moves from the mass \( 2m \) towards the mass \( m \). We will use the concept of gravitational force and acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the Forces**:
The gravitational force \( F \) between two masses \( M_1 \) and \( M_2 \) separated by a distance \( r \) is given by Newton's law of gravitation:
\[ F = \frac{G M_1 M_2}{r^2} \]
In our case, the forces acting on the particle of mass \( \frac{m}{5} \) are due to the masses \( m \) and \( 2m \).

2. **Calculate the Force from Mass \( m \)**:
The distance from the particle to mass \( m \) is \( d - r \), where \( r \) is the distance traveled by the particle towards \( m \). The gravitational force \( F_1 \) exerted by mass \( m \) on the particle is:
\[ F_1 = \frac{G \cdot m \cdot \frac{m}{5}}{(d - r)^2} \]

3. **Calculate the Force from Mass \( 2m \)**:
The distance from the particle to mass \( 2m \) is \( r \). The gravitational force \( F_2 \) exerted by mass \( 2m \) on the particle is:
\[ F_2 = \frac{G \cdot 2m \cdot \frac{m}{5}}{r^2} \]

4. **Determine the Net Force**:
The net force \( F_{net} \) acting on the particle is the difference between these two forces:
\[ F_{net} = F_2 - F_1 = \frac{G \cdot 2m \cdot \frac{m}{5}}{r^2} - \frac{G \cdot m \cdot \frac{m}{5}}{(d - r)^2} \]

5. **Condition for Motion**:
If the particle is projected with an initial velocity \( v_0 \), it will move towards mass \( m \). The particle will continue to move towards \( m \) if its kinetic energy is sufficient to overcome the gravitational pull from \( 2m \). If \( v_0 \) is less than the escape velocity at point \( P \) (the point where the forces balance), the particle will retrace its path.

6. **Conclusion**:
Therefore, the correct statement is that if \( v_0 \) is less than the required value to cross point \( P \), the particle will retrace its path; otherwise, it will continue moving towards mass \( m \).