A particle of mass `'m'` moves on a horizont smooth line `AB` of length `'a'` such that when particle is at any general point `P` on the line two force act on it. A force `(mg(AP))/(a)` towards `A` and another force `(2mg(BP))/(a)` towards `B`.
If the force acting towards A stops acting when the particle is nearest to B then find the velocity with which it crosses point B.

Options

A. `(sqrt(2ga))/(3)`
B. `(sqrt(2ga))/(6)`
C. `(sqrt(2ga))/(5)`
D. `(sqrt(ga))/(3)`

Explanation

To solve the problem, we need to analyze the forces acting on the particle and how they contribute to its motion. Let's break it down step by step.

### Step 1: Understand the Forces Acting on the Particle
When the particle is at a point \( P \) on the line \( AB \):
- The force acting towards point \( A \) is given by \( F_A = \frac{mg(AP)}{a} \).
- The force acting towards point \( B \) is given by \( F_B = \frac{2mg(BP)}{a} \).

Here, \( AP \) is the distance from point \( A \) to point \( P \), and \( BP \) is the distance from point \( B \) to point \( P \).

### Step 2: Express the Forces in Terms of Position
Let \( x \) be the distance from point \( A \) to point \( P \). Therefore:
- \( AP = x \)
- \( BP = a - x \)

Now, we can rewrite the forces:
- \( F_A = \frac{mgx}{a} \)
- \( F_B = \frac{2mg(a - x)}{a} \)

### Step 3: Find the Net Force Acting on the Particle
The net force \( F \) acting on the particle at point \( P \) is:
\[ F = F_B - F_A = \frac{2mg(a - x)}{a} - \frac{mgx}{a} \]
\[ F = \frac{2mg(a - x) - mgx}{a} = \frac{2mga - 3mgx}{a} \]
\[ F = \frac{mg(2a - 3x)}{a} \]

### Step 4: Relate Force to Acceleration
Using Newton's second law, \( F = ma \), we can relate the net force to acceleration:
\[ ma = \frac{mg(2a - 3x)}{a} \]
Dividing both sides by \( m \):
\[ a = \frac{g(2a - 3x)}{a} \]

### Step 5: Identify the Motion Type
The equation we derived resembles the equation of simple harmonic motion (SHM):
\[ a = -\omega^2 x \]
Comparing, we can identify:
\[ \omega^2 = \frac{3g}{a} \]
Thus, \( \omega = \sqrt{\frac{3g}{a}} \).

### Step 6: Find the Time Period of the Motion
The time period \( T \) of the SHM can be calculated as:
\[ T = 2\pi \sqrt{\frac{a}{3g}} \]

### Step 7: Calculate the Maximum Velocity
The maximum velocity \( v_{max} \) in SHM is given by:
\[ v_{max} = \omega A \]
where \( A \) is the amplitude of the motion. The amplitude can be determined from the initial conditions.

Assuming the particle starts from rest at point \( A \) and moves towards \( B \), the maximum displacement from the equilibrium position (which is at \( \frac{2a}{3} \)) can be taken as \( A = \frac{a}{3} \).

Thus,
\[ v_{max} = \sqrt{\frac{3g}{a}} \cdot \frac{a}{3} = \sqrt{\frac{3ga}{3}} = \sqrt{ga} \]

### Step 8: Find the Velocity at Point B
When the force towards \( A \) stops acting as the particle approaches point \( B \), the velocity at point \( B \) can be calculated using energy conservation principles or directly from the maximum velocity derived above.

Thus, the velocity with which the particle crosses point \( B \) is:
\[ v_B = \sqrt{\frac{2g a}{6}} = \sqrt{\frac{ga}{3}} \]

### Final Answer
The velocity with which the particle crosses point \( B \) is:
\[ v_B = \sqrt{\frac{2g a}{6}} = \sqrt{\frac{ga}{3}} \]