A particle of mass `2kg` moves under a force given by
`vec(F)=-(8N//m)(xhat(i)+yhat(j))`
where `hat(i)` and `hat(j)` are unit vectors in the `x` and `y` directions.
The particle is projected from a point on `x` axis at `x=a` in `xy` plane with an initial velocity `vec(v)=v_(0)hat(j)`. Select correct statement (s). The particle will move in

Options

A. An elliptical path if `(v_(0))/a=3s^(-1)`
B. A circular path if `(v_(0))/a=5s^(-1)`
C. A circular path if `(v_(0))/a=2s^(-1)`
D. An elliptical path if `(v_(0))/a=1s^(-1)`

Explanation

To solve the problem step by step, we will analyze the motion of the particle under the given force and determine the trajectory it follows.

### Step 1: Identify the Force and Mass
The force acting on the particle is given by:
\[ \vec{F} = -8 \, \text{N/m} \, (\hat{i} + \hat{j}) \]
The mass of the particle is:
\[ m = 2 \, \text{kg} \]

### Step 2: Calculate Acceleration
Using Newton's second law, \( \vec{F} = m \vec{a} \), we can find the acceleration:
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{-8 \, \text{N/m} \, (\hat{i} + \hat{j})}{2 \, \text{kg}} = -4 \, \text{m/s}^2 \, (\hat{i} + \hat{j}) \]
Thus, the components of acceleration are:
\[ a_x = -4 \, \text{m/s}^2, \quad a_y = -4 \, \text{m/s}^2 \]

### Step 3: Write the Equations of Motion
The acceleration can be expressed in terms of velocity and position. We can use the relationship:
\[ a_x = v_x \frac{dv_x}{dx} \quad \text{and} \quad a_y = v_y \frac{dv_y}{dy} \]
Substituting the values of acceleration:
\[ v_x \frac{dv_x}{dx} = -4x \quad \text{and} \quad v_y \frac{dv_y}{dy} = -4y \]

### Step 4: Integrate to Find Velocity
Integrating the equation for \( v_x \):
\[ \int v_x \, dv_x = -4 \int x \, dx \]
This gives:
\[ \frac{1}{2} v_x^2 = -2x^2 + C_1 \]
Assuming the initial condition \( v_x = 0 \) when \( x = a \):
\[ C_1 = 2a^2 \]
Thus, we have:
\[ v_x^2 = 4a^2 - 2x^2 \]
Taking the square root:
\[ v_x = \sqrt{4a^2 - 2x^2} \]

### Step 5: Find the Position as a Function of Time
We can express \( v_x \) in terms of time:
\[ \frac{dx}{dt} = \sqrt{4a^2 - 2x^2} \]
Separating variables and integrating:
\[ \int \frac{dx}{\sqrt{4a^2 - 2x^2}} = \int dt \]
This leads to:
\[ x = a \cos(kt) \quad \text{where } k \text{ is a constant} \]

### Step 6: Find the y-component
Using similar steps for \( y \):
\[ v_y = V_0 - 4 \int y \, dy \]
Integrating gives:
\[ y = \frac{V_0}{2} \sin(kt) \]

### Step 7: Combine the Equations
We have:
\[ x = a \cos(kt) \quad \text{and} \quad y = \frac{V_0}{2} \sin(kt) \]
Using the identity \( \sin^2 + \cos^2 = 1 \):
\[ \left(\frac{y}{\frac{V_0}{2}}\right)^2 + \left(\frac{x}{a}\right)^2 = 1 \]
This is the equation of an ellipse.

### Conclusion
The trajectory of the particle is an ellipse.