Two particles of masses m and 2 m has initial velocity `vecu_(1)=2hati+3hatj(m)/(s)` and `vecu_(2)=-4hati+3hatj(m)/(s)` Respectivley. These particles have constant acceleration `veca_(1)=4hati+3hatj((m)/(s^(2)))` and `veca_(2)=-4hati-2hatj((m)/(s^(2)))` Respectively. Path of the centre of the centre of mass of this two particle system will be:

Options

A. Straight line
B. Circular
C. Parabolic
D. Elliptical

Explanation

To find the path of the center of mass of the two-particle system, we will follow these steps:

### Step 1: Calculate the initial velocities of the particles
Given:
- Mass of particle 1, \( m_1 = m \) with initial velocity \( \vec{u_1} = 2 \hat{i} + 3 \hat{j} \, \text{m/s} \)
- Mass of particle 2, \( m_2 = 2m \) with initial velocity \( \vec{u_2} = -4 \hat{i} + 3 \hat{j} \, \text{m/s} \)

### Step 2: Calculate the velocity of the center of mass
The formula for the velocity of the center of mass \( \vec{V_{cm}} \) is given by:
\[ \vec{V_{cm}} = \frac{m_1 \vec{u_1} + m_2 \vec{u_2}}{m_1 + m_2} \]
Substituting the values:
\[ \vec{V_{cm}} = \frac{m(2 \hat{i} + 3 \hat{j}) + 2m(-4 \hat{i} + 3 \hat{j})}{m + 2m} \]
\[ = \frac{m(2 \hat{i} + 3 \hat{j}) - 8m \hat{i} + 6m \hat{j}}{3m} \]
\[ = \frac{(2 - 8) m \hat{i} + (3 + 6) m \hat{j}}{3m} \]
\[ = \frac{-6m \hat{i} + 9m \hat{j}}{3m} \]
\[ = -2 \hat{i} + 3 \hat{j} \, \text{m/s} \]

### Step 3: Calculate the accelerations of the particles
Given:
- Acceleration of particle 1, \( \vec{a_1} = 4 \hat{i} + 3 \hat{j} \, \text{m/s}^2 \)
- Acceleration of particle 2, \( \vec{a_2} = -4 \hat{i} - 2 \hat{j} \, \text{m/s}^2 \)

### Step 4: Calculate the acceleration of the center of mass
The formula for the acceleration of the center of mass \( \vec{A_{cm}} \) is given by:
\[ \vec{A_{cm}} = \frac{m_1 \vec{a_1} + m_2 \vec{a_2}}{m_1 + m_2} \]
Substituting the values:
\[ \vec{A_{cm}} = \frac{m(4 \hat{i} + 3 \hat{j}) + 2m(-4 \hat{i} - 2 \hat{j})}{m + 2m} \]
\[ = \frac{m(4 \hat{i} + 3 \hat{j}) - 8m \hat{i} - 4m \hat{j}}{3m} \]
\[ = \frac{(4 - 8) m \hat{i} + (3 - 4) m \hat{j}}{3m} \]
\[ = \frac{-4m \hat{i} - 1m \hat{j}}{3m} \]
\[ = -\frac{4}{3} \hat{i} - \frac{1}{3} \hat{j} \, \text{m/s}^2 \]

### Step 5: Determine the path of the center of mass
Since the velocity and acceleration of the center of mass are not parallel and the acceleration is constant, the path of the center of mass will be parabolic.

### Final Answer
The path of the center of mass of this two-particle system is parabolic.

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