Two particles of equal masses have velocities `vec v_1 = 4hati ms^(-1)` and `vec v_2 = 4hatj ms^(-1)`. First particle has an acceleration `vec a_1 = (5 hat i + 5 hat j) ms^(-2)`, while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

Options

A. Straight line
B. Parabola
C. Circle
D. Ellipse

Answer

To solve the problem, we need to find the path of the center of mass of two particles with given velocities and accelerations.

### Step-by-Step Solution:

1. **Identify Given Data:**
- For Particle 1:
- Velocity: \(\vec{v_1} = 4 \hat{i} \, \text{m/s}\)
- Acceleration: \(\vec{a_1} = 5 \hat{i} + 5 \hat{j} \, \text{m/s}^2\)
- For Particle 2:
- Velocity: \(\vec{v_2} = 4 \hat{j} \, \text{m/s}\)
- Acceleration: \(\vec{a_2} = 0 \, \text{m/s}^2\)

2. **Calculate the Center of Mass Velocity (\(\vec{v_{cm}}\)):**
- Since both particles have equal masses, we can use the formula for the velocity of the center of mass:
\[ \vec{v_{cm}} = \frac{m_1 \vec{v_1} + m_2 \vec{v_2}}{m_1 + m_2} \]
- Let \(m_1 = m_2 = m\):
\[ \vec{v_{cm}} = \frac{m \vec{v_1} + m \vec{v_2}}{m + m} = \frac{\vec{v_1} + \vec{v_2}}{2} \]
- Substitute the values:
\[ \vec{v_{cm}} = \frac{4 \hat{i} + 4 \hat{j}}{2} = 2 \hat{i} + 2 \hat{j} \, \text{m/s} \]

3. **Calculate the Center of Mass Acceleration (\(\vec{a_{cm}}\)):**
- The acceleration of the center of mass is given by:
\[ \vec{a_{cm}} = \frac{\vec{a_1} + \vec{a_2}}{2} \]
- Substitute the values:
\[ \vec{a_{cm}} = \frac{(5 \hat{i} + 5 \hat{j}) + 0}{2} = \frac{5 \hat{i} + 5 \hat{j}}{2} = \frac{5}{2} \hat{i} + \frac{5}{2} \hat{j} \, \text{m/s}^2 \]

4. **Determine the Path of the Center of Mass:**
- Since the center of mass has a constant acceleration \(\vec{a_{cm}}\) and its velocity \(\vec{v_{cm}}\) is also constant, the center of mass will move in a straight line.
- The direction of the motion can be determined from the acceleration vector, which is in the direction of \(\hat{i} + \hat{j}\).

5. **Conclusion:**
- The center of mass of the two particles moves in a straight line.