Trajectory of Centre of Mass for Two Moving Bodies

To determine the nature of the path followed by the centre of mass of the two bodies, we calculate the initial velocity and acceleration of the centre of mass.

1. Velocity of the Centre of Mass ($\vec{v}_{\text{cm}}$): Given that both bodies have equal mass ($m_A = m_B = m$): $\vec{v}_{\text{cm}} = \frac{m \vec{v}_1 + m \vec{v}_2}{m + m} = \frac{\vec{v}_1 + \vec{v}_2}{2}$

Substituting the given initial velocities $\vec{v}_1 = 4\hat{i} \text{ m/s}$ and $\vec{v}_2 = 4\hat{j} \text{ m/s}$: $\vec{v}_{\text{cm}} = \frac{4\hat{i} + 4\hat{j}}{2} = (2\hat{i} + 2\hat{j}) \text{ m/s} = 2(\hat{i} + \hat{j}) \text{ m/s}$

2. Acceleration of the Centre of Mass ($\vec{a}_{\text{cm}}$): $\vec{a}_{\text{cm}} = \frac{m \vec{a}_1 + m \vec{a}_2}{m + m} = \frac{\vec{a}_1 + \vec{a}_2}{2}$

Substituting the given accelerations $\vec{a}_1 = (6\hat{i} + 6\hat{j}) \text{ m/s}^2$ and $\vec{a}_2 = 0$: $\vec{a}_{\text{cm}} = \frac{(6\hat{i} + 6\hat{j}) + 0}{2} = (3\hat{i} + 3\hat{j}) \text{ m/s}^2 = 3(\hat{i} + \hat{j}) \text{ m/s}^2$

3. Trajectory Analysis: Notice that the initial velocity vector $\vec{v}_{\text{cm}}$ and the acceleration vector $\vec{a}_{\text{cm}}$ are parallel to each other: $\vec{v}_{\text{cm}} \times \vec{a}_{\text{cm}} = 0$

The position of the centre of mass as a function of time $t$ (assuming initial position at origin) is given by: $\vec{r}_{\text{cm}}(t) = \vec{v}_{\text{cm}}t + \frac{1}{2}\vec{a}_{\text{cm}}t^2 = \left(2t + \frac{3}{2}t^2\right)\hat{i} + \left(2t + \frac{3}{2}t^2\right)\hat{j}$

Equating the $x$ and $y$ components: $x(t) = 2t + \frac{3}{2}t^2$ $y(t) = 2t + \frac{3}{2}t^2$

Thus, the trajectory equation is: $y = x$

Since the relationship between the coordinates is linear, the centre of mass moves in a straight line.

Correct Option: C