Two identical bodies A and B of equal masses have initial...

Velocity of the centre of mass is given by:

$\vec{v}_{cm} = \dfrac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$

Since $m_1 = m_2 = m$, we get:

$\vec{v}_{cm} = \dfrac{m(4\hat{i}) + m(4\hat{j})}{2m} = 2\hat{i} + 2\hat{j}$ m/s

Acceleration of the centre of mass is given by:

$\vec{a}_{cm} = \dfrac{m_1\vec{a}_1 + m_2\vec{a}_2}{m_1 + m_2}$

$\vec{a}_{cm} = \dfrac{m(6\hat{i} + 6\hat{j}) + 0}{2m} = 3\hat{i} + 3\hat{j}$ m/s$^2$

Here, $\vec{v}_{cm} = 2(\hat{i} + \hat{j})$ and $\vec{a}_{cm} = 3(\hat{i} + \hat{j})$.

Since the initial velocity and acceleration of the centre of mass are parallel to each other, the centre of mass moves in a straight line path.

Answer: straight line