Two identical bodies \(A\) and \(B\) of equal masses have initial velocities \(\vec{v_1} = 4\hat{i}\) m/s and \(\vec{v_2} = 4\hat{j}\) m/s respectively. The body \(A\) has acceleration \(\vec{a_1} = 6\hat{i} + 6\hat{j}\) m/s\(^2\) while the acceleration of the other body \(B\) is zero. The centre of mass of the two bodies moves in __________ path.

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