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(A) The velocity of the centre of mass $(v_{CM})$ is given by: $v_{CM} = \frac{m_1 \overrightarrow{v}_1 + m_2 \overrightarrow{v}_2}{m_1 + m_2} = \frac

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straight line

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(A) The velocity of the centre of mass $(v_{CM})$ is given by: $v_{CM} = \frac{m_1 \overrightarrow{v}_1 + m_2 \overrightarrow{v}_2}{m_1 + m_2} = \frac{\overrightarrow{v}_1 + \overrightarrow{v}_2}{2}$ (since $m_1 = m_2 = m$). $v_{CM} = \frac{4 \hat{i} + 4 \hat{j}}{2} = (2 \hat{i} + 2 \hat{j}) 	ext{ ms}^{-1}$. The acceleration of the centre of mass $(a_{CM})$ is given by: $a_{CM} = \frac{m_1 \overrightarrow{a}_1 + m_2 \overrightarrow{a}_2}{m_1 + m_2} = \frac{\overrightarrow{a}_1 + 0}{2} = \frac{5 \hat{i} + 5 \hat{j}}{2} = (2.5 \hat{i} + 2.5 \hat{j}) 	ext{ ms}^{-2}$. Since the acceleration vector $\overrightarrow{a}_{CM}$ is constant,the centre of mass moves with constant acceleration. Because the initial velocity $\overrightarrow{v}_{CM}$ and the acceleration $\overrightarrow{a}_{CM}$ are parallel (both are in the direction of $\hat{i} + \hat{j}$),the path of the centre of mass is a straight line.

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