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(B) The velocity of the centre of mass is given by $\overrightarrow{v}_{CM} = \frac{m_1 \overrightarrow{v}_1 + m_2 \overrightarrow{v}_2}{m_1 + m_2}$.S

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parabola

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(B) The velocity of the centre of mass is given by $\overrightarrow{v}_{CM} = \frac{m_1 \overrightarrow{v}_1 + m_2 \overrightarrow{v}_2}{m_1 + m_2}$.Since $m_1 = m_2 = m$,we have $\overrightarrow{v}_{CM} = \frac{\overrightarrow{v}_1 + \overrightarrow{v}_2}{2} = \frac{4 \hat{i} + 4 \hat{j}}{2} = (2 \hat{i} + 2 \hat{j}) 	ext{ ms}^{-1}$.The acceleration of the centre of mass is $\overrightarrow{a}_{CM} = \frac{m_1 \overrightarrow{a}_1 + m_2 \overrightarrow{a}_2}{m_1 + m_2}$.Given $\overrightarrow{a}_1 = (5 \hat{i} + 5 \hat{j}) 	ext{ ms}^{-2}$ and $\overrightarrow{a}_2 = 0$,we get $\overrightarrow{a}_{CM} = \frac{\overrightarrow{a}_1 + 0}{2} = (2.5 \hat{i} + 2.5 \hat{j}) 	ext{ ms}^{-2}$.Since the acceleration $\overrightarrow{a}_{CM}$ is constant and the initial velocity $\overrightarrow{v}_{CM}$ is not parallel to the acceleration $\overrightarrow{a}_{CM}$,the path of the centre of mass is a parabola.

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