Two particles of equal masses have velocity ${\vec{v}}_{1} = 2 \hat{i}$ m/s and $\vec{v_{1}} = 2 \hat{j} m / s$ . The first particle has an acceleration $\vec{a_{1}} = (3 \hat{i} + 3 \hat{j}) m / s^{2}$ while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a:

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ellipse

circle

parabola

straight line

answer is C.

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Detailed Solution

$\vec{v_{cm}} = \frac{m 2 \hat{i} + m 2 \hat{j}}{m + m} = \hat{i} + \hat{j} m / s$

$\vec{a_{cm}} = \frac{m (3 \hat{i} + 3 \hat{j}) + m \times 0}{m + m} = \frac{3}{2} \hat{i} + \frac{3}{2} \hat{j}$

Since acceleration and velocity are in same direction, so path should be straight line.

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