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 $\vec{a_1} = 6\hat{i} + 6\hat{j}\ \text{m/s}^2$ while the acceleration of the other body B is zero. The centre of mass of the two bodies moves in ______ path.
Answer: (C) straight line
Concept. The centre of mass moves like a single particle with velocity $\vec v_{cm}=\dfrac{\sum m\vec v}{\sum m}$ and acceleration $\vec a_{cm}=\dfrac{\sum m\vec a}{\sum m}$. If $\vec a_{cm}$ is constant and parallel to $\vec v_{cm}$, the path is a straight line; if not parallel, a parabola.
Step 1. Initial velocity of the centre of mass (equal masses, so a plain average):
$$\vec v_{cm}=\frac{4\hat i+4\hat j}{2}=2\hat i+2\hat j$$
Step 2. Acceleration of the centre of mass:
$$\vec a_{cm}=\frac{(6\hat i+6\hat j)+\vec 0}{2}=3\hat i+3\hat j$$
Step 3. Compare directions. Both vectors point along $\hat i+\hat j$. A constant acceleration along the velocity only changes the speed, not the direction.
Answer. Straight line (Option C).
Solution by Sreeraj P, M.Sc Physics
