Q. Two particles of equal mass have velocities $\text{v}_{\text{1}}$ =4 $\widehat{\text{i}}\mathbf{\ }$and $\text{v}_{\text{2}}$=4$\widehat{\text{j}}$ $\text{ms}^{\text{-1}}$. First particle has an acceleration $\text{a}_{\text{1}}$= (5$\widehat{\text{i}}$ + 5$\widehat{\text{j}}$)$\ \text{ms}^{\text{-2}}$, while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of
A) Straight line
B) Parabola
C) Circle
D) Ellipse
Explanation
(a) $\mathbf{v}_{CM} = \frac{m_{1}\mathbf{v}_{1} + m_{2}\mathbf{v}_{2}}{m_{1} + m_{2}} = \frac{\mathbf{v}_{\mathbf{1}} + \mathbf{v}_{2}}{2}$ (as $m_{1} = m_{2}$)
and $\mathbf{a}_{CM} = \frac{m_{1}\mathbf{a}_{1} + m_{2}\mathbf{a}_{2}}{m_{1} + m_{2}}$
$= \frac{\mathbf{a}_{1} + 0}{2} = \frac{\mathbf{a}_{1}}{2}$
The centre of mass of two particles will move with the mean velocity of two particles having common acceleration $\frac{\mathbf{a}}{2}$.
Hence, path of CM will be a straight line.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.