Question

Two particles of equal masses have velocities ${\overrightarrow{v}}_1=\left(2\hat{i}\right)\text{m/s}$ and ${\overrightarrow{v}}_2=\left(2\hat{j}\right)\text{m/s}$. The first particle has an acceleration of ${\overrightarrow{a}}_1=(3\hat{i}+3\hat{j}){\text{m/s}}^2$ while the acceleration of the other particle is zero. The path of the centre of mass of the two particles system is a/an

• A. circle
• B. parabola
• C. straight line
• D. ellipse

Answer 1

The correct option is C straight line

Velocity of centre of mass of the system of particles will be,
${\overrightarrow{v}}_{COM}=\frac{m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2}{m_1+m_2}=\frac{{\overrightarrow{v}}_1+{\overrightarrow{v}}_2}{2}$
$[\because m_1=m_2]$
$\Rightarrow {\overrightarrow{v}}_{COM}=\frac{2\hat{i}+2\hat{j}}{2}=(\hat{i}+\hat{j})\text{m/s}$
Now, the direction of velocity with respect to $+x$ axis,
$\tan \theta =\frac{y}{x}=\frac{1}{1}=1$
$\Rightarrow \theta ={45}^{\circ }$
Acceleration of centre of mass of the system of particles will be,
${\overrightarrow{a}}_{COM}=\frac{m_1{\overrightarrow{a}}_1+m_2{\overrightarrow{a}}_2}{m_1+m_2}=\frac{{\overrightarrow{a}}_1+{\overrightarrow{a}}_2}{2}$
$[\because m_1=m_2]$
$\Rightarrow {\overrightarrow{a}}_{COM}=\frac{3\hat{i}+3\hat{j}+0}{2}=\frac{3}{2}(\hat{i}+\hat{j}){\text{m/s}}^2$
Now, the direction of acceleration with respect to $+x$ axis,
$\tan \theta =\frac{y}{x}=\frac{3/2}{3/2}=1$
$\Rightarrow \theta ={45}^{\circ }$
As the velocity and acceleration of $COM$ of the system are along the same direction, therefore, path of $COM$ will be a straight line.