Question

Two particles of the same mass move in the x - y plane with velocities $2\hat{i}{\text{ms}}^{-1}$ and $3\hat{j}{\text{ms}}^{-1}$. The first particle is given an acceleration $(2\hat{i}+2\hat{j}){\text{ms}}^{-2}$ and the second particle in given an acceleration $(2\hat{i}+4\hat{j}){\text{ms}}^{-2}$. The centre of mass of the system of two particles will move along a

• A. Straight line
• B. Circle
• C. Ellipse
• D. Parabola

Answer 1

The correct option is A Straight line

$v_{cm}=\frac{m{v}_1+m{v}_2}{m+m}$
$=\frac{1}{2}(v_1+v_2)$
$v_{cm}=\frac{1}{2}(2\hat{i}+3\hat{j}){\text{ms}}^{-1}$
$a_{cm}=\frac{m{a}_1+m{a}_2}{m+m}=\frac{1}{2}(a_1+a_2)$
$=\frac{1}{2}\left[\right(2\hat{i}+2\hat{j})+(2\hat{i}+4\hat{j}\left)\right]=(2\hat{i}+3\hat{j}){\text{ms}}^{-2}$
It is clear that $v_{cm}$ and $a_{cm}$ are parallel vectors. Thus the acceleration of the centre of mass is along its velocity. hence it follows a straight line trajectory.