Question

Let there are three equal masses situated at the vertices of an equilateral triangle, as shown in figure. Now the particle A starts with a velocity $v1$ towards line AB, particle B starts with the velocity $v_2$ towards line BC and particle starts with velocity $v_3$ towards line CA. Find the displacement of the centre of mass of the three particles A, B and C after time t. What would it be the displacement of centre of mass if $v_1=v_2=v_3$?

• A. $(v_1+v_2+v_3)t$
• B. $\frac{(v_1+v_2+v_3)t}{2}$
• C. $\frac{(v_1+v_2+v_3)t}{3}$
• D. '0'

Answer 1

The correct option is D '0'

First we write the three velocities in vectorial form, taking right direction as positive x-axis and upwards as positive y-axis.
${\overrightarrow{v}}_1=-\frac{1}{2}{v}_1\hat{i}-\frac{\sqrt{3}}{2}{v}_1\hat{j}$
${\overrightarrow{v}}_2=v_2\hat{i},{\overrightarrow{v}}_3=-\frac{1}{2}{v}_3\hat{i}+\frac{\sqrt{3}}{2}{v}_3\hat{j}$
Thus the velocity of centre of mass of the system is
${\overrightarrow{v}}_{CM}=\frac{{\overrightarrow{v}}_{+}{\overrightarrow{v}}_2+{\overrightarrow{v}}_3}{3}$
$=(v_2-\frac{1}{2}{v}_1-\frac{1}{2}{v}_3)\hat{i}+\frac{\sqrt{3}}{2}(V_3-V_1)\hat{j}$
Which can be written as${\overrightarrow{v}}_{CM}=v_x\hat{i}+v_y\hat{j}$
$\Delta \overrightarrow{r}=v_{x}t\hat{i}+v_{x}t\hat{j}$
If $v_1=v_2=v_3=v$, we have ${\overrightarrow{v}}_{CM}=0$
Therefore, there is no displacement of centre of mass of the system