Question

Three particles (of same mass) are kept at the three vertices of an equiletral triangle. Each starts moving from one vertex to the other vertex next to it with the same speed $V$ in anticlockwise order. Find the displacement of $COM$ after $t\text{s}$.

• A. $Vt$
• B. $\frac{V}{2}t$
• C. $\frac{V}{4}t$
• D. $0$

Answer 1

The correct option is D $0$

Initially, the particles are at points $A,B$ and $C$.


After $t$ seconds, let the positions be as shown:


(Consider given $x$ and $y$ direction as positive)
For $m_1,v_{1x}=v,v_{1y}=0$
For $m_2$, $v_{2x}=-v\cos {60}^{\circ },v_{2y}=v\cos {30}^{\circ }$
For $m_3$, $v_{3x}=-v\cos {60}^{\circ },v_{3y}=-v\cos {30}^{\circ }$

After $t$ $\text{s}$, $V_{com}=\frac{v_1{m}_1+v_2{m}_2+v_3{m}_3}{m_1+m_2+m_3}$
$(V_{com}{)}_x=\frac{v\times m+(-v\cos {60}^{\circ })\times m+m(-v\cos {60}^{\circ })}{m+m+m}=0$
and $(V_{com}{)}_y=\frac{m\times 0+m\left(v\cos {30}^{\circ }\right)+m(-v\cos {30}^{\circ })}{m+m+m}=0$

Hence, $V_{com}$ is zero, so displacement of $COM$ will be zero, even after $t\text{s}$.