Question

Three particles (of same mass) are kept at the three vertices of a 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 $t\text{s}$.