Question

Three particles of masses $1\text{Kg}$, $2\text{Kg}$ and $3\text{Kg}$ are situated at the corners of an equilateral triangle.
The speed of each particle is shown in the figure. Each particle maintains a direction towards the particle at the next corner symmetrically. The velocity of centre of mass of the system at this instant, if initially the system starts from rest, is:

• A. $3{\text{ms}}^{-1}$
• B. $5{\text{ms}}^{-1}$
• C. $6{\text{ms}}^{-1}$
• D. zero

Answer 1

The correct option is D zero

Velocity of centre of mass is given by,

${\overrightarrow{v}}_{com}=\frac{m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2+m_3{\overrightarrow{v}}_3}{m_1+m_2+m_3}$

$\Rightarrow {\overrightarrow{v}}_{com}=\frac{\text{Total momentum}}{\text{Total mass}}$

In this case, the total momentum is zero, because magnitude of momentum of each particle is same and they are symmetrically oriented.

So, ${\overrightarrow{p}}_1+{\overrightarrow{p}}_2+{\overrightarrow{p}}_3=0$

So, ${\overrightarrow{v}}_{com}=0$

Hence, option $\left(D\right)$ is the correct answer.

Why this Question? Note: Velocity of centre of mass of a system of particle is given by, ${\overrightarrow{v}}_{com}=\frac{m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2+m_3{\overrightarrow{v}}_3}{m_1+m_2+m_3}$