Question

Four particles of masses $1\text{kg},2\text{kg},3\text{kg}$ and $4\text{kg}$ situated at the corners of a square start moving at speeds of $8\text{m/s},6\text{m/s},4\text{m/s}$ and $2\text{m/s}$ respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find the magnitude of velocity of the center of mass of the system at this instant.

• A. $\frac{2}{5\sqrt{2}}\text{m/s}$
• B. $\frac{5\sqrt{2}}{2}\text{m/s}$
• C. $\frac{2\sqrt{2}}{5}\text{m/s}$
• D. $\frac{\sqrt{2}}{5}\text{m/s}$

Answer 1

The correct option is C $\frac{2\sqrt{2}}{5}\text{m/s}$

Given masses, $m_1=1\text{kg},m_2=2\text{kg},m_3=3\text{kg},m_4=4\text{kg}$
From the figure, velocities of masses $m_1,m_2,m_3$ and $m_4$ are
$v_1=-8\hat{j},v_2=6\hat{i},v_3=4\hat{j},v_4=-2\hat{i}$

Velocity of COM of the system is given by
${\overrightarrow{v}}_{CM}=\frac{m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}+m_3\overrightarrow{v_3}+m_4\overrightarrow{v_4}}{m_1+m_2+m_3+m_4}$
$=\frac{1(-8\hat{j})+2\left(6\hat{i}\right)+3\left(4\hat{j}\right)+4(-2\hat{i})}{1+2+3+4}$
$=\frac{-8\hat{j}+12\hat{i}+12\hat{j}-8\hat{i}}{10}$
$=\frac{1}{10}(4\hat{i}+4\hat{j})\text{m/s}$

Therefore, magnitude of velocity of COM,
$\left|{\overrightarrow{v}}_{CM}\right|=\frac{1}{10}\sqrt{(4{)}^2+(4{)}^2}=\frac{4\sqrt{2}}{10}$
$\left|{\overrightarrow{v}}_{CM}\right|=\frac{2\sqrt{2}}{5}\text{m/s}$