Question

Three particles of masses 1 kg, 2 kg and 3 kg are situated at the corners of the equilateral triangle move at speed 6 ms${}^{-1}$, 3 ms${}^{-1}$ and 2 ms${}^{-1}$ respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of COM of the system at this instant :

• A. $3m{s}^{-1}$
• B. $5m{s}^{-1}$
• C. $6m{s}^{-1}$
• D. $zero$

Answer 1

The correct option is D $zero$

Given : $m_1=1kg$ $m_2=2kg$ $m_3=3kg$

$v_1=6$ $m/s$ $v_2=3$ $m/s$ $v_3=2$ $m/s$

Let velocity of their centre of mass be $\overrightarrow{V_{CM}}$

Using $\overrightarrow{V_{CM}}=\frac{m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}+m_3\overrightarrow{v_3}}{m_1+m_2+m_3}$

$\overrightarrow{V_{CM}}=\frac{1(-6sin{30}^o\hat{i}-6cos{30}^o\hat{j})+2\left(3\hat{i}\right)+3(-2cos{60}^o\hat{i}+2sin{60}^o\hat{j})}{1+2+3}$

$\overrightarrow{V_{CM}}=\frac{1(-6\times \frac{1}{2}\hat{i}-6\times \frac{\sqrt{3}}{2}\hat{j})+2\left(3\hat{i}\right)+3(-2\times \frac{1}{2}\hat{i}+2\times \frac{\sqrt{3}}{2}\hat{j})}{6}$

$⟹\overrightarrow{V_{CM}}=0$