Question

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

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

Answer 1

Answer: (D)

${\vec{v}}_{cm}=\frac{m_1{\vec{v}}_1+m_2{\vec{v}}_2+m_3{\vec{v}}_3}{m_1+m_2+m_3}$
$\Rightarrow {\vec{v}}_{cm}=\frac{\text{ Total momentum }}{\text{ Total mass }}$
Here total momentum of system is zero, because momentum of each particle is same in magnitude and they are symmetrically oriented as shown.
So ${\vec{p}}_1+{\vec{p}}_2+{\vec{p}}_3=0$
So, velocity of $CM$ of the system will be zero