question_answer

Three particles of masses 1 kg, 2 kg and 3 kg are situated at the comers of an equilateral triangle move at speed \[6\,m{{s}^{-1}}\], \[3\,m{{s}^{-1}}\] and \[2\,m{{s}^{-1}}\] respectively. Each particle  maintains a direction towards the particle at the next comer symmetrically. Find velocity of CM of the system at this 2kg 4 instant

A) \[3\,m{{s}^{-1}}\]                  

B) \[5\,m{{s}^{-1}}\]

C) \[6\,m{{s}^{-1}}\]      

D) Zero

Correct Answer: D

Solution :

[d] \[{{\vec{v}}_{cm}}=\frac{{{m}_{1}}{{{\vec{v}}}_{1}}+{{m}_{2}}{{{\vec{v}}}_{2}}+{{m}_{3}}{{{\vec{v}}}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[\Rightarrow {{\vec{v}}_{cm}}=\frac{Total\,momentum}{Total\,mass}\] 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