Rotational Motion Question 104

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

Options:

A) $3m{{s}^{-1}}$

B) $5m{{s}^{-1}}$

C) $6m{{s}^{-1}}$

D) Zero

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Answer:

Correct Answer: D

Solution:

[d]

${\vec{v} _{cm}}=\frac{m _{1}{\vec{v} _{1}}+m _{2}{\vec{v} _{2}}+m _{3}{\vec{v} _{3}}}{m _{1}+m _{2}+m _{3}}$

$\Rightarrow {\vec{v} _{cm}}=\frac{Totalmomentum}{Totalmass}$

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



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