SATHEE: Rotational Motion Question 104

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 $6 m s^{- 1}$ , $3 m s^{- 1}$ and $2 m s^{- 1}$ 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) $3 m s^{- 1}$

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

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

D) Zero

Show Answer

Answer:

Correct Answer: D

Solution:

[d]

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

$\Rightarrow {\vec{v}}_{c m} = \frac{T o t a l m o m e n t u m}{T o t a l m a s s}$

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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Rotational Motion Question 104

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