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
Show Answer
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
BETA
