Two blocks of masses 2 ,kg and 1 ,kg are tied | vedclass

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(B) Let the masses be $m_1 = 2 \,kg$ and $m_2 = 1 \,kg$. The acceleration of the system is given by $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(2 - 1

Vedclass · Accepted Answer

$2.22$

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(B) Let the masses be $m_1 = 2 \,kg$ and $m_2 = 1 \,kg$. The acceleration of the system is given by $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(2 - 1)g}{2 + 1} = \frac{g}{3}$.Taking the downward direction as positive for $m_1$ and upward for $m_2$,the accelerations are $a_1 = \frac{g}{3}$ (downward) and $a_2 = -\frac{g}{3}$ (upward).The acceleration of the centre of mass is $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = \frac{2(g/3) + 1(-g/3)}{2 + 1} = \frac{g/3}{3} = \frac{g}{9}$.The distance traversed by the centre of mass in $t = 2 \,s$ starting from rest is $S = \frac{1}{2} a_{cm} t^2$.Substituting the values: $S = \frac{1}{2} 	imes \frac{10}{9} 	imes (2)^2 = \frac{1}{2} 	imes \frac{10}{9} 	imes 4 = \frac{20}{9} \approx 2.22 \,m$.

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