Two blocks of masses 10 ; kg and 30 ; kg are | vedclass

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(C) The position of the centre of mass $X_{CM}$ is given by $X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.To keep the centre of mass unchanged,the ch

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$2 \; cm$ towards the $10 \; kg$ block

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(C) The position of the centre of mass $X_{CM}$ is given by $X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.To keep the centre of mass unchanged,the change in the position of the centre of mass must be zero,i.e.,$\Delta X_{CM} = 0$.This implies $\Delta X_{CM} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2} = 0$.Given $m_1 = 10 \; kg$,$m_2 = 30 \; kg$,and the displacement of the first block $\Delta x_1 = +6 \; cm$ (towards the second block).Substituting the values: $0 = \frac{10 	imes 6 + 30 	imes \Delta x_2}{10 + 30}$.$0 = 60 + 30 \Delta x_2$.$30 \Delta x_2 = -60$.$\Delta x_2 = -2 \; cm$.The negative sign indicates that the $30 \; kg$ block must move in the opposite direction of the $10 \; kg$ block's displacement,which means it must move towards the $10 \; kg$ block by $2 \; cm$.

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