Two particles of masses in the ratio 1:2 are | vedclass

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(A) Let the masses of the particles be $m_1 = m$ and $m_2 = 2m$. The displacement of the centre of mass $\Delta Y_{cm}$ is given by the formula:$\Delt

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moved $1.5 \ cm$ downward

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(A) Let the masses of the particles be $m_1 = m$ and $m_2 = 2m$. The displacement of the centre of mass $\Delta Y_{cm}$ is given by the formula:$\Delta Y_{cm} = \frac{m_1 \Delta y_1 + m_2 \Delta y_2}{m_1 + m_2}$Given that $\Delta Y_{cm} = 2 \ cm$,$\Delta y_1 = 9 \ cm$,$m_1 = m$,and $m_2 = 2m$,we substitute these values into the equation:$2 = \frac{m(9) + 2m(\Delta y_2)}{m + 2m}$$2 = \frac{9m + 2m(\Delta y_2)}{3m}$$2 = \frac{9 + 2\Delta y_2}{3}$$6 = 9 + 2\Delta y_2$$2\Delta y_2 = 6 - 9$$2\Delta y_2 = -3$$\Delta y_2 = -1.5 \ cm$The negative sign indicates that the heavier particle must be moved $1.5 \ cm$ downward.

Two particles of masses in the ratio $1:2$ are placed along a vertical line. The lighter particle is raised through a height of $9 \ cm$. To raise the centre of mass of the system by $2 \ cm$,the heavier particle should be

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