Question 43 of 130

Two blocks of masses $10kg$ and $30kg$ are placed on the same straight line with coordinates $\left(0,0\right)cm$ and $\left(x,0\right)cm$ respectively. The block of $10kg$ is moved on the same line through a distance of $6cm$ towards the other block. The distance through which the block of $30kg$ must be moved to keep the position of centre of mass of the system unchanged is

A$4cm$ towards the $10kg$ blockB$2cm$ away from the $10kg$ blockC$2cm$ towards the $10kg$ blockD$4cm$ away from the $10kg$ block

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Answer: (C) $2cm$ towards the $10kg$ block

Initial position of centre of mass: ${x}_{CM}=\dfrac{10\left(0\right)+30\times x}{40}=\dfrac{3x}{4}$
Final position of the centre of mass: ${{x}^{'}}_{CM}=\dfrac{10\left(6\right)+30\times \left(x-y\right)}{40}=\dfrac{6+3\left(x-y\right)}{4}$
As position of centre of mass remains same,
${x}_{CM}={{x}^{'}}_{CM}\Rightarrow \dfrac{3x}{4}=\dfrac{6+3\left(x-y\right)}{4}\Rightarrow 3x=6+3x-3y$ $\Rightarrow y=2cm$
Alternate method, $∆{x}_{COM}=\dfrac{{m}_{1}∆{x}_{1}+{m}_{2}∆{x}_{2}}{\left({m}_{1}+{m}_{2}\right)}\Rightarrow 0=10\times 6+30y\Rightarrow y=-2cm$