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

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Solution

For $C O M$ to remain unchanged,
$m_{1} x_{1} = m_{2} x_{2}$
$\Rightarrow 10 \times 6 = 30 \times x_{2}$
$\Rightarrow x_{2} = 2 c m$ towards $10 k g$ block.

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10

kg and

30

kg are placed along a vertical line the first block of mass

10

kg is raised through a height of

7

cm by what distance should the second block of mass

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For COM to remain unchanged, m1x1=m2x2 ⇒10×6=30×x2 ⇒x2=2 cm towards 10 kg block.

For $C O M$ to remain unchanged,
$m_{1} x_{1} = m_{2} x_{2}$
$\Rightarrow 10 \times 6 = 30 \times x_{2}$
$\Rightarrow x_{2} = 2 c m$ towards $10 k g$ block.