Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and (x, 0) cm respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is :
Solution
<p>For COM to remain unchanged,</p>
<p>m<sub>1</sub>x<sub>1</sub> = m<sub>2</sub>x<sub>2</sub></p>
<p>$\Rightarrow$ 10 $\times$ 6 = 30 $\times$ x<sub>2</sub></p>
<p>$\Rightarrow$ x<sub>2</sub> = 2 cm towards 10 kg block.</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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