Question

Two blocks of masses 10 kg and 30 kg are placed along a vertical line. The first block is raised through a height of 7 cm. By what distance should the second mass be moved to raise the centre of mass by 1 cm?

• A. 1 cm upwards
• B. 1 cm downwards
• C. 2 cm upwards
• D. 2 cm downwards

Answer 1

The correct option is B

1 cm downwards

We already know ,
$X_{CM}$ = $\frac{X_1{m}_1+x_2{m}_2}{m_1+m_2}$
From this we can say ,
($m_1$ + $m_2$) $\frac{△X_{cm}}{△t}$ = $\frac{m_1△X_1}{△t}$ + $m_2\frac{△X_2}{△t}$ )
or
($m_1$ + $m_2$) $△$ $X_{CM}$ = $m_1$ $△$$X_1$ + $m_2$ $△$$X_2$
Assume upwards to be +ve & downwards as - ve
(40 $\times$ 1 ) = (10 $△$ 7 ) + 30 $△$ $X_2$
$△$ $X_2$ = -1 cm
Which implies that in order to more $X_{CM}$ upwards by 1 cm , we nned to move the 10 kg block upwards by 7 cm which is given & 30 kg block downward by 1 cm.
Let the centre of the system be at $C_i$ as shown in figure. Now in accordance with the question.