The centre of mass of two masses m and m' mov | vedclass

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Question

(B) The displacement of the centre of mass $\Delta R_{cm}$ is given by the formula $\Delta R_{cm} = \frac{m_1 \Delta r_1 + m_2 \Delta r_2}{m_1 + m_2}$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The displacement of the centre of mass $\Delta R_{cm}$ is given by the formula $\Delta R_{cm} = \frac{m_1 \Delta r_1 + m_2 \Delta r_2}{m_1 + m_2}$.Given,$\Delta R_{cm} = \frac{x}{5}$,$\Delta r_1 = x$,and $\Delta r_2 = 0$ (since $m'$ is fixed).Substituting these values into the formula:$\frac{x}{5} = \frac{m(x) + m'(0)}{m + m'}$$\frac{x}{5} = \frac{mx}{m + m'}$Dividing both sides by $x$:$\frac{1}{5} = \frac{m}{m + m'}$$m + m' = 5m$$m' = 4m$Therefore,the ratio $\frac{m'}{m} = 4$.

The centre of mass of two masses $m$ and $m'$ moves by a distance $\frac{x}{5}$ when mass $m$ is moved by a distance $x$ and $m'$ is kept fixed. The ratio $\frac{m'}{m}$ is

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