Two particles of masses m_1 and m_2 are separ | vedclass

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(A) Let the positions of the two particles be $x_1 = 0$ and $x_2 = d$. The initial centre of mass $X_{cm,i}$ is given by:$X_{cm,i} = \frac{m_1(0) + m_

Vedclass · Accepted Answer

$\frac{m_1 - m_2}{m_1 + m_2} d$

Vedclass · Answer

(A) Let the positions of the two particles be $x_1 = 0$ and $x_2 = d$. The initial centre of mass $X_{cm,i}$ is given by:$X_{cm,i} = \frac{m_1(0) + m_2(d)}{m_1 + m_2} = \frac{m_2 d}{m_1 + m_2}$When the particles are interchanged,their new positions are $x_1' = d$ and $x_2' = 0$. The new centre of mass $X_{cm,f}$ is:$X_{cm,f} = \frac{m_1(d) + m_2(0)}{m_1 + m_2} = \frac{m_1 d}{m_1 + m_2}$The shift in the centre of mass $\Delta X_{cm}$ is:$\Delta X_{cm} = X_{cm,f} - X_{cm,i} = \frac{m_1 d}{m_1 + m_2} - \frac{m_2 d}{m_1 + m_2} = \frac{(m_1 - m_2) d}{m_1 + m_2}$Thus,the correct option is $A$.

Two particles of masses $m_1$ and $m_2$ are separated by a distance $d$. The shift in the centre of mass when the two particles are interchanged is:

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