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(A) Let the position of the $5 \,g$ particle be at $x_1 = 0 \,cm$ and the position of the $3 \,g$ particle be at $x_2 = 40 \,cm$.The formula for the c

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lies at a distance of $15 \,cm$ from $5 \,g$ particle

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(A) Let the position of the $5 \,g$ particle be at $x_1 = 0 \,cm$ and the position of the $3 \,g$ particle be at $x_2 = 40 \,cm$.The formula for the centre of mass $x_{cm}$ is given by:$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$Substituting the given values:$x_{cm} = \frac{5 	imes 0 + 3 	imes 40}{5 + 3}$$x_{cm} = \frac{120}{8} = 15 \,cm$This means the centre of mass is at a distance of $15 \,cm$ from the $5 \,g$ particle and at a distance of $40 - 15 = 25 \,cm$ from the $3 \,g$ particle.Therefore,the centre of mass lies at a distance of $15 \,cm$ from the $5 \,g$ particle.

Two particles of masses $5 \,g$ and $3 \,g$ are separated by a distance of $40 \,cm$. The centre of mass of the system of these two particles

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