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Two particles of masses 10 kg and 30 kg are lying on straight line. The 10 kg mass shifted towards 30 kg mass by a distance of 2cm. By what distance should the 30 kg mass be shifted so that the position of their centre of mass does not change

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$\frac{4}{3}$ cm towards 10kg mass

$\frac{2}{3}$ cm towards 10kg mass

$\frac{3}{2}$ cm towards 10kg mass

If $Δ x_{1} = + 2 c m$ then $Δ x_{2} = - \frac{2}{3} c m$

answer is D, B.

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Detailed Solution

$Δ X_{C_{W}} = \frac{m_{1} {Δx}_{1} + m_{2} {Δx}_{2}}{m_{1} + m_{2}}$

$\begin{array}{l} m_{1} {Δx}_{1} = m_{2} {Δx}_{2} \\ 10 (2) = - 30 {Δx}_{2} \\ {Δx}_{2} = \frac{- 20}{30} \\ {Δx}_{2} = - 2 / 3 c m \end{array}$

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