Consider a two-particle system with the particles having masses $m_{1}$ and $m_{2}$ . If the first particle is pushed towards the centre of mass by a distance 'd', by what distance should the second particle be moved so as to keep the centre of mass at the same position?

( $\frac{m_{1} m_{2}}{m_{1} + m_{2}}$ )d

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$\frac{m_{2}}{m_{1}}$ d

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$\frac{m_{1}}{m_{2}}$ d

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( $\frac{m_{1} - m_{2}}{m_{1} + m_{2}}$ )d

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Solution

The correct option is C
$\frac{m_{1}}{m_{2}}$ d

Again, as we aleady know ,
$△$ $X_{C M}$ = $\frac{m_{1} △ X_{1} + m_{2} △ X_{2}}{m_{1} + m_{2}}$
$△$ $X_{C M}$ = 0
Assuming leftards as negative and rightwards as potive
$0 = \frac{m_{1} d + m_{2} d^{'}}{m_{1} + m_{2}}$
d' = $\frac{- m_{1} d}{m_{2}}$
The negative sign indicates that $m_{2}$ should be shifted by
( $\frac{m_{1}}{m_{2}}$ ) d leftwards in order to keep the COM stationary.

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Consider a two-particle system with the particles having masses m1 and m2. If the first particle is pushed towards the centre of mass by a distance 'd', by what distance should the second particle be moved so as to keep the centre of mass at the same position?

The correct option is C m1m2d Again, as we aleady know , △XCM = m1△X1+m2△X2m1+m2 △XCM = 0 Assuming leftards as negative and rightwards as potive 0=m1d+m2d′m1+m2 d' = −m1dm2 The negative sign indicates that m2 should be shifted by (m1m2) d leftwards in order to keep the COM stationary.

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