Figure below shows the system is at rest initially with x = 0. A man and a woman both are initially at the extreme ends of the platform. The man and the woman start to move towards each other. Obtain an expression for the displacement 's' of the platform when the two meet in terms of the displacement ' $x_{1}$ ' of the man relative to the platform.

\frac{m_{1} x_{1} + m_{2} l}{m_{1} + m_{2} + m_{0}}

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\frac{m_{1} x_{1} - m_{2} l}{m_{1} + m_{2} + m_{0}}

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\frac{m_{1} x_{1} + m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}

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\frac{m_{1} x_{1} - m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}

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Solution

The correct option is D $\frac{m_{1} x_{1} - m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}$
As no external force is acting on the system, the displacement of the centre of mass should be zero. Let the displacement of the platform toward right be s,
$Δ {\to x}_{1} = Δ {\to x}_{1, p} + Δ {\to x}_{p} = (- x_{1} + s)$
$Δ {\to x}_{2} = Δ {\to x}_{2, p} + Δ {\to x}_{p} = (l - x_{1} + s)$
Hence,
$0 = \frac{m_{1} (- x_{1} + s) + m_{2} (l - x_{1} + s) + m_{0} s}{m_{1} + m_{2} + m_{0}}$
$s (m_{1} + m_{2} + m_{0}) - m_{1} x_{1} + m_{2} (l - x_{1}) = 0$
$s = \frac{m_{1} x_{1} - m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}$

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