Figure shows position and velocities of two particles moving under mutual gravitational attraction in space at time t=0.
The position of centre of mass after one second is

x = 4 m

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x = 6 m

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x = 8 m

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x = 10 m

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Solution

The correct option is A $x = 10 m$
mutual force with not affect motion
the motion of center of mass can be supposed to be uniform
So initial velocity of center of mass
$u = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = \frac{2 k g (8) + 3 k g (- 2)}{2 + 3}$
$\Rightarrow u = \frac{10}{5} = 2 m / s$ $+ v e$
$⇓$
towards increasing $x$
distance in second will be $x = x t = 2 x = 2 m$
$\Rightarrow$ New position $= o l d p o s i t i o n + 2$
or new position $= \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} + 2$
$x = \frac{2 \times 2 + 3 \times 12}{3} + 2$
$x = 10$ meters

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Figure shows position and velocities of two particles moving under mutual gravitational attraction in space at time t=0.The position of centre of mass after one second is

Figure shows position and velocities of two particles moving under mutual gravitational attraction in space at time t=0.
The position of centre of mass after one second is