Two particles are kept on a smooth surface, move towards each other due to mutual attraction. If the position and velocities of the particles at $t = 0$ is as shown in the figure. Then position of the center of mass at $t = 2 s e c$ is

$x = 7 m$

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$x = 5 m$

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$x = 3 m$

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$x = 2 m$

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Solution

The correct option is A
$x = 7 m$

Initial position of centre of mass
$x_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$
$x_{C M} = \frac{1 \times 0 + 1 \times 10}{2} = 5 m$
Velocity of centre of mass
$v_{C M} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

$v_{C M} = \frac{1 \times 5 - 1 \times 3}{2} = 1 m / s$

Displacement of $C M$ in $2 s e c = v_{C M} \times t = 2 m$

So location of $C M = 5 + 2 = 7 m$

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Two particles are kept on a smooth surface, move towards each other due to mutual attraction. If the position and velocities of the particles at t=0 is as shown in the figure. Then position of the center of mass at t=2 sec is

The correct option is A x=7 m Initial position of centre of mass xCM=m1x1+m2x2m1+m2 xCM=1×0+1×102=5 m Velocity of centre of mass vCM=m1v1+m2v2m1+m2 vCM=1×5−1×32=1 m/s Displacement of CM in 2 sec=vCM×t=2 m So location of CM =5+2=7 m

Two particles are kept on a smooth surface, move towards each other due to mutual attraction. If the position and velocities of the particles at $t = 0$ is as shown in the figure. Then position of the center of mass at $t = 2 s e c$ is