The figure shows the positions and velocities of two particles. If the particles move under the mutual attraction of each other, then the position of the centre of mass at $t = 1 s$ is ______ $c m$ .

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Solution

$∵$ Distance between bodies $= 6 m$
$V_{r e l} = 8 m / s$
Hence after $1 s e c$ balls collide
Hence CM lies at point of contact
Now, point of contact $= \frac{6}{8} \times 5 = \frac{30}{8} = \frac{15}{4} = 3.75 m$

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