Two masses ${\mathrm{m}}_1$ = 1 kg and ${\mathrm{m}}_2 =$ 2 kg arc connected by a light inextensible string and suspended by means of a weightless pulley as shown in the figure. Assuming that both the masses start from rest, the distance travelled by the centre of mass in two seconds is (Take g = 10 ${\mathrm{ms}}^{-2}$)

Here,${\mathrm{m}}_1$ = 1 kg, ${\mathrm{m}}_2$ = 2kg

The acceleration of the system is

$\mathrm{a} = \frac{({\mathrm{m}}_2-{\mathrm{m}}_1)\mathrm{g}}{{\mathrm{m}}_1+{\mathrm{m}}_2} = \frac{(2-1)\mathrm{g}}{(1+2)} = \frac{\mathrm{g}}{3} = \frac{10}{3}$

Acceleration of the centre of mass is

${\mathrm{a}}_{\mathrm{cm}} = \frac{{\mathrm{m}}_1{\mathrm{a}}_1+{\mathrm{m}}_2{\mathrm{a}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2} = \frac{1(-\mathrm{a})+2\left(\mathrm{a}\right)}{1+2}$

        = $\frac{\left({\displaystyle \frac{-\mathrm{g}}{3}}\right)+2\left({\displaystyle \frac{\mathrm{g}}{3}}\right)}{3} =\frac{\mathrm{g}}{9} = \frac{10}{9}$

The distance travelled by the centre of mass in two seconds is

$\mathrm{S} = \frac{1}{2}{\mathrm{a}}_{\mathrm{cm}}{\mathrm{t}}^2 = \frac{1}{2}\times \frac{10}{9}\times {\left(2\right)}^2 = \frac{20}{9}\mathrm{m}$