Two masses $m_{1} = 1 kg and m_{2} = 2 kg$ are 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 $(take g = 10 {ms}^{- 2})$

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$\frac{40}{9} m$

$\frac{2}{3} m$

$\frac{1}{3} m$

$\frac{20}{9} m$

answer is A.

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Detailed Solution

Here, $m_{1} = 1 kg, m_{2} = 2 kg$
The acceleration of the system is $a = \frac{(m_{2} - m_{1}) g}{m_{1} + m_{2}} = \frac{(2 - 1) g}{1 + 2} = \frac{g}{3} = \frac{10}{3}$
Acceleration of the center of mass is $a_{cm} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}} = \frac{1 (- a) + 2 (a)}{1 + 2}$
$= \frac{1 (\frac{- g}{3}) + 2 (\frac{g}{3})}{3} = \frac{g}{9} = \frac{10}{9}$

The distance travelled by the center of mass in two second is $S = \frac{1}{2} a_{cm} t^{2} = \frac{1}{2} \times \frac{10}{9} \times (2)^{2} = \frac{20}{9} m$