Two blocks of masses 2 kg and 1 kg respectively, are tied to the...

The acceleration of the blocks in the Atwood machine is given by:

$a = \dfrac{m_1 - m_2}{m_1 + m_2} g$

Substituting the given values ($m_1 = 2$ kg, $m_2 = 1$ kg):

$a = \dfrac{2 - 1}{2 + 1} g = \dfrac{g}{3}$

Taking the downward direction as positive, the acceleration of the $2$ kg block is $a_1 = \dfrac{g}{3}$ and the acceleration of the $1$ kg block is $a_2 = -\dfrac{g}{3}$.

The acceleration of the centre of mass is:

$a_{cm} = \dfrac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$

$a_{cm} = \dfrac{2 \left(\dfrac{g}{3}\right) + 1 \left(-\dfrac{g}{3}\right)}{2 + 1} = \dfrac{\dfrac{g}{3}}{3} = \dfrac{g}{9}$

Since the system is released from rest, the initial velocity of the centre of mass is zero.

The distance traversed by the centre of mass in $t = 2$ s is:

$S_{cm} = \dfrac{1}{2} a_{cm} t^2$

$S_{cm} = \dfrac{1}{2} \left(\dfrac{10}{9}\right) (2)^2 = \dfrac{20}{9} \approx 2.22 \text{ m}$

Answer: $2.22$