Two blocks of masses $2$ kg and $1$ kg respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in $2$ s is ______ m. (Take $g=10\ \text{m/s}^2$)
Answer: (C) $2.22$
Acceleration of each block: $a=\dfrac{(m_1-m_2)g}{m_1+m_2}=\dfrac{(2-1)\times10}{3}=\dfrac{10}{3}\ \text{m/s}^2$ (2 kg down, 1 kg up).
Acceleration of the centre of mass (taking down as positive):
$$a_{cm}=\frac{m_1a-m_2a}{m_1+m_2}=\frac{2-1}{3}\times\frac{10}{3}=\frac{10}{9}\ \text{m/s}^2\ \text{(downward)}$$
Starting from rest, $s_{cm}=\tfrac12a_{cm}t^2=\tfrac12\times\tfrac{10}{9}\times4=\tfrac{20}{9}\approx2.22$ m.
Solution by Sreeraj P, M.Sc Physics