Question

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. The masses are held at rest at the same horizontal level and then released. The distance traversed by centre of mass in 2 s is $(g=10\frac{m}{s^2})$

• A. 1.42 m
• B. 2.22 m
• C. 3.12 m
• D. 3.33 m

Answer 1

The correct option is B 2.22 m

The masses of two blocks are 2 kg and 1 kg, the time is 2 sec.
Acceleration of the blocks is given by
$a=\frac{(m_2-m_1)g}{(m_2+m_1)}$
Putting the values, we get $a=g/3\frac{m}{s^2}$

Centre of mass of the system is
$a_{com}=\frac{(m_2-m_1)g/3}{(m_2+m_1)}$

Putting the values, we get $a_{com}=g/9\frac{m}{s^2}$ $S_{com}=\frac{1}{2}{a}_{com}{t}^2\Rightarrow 2.22\text{m}$