Question

Two masses $m_1=1kg$ and $m_2=2kg$ are connected by means of a light inextensible string which passes over a weightless pulley as depicted in the figure. If both the masses start from rest, the distance travelled by the center of mass in two seconds will be(Take $g=10m{s}^{-2}$ )

• A. $\frac{20}{9}m$
• B. $\frac{40}{9}m$
• C. $\frac{2}{3}m$
• D. $\frac{1}{3}m$

Answer 1

Answer: (A)

Here, $m_1=1kg,m_2=2kg$
The acceleration of the system
$a=\frac{(m_2-m_1)g}{m_1+m_2}=\frac{(2-1)g}{(1+2)}=\frac{g}{3}=\frac{10}{3}$
Acceleration of the centre of mass is
$a_{CM}=\frac{1\left(\frac{-g}{3}\right)+2\left(\frac{g}{3}\right)}{3}=\frac{g}{9}=\frac{10}{9}$
The distance travelled by the centre of mass in two seconds is
$s=\frac{1}{2}{a}_{Cm}{t}^2=\frac{1}{2}\times \frac{10}{9}\times {\left(2\right)}^2=\frac{20}{9}m$