Question

Two masses $m_1=1kg$ and $m_2=2kg$ 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 $g=10m{s}^{-2}$)

• 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 is
$a=\frac{\left(m_2-m_1\right)g}{m_1+m_2}=\frac{\left(2-1\right)g}{\left(1+2\right)}$
$=\frac{g}{3}=\frac{10}{3}$
Acceleration of the centre of mass is
$a_{cm}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}=\frac{1\left(-a\right)+2\left(a\right)}{1+2}$
$=\frac{1\left(\frac{-g}{3}\right)+2\left(\frac{g}{3}\right)}{3}$
$=\frac{g}{9}=\frac{10}{9}$
The distance travelled by the centre of mass in two seconds is
$S=\frac{1}{2}{a}_{cm}{t}^2$
$=\frac{1}{2}\times \frac{10}{9}\times {\left(2\right)}^2$
$=\frac{20}{9}m$