Question

Two masses m1= 1 kg and m2 = 2 kg 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= 10 m 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)

$\frac{20}{9}$m

Answer 2

Here, $m_{1} = 1kg,m_{2} = 2kg$

The accelerations of the system is

$a = \frac{(m_{2} + m_{1})g}{m_{1} + m_{2}} = \frac{(2 - 1)g}{(1 + 2)} = \frac{g}{3} = \frac{10}{3}$

Accelerations of the centre of mass is

$a_{cm} = \frac{m_{1}a_{1} + m_{2}a_{2}}{m_{1} + m_{2}} = \frac{1( - a) + 2(a)}{1 + 2}$

$= \frac{1\left( \frac{- g}{3} \right) + 2\left( \frac{g}{3} \right)}{3} = \frac{g}{9} = \frac{10}{9}$

The distance travelled by the centre of mass in two seconds is

$S = \frac{1}{2}a_{cm}t^{2}\frac{1}{2} \times \frac{10}{9} \times (2)^{2} = \frac{20}{9}m$