Two masses `m_1 = 1 kg and m_2 = 2 kg` are connected by a light inextensible string and suspended by means of a weightness 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 ms^-2`.
.

Options

A. `(20)/(9)m`
B. `(40)/(9)m`
C. `(2)/(3)m`
D. `(1)/(3)m`

Answer

A

Explanation

Here, `m_(1)= 1kg, m_(2)= 2kg`
The acceleration of the system is `a= ((m_(2)- m_(1))g)/(m_(1) + m_(2))= ((2-1)g)/(1+ 2)= (g)/(3)= (10)/(3)`
Acceleration of the centre of mass is `a_(cm)= (m_(1) a_(1) + m_(2) a_(2))/(m_(1) + m_(2))= (1 (-a) + 2(a))/(1+ 2)`
`=(1 ((-g)/(3)) + 2 ((g)/(3)))/(3)= (g)/(9)= (10)/(9)`
The distance travelled by the centre of mass in two seconds is
`S= (1)/(2) a_(cm) t^(2)= (1)/(2) xx (10)/(9) xx (2)^(2)= (20)/(9)m`