Two masses `m_(1)` = 1 kg and `m_(2)` = 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 1 kg seconds is (Take g= `10 m s^(-2)`)

Options

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

Answer

A

Explanation

Here `m_(1) = 1` kg , `m_(2) = 2` kg
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`