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)=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`