Question

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

Answer 1

Correct Answer - a
Here, `m_(1)=1 kg, 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 center of mass is ltbr. `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 center of mass in two seconds is
`S=1/2a_(CM)t^(2)= 1/2 xx 10/9 xx (2)^(2)= 20/9` m