two uniform rods A and B of length 5m and 3m are placed end to end. if their linear densities are 3kg/m and 2kg/m the position of thier centre of mass from the interface is

Dear student, 

Consider the interface of the system to be at the origin. The rod having length 5 m be to the left of the origin and the rod having a length 3 m be on the right side of the origin. Since the rods are uniform, the center of mass of the two rods rods will me
 $$\begin{gathered} x_1=-2.5 \mathrm{m} \\ {x}_2=-1.5 \mathrm{m} \end{gathered}$$

The masses of the two rods are:
$$\begin{gathered} \mathrm{Mass}=\mathrm{Mass} \mathrm{density}\times \mathrm{length} \\ {m}_1=3\times 5=15 \mathrm{kg} \\ {m}_2=2\times 3=6 \mathrm{kg} \end{gathered}$$

Then center of mass can is calculated as:

$$\begin{gathered} x_{cm}=\frac{x_1{m}_1+x_2{m}_2}{m_1+m_2}=\frac{(-2.5)\times 15+1.5\times 6}{15+6}=\frac{-37.5+9}{21}=\frac{-28.5}{21} \\ \Rightarrow x_{cm}=-1.36 \mathrm{m} \end{gathered}$$

Hence, the center of mass is at a distance of 1.36 m on the right of the interface.

Regards