Question

Four point masses are placed in a plane so that their centre of mass is at $(1,1)$. Three of them are of mass $m$ each and are placed at $(0,0),(2,0)$ and $(0,2)$ respectively. The fourth point of mass $2m$ is displaced from its initial position such that centre of mass of the system moves to $(2,1)$. Then, the displacement of the fourth point mass is

• A. Parallel to X axis
• B. Inclined at an angle ${45}^{\circ }$ with X axis
• C. Of magnitude $\frac{5}{2}units$
• D. Of magnitude $5units$

Answer 1

The correct option is C Of magnitude $\frac{5}{2}units$

The three masses $m$ can be thought of as a $single$ mass $3m$ placed at $x_1=\frac{m(0+2+0)}{m+m+m}=\frac{2}{3}$ and $y_1=\frac{1}{3}(0+0+2)=\frac{2}{3}$

the fourth mass is supposed to be a $second$ mass placed at $x,y$ so the center of mass of the system in old case

will be at $1=\frac{3m\times \frac{2}{3}+2m\times x}{2m+3m}$ so $x=1.5$ similarly $y=1.5$

to make the center of mass at $(2,1)$ the $x$ will change to $x_0$, given by $2=\frac{3m\times \frac{2}{3}+2m\times x_0}{2m+3m}$

or $x_0=4$

so the displacement will be $x_0-x=4-1.5=\frac{5}{2}units$ parallel to X axis because $y$ didn't change.