Question

Three equal objects of mass $M$ each are placed on a table at points $(0,0),(6,0)$ and $(0,6)$ respectively . If another object of mass $2M$ is placed at point $(2,7)$, the displacement of $COM$ is

• A. $2\text{units}$
• B. $2\sqrt{2}\text{units}$
• C. $2\sqrt{3}\text{units}$
• D. $0\text{units}$

Answer 1

The correct option is A $2\text{units}$

$COM$ of mass of initial system (three objects)
$X_{COM}=\frac{M_1{X}_1+M_2{X}_2+M_3{X}_3}{M_1+M_2+M_3}$
$X_{COM}=\frac{0\times M+6\times M+0\times M}{M+M+M}=\frac{6M}{3M}=2$

Similarly,
$Y_{COM}=\frac{M_1{Y}_1+M_2{Y}_2+M_3{Y}_3}{M_1+M_2+M_3}$
$Y_{COM}=\frac{0\times M+0\times M\times 6\times M}{M+M+M}=\frac{6M}{3M}=2$
$\Rightarrow$ Initial coordinates of $COM$ are $(2,2)$,

Coordinates of $COM$ of new system.
$X_{COM}^{\prime }=\frac{M_1{X}_1+M_2{X}_2+M_3{X}_3+M_4{X}_4}{M_1+M_2+M_3+M_4}$
$=\frac{0\times M+6\times M+0\times M+2\times \left(2M\right)}{M+M+M+2M}$
$=\frac{10M}{5M}=2$

Similarly,
$Y_{COM}^{\prime }=\frac{Y_1{M}_1+Y_2{M}_2+Y_3{M}_3+Y_4{M}_4}{M_1+M_2+M_3+M_4}$
$=\frac{0\times M+0\times M+6\times M+7\times \left(2M\right)}{M+M+M+2M}$
$=\frac{20M}{5M}=4$
$\Rightarrow$ Final coordinates of $COM$ are $(2,4)$

Displacement of $COM$, $d=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$
$=\sqrt{(2-2{)}^2+(2-4{)}^2}$
$=2\text{units}$