Question

Centre of mass of three particles of masses $1\text{kg},2\text{kg}$ and $3\text{kg}$ lies at the point $(1,2,3)$ and centre of mass of another system of particles $3\text{kg}$ and $2\text{kg}$ lies at the point $(-1,3,-2)$. Where should we put a particle of mass $5\text{kg}$ so that the centre of mass of entire system lies at the centre of mass of $1^{st}$ system?

• A. $(0,0,0)$
• B. $(1,3,2)$
• C. $(-1,2,3)$
• D. $(3,1,8)$

Answer 1

The correct option is D $(3,1,8)$

We can imagine one particle of mass $(1+2+3)\text{kg}$ located at $(1,2,3)$ and another particle of mass $(3+2)\text{kg}$ located at $(-1,3,-2)$. Assume that $3^{rd}$ particle of mass $5\text{kg}$ is placed at $(x_3,y_3,z_3)$.
Hence,
$m_1=6\text{kg};(x_1,y_1,z_1)=(1,2,3)$
$m_2=5\text{kg};(x_2,y_2,z_2)=(-1,3,-2)$
$m_3=5\text{kg};(x_3,y_3,z_3)=?$
Given that $(X_{CM},Y_{CM},Z_{CM})=(1,2,3)$
Now, $X_{CM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$\Rightarrow 1=\frac{6\times 1+5\times (-1)+5x_3}{16}\therefore x_3=3$
$Y_{CM}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$\Rightarrow 2=\frac{6\times 2+5\times 3+5y_3}{16}\therefore y_3=1$
$Z_{CM}=\frac{m_1{z}_1+m_2{z}_2+m_3{z}_3}{m_1+m_2+m_3}$
$3=\frac{6\times 3+5(-2)+5z_3}{16}\therefore z_3=8$
$\therefore (x_3,y_3,z_3)=(3,1,8).$
Hence, the correct answer is option (d).