Question

Centre of mass of a system of three particles of masses $1\text{kg}$, $2\text{kg}$ and $3\text{kg}$ at the point $(1\text{m},2\text{m},3\text{m})$ and centre of mass of another group of two particles of masses $2\text{kg}$ and $3\text{kg}$ is at point $(-1\text{m},-2\text{m})$, where a $10\text{kg}$ particle should be placed, so that the centre of mass of the system of all those six particles shifts to centre of mass of the first system?

• A. $(3\text{m},6\text{m},6\text{m})$
• B. $(3\text{m},6\text{m},4\text{m})$
• C. $(2\text{m},4\text{m},4.5\text{m})$
• D. $(3\text{m},6\text{m},4.5\text{m})$

Answer 1

The correct option is C $(2\text{m},4\text{m},4.5\text{m})$

We can analyze the situation as mass $6\text{kg}$ is at $(1,2,3)$, another $5\text{kg}$ at $(-1,-2,0)$. we have to find the position of other mass of $10\text{kg}$ to be put say $(x,y,z)$ so that the final centre of mass is at $(1,2,3)$
$x-$ coordinate of centre of mass $\left(x_{com}\right):$
$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$1=\frac{6\left(1\right)+5(-1)+10x}{6+5+10}\Rightarrow x=2\text{m}$

$y-$ coordinate of centre of mass $\left(y_{com}\right):$
$y_{com}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$2=\frac{6\times 2+5\times (-2)+10\times y}{6+5+10}$
$\Rightarrow y=4\text{m}$

$z-$ coordinate of COM
$Z_{com}=\frac{m_1{z}_1+m_2{z}_2+m_3{z}_3}{m_1+m_2+m_3}$
$3=\frac{6\times 3+5\times 0+10z}{6+5+10}$
$\Rightarrow z=4.5\text{m}$
Hence, position of new mass of $10\text{kg}$ sholud be $(2\text{m},4\text{m},4.5\text{m})$