Question

Mass centre of a system of three particles of masses $1,2,3kg$ is at the point $(1m,2m,3m)$ and mass centre of another group of two particles of masses $2kg$ and $3kg$ is at point $(-1m,-2m)$. Where a $5kg$ particle should be placed, so that the mass centre of the system of all those six particles shifts to mass centre of the first system?

Answer 1

We can analyze the situation as mass $6kg$ is at $(1,2,3)$, another mass of $5kg$ at $(-1,-2,0).$ We have to find the position of other mass of $5kg$ to be put say (x,y,z) so that the final center of mass is at $(1,2,3)$

$x_{CM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$

$⟹1=\frac{6\times 1+5\times (-1)+5\times x}{16}$

$⟹x=3m$

$y_{CM}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$

$⟹2=\frac{6\times 2+5\times (-2)+5\times y}{16}$

$⟹y=6m$

$z_{CM}=\frac{m_1{z}_1+m_2{z}_2+m_3{z}_3}{m_1+m_2+m_3}$

$⟹3=\frac{6\times 3+5\times 0+5\times z}{16}$

$⟹z=6m$

Hence position of new mass of $5kg$ should be $(3m,6m,6m)$