Question

The co-ordinates of centre of mass of particles of mass $10,20and30gm$ are $(1,1,1)cm$. The position coordinates of mass $40gm$ which when added to the system, the position of combined centre of mass be at $(0,0,0)$ are,

• A. $(3/2,3/2,3/2)$
• B. $(-3/2,-3/2,-3/2)$
• C. $(3/4,3/4,3/4)$
• D. $(-3/4,-3/4,-3/4)$

Answer 1

Answer: (B)

$(-3/2,-3/2,-3/2)$

Total mass of the system in case-1 was $m_1=10+20+30=60gm$

and its position was $(1,1,1)cm$

now the second position is $(0,0,0)$ with added mass as $m_2=40gm$

If $x,y,z$ was the position of $4^{th}$ mass then so we should get

$0=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ or $0=60\times 1+40x$ or $x=-\frac{3}{2}cm$

similarly

$0=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$ or $0=60\times 1+40y$ or $y=-\frac{3}{2}cm$

and $0=\frac{m_1{z}_1+m_2{z}_2}{m_1+m_2}$ or $0=60\times 1+40z$ or $z=-\frac{3}{2}cm$

so the required coordinate was $(-3/2,-3/2,-3/2)$