Question

The centre of mass of three particles of masses $1kg$ , $2kg$ and $3kg$ is at $(2,2,2)$ . The position of the fourth mass of $4kg$ to be placed in the system such that the new centre of mass is at $(0,0,0)$ is

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

Answer 1

Answer: (A)

$m_1=1kg$ , $m_2=2kg$ , $m_3=3kg$
Position of centre of mass $(2,2,2,)$
$m_4=4kg$
New position of centre of mass $(0,0,0)$
For initial position
$X_{CM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$2=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{1+2+3}$
$m_1{x}_1+m_2{x}_2+m_3{x}_3=12$
Similarly, $m_1{y}_1+m_2{y}_2+m_3{y}_3=12$
and $m_1{z}_1+m_2{z}_2+m_3{z}_3=12$
For new position,
$X_{CM}^{\prime }=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$
$0=\frac{12+4\times x_4}{1+2+3+4}$
$4x_4=-12$
$x_4=-3$
Similarly, $y_4=-3$
$z_4=-3$
$\therefore$ Position of fourth mass $(-3,-3,-3)$