Question

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

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

Answer 1

Answer: (A)

$m_{1}=1 \, kg$ , $m_{2}=2 \, kg$ , $m_{3}=3 \, kg$
Position of centre of mass $\left(\right.2, \, 2, \, 2,\left.\right)$
$m_{4}=4 \, kg$
New position of centre of mass $\left(\right.0, \, 0, \, 0\left.\right)$
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,
$X_{CM}^{′}=\frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$0=\frac{12 + 4 \times x_{4}}{1 + 2 + 3 + 4}$
$4x_{4}=-12$
$x_{4}=-3$
Similarly, $y_{4}=-3$
$z_{4}=-3$
$\therefore $ Position of fourth mass $\left(\right.-3, \, -3, \, -3\left.\right)$