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

The masses and positions (in rectangular coordinates) of four particles are as follows: $1 \ kg$ at $(a, a)$,$2 \ kg$ at $(-a, a)$,$3 \ kg$ at $(-a, -a)$,and $4 \ kg$ at $(a, -a)$. The position vector of the centre of mass of the system of four particles is:

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

Three masses are placed on the $x-$axis: $300 \, g$ at the origin,$500 \, g$ at $x = 40 \, cm$ and $400 \, g$ at $x = 70 \, cm$. The distance of the centre of mass from the origin is ....... $cm$.

Where is the centre of mass of two isolated particles of equal mass located?

An isolated particle of mass $m$ is moving in a horizontal $x-y$ plane along the $x-$axis at a certain height above the ground. It suddenly explodes into two fragments of masses $m/4$ and $3m/4$. An instant later,the smaller fragment is at $y = 15 \ cm$. The larger fragment at this instant is at $y = \dots \ cm$.