Where will the centre of mass be located when combining two masses $m$ and $M$ $(M > m)$?

Two particles of masses $M$ and $4M$ are released from a distance of $10 \ m$ from each other. Find the point of collision from the smaller particle in $meters$.

Three particles of masses $1\,kg$,$\frac{3}{2}\,kg$,and $2\,kg$ are located at the vertices of an equilateral triangle of side $a$. The $x, y$ coordinates of the centre of mass are

Three identical spheres each of diameter $2\sqrt{3} \text{ m}$ are kept on a horizontal surface such that each sphere touches the other two spheres. If one of the spheres is removed,then the shift in the position of the centre of mass of the system is

Two particles of masses $1\,kg$ and $3\,kg$ have position vectors $2\hat{i} + 3\hat{j} + 4\hat{k}$ and $-2\hat{i} + 3\hat{j} - 4\hat{k}$ respectively. The position vector of the centre of mass is:

Three identical spheres each of mass $M$ are placed at the corners of a right-angled triangle with mutually perpendicular sides equal to $3\,m$ each. Taking the point of intersection of mutually perpendicular sides as the origin,the magnitude of the position vector of the centre of mass of the system will be $\sqrt{x}\,m$. The value of $x$ is