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 $meters$.

$A$ uniform rod of length $L$ and mass $M$ is placed along the $x$-axis with its one end at the origin. The centre of mass of the rod is located at:

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

Four bodies of masses $2\,kg, 3\,kg, 5\,kg$ and $8\,kg$ are placed at the four corners of a square of side $2\,m$ as shown in the figure. The position of the centre of mass $(CM)$ is:

What do you mean by mass element $dm$?

Centre of mass $(C.M.)$ of three particles of masses $1 \,kg, 2 \,kg$ and $3 \,kg$ lies at the point $(1, 2, 3)$ and $C.M.$ of another system of particles of $3 \,kg$ and $2 \,kg$ lies at the point $(-1, 3, -2)$. Where should we put a particle of mass $5 \,kg$ so that the $C.M.$ of the entire system lies at the $C.M.$ of the first system?