Two bodies $A$ and $B$ have masses $M$ and $m$ respectively,where $M > m$,and they are at a distance $d$ apart. Equal forces are applied to each of them so that they approach each other. The position where they hit each other is

On what does the position of the center of mass of a rigid body depend?

Two uniform plates of the same thickness and area but of different materials,one shaped like an isosceles triangle and the other shaped like a rectangle,are joined together to form a composite body as shown in the figure. If the centre of mass of the composite body is located at the mid-point of their common side,then the ratio between the masses of the triangle to that of the rectangle is:

Three particles of mass $1 \ kg, 2 \ kg$ and $3 \ kg$ are placed at the vertices $A, B$ and $C$ respectively of an equilateral triangle $ABC$ of side $1 \ m$. The centre of mass of the system from vertex $A$ (located at origin) is

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

$A$ thin uniform wire is bent to form the two equal sides $AB$ and $AC$ of triangle $ABC$,where $AB = AC = 5\,cm$. The third side $BC$,of length $6\,cm$,is made from a uniform wire of twice the linear density of the first. The distance of the centre of mass from $A$ is