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:

Consider a two-particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the center of mass through a distance $d$,by what distance should the second particle be moved so as to keep the center of mass at the same position?

The center of mass of a system of particles with masses $1 \, g, 2 \, g$,and $3 \, g$ is at the origin. When a particle of mass $4 \, g$ with position vector $\alpha(\hat{i} + 2\hat{j} + 3\hat{k})$ is added,the center of mass of the system becomes $(1, 2, 3)$. If $\alpha$ is a constant,its value must be:

The centre of mass of a system of two particles divides the distance between them

Three identical spheres,each of mass $2 M$,are placed at the corners of a right-angled triangle with mutually perpendicular sides equal to $4 \ m$ each. Taking the point of intersection of these two sides as the origin,the magnitude of the position vector of the centre of mass of the system is $\frac{4 \sqrt{2}}{x}$,where the value of $x$ is . . . . . .

An isolated particle of mass $m$ is moving in a horizontal plane $(x-y)$ 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