If two bodies of masses $2 \,kg$ and $3 \,kg$ are moving at right angles with velocities $20 \,m \,s^{-1}$ and $10 \,m \,s^{-1}$ respectively, then the velocity of the centre of mass of the system of the two bodies is

Masses $m, (1/2)(m/2), (1/2)^2(m/3), \dots, (1/2)^{N-1}(m/N), \dots \infty$ are placed at $x = 1, 2, 3, \dots, N, \dots \infty$ respectively. If the total mass is $M$,then the centre of mass of the system is:

Infinite numbers of blocks are placed on a table edge as shown in the diagram. What is the minimum value of $L$ so that the blocks are just going to topple over the table?

Three particles of masses $50\, g$,$100\, g$,and $150\, g$ are placed at the vertices of an equilateral triangle of side $1\, m$ (as shown in the figure). The $(x, y)$ coordinates of the centre of mass will be

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square of side $a$. Let the corners be $(0,0)$,$(a,0)$,$(a,a)$,and $(0,a)$ respectively. What should be the value of $m_4$ so that the centre of mass of the system is at the centre of the square,i.e.,at $(\frac{a}{2}, \frac{a}{2})$?

The distance between the carbon and oxygen atoms in a carbon monoxide molecule is $1.1 \, \mathring A$. Given that the mass of the carbon atom is $12 \, a.m.u.$ and the mass of the oxygen atom is $16 \, a.m.u.$,find the position of the center of mass of the carbon monoxide molecule.