Particles of masses 1kg and 2kg are at `(i+2j+3k)` m and `(-2i+4j+2k)m` then instantaneous position of their centre of mass is

Particles of masses 1 kg and 3 kg at (2i+5j+13 k) m and (-6i+4j-2k)m then instantaneous position of their centre of mass is

Two particles of masses 1kg and 2kg are located at x=0 and x=3m . Find the position of their centre of mass.

Two particles of masses 1kg and 2kg are located at x=0 and x=3m . Find the position of their centre of mass.

Particles of masses 1kg and 3kg are moving with the velocities (10i-7j-3k)ms^(-1) and (7i-9j+6k)ms^(-1) then the instantaneous velocity of their centre of mass is

The coordinates of a triangle ABC are A(1,2) , B(4,6) and C(-3,-2) . Three particles of masses 1kg , 2kg and m kg are placed the vertices of the triangle. If the coordinate of the centre of mass are ((3)/(5),2) , calculate the mass m .

Two bodies of mass 2 kg and 5 kg have position (1 m, 2 m, 1 m) and (3 m, 2 m, -1 m) respectively. The position vector of centre of mass is

The centre of mass of three particles of masses 1 kg, 2 kg and 3 kg is at (2,2, 2). The position of the fourth mass of 4 kg to be placed in the system as that the new centre of mass is at (0, 0, 0) is

Two particles of masses 1 kg and 3 kg have position vectors 2hati+3hatj+4hatk and-2hati+3hatj-4hatk respectively. The centre of mass has a position vector

Two particles of masses 1 kg and 3 kg have position vectors 2hati+3hatj+4hatk and-2hati+3hatj-4hatk respectively. The centre of mass has a position vector

Consider a system of two particles having masses 2kg and 5 kg the particle of mass 2 kg is pushed towards the centre of mass of particles through a distance 5 m,by what distance would particle of mass 5kg move so as to keep the centre of mass of particles at the original position?