Two particles of masses 1 kg and 2 kg are initially at rest at position vectors 2 i + 3 j m and - i + 3 j m respectively. They are subjected to constant external forces F ₁ = 2 i + 4 j + 6 k N (on the 1 kg particle) and F ₂ = 4 i - 4 j + 6 k N (on the 2 kg particle). The position vector of the centre of mass of the system at t = 2 s will be:

Step-by-step solution

First, calculate the initial position vector of the centre of mass at t = 0 : r _ com (0) = m₁ r ₁ + m₂ r ₂ m₁ + m₂ r _ com (0) = 1(2 i + 3 j ) + 2(- i + 3 j ) 1 + 2 r _ com (0) = 2 i + 3 j - 2 i + 6 j 3 = 9 j 3 = 3 j m Next, find the net external force acting on the system: F _ net = F ₁ + F ₂ = (2 i + 4 j + 6 k ) + (4 i - 4 j + 6 k ) = 6 i + 12 k N The total mass of the system is M = 1 + 2 = 3 kg . The acceleration of the centre of mass is: a _ com = F _ net M = 6 i + 12 k 3 = 2 i + 4 k m/s ^2 Since the particles