Two masses, `m_(1)` and `m_(2)` , are moving with velocities `v_(1)` and `v_(2)`. Find their total kinetic energy in the reference frame of centre of mass.

Two bodies of masses m_(1) and m_(2) are moving with velocity v_(1) and v_(2) respectively in the same direction. The total momentum of the system in the frame of reference attached to the centre of mass I (v is relative velocity between the masses)

Two bodies of equal masses are moving with uniform velocities v and 2v. Find the ratio of their kinetic energies

Two particles having masses m_(1) and m_(2) are moving with velocities vec(V)_(1) and vec(V)_(2) respectively. vec(V)_(0) is velocity of centre of mass of the system. (a) Prove that the kinetic energy of the system in a reference frame attached to the centre of mass of the system is KE_(cm) = (1)/(2)mu V_(rel)^(2) . Where mu=(m_(1)m_(2))/(m_(1)+m_(2)) and V_(rel) is the relative speed of the two particles. (b) Prove that the kinetic energy of the system in ground frame is given by KE=KE_(cm)+(1)/(2)(m_(1)+m_(2))V_(0)^(2) (c) If the two particles collide head on find the minimum kinetic energy that the system has during collision.

Two objects having same mass 'm' are moving with velocities v and 2v. Find ratio of their kinetic energies.

Two objects having same mass 'm' are moving with velocities v and 2v. Find ratio of their kinetic energies.

Two bodies of masses m_1 and m_2 are moving with velocity v_1 and v_2 respectively in the same direction. The total momentum of the system in the frame of reference attached to the centre of mass is ( v is relative velocity between the masses)