Two bodies having masses `m_(1)` and `m_(2)` and velocities `v_(1)` and `v_(2)` colide and form a composite system. If `m_(1)v_(1) + m_(2)v_(2) = 0(m_(1) ne m_(2)`. The velocity of composite system will be

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 bodies P and Q of masses m_(1) and m_(2) (m_(2) gt m_(1)) are moving with velocity v_(1) and v_(2) force exerted by P on Q during the collision is

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 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.