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.

Answer

(c) `(1)/(2)(m_(1)+m_(2))V_(0)^(2)`