In the reference frame K two particles travel along the x axis, one of mass `m_1` with velocity `v_1`, and the other of mass `m_2` with velocity `v_2`. Find:
(a) the velocity V of the reference frame `K^'` in which the cumulative kinetic energy of these particles is minimum,
(b) the cumulative kinetic energy of these particles in the `K^'` frame.

Explanation

(a) Total kinetic energy in frame `K^'` is
`T=1/2m_1(vecv_1-vecV)^2+1/2m_2(vecv_2-vecV)^2`
This is minimum with respect to variation in `vecV`, when
`(deltaT^')/(deltavecV)=0`, i.e. `m_1(vecv_1-vecV)^2+m_2(vecv_2-vecV)=0`
or `vecV=(m_1vecv_1+m_2vecV_2)/(m_1+m_2)=vecv_c`
Hence, it is the frame of C.M. in which kinetic energy of a system is minimum.
(b) Linear momentum of the particle 1 in the `K^'` or C frame
`overset~vecp_1=m_1(vecv_1-vecv_c)=(m_1m_2)/(m_1+m_2)(vecv_1-vecv_2)`
or, `overset~vecp_1=mu(vecv_1-vecv_2)`, where, `mu=(m_1m_2)/(m_1+m_2)`= reduced mass
Similarly, `overset~vecp_2=mu(vecv_2-vecv_1)`
So, `|overset~vecp_1|=|overset~vecp_2|=overset~p=muv_(rel)` where, `v_(rel)=|vecv_1-vecv_2|` (3)
Now the total kinetic energy of the system in the C frame is
`overset~T=overset~T_1+overset~T_2=(overset~p^2)/(2m_1)+(overset~p^2)/(2m_2)=(overset~p^2)/(2mu)`
Hence `overset~T=1/2muv_(rel)^2=1/2mu|vecv_1-vecv_2|^2`