Case II: when the target particle is at rest i,e u_{2}=0
From equation (8) and (9)
Hence some part of the KE which is transformed into second particle would be
Therefore after collisom first particle moving with initial velocity u_{1} would come to rest and the second particle which was at rest would start moving with the velocity of first particle.Hence in this case when m_{1}=m_{2} transfer of energy is 100%.if m_{1} > m_{2} or
m_{1} < m_{2} ,then energy transformation is not 100%
Therefore when a heavy particle collide with a very light particle at rest ,then the heavy particle keeps on moving with the same velocity and the light particle come in motion with a velocity double that of heavy particle
Velocity of the center of mass would be
Intial velocity of the m_{1} w.r.t center of mass frame of refrence is
Similarly Intial velocity of m_{2} w.r.t center of mass frame of refrence is
So u_{2}^{'}=(m_{1}/m_{2})u_{1}^{'}
Since the collision is elastic,Kinetic energy will be conserved
From which |v_{1}^{'}|=|u_{1}^{'}| and |v_{2}^{'}|=|u_{2}^{'}|
hence after collison velocities of particles remain unchanged in center of mass frame of refrence.If the collision is one dimmension then because of the collsion direction of these would be opposite to that of their intial velocites