A thin ring of radius is rolling without slipping on the horizontal surface AB with velocity. Then the ring strikes the inclined surface BC and starts rolling up the plane. What fraction of initial energy is lost in the process ?

Angular momentum of the ring about the point of impact P is conserved.Just after the impactLPf  =  mVR  +  mR2  VR  =  2mVRJust before the impact, LPi  =  mV0 Rcosϕ +  mR2.  V0R∴    2mVR =  mV0R cosϕ +  mV0R⇒   V=V02   1+ cosϕInitial Kinetic energy,Ki =  12  mV02  +  12  mR2  V0R2  =  mV02Final Kinetic energy,Kf =  12  mV 2  +  12  mR2  VR2  =  mV 2 =  14  mV02   1+  cos6002=  916   mV02∴   Fraction of initial energy lost due to impact=  Ki −  KfKi  = 1−  916  =  716