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 impact

${\left({\mathrm{L}}_{\mathrm{P}}\right)}_{\mathrm{f}}=\mathrm{mVR}+\left({\mathrm{mR}}^2\right)\left(\frac{\mathrm{V}}{\mathrm{R}}\right)=2\mathrm{mVR}$

Just before the impact, ${\left({\mathrm{L}}_{\mathrm{P}}\right)}_{\mathrm{i}}={\mathrm{mV}}_0\mathrm{Rcos\varphi }+{\mathrm{mR}}^2.\left(\frac{{\mathrm{V}}_0}{\mathrm{R}}\right)$

$\therefore 2\mathrm{mVR}={\mathrm{mV}}_0\mathrm{R}\mathrm{cos\varphi }+{\mathrm{mV}}_0\mathrm{R}$

$\Rightarrow \mathrm{V}=\frac{{\mathrm{V}}_0}{2}\left(1+\mathrm{cos\varphi }\right)$

Initial Kinetic energy,

$\mathrm{Ki}=\frac{1}{2}{\mathrm{mV}}_0^2+\frac{1}{2}\left({\mathrm{mR}}^2\right){\left(\frac{{\mathrm{V}}_0}{\mathrm{R}}\right)}^2={\mathrm{mV}}_0^2$

Final Kinetic energy,

${\mathrm{K}}_{\mathrm{f}}=\frac{1}{2}{\mathrm{mV}}_{ }^2+\frac{1}{2}\left({\mathrm{mR}}^2\right){\left(\frac{\mathrm{V}}{\mathrm{R}}\right)}^2={\mathrm{mV}}_{ }^2=\frac{1}{4}{\mathrm{mV}}_0^2{\left(1+\cos {60}^0\right)}^2$

$=\frac{9}{16}{\mathrm{mV}}_0^2$

$\therefore$   Fraction of initial energy lost due to impact

$=\frac{{\mathrm{K}}_{\mathrm{i}}-{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{i}}}=1-\frac{9}{16}=\frac{7}{16}$