A small disc A is placed on an inclined plane forming an angle $\mathrm{\alpha }$ as shown in figure and is imparted an initial velocity ${\mathrm{v}}_{\mathrm{o}}$. If the friction coefficient $\mathrm{\mu }=\tan \mathrm{\alpha }$ and at initial moment $\mathrm{\varphi }=\frac{\mathrm{\pi }}{2}$. Then

${\mathrm{a}}_{\mathrm{t}}=\frac{\mathrm{mg}\mathrm{sin\alpha cos\varphi }-\mathrm{\mu mgcos\alpha }}{\mathrm{m}}=\mathrm{gsin\alpha }[\mathrm{cos\varphi }-1]$

${\mathrm{a}}_{\mathrm{x}}=\frac{-\left(\mathrm{\mu mg}\mathrm{cos\alpha }\right)\mathrm{cos\varphi }+\mathrm{mgsin\alpha }}{\mathrm{m}}=\mathrm{gsin\alpha }[1-\mathrm{cos\varphi }]$
Hence change in velocity along tangential and x direction will be same
Hence, ${\mathrm{v}}_{\mathrm{x}}=\mathrm{v}+\mathrm{c}$
Initially, ${\mathrm{v}}_{\mathrm{x}}=0,\mathrm{v}={\mathrm{v}}_0$
We get $\mathrm{c}=-{\mathrm{v}}_0$
Hence ${\mathrm{v}}_{\mathrm{x}}=\mathrm{v}-{\mathrm{v}}_0$
$$\begin{gathered} {\text{v}}_0\mathrm{cos\alpha }=\mathrm{v}-{\mathrm{v}}_0 \\ {\mathrm{v}}_0=\frac{\mathrm{v}}{1+\mathrm{cos\alpha }} \end{gathered}$$