A small disc P is placed on an inclined plane forming an angle $\theta$ with the horizontal and imparted an initial velocity $v_0$. Find how the velocity of disc depends on the angle $\varphi$  which its velocity vector makes with x axis (see figure). The coefficient of friction is $\mu =\tan \theta$ and   ${\varphi }_{initial}=\frac{\pi }{2}$.
 

Normal reaction on the particle is always $N=mg \cos \theta$  

Friction force on the particle is $f=\mu mg\cos \theta$ and is always opposite of instantaneous velocity. Another force on the particle in the plane of the incline is $mg\sin \theta$ which is always directed along x direction. Force on the particle in tangential direction ( i.e., in direction of its velocity) is 
$F_t=mg\sin \theta .\cos \varphi -\mu mg\cos \theta$
Acceleration in tangential direction is 
$a_t=g\sin \theta (\cos \varphi -1)[\because \mu =\tan \theta ]$    ………….(a)
Similarly, acceleration of disc along x-direction is 
$a_x=g\sin \theta -\left(\mu g\cos \theta \right)\cos \varphi$
$=g\sin \theta (1-\cos \varphi )$   ………..(b)
$\therefore a_t+a_x=0$                      [Adding (a) and (b) ]
Integrating, we get
$v+v_x=c$                           [c is constant]
$But{v}_x=v\cos \varphi$ 
$\therefore v+v\cos \varphi =c$
Initially  $\varphi =\frac{\pi }{2};v=v_0$
$\begin{array}{l}\therefore c=v_0\\ \therefore v=\frac{v_0}{1+\cos \varphi }\end{array}$