Spinning ball colliding inclined plane 
[![Solution][2]][2]
"A solid uniform ball of mass M and radius R collides elastically with a rough fixed with a rough inclined surface as shown. Velocity and angular velocity of the ball just before collision are $v_0$ and $\omega_0=\frac{5v_0}{R}$. Coefficient of friction is 0.75. Find the velocity and the angular velocity just after collision."
How do I approach this problem?
 A: You have three unknowns: two velocity components and the rotation rate.  So you need three equations to determine them.  Here are my recommendations:

*

*Conservation of energy (there is translational and rotational, and the total can't change.)


*Conservation of angular momentum about the contact point. (This one has not been mentioned in the other answers. There is angular momentum both before and after the collision due to both the CM motion and the rotation about the CM, and the total can't change because the contact force provides no torque about the contact point.)


*Normal force does not depend on friction force. Or in other words, the component of the impulse in a direction perpendicular to the ramp surface will be the same regardless of the rotation rate. This seems reasonable, and it is the assumption underlying @John Wick's Step 2. It implies that component of the linear momentum of the ball in the direction perpendicular to the ramp changes sign from before to after the collision, but keeps the same magnitude.  This seems reasonable, but it is more of an assumption than any kind of law--you could imagine friction mechanisms that would violate this assumption.
A: STEP 1 :  Make the components of the velocity, along the incline perpendicular to it.
those will be: along incline = $\frac{3v_{o}}{5}$ and perpendicular velocity: $\frac{4v_{o}}{5}$
STEP 2: Since it is colliding elastically, e=1, so velocity just after collision perpendicular to the incline will be $\frac{4v_{o}}{5}$ but opposite to the original direction
STEP 3 Change in momentum = $\left|2m\cdot\frac{4}{5}v_{o}\right|$ = $\int_{ }^{ }Ndt$
STEP 4 Now impulse due to friction (along incline, of course)  will be $μ\int_{ }^{ }Ndt$
Now try to solve!
A: Taking all the forces in the x direction:
$$\sum{F_{netx}}=mg\sin(37^\circ)-F_{friction}=\frac{3mg}{5}-\mu N=\frac{3mg}{5}-\frac{3mg}{5}=0$$
Hence momentum is conserved in the x direction.
$$\begin{bmatrix} -\frac{3v_0}{5} \cr -\frac{4v_0}{5} \end{bmatrix}\rightarrow\begin{bmatrix} -\frac{3v_0}{5} \cr \frac{4v_0}{5} \end{bmatrix}$$
Impulse along y axis;
$$N\Delta t=\Delta p=\frac{8mv_0}{5}\Rightarrow\Delta t=\frac{8mv_0}{5N}$$
Now for the angular velocity, about the centre of mass about the solid ball
$$\sum{\tau}=r\mu N=I\alpha \Rightarrow \alpha = -\frac{R \mu N}{I}$$
$$\Delta \omega=\alpha \Delta t=-\frac{R\mu 8mv_0}{5I}=-\frac{R\frac{3}{4} 8mv_0}{5\frac{2mR^2}{5}}=-\frac{3v_0}{R}$$
$$\Rightarrow \omega=\frac{2v_0}{R}$$

A: Assumptions:  If the collision is elastic then: (1/2)m${v_o}^2$ + (1/2)I${ω_o}^2$  =  (1/2)m${v_f}^2$ + (1/2)I${ω_f}^2$,  and the friction is entirely static.  The x axis is up the incline and y is up (but leaning from the vertical).    $ ω_o$ as a vector is toward -z (into the sketch). With static contact $v_{fx}$ = - R$ω_f $.  (Where R is the radius of the ball and I is its rotational inertia.)  The  contact time is short enough to ignore the effect of gravity.  The angular impulse (from friction) equals R times the linear impulse in the x direction:  I(Δω) = Rm(Δ$v_x$).
Expanding and solving the last equation gives:   $ω_f $ = [I${ω_o}$  - mR$v_{ox}$]/[I – m$R^2$].  Knowing  $ω_f $  lets you find  $v_{fx}$ and putting both in the original energy equation yields $v_{fy}$.
