Angular Velocity after a frictional impulse I am modelling 2D physics collision into simulations. In Physics for Game Programmers, Grant Palmer book,

the velocity Vn1 after collision is mentioned to be independent of the friction coeff. between the surface. 
for a sphere Vn1=5/7Vn0. But this holds to true only when the sphere is assumed to be  in pure rolling after the impact.
How can i determine the velocity of the ball for a sliding case as well. I have to use it in a physics simulation and the duration of impulse is not known. Is there any other way to determine it?
 A: For this post I'm using the subscripts $_0$, $_1$, $_p$, and $_n$ to denote pre-impact, post-impact, normal to the impact surface, and tangential to the impact surface as per the OP's diagrams.
Set Up
If the ball is spherical and of uniform density, $I=\frac25m\,r^2$
The force acting on the ball over time can be integrated into an impulse. The impulse must act at the point of contact which is $r$ away from the center of mass. The $p$ component will only affect the $p$ component of velocity and the maximum $n$ force through friction. The $n$ component will affect the $n$ velocity and the rotational velocity.
$$V_{n1}=V_{n0}+\frac{J_n}{m}$$
$$\omega_1=\omega_0+\frac{J_n\,r}{I}$$
(The sign on the last term depends on your choice of coordinate system)
Rolling Case
If the ball has enough friction that by the end of the impact it is rolling along the surface, $V_{n1}=-\omega_1\,r$
This would be the case when there are high friction surfaces and the ball is not very torsionally elastic. I would imagine this to be the case for basketballs. (Note though that as basketballs are hollow $I\approx\frac23m\,r^2$ so the calculations below would need to be redone for that value of $I$)
This is now a system of three equations with three unknowns.
Solving yields:
$$J_n=-\frac27 m(r\,\omega_0+V_{n0})$$
$$\omega_1=\frac{2\,\omega_0}7-\frac{5V_{n0}}{7r}$$
$$V_{n1}=-\frac{2r\,\omega_0}7+\frac{5V_{n0}}{7}$$
Which if you set $\omega_0=0$ yields $V_{n1}=\frac57V_{n0}$ just as stated in the problem statement. Note that there was no conservation of energy used as in this case the friction would consume some of the energy.
Sliding Impact
Now, if you want to modify this system to one where the impact finishes before the slip velocity becomes zero, then the ball would be slipping the entire duration of the impact, which means that $F_n=\mu\,F_p$ for the entire duration of impact so therefore $J_n=\mu\,J_p$ So now we have:
$$V_{n1}=V_{n0}+\frac{J_n}{m}$$
$$\omega_1=\omega_0+\frac{J_n\,r}{I}$$
$$V_{p1}=V_{p0}+\frac{J_p}{m}=-V_{p0}\,COR$$
$$J_n=\mu\,J_p$$
Where $COR$ is the coefficient of restitution for the velocity perpendicular to the surface.
Solving directly yields some long ugly equations, but if I were programming this, I would probably calculate values as follows:
$$maxFriction=\mu\,(COR+1)$$
$$V_{p1}=-V_{p0}\,COR$$
$$V_{slide}=r\,\omega_0+V_{n0}$$
$$\frac{J_n}{m}=-sign(V_{slide})\,min(maxFriction\,|V_{p0}|,\frac27 |V_{slide}|)$$
$$V_{n1}=V_{n0}+\frac{J_n}{m}$$
$$\omega_1=\omega_0+\frac52\frac{J_n}{m}$$
Torsionally Elastic
What happens when you get something like a super ball, or lacrosse ball, that can deform and store energy torsionally?
In this case the impulse can actually be stronger than that required to bring the sliding velocity to zero. This is similar to how in elastic collisions the difference in velocity between two objects isn't just brought to zero, it actually reverses direction. As it turns out, the maximum impulse that doesn't produce net energy, reverses the direction of the sliding velocity. However, I believe there is an interaction between the perpendicular velocity and how much the sliding velocity is changed. So all I can say for now is that $|J_n| < \frac47m(r\,\omega_0+V_{n0})$ and $|J_n| < \mu\,m\,(COR+1)V_{p0}$
A: $F_r = \mu N$
$\mu$ is the coefficient of friction.
N is the normal force ( the portion of the gravitational force perpendicular to the surface).
If it is partially slipping then the amount of work done is not F*length-of-slope - don't make this mistake.
A: This is a pretty long answer - skip to the bottom if you just want to see the solution.
It would appear that you are assuming the collision is sufficiently inelastic that the ball does not bounce off the surface, correct?  Seems a bit unusual for a ball to be inelastic enough to stick to a surface, but stiff enough to roll smoothly afterwards, but we'll go with it.
We can divide the initial kinetic energy into two parts:  $KE_p$ and $KE_n$  I'm using your dimension labeling system of "p" and "n".
Since the ball apparently loses all its velocity in the "p" direction, and I don't see evidence of any force partially in both the "p" and "n" directions that could facilitate a transfer of kinetic energy between the "p" and "n" directions, I am led to the conclusion that 100% of $KE_p$ must be dissipated as heat.
That leaves $KE_n$ to divide up between the final translational motion in the "n" direction and rotational motion.
I will take a moment to solve the case where the ball is rolling without slipping.  In that case, the total energy available is $KE_n$
$$KE_n = \frac{1}{2} m ~V_{n0}$$
That energy is distributed between translational kinetic energy, rotational kinetic energy, and possibly heat.
$$ KE_{n0} = KE_{n1} + KE_{rot} + Q$$
$$ \frac{1}{2} m ~V_{n0}^2 = \frac{1}{2} m ~V_{n1}^2 + \frac{1}{2} I ~\omega^2 + Q$$
For a uniform sphere, $I = \frac{2}{5} m r^2$.  If your ball begins rolling without slipping immediately upon impact, the rotational velocity and the translational velocity must be related via $\omega = \frac{V_{n1}}{r}$.  Substituting those in and simplifying, we get
$$ V_{n0}^2 = \frac{7}{5} V_{n1}^2 + Q$$
Now, if we assume (probably reasonably) that negligible heat is generated in the impact from the ball's initial velocity in the "n" direction, then we get
$$V_{n1} = \sqrt{\frac{5}{7}} V_{n0}$$
Note the square root!  It would appear that you or your source forgot a square root in there somewhere.
Now, your question was about the behavior if the frictional force is not strong enough to cause "pure rolling", or "rolling without slipping" as I was taught to call it.
When the ball first hits, it is not rotating at all.  Thus, in the first instant, it must be purely sliding.  In this case, there is no rotational energy, and the energy is divided up thusly:
$$ \frac{1}{2} m ~V_{n0}^2 = \frac{1}{2} m ~V_{n1}^2$$
leaving us with
$$V_{n1} = V_{n0}$$
Pretty simple answer.  Now, it's not entirely clear if you want this, but here's how the speed will evolve:
The forces in the "n" direction will be  $F_n = F_{gn} + F_f$, where F_{gn} is the component of the gravitational force in the "n" direction.  The frictional force must be $\mu F_N$ where $F_N$, the normal force, must be $m g cos(\theta)$.
$$F_n = - m g sin(\theta) - \mu m g cos(\theta)$$
Since this is a constant force, we can divide by mass, get the acceleration, and predict the translational velocity over time:
$$V_{n1}(t) = V_{n0} - (sin(\theta) + \mu cos(\theta)) g t$$
The rotational velocity can be gotten via the torque.  The only force that produces a torque is the frictional force, so
$$\tau = \mu m g cos(\theta) r$$
Which, if we assume the force of kinetic friction is constant, similarly allows us to find
$$\omega(t) = \frac{\tau}{I} t$$
$$\omega(t) = \frac{\mu m g cos(\theta) r}{\frac{2}{5} m r^2} t$$
$$\omega(t) = \frac{5}{2} \frac{\mu g cos(\theta)}{r} t$$
So my final answer is that $V_{n1}(t)$ is given by
$$V_{n1}(t) = V_{n0} - (sin(\theta) + \mu cos(\theta)) g t$$
Where $t=0$ is the time of the impact.  This answer is valid up to the point when the ball stops moving, or until the ball begins rolling without slipping (since static friction takes over at that point).
