Assuming it is an elastic collision, you will have, right after it,
a ball with the same rolling motion (anticlockwise) but translating in the opposite direction at $v_0$. Now there is friction because the ball is "sliding", and the new equilibrium movement will be when both are moving without sliding again at $v_k$. Let me know if you do not know how to solve this last problem.
In short, the effect of the collision is only to change the direction of $v_0$. The friction between the wall and the ball should have no effect at all (as the collision is infinitesimally short)
UPDATE:
I assume that the ball reverses $v_0$ and friction starts until the ball stops sliding. Thus the final angular speed will be $v_f/r$.
We have
$v_f=v_0-at_s=v_0-\mu gt_s$ (1)
where $t_s$ is the time it takes to stop sliding. You can obtain $t_s$ from the torque:
$\tau.t_s=\Delta L=\frac{2}{5} m r^2(\frac{v_f}{r}+\frac{v_0}{r})=\mu mgrt_s$ (2)
from here you obtain $t_s$ and replace in (1) to obtain:
$v_f=\frac{3}{7}v_0$
UPDATE 2:
if we accept the explanation that the ball will roll up until it stops sliding, reaching $w_2$, then we need to change, in eq. (2), the initial angular speed in the previous solution from $\frac{v_0}{r}$ to $\frac{2v_0}{7r}$.
In such a case we get:
$v_f=\frac{31}{49}v_0$
Exert a constant force $F$ pointing from the origin in $(x,y)$ to $(x_1,y_1)$ at angle $\theta$, for a period of time $\Delta t$, so that:
$$F=ma,$$
and:
$$v_1=\frac{F}{m}\Delta t$$
Choose the angle $\theta$ so that $\vec{v_1}$ is tangential at $(x_1,y_1)$ to your trajectory $y=f(x)$ (I assume other forces will then take over, hence the curvature).
If it makes things easier, you can later skip from one coordinate system to the other, e.g.:
$$x'=x-x_1$$
$$y'=y-y_1$$
Edit: (in response to OP's edit to the question)
The initial velocity and angle of launch of the ball could be simply modelled as follows, see diagram below:
Some assumptions need to be made:
- The kicker hits the ball at height above the horizontal $h (<R)$ and the speed vector $\vec{v}$ of the shoe is purely horizontal with magnitude $v$.
- No momentum is lost by the shoe: we consider it much more massive than the ball itself.
- The collision is free of friction.
Due to assumption 2. the situation now becomes equivalent (simply change frame of reference) to the ball falling onto a stationary shoe with speed $v_b$ and rebounding.
$v_b$ is the radial component of $\vec{v}$ and is simply (in scalar form):
$$v_b=v\cos\theta,$$
and from simple trigonometry we can derive:
$$\tan\theta=\frac{R-h}{\sqrt{R^2-(R-h)^2}}.$$
If a restitution coefficient $r$ needs to be taken into account, then:
$$v_b=vr\cos\theta$$
Note also that with $y=f(x)$ the trajectory's function, then:
$$\tan\theta=\bigg(\frac{dy}{dx}\bigg)_{x_1}=y'(x_1).$$
Best Answer
Given the mass m of the ball, the incident normal speed v, and the coefficient of restitution $\rho$, Then the integral of F over the duration of the collision $\Delta t$ is $$\int_0^{\Delta t} F dt = \frac{m(1 + \rho)v }{\Delta t}$$ assuning no rotational effects are incurred.
This follows from the fact that at any instant the acceleration of the ball away from the wall is F/m, so the integrated acceleration over the duration of the collision is the total change in normal velocity of the ball. Since the post-bounce velocity is simply the approach velocity times the coefficient of restitution, the total velocity change is as indicated.
If you want to simplify the force profile to assume a constant force FB during the collision (which is clearly not accurate) then $$\frac{(FB)(\Delta t)}{m} = (1 + \rho)v $$
The collision clearly does not produce a uniform force level over the duration of the collision, because the elastic forces on the ball will vary with the amount of deformation of the ball.