Before you try to analyze a situation you have to think of the appropriate assumptions that would allow you do idealize a situation and make it solvable, or at least describable.
When it comes to collisions there are three main assumptions that are typically used:
The collision happens near instantaneously. This means we can talk about before and after a collision, but any finite force (such as gravity) acting on the bodies will have no effect on their motion during the collision. Only the collision forces matter, which are said to happen for an infinitely short time and infinite large magnitude. But the end result is a step change in velocity on each body at the moment of impact.
Total momentum before and after the collision is preserved as no external forces act during the collision (see #1). This is handled with an exchange of momentum between the bodies, such that however much momentum is reduced on one body is added to the other body. The unit of currency of momentum that is exchanged is called an impulse. This impulse is a vector and acts in an equal and opposite fashion between the colliding bodies (Newton's 3rd law). Typically the direction of the impulse is known, but the magnitude isn't.
The relative speed of the two bodies after the collision is a fraction of the relative speed of the two bodies before the collision. The proportionality ratio is called the "coefficient of restitution". Also, the relative velocity before acts in the opposite sense from the relative velocity before.
If finite friction is to be considered then an additional assumption is needed
- The total impulse has two components, one along the normal direction of the contact and one tangent to the contact due to friction. Dynamic friction between the bodies limits the magnitude of the tangent component proportionally to the normal component.
So consider an example with a general object that is rotating hits the ground, with the coefficient of friction $\mu$. In the diagram below I have separated the object from the ground for clarity but it is assumed they are in contact at this moment. The location of the contact point A from the center of mass is $\vec{r}_A$
Before the contact, the motion of the center of mass is described by the velocity vector $\vec{v}$ and rotational velocity $\vec{\omega}$. The body has mass $m$ and mass moment of inertia tensor ${\rm I}$.
At the point of contact A the velocity of the object is
$$\vec{v}_A = \vec{v} + \vec{\omega} \times \vec{r} \tag{1} $$
where $\times$ is the vector cross product.
At the moment of impact, the contact normal direction is $\hat{n}$ and the contact tangent direction is $\hat{t}$. For this example, these are perpendicular and parallel to the floor.
This means that the impact speed along the contact normal is
$$v_{\rm imp} = \hat{n} \cdot \vec{v}_A \tag{2}$$
where $\cdot$ is the vector dot product.
As a result of the impact, a normal impulse $J_n$ and a tangential impulse $J_t$ develop at the point of contact and act in an equal and opposite fashion on the body and the floor. Thus the total impulse acting on the body is the combination of those two impulses.
$$ \vec{J} = J_n \hat{n} + J_t \hat{t} \tag{3}$$
We will find the impulse magnitudes later, but the important point here is that once $\vec{J}$ is known, then the response of the object to the collision is known. Each motion vector will suddenly change values according to the following changes
$$ \begin{aligned}
\Delta \vec{v} &= \tfrac{1}{m} \vec{J} \\
\Delta \vec{\omega} &= {\rm I}^{-1} (\vec{r}\times \vec{J})
\end{aligned} \tag{4}$$
This means the rotational velocity after the impact would be $\vec{\omega} + \Delta \vec{\omega}$ for example.
This also means the change in velocity of the body at the point of contact A is
$$ \Delta \vec{v}_A = \Delta \vec{v} + \Delta \vec{\omega} \times \vec{r} \tag{5}$$
which is the same velocity transformation as (1), but after the collision.
Now how to calculate the impulse component magnitudes $J_n$ and $J_t$?
Use the law of contact to project the impact velocity along the contact normal
$$ \underbrace{ \hat{n} \cdot ( \vec{v}_A + \Delta \vec{v}_A)}_\text{bounce} =- \epsilon\; \underbrace{\hat{n} \cdot \vec{v}_A}_\text{impact} = \tag{6}$$
where $\epsilon$ is the coefficient of restitution.
The above can be re-arranged as $\hat{n} \cdot \Delta \vec{v}_A =-(1+\epsilon)\,v_{\rm imp}$ and use (5) to relate the change in velocity at the point of contact to the change in velocity at the center of mass. This is $\hat{n} \cdot (\Delta\vec{v}-\vec{r}\times\Delta\vec{\omega}) =-(1+\epsilon)\,v_{\rm imp}$ and now use (4) to relate to the impulse magnitudes
$$ \hat{n}\cdot\left(\tfrac{1}{m}\vec{J}-\vec{r}\times{\rm I}^{-1}\left(\vec{r}\times\vec{J}\right)\right)=-(1+\epsilon)\,v_{{\rm imp}} \tag{7a}$$
The nature of friction is such that the surfaces either stick together or slide apart. This means that along the contact tangent, the effective coefficient of restitution is zero. This is not always the case, as with super bouncy balls, which allow bouncing along the tangential direction as well as the normal direction. But for this example, I am ignoring this and formulating an expression similar to (7) but projected along the tangential direction
$$ \hat{t}\cdot\left(\tfrac{1}{m}\vec{J}-\vec{r}\times{\rm I}^{-1}\left(\vec{r}\times\vec{J}\right)\right)=-(1+\epsilon)\left(\hat{t}\cdot\vec{v}_{A}\right) \tag{7b}$$
Considering the directions $\hat{n}$ and $\hat{t}$ are unit vectors and perpendicular to each other when the impulse vector $\vec{J}$ is decomposed using (3) some simplifications occur to arrive at the following 2×2 system of equations in terms of $J_n$ and $J_t$.
$$ \begin{aligned}\left[\tfrac{1}{m}+\left(\hat{n}\times\vec{r}\right)\cdot{\rm I}^{-1}\left(\hat{n}\times\vec{r}\right)\right]J_{n}+\left[\left(\hat{n}\times\vec{r}\right)\cdot{\rm I}^{-1}\left(\hat{t}\times\vec{r}\right)\right]J_{t} & =-(1+\epsilon)\,\left(\hat{n}\cdot\vec{v}_{A}\right)\\
\left[\left(\hat{t}\times\vec{r}\right)\cdot{\rm I}^{-1}\left(\hat{n}\times\vec{r}\right)\right]J_{n}+\left[\tfrac{1}{m}+\left(\hat{t}\times\vec{r}\right)\cdot{\rm I}^{-1}\left(\hat{t}\times\vec{r}\right)\right]J_{t} & =-\left(\hat{t}\cdot\vec{v}_{A}\right)
\end{aligned} \tag{8} $$
If it so happens that $\vec{r}$ and $\hat{n}$ are parallel, then the above simplifies to
$$\begin{aligned}\left[\tfrac{1}{m}\right]J_{n} & =-(1+\epsilon)\,\left(\hat{n}\cdot\vec{v}_{A}\right)\\
\left[\tfrac{1}{m}+\left(\hat{t}\times\vec{r}\right)\cdot{\rm I}^{-1}\left(\hat{t}\times\vec{r}\right)\right]J_{t} & =-\left(\hat{t}\cdot\vec{v}_{A}\right)
\end{aligned}$$
which is directly solvable.
Solve the system to get $J_n$ and $J_t$ assuming the contact sticks. But if finite friction is available then $| J_t | \leq \mu | J_n |$ which means $J_t$ needs to be capped at this value
$$ J_t = {\rm sign}(J_t) \; {\rm arg min}( | J_t|, \; \mu |J_n| ) \tag{9}$$
where ${\rm sign}(x)$ returns -1 or +1 depending if the argument $x$ is negative or positive, and ${\rm argmin}(x,y)$ return the minimum of $x$ or $y$.
Once the tangential impulse is capped (if needed) then use (3) to find the change in motion
$$ \begin{aligned}\Delta\vec{v} & =\tfrac{1}{m}\left(J_{n}\hat{n}+J_{t}\hat{t}\right)\\
\Delta\vec{\omega} & ={\rm I}^{-1}\left(\vec{r}\times\left(J_{n}\hat{n}+J_{t}\hat{t}\right)\right)
\end{aligned} \tag{10}$$
Note that the above analysis is entirely impulse-based, and forces or time do not play a role here. But if you had an estimate for the time slice $\Delta t$ in which the collision occurs, then you can do a very crude approximation of the forces involved.
Assuming the contact force is almost proportional to the deformation, then the shape of the force is going to be that of half a cosine with time
$$ F = F_{\rm max} \cos \left( \frac{ \pi t}{\Delta t} \right) $$
The above function provides a positive force from the time between $-\Delta t/2 $ to $\Delta t$.
The impulse along the contact normal can be equated to the area under this curve
$$ J_n = \int_{-\Delta t/2}^{\Delta t/2} F_{\rm max} \cos \left( \frac{ \pi t}{\Delta t} \right) \,{\rm d}t = F_{\rm max} \frac{2 \Delta t}{\pi} $$
In reverse, the estimated peak force is then
$$ F_{\rm max} = \frac{\pi J_n}{2 \Delta t} \tag{11}$$
Appendix I
For the 2D case where $\vec{r}=(r_x,\,r_y)$, $\hat{n} = (n_x,\,n_y)$ and $\hat{t} = (-n_y,\,n_n)$ the 2×2 system in (8) is states in matrix form as
$$\small \begin{vmatrix}
\tfrac{1}{m} + \tfrac{(n_x r_y - n_y r_x)^2}{I} & \tfrac{ (n_x r_x + n_y r_y) ( n_y r_x - n_x r_y)}{I} \\
\tfrac{ (n_x r_x + n_y r_y) ( n_y r_x - n_x r_y)}{I} & \tfrac{1}{m} + \tfrac{(n_x r_x + n_y r_y)^2}{I} \end{vmatrix}
\begin{vmatrix} J_n \\ J_t \end{vmatrix} = - \begin{vmatrix} (1+\epsilon)( n_y (v_y + \omega r_x)-n_x ( -v_x + \omega r_y)\\
n_x ( v_y + \omega r_x) + n_y (-v_x + \omega r_y) \end{vmatrix} $$
where $\vec{v} = (v_x,\,v_y)$
