At the moment of impact, this is the picture.
All vectors on body A are shown in a negative sense, and vectors on body B in a positive sense.
Conservation of momentum, linear and angular, is achieved by applying the same impulse $\color{magenta} J$ to both bodies in equal and opposite sense, on the same line of action.
As a result, the response is a change in velocity, both linear and angular, for both bodies $\color{blue}{\Delta v_A}$, $\color{blue}{\Delta \omega_A}$, $\color{blue}{\Delta v_B}$, and $\color{blue}{\Delta \omega_B}$
The four equations (two bodies, linear and angular) that describe the response are summarized below
$$ \left.\begin{aligned}m_{A}\Delta v_{A} & =-J\\
m_{B}\Delta v_{B} & =+J\\
I_{A}\Delta\omega_{A} & =-(r_{A}\cos\alpha)J\\
I_{B}\Delta\omega_{B} & =-(r_{B})J
\end{aligned}
\right\} \begin{aligned}\Delta v_{A} & =-\tfrac{1}{m_{A}}J\\
\Delta v_{B} & =+\tfrac{1}{m_{B}}J\\
\Delta\omega_{A} & =-\tfrac{1}{I_{A}}(r_{A}\cos\alpha)J\\
\Delta\omega_{B} & =-\tfrac{1}{I_{B}}(r_{B})J
\end{aligned}
\tag{1,2,3,4}$$
Yet the above is not sufficient to solve the problem. You need to describe what type of contact this is (plastic, elastic, fully elastic) and describe the law of collision relating to the relative speeds before and after impact.
$$ (v_A^\star + \Delta v_A^\star) – (v_B^\star + \Delta v_B^\star) = – \epsilon \, (v_A^\star – v_B^\star) \tag{5}$$
where $\epsilon = 0 \ldots 1$ is the coefficient of restitution, and the star superscript denotes velocities resolved at the point of contact and along the contact normal direction for each body
$$ \begin{aligned}
v_A^\star & = v_A + (r_A \cos \alpha) \omega_A \\
v_B^\star & = v_B – (r_B) \omega_B
\end{aligned} \tag{6,7}$$
and
$$ \begin{aligned}
\Delta v_A^\star & = \Delta v_A + (r_A \cos \alpha) \Delta \omega_A \\
\Delta v_B^\star & = \Delta v_B – (r_B) \Delta \omega_B
\end{aligned} \tag{8,9}$$
which makes equation (5) after some re-arranging equal to
$$\left(\Delta v_{A}+(r_{A}\cos\alpha)\Delta\omega_{A}\right)-\left(\Delta v_{B}-(r_{B})\Delta\omega_{B}\right)=-\left(1+\epsilon\right)\,v_{{\rm imp}} \tag{10}$$
where $$v_{{\rm imp}}=\left(v_{A}+(r_{A}\cos\alpha)\omega_{A}\right)-\left(v_{B}-(r_{B})\omega_{B}\right) \tag{11}$$ is the relative impact speed.
Now use equations (1),(2),(3) and (4) above to find a single equation in terms of the unknown impulse $J$
$$\boxed{\vphantom{\begin{array}{c}
\\
\\
\\
\end{array}}J=\frac{\left(1+\epsilon\right)\,v_{{\rm imp}}}{\tfrac{1}{m_{A}}+\tfrac{1}{I_{A}}(r_{A}\cos\alpha)^{2}+\tfrac{1}{m_{B}}+\tfrac{1}{I_{B}}(r_{B})^{2}}} \tag{12}$$
And once $J$ is calculated, then (1),(2),(3),(4) are used to find the response.
Note that the above is complicated and gets even worse with 2D contacts involving vectors. Worst yet are 3D contacts with inertia tensors and other complexities beyond this question’s scope.
But there is a general shortcut that can be applied in most cases
Notice the solutions has the form of $$ J = \left(1+\epsilon\right)\,m_{\rm eff}\,v_{{\rm imp}} $$ where $m_{\rm eff}$ is the effective reduced mass of the system.
Given two bodies A, B and their perpendicular distances to the impulse line of action from the respective centers of mass as $d_A$ and $d_B$, then the effective mass is evaluated as
$$ m_{\rm eff} = \frac{1}{ \tfrac{1}{m_A} + \tfrac{d_A^2}{I_A} + \tfrac{1}{m_B} + \tfrac{d_B^2}{I_B}} $$
you can see with the special case of two spheres colliding that have $d_A=0$ and $d_B=0$ the above is $m_{\rm eff} = \frac{1}{\tfrac{1}{m_A} + \tfrac{1}{m_B}} = \frac{m_A m_B}{m_A + m_B}$ which is a familiar expression for reduced mass.
To evaluate the effective mass you would feel if you hit the rod at the impact point, you consider only their respective parts.
$$\begin{aligned}
m_{\rm eff}^A & = \frac{1}{ \tfrac{1}{m_A} + \tfrac{d_A^2}{I_A} } \\
m_{\rm eff}^B & = \frac{1}{ \tfrac{1}{m_B} + \tfrac{d_B^2}{I_B} } \\
\end{aligned}$$
Notice that the values are maximum when the line of action goes through the center of mass and the perpendicular distances are zero. The further away the point of impact is, the less the effective mass is.
