The angular momentum acquired after the impact is a result of the impulse exchange $J$ at a distance $c = d \frac{m_1}{m_1+m_2}$ from the center of mass (distance between point of impact and COM). So the equation for the change in rotation is

$$ \Delta \omega = \mathrm{I}_C^{-1} c J $$ where $\mathrm{I}_C = \left( \frac{m_1 m_2}{m_1 + m_2} \right) d^2$ is the mass moment of inertia of the system of two masses. This simplifies to $$ \Delta \omega = \frac{1}{m_2 d} J $$

Now you need to find the impulse $J$ (linear momentum exchange). One easy way of doing this is finding the reduced mass $\mu$ of the exchange and stating the law of impact $ J = (1+\epsilon) \mu\, v_0$ with a coefficient of restitution $\epsilon =0$.

The reduced mass between a particle with mass $m_0$ and a rigid body with properties stated above is

$$ \mu = \frac{1}{ \frac{1}{m_0} + \frac{c^2}{\mathrm{I}_C} } = \frac{1}{ \frac{1}{m_0} + \frac{m_1}{m_2 (m_1+m_2)} } = \frac{m_0 m_2 (m_1+m_2)}{m_0 m_1 + m_2 (m_1 + m_2)} $$

Putting together the impulse is $$ J = \frac{m_0 m_2 (m_1+m_2)}{m_0 m_1 + m_2 (m_1 + m_2)} v_0 $$ and thus the change in rotational speed

$$ \boxed{ \Delta \omega = \frac{m_0 (m_1+m_2)}{m_0 m_1 + m_2 (m_1 + m_2)} \frac{v_0}{d} } $$

And yes, the result does depend on $m_0$.

_{Link to another similar post with the same method.}

You are correct.

The angular momentum acquired after the impact is a result of the impulse exchange $J$ at a distance $c = d \frac{m_1}{m_1+m_2}$ from the center of mass (distance between point of impact and COM). So the equation for the change in rotation is

$$ \Delta \omega = {I}_C^{-1} c J $$ where ${I}_C = \left( \frac{m_1 m_2}{m_1 + m_2} \right) d^2$ is the mass moment of inertia of the system of two masses. This simplifies to $$ \Delta \omega = \frac{1}{m_2 d} J $$

Now you need to find the impulse $J$ (linear momentum exchange). One easy way of doing this is finding the reduced mass $\mu$ of the exchange and stating the law of impact $ J = (1+\epsilon) \mu\, v_0$ with a coefficient of restitution $\epsilon =0$.

The reduced mass between a particle with mass $m_0$ and a rigid body with properties stated above is

$$ \mu = \frac{1}{ \frac{1}{m_0} + \frac{1}{m_1+m_2} + \frac{c^2}{\mathrm{I}_C} } = \frac{m_0 m_2}{m_0+m_2} $$

Putting together the impulse is $$ J = \frac{m_0 m_2}{m_0 + m_2} v_0 $$ and thus the change in rotational speed

$$ \boxed{ \Delta \omega = \frac{m_0}{m_0 + m_2 } \frac{v_0}{d} } $$

And yes, the result does not depend on $m_1$. This is a direct result of $m_1$ simplifying out of the reduced mass calculation. This is probably because the resulting center of rotation is at $m_1$ which means it has zero motion and zero contribution to the kinetic energy.

The relationship between the impact axis (movement line of $m_0$) and the instant pivot point is called the pole-polar relationship. The location of the pole is $$ \ell = c + \frac{I_C}{(m_1+m_2) c} = \frac{m_1}{m_1+m_2} d + \frac{ \frac{m_1 m_2}{m_1+m_2} d^2 }{ (m_1+m_2) \frac{m_1}{m_1+m_2} d } = d$$

This confirms that the impact point is on the percussion axis when pivoting about $m_1$.

_{Link to another similar post with the same method.}