What is the initial angular momentum of a rigid body given an offset impulsed force? What is the imparted angular momentum to a rigid body if the impulse force is offset by a distance $h$ from the center of mass and the imparted momentum from the center of mass is $mv$?
For a homogeneous sphere I said the imparted angular momentum is $L=mvh$, but I am not sure if that is correct. 
 A: In general if the imparted momentum vector $\vec{J}$ goes through a point $\vec{r}$ relative to the center of mass then the change in speed of the center of mass is
$$ \begin{aligned}
  \Delta \vec{v} &= \frac{1}{m} \vec{J} \\
  \Delta \vec{\omega} & = I^{-1} (\vec{r} \times \vec{J})
\end{aligned} $$
where $\times$ is the vector cross product. In 2D if the impact momentum is $J$ at a distance $h$ from the center of mass, then angular momentum is $\vec{r} \times \vec{J} = (0,0,J h)$
The change in speed of the point of impact A is thus
$$ \begin{aligned}
  \Delta \vec{v}_A &= \Delta \vec{v} -\vec{r} \times \Delta \vec{\omega}  \\
& = \frac{1}{m}\vec{J} - \vec{r} \times I^{-1} (\vec{r} \times \vec{J}) 
\end{aligned}$$
Making this into a 2D problem with $\vec{v}_A =(0,v_{impact},0)$, $\vec{J}=(0,J,0)$ and $\vec{r}=(h,0,0)$ you have
$$ \left. v_{impact} = \frac{J}{m} + \frac{J h^2}{I} \right\} J = \frac{1}{\frac{1}{m}+
\frac{h^2}{I}} v_{impact} $$
So the reduced mass of the system is $J=m_{reduced} v_{impact}$ with $m_{reduced} = \frac{1}{\frac{1}{m}+
\frac{h^2}{I}} $
A lot of insight comes from transforming the problem from a rigid body impact to a equivalent particle impact with reduced mass.
A: When you have an impulse $F\Delta t$ (I prefer that notation over $m\Delta v$ because it allows impulse to be imparted without worrying about the mass of the thing giving the impulse), then


*

*The momentum of the center of mass changes as though the impulse was applied there, so $$m\Delta v = F\Delta t$$

*The angular momentum changes according to the torque imparted $$\Delta \vec{L} = \vec{F}\Delta t \times \vec{h} $$


So yes, you got it right.
