Scenario: a ball of mass $m_B$ collides elastically collides with a door. The door is a uniform rectangle of mass $m_D$, moment of inertia $I$, length $R$, and height $h$, and it is free to rotate about an axis through its left edge. The ball impacts the door perpendicularly with an initial velocity $v_i$, and bounces away from the door with a final velocity $v_f$ that is also perpendicular to the door. The point of impact is a distance $d$ away from the left edge of the door. The collision imparts the door with an angular velocity $\omega$. In terms of the initial conditions, what are $v_f$ and $\omega$?
My approach: I wrote a conservation of energy equation and a conservation of momentum equation, which together would let me solve for the two unknowns. Since I was presented this problem before I knew about angular momentum, I tried to linearize the angular momentum of the door instead of trying to angularize the linear momentum of the ball. To do this, I reasoned that, at the moment of collision, I each vertical strip of the door would gain an instantaneous tangential velocity $v_T$, which has the same direction as $v_i$. If $r$ is the distance from the strip to the left edge, then we would have $v_T=\omega r$. To get a momentum from this, I would just have to multiply $v_T$ times the mass of the strip. Since the door is uniform, this mass would be the mass density of the door, $\frac{m_D}{Rh}$, times the the area of the strip, $A$. Integrating over all strips, I found that the total linear momentum of the door was $\int\limits_0^R\left(\omega r\right)\left(\frac{m_D}{Rh}\cdot h \; \mathrm{d}r\right)$. So in all I got that
\begin{align*} \text{Conservation of Energy: } \frac{1}{2}m_Bv_i^2 &= \frac{1}{2}m_Bv_f^2 + \frac{1}{2}\omega^2I \\ \text{Conservation of Momentum: } m_Bv_i &= m_Bv_f + \int\limits_0^R\left(\omega r\right)\left(\frac{m_D}{Rh} \; \mathrm{d}A\right) = m_Bv_f + \int\limits_0^R\left(\omega r\right)\left(\frac{m_D}{Rh}\cdot h \; \mathrm{d}r\right)\end{align*}
but this is clearly wrong since it implies that $v_f$ and $\omega$ don't depend on $d$, which they obviously do.
After learning a bit more about angular momentum I rewrote the second equation as $m_bv_id=m_bv_fd+\omega I$, which after solving gave me a completely reasonable answer, but I'm still not sure why the my first approach isn't valid. Is there a way to linearize the door's angular momentum in a way that depends on $d$, or is it necessarily a hopeless endeavor?
Let me know if the question is unclear, and thank you for your responses.
