A small, uniform disk of mass $m_1=2kg$ travels with a velocity $v_{1i}=3m s^{−1}$ on a frictionless surface in the xy-plane. It strikes the end of a flat, narrow, uniform rod of length, $l=4m$ and mass $m_2=1kg$. The moment of inertia of the rod about its centre of mass, $I_2=ml^2/12$ The collision is elastic and the disk leaves the collision with velocity $v_{1f}=2.35ms^{−1}$ . By considering the conservation of momentum and the conservation of angular momentum in the collision, find the translational and angular velocity of the rod after the collision. [The disk can be considered as a point particle.]
What I did was say $m(v_{1i}-v_{1f})r=I\omega$ and solve for $\omega$, which gave me $1.95s^{-1}$
To work out the final velocity I just said $mv_{1i}=mv_{1f}+mv_{2f}$ and solve for $v_{2f}$. which gave me $1.3ms^{-1}$
To check my answer, I checked if energy was conserved, I found that it wasn't.
$\frac12mv_{1i}^2=\frac12I\omega^2+\frac12mv_{2f}^2+\frac12mv_{1f}^2$
I got $LHS=9J$ and $RHS=8.9J$, surely these should be equal, as I didn't round at all?
Just to sum up, we have the following system:
\begin{equation} \begin{cases} m_1(v_{1i}-v_{1f})\frac l2=I\omega\\m_1v_{1i}=m_1v_{1f}+m_2v_{2fG}\\\frac12m_1v_{1i}^2=\frac12I\omega^2+\frac12m_2v_{2fG}^2+\frac12m_1v_{1f}^2 \end{cases} \end{equation}
where $G$ is the center of mass of the rod. In this system we have two variables: $\omega,v_{2fG}$ that they can be obtained by the first two equations; the third equation can be satisfied or not. I think that this system is impossible with the data of the problem, so there must be something wrong in the formulation.
Please check it.