I am currently studying for the GRE Physics subject test by working through published past tests. My question is about problem 44 from the test GR8677:
$44.$ A uniform stick of length $L$ and mass $M$ lies on a frictionless horizontal surface. A point particle of mass $m$ approaches the stick with speed $v$ on a straight line perpendicular to the stick that intersects the stick at one end, as shown above. After the collision, which is elastic, the particle is at rest. The speed $V$ of the center of mass of the stick after the collision is
(A) $\frac{m}{M}v$
(B) $\frac{m}{M+m}v$
(C) $\sqrt{\frac{m}{M}}v$
(D) $\sqrt{\frac{m}{M+m}}v$
(E) $\frac{3m}{M}v$
My approach was to first write down expressions for the conservation of energy, linear momentum, and angular momentum. The problem states that the particle is at rest after the collision, which simplifies the equations:
$$\frac{1}{2} m v^2 = \frac{1}{2} M V^2 + \frac{1}{2} I \omega^2$$
$$m v = M V$$
$$\frac{1}{2} L m v = I \omega$$
where $I=ML^2/12$ is the moment of inertia of the stick about its CM and $\omega$ is the stick's angular velocity. The most natural next step is to solve the linear momentum equation, giving us the correct answer (A). This is the solution used here and here.
However, adhering to my penchant for valuing understanding above efficiency, I attempted to verify this answer by combining the other two conservation equations. I solved the angular momentum equation for $\omega$ to obtain $$\omega = \frac{L m v}{2 I}.$$ I then solved the energy equation for $V$ and substituted in this result: $$V^2 = \frac{1}{M}(m v^2 - I \omega^2)$$ $$= \frac{1}{M}\left( m v^2 - I \left( \frac{L m v}{2 I} \right)^2 \right)$$ $$= \frac{1}{M}\left( m v^2 - \frac{(L m v)^2}{4 I} \right)$$ $$= \frac{m v^2}{M}\left( 1 - \frac{L^2 m}{4 (M L^2 / 12)} \right)$$ $$= \frac{m v^2}{M} \left( 1 - 3\frac{m}{M} \right)$$ $$\Longrightarrow V = v \sqrt{ \frac{m}{M} \left( 1 - 3\frac{m}{M} \right) }$$ $$= v \frac{m}{M} \sqrt{ \frac{M}{m} - 3 }$$
I see several problems with this result. First, it does not immediately resemble any of the answers, though it can be made to match either (A) or (E) with a choice of $M/m=4$ or $M/m=12$, respectively. Second, if $M/m < 3$, the velocity of the stick becomes imaginary which (to me) does not have an obvious physical interpretation. I also do not see why there should be any restriction on the mass ratio in the first place.
Is this approach illegitimate or does it contain an error? If so, why/where? If not, why does the result not coincide with the first approach?
In the first solution link above, there are many comments going back-and-forth on the reasons for the discrepancy but, frustratingly, no conclusion. Any insights would be greatly appreciated.
