newtonian mechanics - Violation of energy conservation during collisions of a particle with different sections of a rod

Imagine a homogenous rod with total mass $M$ and length $l$ floating in free space without any force or constraint acting on it. Then, think about two possible scenarios. In the first, a particle with lineal momentum $mv$ impacts right at the center of the rod. In the second, the same particle with same lineal momentum impacts right at one of the ends of the rod.

First case

In the first case there will be no rotation, only translation. We apply conservation of momentum (where I think there are two cases, one with the particle having final lineal momentum $v_f$ and the other with $-v_f$, but i will omit the first one).

$$mv = MV - mv_f \quad \quad V= \frac{m}{M}(v+v_f)$$

Then the final energy is:

$$E_f = \frac{1}{2} MV^2 + \frac{1}{2} mv_f^2 = \frac{m^2}{2M} (v+v_f)^2 + \frac{1}{2} mv_f^2$$

Second case

In the second case there will only be rotation (or so I think, but I'm getting really confused so that's why I've come up with this scenarios to try to prove it) without translation. Again we apply conservation of momentum:

$$mv = MV - mv_f \quad \quad V= \frac{m}{M}(v+v_f)$$

Let me explain what $V$ represents in this case. Since the rod rotates all the points in it don't have the same velocity, but, since the velocity increases linearly with the radius and the rod is homogenous, if the velocity of the ends is $2V$ (which I compell it to be this way) then the average speed of thte whole rod is $V$. Then we proceed (the cross product of $L$ doesn't matter):

$$I \omega = L = pr = MV l/4 = \frac{ml(v+v_f)}{4}$$

So the final energy of the whole system is (having $I=\frac{1}{12}Ml^2$):

$$E_f= \frac{1}{2} \frac{(I\omega)^2}{I} + \frac{1}{2} mv_f^2 = \frac{3m^2}{8M} (v+v_f)^2 + \frac{1}{2} mv_f^2$$

So comparing the two cases, there's a difference in the final energy, specifically a difference of $\frac{m^2}{8M} (v+v_f)^2$, which doesn't make sense.

Conclusions

This difference can mean various things, and I don't know which of this ones is, or even if it implies something else I haven't thought about. But before, there's something I don't understand and then there would be no need of other explanations.

As I've said the average velocity of the rod is $V$ so why should it have different speed when rotating than if the whole rod is advancing? In the end, the product mass times velocity is the same. So that's one thing I don't understand.

If that's not the case and the kinetic energy is different for some reason here's the explanations I've thought about for this difference in the energies:

1. In the second case there's also kinetic energy in the form of translation of the rod, and not only rotation, specifically with a translational kinetic energy of $\frac{m^2}{8M} (v+v_f)^2$, but why this specific quantity?


2. Actually, apart from the rotation, the rod has the same translational kinetic energy as in the first case, and $v_f$ in the second case is what causes the difference, being much smaller, so in this case the rod would have much more kinetic energy in the end when having the particle impact at one of the ends than having it impact at the very center.


3. There's no translation, which would agree with my hypothesis, and $v_f$ would be bigger in the second case than in the first.

Answer

In both cases if you apply the conservation of linear momentum then the relationship between the velocity of the centre of mass of the rod $V$ and the final velocity of the particle $(-) v_{\rm f}$ is $V = \frac mM (v+v_{\rm f})$.

However the final velocities of the particle $v_{\rm f1}$ and $v_{\rm f2}$ and the rod $V_1$ and $V_2$ in the two cases are not necessarily the same.

As there are no external torques acting on the rod and the particle, the conservation of angular momentum must be satisfied which when applied about the centre of mass in the second case gives $mv\frac L2= mv_{\rm f2} \frac L2 + I_{\rm c}\omega$ where $I_{\rm c}$ is the moment of inertia of the rod about its centre of mass and $\omega$ is its angular speed.

The final kinetic energy for the second case is $\frac 12 MV_2^2+ \frac 12 m v_{f2}^2 + \frac 12 I_{\rm c}\omega^2$ whereas in the first case it was $\frac 12 MV_1^2+ \frac 12 m v_{f1}^2$.

If you want kinetic energy to be conserved during the collision then the final translational velocities of the rod and the particle are less in the second case as compared with the first case.