A consequence of Noether's theorem is that the energy of a Hamiltonian system is conserved if and only if the Hamiltonian is time-translation invariant. However, to my surprise I found some pathological systems in which this correspondence doesn't seem to hold.
In the following examples, what exactly are the conditions that fail to hold for the proof of Noether's theorem to hold (pertaining to the link between energy conservation and time-translation invariance)? Can anyone pinpoint exactly what breaks down?
If more details or reasoning needs to be provided, please let me know.
Example 1
Jon Pérez Laraudogoita's "Beautiful Supertask" (described here and here) involves a row of infinitely many pool balls (we'll abstract them as point particles) of equal mass positioned at $x_{0} = -1, x_{1} = -1/2, x_{2} = -1/4$, and so on (with no final ball). A ball (labeled $-1$) from the $-x$-direction comes in with velocity $v_{0} > 0$ and a chain reaction of elastic collisions take place.
Particle $-1$ ends up stationary at the initial position of particle $0$, particle $0$ ends up stationary at the initial position of particle $1$, particle $1$ ends up stationary at the initial position of particle $2$, and so on. All this takes place within a finite time span, and the result is that every particle is stationary.
Please read the links to understand the scenario in better detail. The Hamiltonian for the system, I believe could be written as
$$ H = \sum_{k=-1}^{\infty} \frac{1}{2}m\dot{x}_{k}^{2} + \sum_{i<j} c\delta(|x_{i}-x_{j}|) $$
where $c > 0$ is some constant. Here $H$ has no explicit time-dependent, and yet in our scenario we started with total energy $E = \frac{1}{2}mv_{0}^{2}$ and ended up with $E = 0$ after a finite time interval.
Example 2
I was thinking perhaps the use of the Dirac-delta function could have been problematic, so I considered a similar system with the Dirac-deltas taken out. Here we take the initial positions and velocities of all the particles to be exactly the same as in Example 1, but now we have
$$ H = \sum_{k=-1}^{\infty} \frac{1}{2}m\dot{x}_{k}^{2} + \sum_{i < j} U_{ij}(x_{i}, x_{j}) $$
where
\begin{align} U_{ij}(x_{i}, x_{j}) = \begin{cases} A_{ij}(1 - B_{ij}|x_{i}-x_{j}|), & |x_{i}-x_{j}|\in[-\frac{1}{B_{ij}}, \frac{1}{B_{ij}}], \\ 0, & \text{else} \end{cases} \end{align}
with $A_{ij} = \frac{1}{2}mv_{0}^{2}$ and $B_{ij} = 2^{2^{j}}$ ($i < j$). With initial condition
$$ x_{-1}(0) = -2 \qquad\text{and}\qquad x_{k}(0) = -1/2^{k} \quad (k\ge 0) $$
and
$$ \dot{x}_{-1}(0) = v_{0} \qquad\text{and}\qquad \dot{x}_{k}(0) = 0 \quad (k\ge 0) $$
we obtain the same scenario playing out as in Example 1 where each ball transfers its kinetic energy to the next in a chain reaction until it "disappear" after a time $T = 2/v_{0}$.
Note that the fact that I used a non-differentiable functions for the potential is not an issue, because they can be approximated by smooth / infinitely differentiable functions to arbitrary precision.
The same situation plays out as in Example 1, even though $H$ is time-independent, so I am wondering what assumption for Noether's theorem is broken here. Note we can treat this as a system of infinitely many coupled differential equations.