There are no irresolvable contradictions in this scenario. At least none that have been pointed out.
The observers disagree on the relative shape of the rod during the collision.
So what? Shape is relative
Wrong, we aren't talking about the shape of the rod in the train car frame as viewed from the platform frame.
We are talking about the shape of the rod in the platform frame as measured in the platform frame vs. the shape of the rod in the train car frame as measured in the train car frame. We are talking about a comparison of the
proper shapes of the rod.
And no one disagrees about the shape of the rod measured in its own frame.
Actually, according to the discussion, they would, IF the rod were to be measured to fall at a non-zero angle in the platform frame. Turns out that I can prove that is not the case.
There is no disagreement about the absolute measures such as proper length.
Actually you have been disagreeing on that by virtue of disagreeing on the angle of fall. You may not realize or admit it but this is the case.
I haven't asked those exact questions before.
All you've said is that the scenario leads to a contradiction (it doesn't), that the two observers disagree on the outcome of the experiment (they only disagree on relative measures such as length and simultaneity) therefore you conclude it must be ill-formed.
But you haven't attempted to discern what exactly is ill-formed about it.
I have, several times. But you won't admit to it.
Try this, the scenario is posed a little more formally, and leaves it to you to determine the outcome:
A train is at rest in frame S(x,y,t).
The floor of the train is at y=0 (ie along the x-axis).
A rod of negligible mass is above the floor, parallel to the floor, moving toward the floor at velocity $$\vec{v_r}(x,y) = (0, -v_r)$$
At t=0, the rod collides inelastically with the floor.
Frame S' is moving relative to S at $$\vec{v}(x,y) = (v, 0)$$
What happens to the rod in frame S'?
I am glad that you are attempting to formalize a little the dumbed down version of the problem. Since you are so intent on solving the dumbed down version instead of the realistic one, I'll solve this version for you.
First off, there are THREE frames of reference that you must consider, not TWO:
$$S"=$$ the frame of the train car
$$S'=$$ the frame of the falling rod
$$S=$$ the frame of the platform
There are two speeds that we need to consider:
$$v'=(0,-u)$$ the speed between $$S'$$ and $$S"$$
$$v=(V,0)$$ the speed between $$S$$ and $$S'$$
This is a classical problem of
Thomas precession.
The angle made by the rod with the floor of the car , in the frame of the car $$S"$$ is $$\theta"$$.
The angle made by the rod with the platform , in the frame of the platform is $$\theta$$.
Form the description of Thomas precession:
$$tan(\theta)=\frac{v'}{v \gamma(v)}$$
$$tan(\theta")=\frac{v'\gamma(v')}{v}$$
So:
$$\tan(\theta")=tan(\theta) \gamma(v)\gamma(v')$$
Since $$\tan(\theta")=0$$ it follows $$tan(\theta)=0$$.
In other words:
$$\theta"=0 => \theta=0$$
QED.
Now, having solved the dumbed down version of the problem, I'll let you ponder on how to solve the much more difficult problem in the realistic case of accelerated motion in the presence of the gravitational field. Hint: you will no longer be able to use the Lorentz transforms, you will need to use the equations of accelerated motion (Rindler coordinates).