When an electron hits a fluorescent screen mounted on a spring, why can't we get both position and momentum? From the Heisenberg's Uncertainty Principle for position and momentum we know that, $x$ and $p$ of a particle cannot be measured simultaneously with arbitrary accuracy.
$$\Delta x \Delta p \geq \hbar/2 $$
How does the uncertainty principle work in the following experiment?
Consider an electron hitting a fluorescent screen which is mounted on a very sensitive (hypothetical) spring. When the electron hits the screen, we know its position at the instant of it hitting the screen. Besides this, it imparts momentum at the instant of hitting the sunscreen which can be calculated from the compression of the spring. Thus getting both position and momentum simultaneously.
What part of the above experiment am I confused about that gives me this wrong conclusion?
 A: Thomas Fritsch's answer covers the fact that your experiment doesn't probe the essential problem. So what is the essential problem?
Your measurement system can record whatever it records to whatever precision it's capable of. The HUP constrains your ability to predict what it will record. If you prepare many electrons in exactly the same way, so their positions and momenta should be the same for this purpose, your measurement system will record a distribution of positions and momenta whose width is at least as much as the HUP demands.
A: When an electron hits the screen, it will give us a spot of diameter $\Delta x$, whereas the compression of the spring is also measured with a certain precision, which will limit the precision of measuring momentum to $\Delta p$. The Heisenberg uncertainty principle then says that $$\Delta x \Delta p \geq \frac{\hbar}{2}.$$
There is nothing special about this type of measurement as compared to other kinds of measurements that one could envisage, however it is worth adding a few remarks:

*

*The material of the screen and the spring are also made of atoms. Having precision of measurement sufficiently high to try to break the HUP would require treating these as quantum, which would make the survival of HUP in this experiment far less mysterious.

*Experiments of this type have been actively tried (and corresponding calculations have been performed) in the last decade in the context of nanomechanics - see the references in this Wikipedia article. There is a lot of interesting stuff going there, but, unsurprisingly, no breaking of the fundamental QM principles.

A: You are assuming that the screen is a classical object with a perfectly well-defined position, a perfectly well defined distribution of detection points, and that the spring has a perfectly well defined length. In reality, the screen and the spring are made of electrons and ions, all of which have imperfectly defined positions and momenta in accordance with the HUP. So the HUP imposes a limit on how well you can define the position and momentum of your detecting equipment, which in turn prevents you from pinning down the exact position and momentum of the electron, even supposing it had an exact position and momentum, which, of course, it doesn't.
The HUP not only means that the electron's position and momentum can't be measured exactly together, it goes further and says that the electron cannot have an exact momentum and position at the same time.
A: If I understand you correctly you have an experimental
setup like this.

(image partially taken from The Physics of Springs)
You measure two things:

*

*the $x$ position
given by the position where the fluorescent spot appears on the screen

*$p_y$, the $y$-component of momentum
given by the compression of the spring when hit by the particle

You claim there should be no restriction between these two measurements.
And you are right. Let us see how this is no contradiction to the uncertainty
relation.
Heisenberg's uncertainty relation says
$$\Delta x\ \Delta p_x \geq \frac{\hbar}{2}$$
However, this relation doesn't apply to your experiment because
you measure $p_y$, but not $p_x$.
There is the more general Robertson uncertainty relation between two
arbitrary observables $A$ and $B$:
$$\Delta A\ \Delta B \geq \frac{1}{2} \left|\left<[A,B]\right>\right|$$
where $[A,B]$ is the commutator between $A$ and $B$.
(Heisenberg's uncertainty relation obviously is a special case
of this because $[x,p_x]=i\hbar$.)
Now the commutator between your two measured observables is
$$[x,p_y]=0$$
and therefore you have the uncertainty relation
$$\Delta x\ \Delta p_y \geq 0.$$
