You can simplify this thought-experiment a lot by talking about the X and Z axis of entangled qubits instead of position and velocity. Position and velocity are very complicated in comparison to qubits (because modelling qubits doesn't require any calculus, qubits don't live in vector spaces with infinitely many dimensions, and qubits don't drag along as much incorrect-intuition baggage).
A qubit can have a well-defined value along its X axis, and a qubit can have a well-defined value along its Z axis, but a qubit can't have both at the same time. This comes down to the fact that the Z axis up/down states are ##\left| 0 \right\rangle## and ##\left| 1 \right\rangle## while the X axis up/down states are ##\frac{1}{\sqrt{2}} (\left| 0 \right\rangle + \left| 1 \right\rangle)## and ##\frac{1}{\sqrt{2}} (\left| 0 \right\rangle - \left| 1 \right\rangle)##. The X value is not independent of the Z value, it's the same thing expressed in a different basis. Asking for both X and Z to have a guaranteed well-defined value is like trying to find an ##a## and ##b## that satisfy the constraints ##a \cdot b = 0## and ##a+b=1## and ##a-b=0## or like asking for a line that's both diagonal and axis-aligned. It's just not going to happen.
A mathematically elegant way to summarize the fact that the X and Z values are incompatible is that their commutator ##[X, Z] = XZ - ZX = -2iY## is not equal to zero. (See:
Pauli Matrices.) But here's an interesting fact: the commutator ##[X \otimes X, Z \otimes Z]##
is equal to zero:
##[X \otimes X, Z \otimes Z]##
##= (X \otimes X) \cdot (Z \otimes Z) - (Z \otimes Z) \cdot (X \otimes X)##
##= (X \cdot Z) \otimes (X \cdot Z) - (Z \cdot X) \otimes (Z \cdot X)##
##= (-iY) \otimes (-iY) - (iY) \otimes (iY)##
##= ((-i)^2) (Y \otimes Y) - (i^2) (Y \otimes Y)##
##= ((-i)^2 - i^2) (Y \otimes Y)##
##= ((-1) - (-1)) (Y \otimes Y)##
##= 0 \cdot (Y \otimes Y)##
##= 0##
A
pair of qubits can have a well-defined
joint measurement for both their X axes and their Z axes! If I tell you the two qubits are in a state where their X axes
agree (or, alternatively, disagree), the fact of whether their Z axes agree or disagree is an independent additional fact. Even though the uncertainty principle prevents the individual axes from having simultaneously well-defined values, it doesn't prevent the combination of axes from having simultaneously well-defined agree/disagree values.
(Practical consequence: this is why
superdense coding works. You can independently toggle the X-agreement and the Z-agreement of an EPR pair using only one of the involved qubits, then send that single qubit to the other side, and do a joint measurement of both ##X \otimes X## and ##Z \otimes Z## to recover the two encoded bits.)
But beware! Don't think you can use well defined agreement/disagreement of a pair of qubits to measure both axes of a single qubit. The Z-axis agreement observable ##Z \otimes Z## doesn't commute with the individual X axis observables (##X \otimes I## and ##I \otimes X## for each qubit respectively). Specifying that the Z axes agree is incompatible with the individual X axes having well-defined values. Conversely, if you know that the X axes agree (or disagree), then you can't know the individual Z values.
In an EPR pair, we know whether the X axes and the Z axes agree or disagree. In the case of the EPR
singlet state, we know the X and Z values must both respectively disagree. Since the singlet state has well defined ##Z \otimes Z## and ##X\otimes X## values, it can't have well defined ##Z \otimes I## or ##X \otimes I## or ##I \otimes Z## or ##I \otimes X## values. Neither qubit has a well defined X axis value or Z axis value. We've made the problem worse instead of better! At the individual qubit level we now have zero well defined axes instead of one well defined axis!
If you try to get the X and Z axis values of a single qubit by creating a single state and measuring the X axis of one qubit and the Z axis of the other qubit, you will not have done what you set out to do. Measuring the X axis of one qubit destroyed the Z-agreement, and measuring the Z axis of the other qubit destroyed the X-agreement. The qubit you measured along the Z axis doesn't have an X value that agrees with the measurement result from the other qubit, precisely because of the Z axis measurement.
I suspect the same thing happens with entangled position and velocity. You can know a pair of particles agrees in both position and velocity, but that agreement forces the individual positions and velocities to be uncertain. Furthermore, measuring the position of one is incompatible with maintaining the "velocities agree" information. Measuring position on one side and velocity on the other side will destroy the agreement you wanted to use to bypass the uncertainty principle.