Why can't I assume the quarks inside a hadron move together?

Question

Looking at the following Feynman diagram:

Using conservation of energy, we can see that in the rest frame of $D^0$, the energy of $K^-$ is higher than its rest energy. Meaning, it is in motion. I expected that since the particles in the hadron only exist as a set, that means that also the resulting $\bar{u}$ is in motion. But clearly by the same assumption, since $D^0$ is at rest in its own rest frame, also the initial $\bar{u}$ must be at rest. During the entire decay, $\bar{u}$ doesn't partake in any interaction, meaning that at the end, it should still be at rest, which is a contradiction.

I see two possible options to resolve this discrepancy: Either the different quarks that compose a Hadron can move at different speeds, or $\bar{u}$ somehow changes its speed without participating in any interaction. Neither of these seems possible.

Can someone please explain to me what I'm missing? Thanks.

Edit: After reading both anna's and nu's answers, I now understand, but I don't think I would have understood after reading either answer individually, so I'm not sure which answer I should select as correct. For now, I'm not selecting any answer, as I think ideally people visiting this question should read both.

Answer 1

In the given diagram, it is $D^0$ that is at rest, not the individual quarks composing it. Within energy and momentum conservation, (even if there were not a sea of quark antiquark and gluons in hadrons, example,) since the rest mass energy of the $D^0$ is larger than the masses of the two quarks , they are not at rest. In a funny Bohr like model the kinematics allows them to be orbiting around each other.

This is the wrong statement:

But clearly by the same assumption, since D0 is at rest in its own rest frame, also the initial u¯ must be at rest

The antiup is not at rest , as explained above.

Answer 2

Quarks cannot exist on their own (at least at any reasonable temperature), they only occur in color-neutral bound states, in which all bound particles move at the same speed.

The Feynman diagram, however, only captures a short moment in time, at which the $D^0$ decays into a $K^-$. What happens thereafter is not included. It could be, that the strange quark has kinetic energy and exchanges a gluon with the $\bar u$ to transfer some of it, which would cause the whole $K^-$ to move. On the other hand, you also get that $\pi^+$-Meson, that can carry away kinetic energy, too, which it got from the $W$-Boson.

In fact, due to conservation of momentum, in the rest frame of the $D$-Meson, which becomes the reference frame of the center of mass after the decay, both the $K^-$ and the $\pi^+$ will have some kinetic energy and momenta in opposite directions.

Edit:

As stated above, quarks cannot exist as single particles and likewise cannot just drift apart, because they are bound by the strong interaction, which is very strong (hence the name). If quarks have a lot of energy, their distance can increase and the gluons (the gauge bosons of the strong force) between them can decay into more quarks. This process is called hadronisation and typically takes place in hadron colliders at high energies. Theoretically, there could be no further interaction, but that is extremely unlikely (so unlikely, that it will not be observed for several times the current age of the universe) to happen, because due to the strength of the strong force, any hadronic bound state has a much lower energy than a single quark and consequentially a much higher probability to be realised.