Looking at the following Feynman diagram:

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/x89yd.png

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.