Maximum electron momentum in $\beta^-$-decay This should be easy, but I think I have a mind-block... 
For $\beta^-$-decay, what is the maximum possible momentum for the electron? The two equations I can use are conservation of energy and conservation of momentum, but I have three unknowns: Momentum of electron, nucleus and anti-neutrino, so what am I missing?
Could I just set the kinetic energy of the nucleus to zero and work from there?
 A: Answer of Zassounotsukushi is correct, I'd like to expand it
consider center of mass frame, a resting neutron for simplicity. four-momentum is $\underline{p}_n=\binom{m_n}{\vec{0}}$.
neutron decays into proton, electron, antineutrino: $n \rightarrow p^+ e^- \bar{\nu}$ 
$$
\binom{m_n}{\vec{0}}=\binom{E_p}{\vec{p}_p}+\binom{E_e}{\vec{p}_e}+\binom{E_\nu}{\vec{p}_\nu}
$$
neutrino mass $(m_\nu < 0.2eV)$ is negligible $\Rightarrow E_\nu=|\vec{p}_\nu|$, to maximize electron energy neutrino has to be left without any  $\Rightarrow E_\nu=0$
$$
\Rightarrow\binom{m_n}{\vec{0}}=\binom{E_p}{\vec{p}_p}+\binom{E_e}{\vec{p}_e}
$$
for momentum conservation to be satisfied , $\vec{p}_p+\vec{p}_e=\vec0 $, proton momentum has to be $\vec{p}_p=-\vec{p}_e$
$$
\Rightarrow\binom{m_n}{\vec{0}}=\binom{E_p}{-\vec{p}_e}+\binom{E_e}{\vec{p}_e}
$$
Energy-momentum relation can be used: $E_p=\sqrt{p_p^2+m_p^2}=\sqrt{p_e^2+m_p^2}$, $E_e=\sqrt{p_e^2+m_e^2}$
energy conservation:
$$
\Rightarrow m_n = \sqrt{p_e^2 + m_p^2} + E_e
$$
$$
\Rightarrow m_n = \sqrt{E_e^2 - m_e^2 + m_p^2} + E_e
$$
$$
\Rightarrow (m_n - E_e)^2  = E_e^2 - m_e^2 + m_p^2
$$
$$
\Rightarrow E_{e,max} = \frac{m_n^2 - m_p^2 + m_e^2}{2m_n}
$$
$$
\Rightarrow p_{e,max} = \sqrt{\left(\frac{m_n^2 - m_p^2 + m_e^2}{2m_n}\right)^2 - m_e^2}
$$
the "mass defect" in transition $n\rightarrow p$ is in the term $m_n^2 - m_p^2$ included
I assumed $c=\hbar=1$

Edit after the question by AAAAAA: 
if the proton is at rest, similar consideration about the momentum conservation leads to:
$$
\Rightarrow\binom{m_n}{\vec{0}}=\binom{m_p}{0}+\binom{E_\nu}{\vec{p}_\nu}+\binom{E_e}{\vec{p}_e}
$$
Now the $E_\nu$ has to be substituted approximately by $E_\nu\approx|\vec{p}_\nu|$ and thus $E_\nu\approx|\vec{p}_\nu| = |-\vec{p}_e| = \sqrt{E_e^2 - m_e^2}$
Energy conservation gives:
$$
\Rightarrow m_n = m_p + \sqrt{E_e^2 - m_e^2} + E_e
$$
$$
\Rightarrow (m_n - m_p) - E_e =  + \sqrt{E_e^2 - m_e^2}
$$
with $\Delta m:=(m_n - m_p)$
$$
\Rightarrow (\Delta m - E_e)^2 =  E_e^2 - m_e^2
$$
$$
\Rightarrow -2\Delta m E_e + \Delta m^2 =  - m_e^2
$$
$$
\Rightarrow E_e = \frac{m_e^2 + \Delta m^2}{2\Delta m}
$$
This results in a smaller electron energy (and momentum):
$E_{e, \textrm{proton at rest}} = 748 keV$ while $E_{e, \textrm{neutrino at rest}} = 1.293 MeV$ above.
Maybe it's in limits valid to imagine that it's easier for the electron to "bounce off" a massive proton that off a light neutrino.
A: The answer to your question is that the beta endpoint energy is equal to the total Q value of the reaction.  Here is a graph of the spectrum from a beta emission, which is why we use internal conversion instead when looking for a monoenergetic source of beta particles.  This graph is used in the German Wikipedia, the y-axis is counts per unit energy and the x-axis is energy (of the electron).

You asked for momentum instead, so use $E^2=(m_e c^2)^2 + (pc)^2$.  Do not use the classical analog because it is extremely common that beta decay gives relativistic electrons.
