Speed of an antimuon in a pi meson decay

[Je m'appelle]

Homework Statement

I'm trying to confirm the speed of an antimuon in the \pi^+ \rightarrow \mu^+ \nu_{\mu} decay through the laws of conservation but it doesn't add up.

Homework Equations

[/B]
1.Energy-momentum relation:

E^2 = (pc)^2 + (mc^2)^2

2. Rest masses:

m_{\pi} = 139.6 \ \frac{MeV}{c^2}
m_{\mu} = 105.7 \ \frac{MeV}{c^2}
m_{\nu} \approx 0 \frac{MeV}{c^2}

3. Relativistic kinetic energy formula:

E_k =m_{\mu}c^2 \left( \frac{1}{\sqrt{1 - \frac{v_{\mu}^2}{c^2}}} - 1 \right)

The Attempt at a Solution

By the way, the pi meson decays at rest, so p_{\pi}=0.

I'm considering the difference of mass, before and after the decay, as pure kinetic energy, so around (m_{\pi} - m_{\mu})c^2 = 33.9 MeV.

m_{\mu}c^2 \left( \frac{1}{\sqrt{1 - \frac{v_{\mu}^2}{c^2}}} - 1 \right) = 33.9 \ MeV

Carrying this out yields v_{\mu}=0.65c when in fact it should be 0.27c.

What am I doing wrong?

 

[fzero]

In order for momentum to be conserved, the neutrino must have momentum, so not all of that energy is available to the muon.