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.