$$\gamma_p m_p v_p =\gamma_M M_p V_N $$ because the momentum of an inelastic collision is conserved.

The relativistic mass of the new particle
$$M=2m_p \gamma_N $$

$$MV_N \gamma_N = (2m_p \gamma_N) V_N \gamma_N=m_p v_p \gamma_p$$

$$V_N = \frac {m_p v_p \gamma_p} {2m_p \gamma_N^2} = \frac {v_p \gamma_p} {2\gamma_N^2}$$

If the relativistic kinetic energy is
$$K = (\gamma -1) mc^2$$
where gamma is gamma_N and the mass is 2m_p*gamma_N
how do i manipulate the equation to get
$$Mc^2 = 2m_p c^2 \sqrt{ 1+ {K \over 2m_p c^2} }$$

Question

$$\gamma_p m_p v_p =\gamma_M M_p V_N $$ because the momentum of an inelastic collision is conserved.

The relativistic mass of the new particle
$$M=2m_p \gamma_N  $$

$$MV_N \gamma_N = (2m_p \gamma_N) V_N \gamma_N=m_p v_p \gamma_p$$

$$V_N = \frac {m_p v_p \gamma_p} {2m_p \gamma_N^2} = \frac {v_p \gamma_p} {2\gamma_N^2}$$ 

If the relativistic kinetic energy is
$$K = (\gamma -1) mc^2$$
where gamma is gamma_N and the mass is 2m_p*gamma_N
how do i manipulate the equation to get 
$$Mc^2 = 2m_p c^2 \sqrt{ 1+ {K \over 2m_p c^2}  }$$

roadjester · Answer

oops, I screwed up the first equation
$$\gamma_p m_p v_p = \gamma_N M V_N$$

roadjester · Answer

@LastDayWork 
@agent0smith

roadjester · Answer

This time there's no crazy square roots since i didn't conserve the energy