System of equations of non-relativistic scattering in the laboratory system 
Considering the system of equations of non-relativistic scattering in the laboratory system:
$$\begin{cases} \dfrac{1}{2} m_{1} v_{1}^{2} &=\dfrac{1}{2} m_{1} v_{2}^{2}+T_2 \,,\\
m_{1} v_{1} &=m_{1} v_2 \cos  \psi_{1}+p_{2} \cos \psi_{2} \,,\\
0 &=m_{1}v_{2} \sin \psi_{1}+p_{2} \sin \psi_{2} \,,\\
T_2 &= \dfrac{p_2^2}{2 m_2}\,. \end{cases}$$
Prove that
$$T_2= 2 \frac{m_1^2 m_2}{(m_1+m_2)^2}v_1^2 \cos^2 \psi_2\,.$$

Is there a way to find the solution without going into the centre-of-mass reference system, i.e., considering only the 4 equations above?
 A: I assume that we are to solve for $T_2$ in terms of $m_1$, $m_2$, $v_1$, and $\psi_2$.    I also assume that, in the first equation of the given system of equations, there is a typo and the correct equation should be $$\frac{1}{2}m_1v_1^2=\frac12m_1v_2^2+T_2\,.$$
From the second and the third equations in the given system of equations, we have
$$m_1v_2\cos(\psi_1)=m_1v_1-p_2\cos(\psi_2)$$
and
$$m_1v_2\sin(\psi_1)=-p_2\sin(\psi_2)\,.$$
By squaring the two equations above and then adding them, we obtain
$$m_1^2v_2^2=\big(m_1v_1-p_2\cos(\psi_2)\big)^2+\big(-p_2\sin(\psi_2)\big)^2=m_1^2v_1^2-2m_1v_1p_2\cos(\psi_2)+p_2^2\,.$$
Hence,
$$m_1^2(v_1^2-v_2^2)-2m_1v_1p_2\cos(\psi_2)+p_2^2=0\,.\tag{*}$$
From the first equation in the given system of equations, we get
$$2T_2=m_1(v_1^2-v_2^2)\,.\tag{#}$$
From the fourt equation in the given system of equations, we have
$$2T_2=\frac{p_2^2}{m_2}\,.\tag{@}$$
Plugging (#) and (@) into (*) yields
$$\frac{m_1}{m_2}\,p_2^2-2m_1v_1p_2\cos(\psi_2)+p_2^2=0\,.$$
Thus, $p_2=0$ or
$$\frac{m_1}{m_2}\,p_2-2m_1v_1\cos(\psi_2)+p_2=0\,.\tag{%}$$
The case $p_2=0$ corresponds to the initial condition, so we eliminate it.  From (%), we get
$$p_2=\frac{2m_1m_2}{m_1+m_2}\,v_1\cos(\psi_2)\,.$$
That is,
$$T_2=\frac{p_2^2}{2m_2}=\frac{2m_1^2m_2}{(m_1+m_2)^2}\,v_1^2\cos^2(\psi_2)\,.$$
