# Elastic collisions

How do I derive the energy transfer in an elastic collision between two bodies of mass $m_1$ and $m_2$ respectively. $E_0$ is the incident energy of the incoming body $m_1$ and $\theta$ is the angle of scattering of the incoming particle.

I started with the conservation relations:

$$\frac{1}{2} m_1 v_0^2 = \frac 1 2 m_1 v^2 + \frac 1 2 m_2 V^2$$ $$m_1 v \cos(\phi)=m_1 v_0 -m_2 V \cos(\theta)$$ $$m_1 v \sin(\phi)=-m_2 V \sin(\theta)$$ but I could not solve for the final velocities $v$ and $V$ respectively which goes like, $$v^2=v_0^2 \left(1- \frac{4 m_1 m_2 \cos^2(\theta)}{(m_1+m_2)^2} \right)$$ and $$V=2 v_0 \frac{m_1 \cos(\theta)}{m_1+m_2}$$ I do not understand how the $\cos^2(\theta)$ term enters the above expression of $v^2$. Can anyone please help me to figure it out.?