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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.?

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If you first convert to the center of mass frame (that is, the frame of reference in which the net momentum before and after the collision is zero), you will find the equations become much easier to solve. You then add the velocity of the frame of reference again at the end, to get the solution in the lab frame. Try that and see if you can get the result you need.

Possible questions / answers to help you:

Elastic collision between two circles

Determine resultant velocity of an elastic particle-particle collision in 3d space

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  • $\begingroup$ I used the relation, (m+M)Vcm =mv_0 to get Vcm but I cannot write the exact transformation equations because there are both x and y components.@Floris $\endgroup$ – Payel Sep 2 '16 at 17:50
  • $\begingroup$ Note that in the com frame, the (magnitude of the) velocity is the same before and after the collision. This means you can draw a simple vector diagram, and the result follows from trigonometry. I will update this later if you haven't figured it out by then. $\endgroup$ – Floris Sep 2 '16 at 19:51

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