- #1
Payel Das
- 1
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How do I derive the energy transfer equation in an elastic collision of two bodies of masses m and M respectively,using the energy and momentum conservation relations in the laboratory frame?
$$\frac{1}{2} m_1 v_0^2 = \frac 1 2 m_1 v^2 + \frac 1 2 m_2 V^2$$
$$m_1 v \cos(\phi)=m_1 v_0 -m_2 V \cos(\theta)$$
$$m_1 v \sin(\phi)=-m_2 V \sin(\theta)$$
but I could not solve for the final velocities $v$ and $V$ respectively
$$\frac{1}{2} m_1 v_0^2 = \frac 1 2 m_1 v^2 + \frac 1 2 m_2 V^2$$
$$m_1 v \cos(\phi)=m_1 v_0 -m_2 V \cos(\theta)$$
$$m_1 v \sin(\phi)=-m_2 V \sin(\theta)$$
but I could not solve for the final velocities $v$ and $V$ respectively