Energy transfer in elastic collision.

[Payel Das]

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

 

[Chandra Prayaga]

You cannot expect to solve for the unknown quantities here. There are four unknowns, the two final speeds and the two angles, and only three equations. If it is a head-on collision, then it is just a 1-dimensional problem with two unknowns, the velocities after collision. There are two equations, the conservation equations, and the problem can be solved. In the general case, you have to specify one of the four unknowns. In Nuclear physics problems, such as the collision between an α-particle and a gold nucleus, the direction of the α-particle, after collision, is measured, and the corresponding values of the other three quantities can be calculated.