What is the energy loss of a solid sphere hitting a wall obliquely?

[physmurf]

So, I am reading about a compton scattering problem, and I don't understand part of the derivation of a formula. I will explain my confusion.

If a gamma photon with energy $$E_{\gamma}$$, undergoes compton scattering with an electron which is at rest, how does one arrive at the following expression?

$$E^{'}_{\gamma}=\frac{E_{\gamma}}{1+(2E_{\gamma}/m_{o}c^{2})}$$

So far it says we start with the conservation of energy and momentum:
$$E_{\gamma}=E^{'}_{\gamma}+E_{e} \ \ (eqn 1)$$
$$\frac{E_{\gamma}}c=P_{e}-\frac{E^{'}_{\gamma}}{c}\ \ (eqn 2)$$

From eqn 2 we get:
$$E_{\gamma}+E^{'}_{\gamma}=p_{e}c=\sqrt{(E_{e}+m_{o}c^{2})^{2}-(m_{o}c^{2})^{2}}$$

This is where I am confused. I don't understand where the term inside of the radical comes from. Any ideas?

 

[jtbell]

$$E_{\gamma}+E^{'}_{\gamma}=p_{e}c=\sqrt{(E_{e}+m_{o}c^{2})^{2}-(m_{o}c^{2})^{2}}$$

The $E_e$ in the square root on the right is the electron's kinetic energy, which most books call $K_e$. $E_e$ usually means the total energy:

$$E_e = K_e + m_0 c^2$$

Using this notation, the relationship between energy, momentum and mass is

$$E^2 = (pc)^2 + (m_0 c^2)^2$$

so

$$p_e c = \sqrt {E_e^2 - (m_0 c^2)^2} = \sqrt {(K_e + m_0 c^2)^2 - (m_0 c^2)^2}$$

 

[ranaroy]

hi all,

a similar post for me. pls help me.
i need to calculate the energy lost by a solid sphere (~ 0.5 mm dia) on hitting the solid wall obliquely.
i know the tangetial and normal velocities and the restitution coefficients. what other parameters do i need to know ??

help with the expression or a reference to articles in the web would be of great help.

thanks you.