Question

Photons from space are bombarding your laboratory and smashing massive objects to pieces! Your detectors indicate that two fragments each of mass m₀ depart such a collision moving at 0.6c at 60^o to the photon’s original direction of motion. In terms of m₀ what are the energy of the cosmic ray photon and the massMof the particle being struck (assumed stationary initially).

Short answer

The energy of the cosmic ray photon $=\frac{3}{4}{m}_0{c}^2$

The mass M of the particle has been struck $=\frac{7}{4}{m}_0$.

Step-by-step solution

Given data

The final speed of each mass m₀ is u_f= 0.6c at 60^o to the photon's original direction of motion

The wavelength of the incoming photon $\lambda$.

Concept  used

The Lorentz factor is given by

$\mathit{\gamma }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\left({\displaystyle \frac{u}{c}}\right)}^{\mathbf{2}}}}$

Law of conservation of momentum

First, calculate the$\gamma$ value

$\begin{array}{rcl}\gamma & =& \frac{1}{\sqrt{1-{\left({\displaystyle \frac{u_f}{c}}\right)}^2}}\\ & =& \frac{1}{\sqrt{1-{\left({\displaystyle \frac{0.6c}{c}}\right)}^2}}\\ & =& \frac{5}{4}\end{array}$

By Law of conservation of momentum:

$\begin{array}{rcl}\frac{h}{\lambda }& =& 2\gamma m_0{u}_f\cos {60}^o\\ & =& 2\times \frac{5}{4}{m}_0\times \frac{6}{10}c\times \frac{1}{2}\\ & =& \frac{3}{4}{m}_{0}c\end{array}$

Calculate the energy of the photon

Energy of the photon can be written as

$E=\frac{hc}{\lambda }$

Therefore, by substituting

$E=\frac{hc}{\lambda }=\frac{3}{4}{m}_0{c}^2$

Law of energy conservation

By Law of energy conservation

$\begin{array}{rcl}E+M{c}^2& =& 2\gamma m_0{c}^2\\ \frac{3}{4}{m}_0{c}^2+M{c}^2& =& 2\times \left(\frac{5}{4}{m}_0{c}^2\right)\\ M{c}^2& =& \frac{7}{4}{m}_0{c}^2\\ M& =& \frac{7}{4}{m}_0\end{array}$