Chapter 3: Q55CE (page 96)

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

Expert verified

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 $λ$ .

Concept used

The Lorentz factor is given by

$γ = \frac{1}{\sqrt{1 - {(\frac{u}{c})}^{2}}}$

Law of conservation of momentum

First, calculate the $γ$ value

\begin{array}{rcl} γ & = & \frac{1}{\sqrt{1 - {(\frac{u_{f}}{c})}^{2}}} \\ = & \frac{1}{\sqrt{1 - {(\frac{0.6 c}{c})}^{2}}} \\ = & \frac{5}{4} \end{array}

By Law of conservation of momentum:

$\begin{array}{rcl} \frac{h}{λ} & = & 2 γ m_{0} u_{f} \cos 60^{o} \\ = & 2 \times \frac{5}{4} m_{0} \times \frac{6}{10} c \times \frac{1}{2} \\ = & \frac{3}{4} m_{0} c \end{array}$

Calculate the energy of the photon

Energy of the photon can be written as

$E = \frac{h c}{λ}$

Therefore, by substituting

$E = \frac{h c}{λ} = \frac{3}{4} m_{0} c^{2}$

Law of energy conservation

By Law of energy conservation

$\begin{array}{rcl} E + M c^{2} & = & 2 γ m_{0} c^{2} \\ \frac{3}{4} m_{0} c^{2} + M c^{2} & = & 2 \times (\frac{5}{4} m_{0} c^{2}) \\ M c^{2} & = & \frac{7}{4} m_{0} c^{2} \\ M & = & \frac{7}{4} m_{0} \end{array}$

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Short Answer

Step by step solution

Given data

Concept used

Law of conservation of momentum

Calculate the energy of the photon

Law of energy conservation

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