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Chapter 20: Problem 34

Calculate the speed of a particle \((a)\) whose kinetic energy is equal to twice its rest energy and \((b)\) whose total energy is equal to twice its rest energy.

Short Answer

Expert verified
The velocity of the particle when its kinetic energy is twice its rest energy is \(v=\sqrt{8}/3 c\). The velocity of the particle when its total energy is twice its rest energy is \(v=\sqrt{3}/2 c\).

Step by step solution

01

Define the known and find equations

The problem involves principles of relativity and energy. Let's denote the rest energy as \(E_0=m_0c^2\), where \(m_0\) is the rest mass of the particle and \(c\) is the speed of light. Kinetic energy is \(KE=E_{total}-E_0=((\gamma-1)m_0)c^2\), where \(\gamma\) is the Lorentz factor and it equals \(1/\sqrt{1-(v/c)^2}\). Total energy is \(E_{total}=\gamma m_0 c^2\). The term 'twice' in the problem indicates we need to set kinetic energy equal to \(2E_0\) and total energy equal to \(2E_0\) to solve for \(v\).
02

Solve part (a)

To find the velocity of the particle when kinetic energy is twice the rest energy, we set the equation as \(KE=2E_0\), which gives us \((\gamma-1)m_0c^2=2m_0c^2\). Solving for \(\gamma\) gives \(\gamma=3\). Substituting it into the formula for \(\gamma\), \(1/\sqrt{1-(v/c)^2}\)=3, we can solve for \(v\), which gives us \(v=\sqrt{8}/3 c\).
03

Solve part (b)

To find the velocity when the total energy is twice rest energy, we set \(E_{total}=2E_0\), which gives us \(\gamma m_0c^2=2m_0c^2\). Solving this yields \(\gamma=2\). Substitute \(\gamma=2\) into the formula for \(\gamma\), \(1/\sqrt{1-(v/c)^2}\)=2, solving for \(v\) gives us \(v=\sqrt{3}/2 c\).

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