The relativistic energy of a particle in terms of momentum is given by: a) E = \sqrt{p^2c^2 + m_0^2c^4} b) E = \sqrt{p^2c^2 + m_0^4c^4} c) E = \sqrt{p^2c^2 + m_0^2c^2} d) E = \sqrt{p^2c^4 + m_0^2c^2}

Name: OpenStax College Physics Solution, Chapter 28, Problem 6 (Advanced Placement)
Uploaded: 2021-04-25T18:38:11-07:00
Duration: 49 s
Description: The relativistic energy of a particle in terms of momentum is given by: a) E = \sqrt{p^2c^2 + m_0^2c^4} b) E = \sqrt{p^2c^2 + m_0^4c^4} c) E = \sqrt{p^2c^2 + m_0^2c^2} d) E = \sqrt{p^2c^4 + m_0^2c^2}

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The relativistic energy of a particle in terms of momentum is given by:

$E = \sqrt{p^2c^2 + m_0^2c^4}$
$E = \sqrt{p^2c^2 + m_0^4c^4}$
$E = \sqrt{p^2c^2 + m_0^2c^2}$
$E = \sqrt{p^2c^4 + m_0^2c^2}$

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Final Answer

(a)

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OpenStax College Physics for AP® Courses, Chapter 28, Problem 6 (Test Prep for AP® Courses)

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This is College Physics Answers with Shaun Dychko. The textbook gives us a formula for the total energy in terms of momentum and it says that the energy squared is the momentum times the speed of light squared plus the rest mass energy squared and if we take the square root of both sides here then we solve for energy and the option here, which is the square root of pc squared plus mc squared squared is this one here. If you were to continue along here, we have the square root sign and then we have a p squaredc squared when we multiply this pc by itself and then add to that mc squared squared, this is m squared times c to the fourth. So the answer is (a).

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OpenStax College Physics for AP® Courses, Chapter 28, Problem 6 (Test Prep for AP® Courses)

Chapter 28: Special Relativity

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