Speed of particle before its KE is doubled

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Homework Help Overview

The discussion revolves around determining the speed a particle must reach for its kinetic energy to be double that predicted by the classical nonrelativistic expression, KE=0.5mv^2. The problem is situated within the context of relativistic physics, particularly the comparison between classical and relativistic kinetic energy equations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between classical and relativistic kinetic energy, with one suggesting to equate the relativistic kinetic energy to twice the classical value. Questions arise regarding the derivation of specific terms in the equations, particularly the appearance of the number 5 in a square root expression.

Discussion Status

The discussion is active, with participants providing insights into the equations involved and questioning specific aspects of the derivation. Some guidance has been offered regarding the manipulation of the equations, but no consensus has been reached on the interpretation of certain terms.

Contextual Notes

The original poster notes the absence of numerical data, which may limit the approaches discussed. There is an emphasis on understanding the transition from classical to relativistic kinetic energy without specific values to work with.

Miri
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Homework Statement


What speed must a particle attain before its kinetic energy is double the value predicted by the nonrelativistic expression KE=0.5mv^2?
(there is no data given)

Homework Equations


How should I solve this if I have no numbers to insert...? Should I set this equation equal to the relativist one (e.g. E=KE-rest energy)
In the end v/c=~0.786151
 
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Corrct, but I think you meant to equate the relativistic kenetic energy to twice the classical kinetic energy:

Kinetic energy is (gamma - 1) m c^2, so we have:

(gamma - 1) m c^2 = m v^2 -------->

gamma - 1 = (v/c)^2

v = c sqrt{[sqrt(5) - 1]/2}
 
Thank you..But I don't get it how you have this 5 in the 2nd sqrt...v = c sqrt{[sqrt(5) - 1]/2} The result is correct.
 
Miri said:
Thank you..But I don't get it how you have this 5 in the 2nd sqrt...v = c sqrt{[sqrt(5) - 1]/2} The result is correct.

You just put gamma = 1/sqrt[1-(v/c)^2], move gamma to one side of the equation, square both sides, and solve the quadratic equation.
 

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