Double Speed: Proton Acceleration with 2000MV Potential

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Homework Help Overview

The discussion revolves around the acceleration of a proton through a potential difference of 2000MV, with a focus on the implications of classical mechanics versus relativistic effects on speed. Participants are exploring the relationship between potential difference and speed in the context of particle acceleration.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply classical mechanics to determine the speed of a proton accelerated through a potential difference, questioning why doubling the initial speed calculated for 500MV does not yield the expected result for 2000MV. Some express confusion over the validity of their reasoning and the application of kinetic energy principles.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and interpretations. There is a recognition of differing approaches, including a reference to kinetic energy equations, but no consensus has been reached on the correct method to determine the new speed.

Contextual Notes

Participants are grappling with the transition from classical to relativistic mechanics, indicating a potential misunderstanding of the assumptions underlying their calculations. The original speed of the proton at 500MV is noted, but the implications of increasing the potential to 2000MV are under scrutiny.

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Homework Statement


A proton is accelerated from rest through a potential difference of 500MV. Classical mechanics indicates that quadrupling the potential difference would double the speed. Were a classical analysis valid, what speed would result from a 2000MV potential difference?


Homework Equations



4V->2v

The Attempt at a Solution


We were first asked to find the original speed of the proton going through 500MV which is 0.76c Since 2000MV = 4*500MV and 4V means v is doubled, why can't I just double 0.76c to get the new speed for the proton accelerated through the 2000MV?
 
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channel1 said:

Homework Statement


A proton is accelerated from rest through a potential difference of 500MV. Classical mechanics indicates that quadrupling the potential difference would double the speed. Were a classical analysis valid, what speed would result from a 2000MV potential difference?

Homework Equations



4V->2v

The Attempt at a Solution


We were first asked to find the original speed of the proton going through 500MV which is 0.76c Since 2000MV = 4*500MV and 4V means v is doubled, why can't I just double 0.76c to get the new speed for the proton accelerated through the 2000MV?
I believe that you are asked to do precisely that.
 
Yet the answer is 2.07c :/
 
the book solves it by 4*(the original KE, 8*10^-11 J) = .5 * m * v^2 then solving for v...but I really just don't see why I can't logically just straight-up double the velocity...
 

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