Double Speed: Proton Acceleration with 2000MV Potential

[channel1]

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?

 

[SammyS]

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.

 

[channel1]

Yet the answer is 2.07c :/

 

[channel1]

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...