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What happens to the kinetic energy of a speedy proton

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Dx

Hello,

What happens to the kinetic energy of a speedy proton when its relativistic masss doubles?

a) it doubles b) it more than doubles c) it less than doubles d) it must increase but impossible to say by how much

Now, what kind of question is this? Or should I ask what klind of answers are these, its like really vagues, huh?

I mean whith a steady net force applied to an object of rest mass, the object increases speed. Since its acting over a distance the work done and its KE increases but not over c (speed of light in vacuum). On the other hand the mass of the object not only increases
with increaseing speed. That is to say the work done on an object not only increases its speed but also contributes to increaseing mass. My question now is which answer would you choose?

dx
 
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HallsofIvy

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Mass increases with speed as m= m0/sqrt(1-v2/c2.

Kinetic energy is given by (1/2)mv2.

Put those together and you get ((1/2)m0)(v2/sqrt(1-v2/c2)).

If v is such that 1/sqrt(1-v2/c2= 2 (mass doubles) then the v2 part will have caused the kintic energy to be far more than double.
 
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" Kinetic energy is given by (1/2)mv2 " ?

I'm not sure about that...

Kinetic energy = m0*c^2*[1/sqrt(1-v^2/c^2)-1]...

m1=m0/sqrt(1-v1^2/c^2)...T1=m0*c^2*[1/sqrt(1-v1^2/c^2)-1]...
m2=m0/sqrt(1-v2^2/c^2)...T2=m0*c^2*[1/sqrt(1-v2^2/c^2)-1]...
m2=2*m1...
1/sqrt(1-v2^2/c^2)=2/sqrt(1-v1^2/c^2)...
Let a=1/sqrt(1-v1^2/c^2)...b=1/sqrt(1-v2^2/c^2)...
b=2*a...
T1=m0*c^2*(a-1)...
T2=m0*c^2*(b-1)...
T2/T1=(b-1)/(a-1)=(2*a-1)/(a-1)=[2*(a-1)+1]/(a-1)=2+1/(a-1)...
So... T2/T1>2...
 

Tom Mattson

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Originally posted by bogdan
" Kinetic energy is given by (1/2)mv2 " ?

I'm not sure about that...

Kinetic energy = m0*c^2*[1/sqrt(1-v^2/c^2)-1]...
Yes, you're right. KE=(γ-1)m0c2

The formula KE=(1/2)mv2is not valid at relativistic speeds.
 

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