# Acceleration of Proton (Kinetic Energy & relativity)

## Homework Statement

a)
Calculate the kinetic energy required to accelerate a single proton from a rest position to 0.9999c. The mass of a proton is 1.67*10^-27 Kg.

b)
Find the ratio of kinetic energy to the energy of a proton at rest

## Homework Equations

Ekrest = mc2

Ek= (mc2)/√(1-(v2/c2))-mc2

## The Attempt at a Solution

a)
Ekrest = mc2
Ekrest = (1.67*10^-27)(c)2
Ekrest = 1.5*10^-10

Ek= (1.5*10^-10)/√(1.9999*10^-4)-(1.5*10^-10)
EK=1.047*10^-8J

I am not confident in this answer as that does not seem nearly enough energy to accelerate the proton...

B)
1.047*10^-8/ 1.5*10^-10
=69.66%

I am very confused! Any help is much appreciated.

collinsmark
Homework Helper
Gold Member

## Homework Statement

a)
Calculate the kinetic energy required to accelerate a single proton from a rest position to 0.9999c. The mass of a proton is 1.67*10^-27 Kg.

b)
Find the ratio of kinetic energy to the energy of a proton at rest

## Homework Equations

Ekrest = mc2

Ek= (mc2)/√(1-(v2/c2))-mc2
I think you might find it easier for this problem to first define gamma, $\gamma$ as

$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

and then just define the kinetic energy as

$$E_k = \left( \gamma - 1 \right) mc^2.$$

But of course, it's up to you.

## The Attempt at a Solution

a)
Ekrest = mc2
Ekrest = (1.67*10^-27)(c)2
Ekrest = 1.5*10^-10

Ek= (1.5*10^-10)/√(1.9999*10^-4)-(1.5*10^-10)
EK=1.047*10^-8J
That looks about right to me! Very nice.
I am not confident in this answer as that does not seem nearly enough energy to accelerate the proton...
Protons are pretty small, don't forget.
B)
1.047*10^-8/ 1.5*10^-10
=69.66%
Ignoring a rather minor difference in rounding errors (between your result and mine), why in the world did you throw on a "%" at the end?

(That "%" is throwing you off by two orders of magnitude. )

Thank you very much!! and I honestly have no idea why I put the % haha thank you :)