Acceleration of Proton (Kinetic Energy & relativity)

[Lauren12]

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

Ek_rest = mc²

Ek= (mc²)/√(1-(v²/c²))-mc²

The Attempt at a Solution

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

Ek_rest = 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)
Ek_rest = mc²
Ek_rest = (1.67*10^-27)(c)²
Ek_rest = 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. )

 

[Lauren12]

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