# Kinetic Energy Needed for Proton Acceleration

• fatcats
In summary, the homework statement is that Ek = Etotal - Erest and Ek/Erest. The Attempt at a Solution includes the calculations on a word document that are cross-referenced with others. The question is worth one mark and the student is stuck.
fatcats

## 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 the kinetic energy to the energy of a proton at rest.
c) Explain why no particle accelerator will ever be able to accelerate any particle to the speed of light.

## Homework Equations

a) Ek = Etotal - Erest

Ek = mc^2/sqroot 1 - v^2/c^2 - mc^2

b) Ek/Erest

## The Attempt at a Solution

a) I am including the calculations I have done on my word document page to make it more legible than typing it in here. Please see the attached image.

I think my work is right, but I have cross-referenced this answer with others and the math does not seem to be right... it seems to be correct until at least the step where:

Ek = 1.50 * 10^-10 / sqroot(1.9999*10^-4) - 1.50 *10^-10

After that I am not sure, but the answer I keep getting is 9.11 * 10^-10, and other answers, deemed correct have a totally different number multiplied by 10^-8...
I have been reworking this and trying to figure out where/if my calculation is wrong and I'm stuck.

b) For this question, all I do is divide the two numbers for the ratio, correct? This question is worth three marks and it seems like they want something more than just this? Am I missing something?

c) Not sure how to word this. Here's my best attempt. This question is worth one mark.

No particle accelerator will ever be able to accelerate any particle to the speed of light because according to the denominator of the aforementioned equation, sqroot (1-v^2/c^2), the speed of light acts as a limit to velocity. If velocity were to equal the speed of light, then there would be no difference in resting and kinetic energies.

Please help me understand, I am doing this course online and don't really have a teacher who can help me out. Thanks for your time.

#### Attachments

• Screen Shot 2018-03-21 at 11.02.25 AM.png
8.3 KB · Views: 853
In your penultimate step you have divided by 0.141... instead of 0.0141...
fatcats said:
If velocity were to equal the speed of light, then there would be no difference in resting and kinetic energies.
No - if velocity = c, then KE is infinite.

Oh! Is that my only mistake with calculations? Thank you so much.

Why is KE infinite with velocity = c? Wouldn't that make the denominator 1? I don't understand

fatcats said:
Why is KE infinite with velocity = c? Wouldn't that make the denominator 1? I don't understand
It's hard to read your equations -- consider learning to use LaTeX for your equations when posting at the PF. There is a LaTeX Primer under INFO (at the top of the page), Help/How-To.

When written out as an equation, it's easy to see how the denominator goes to zero if v = c

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/imgrel/rke2.gif

#### Attachments

• rke2.gif
1.5 KB · Views: 608
I usually upload my equations in a neat screenshot, but I will look at Latex

I don't understand why it is infinity.

(c)^2/(c)^2 = 1
1 - (c)^2/(c)^2 = 0
squareroot of 0 = 0

so how is that infinity?

fatcats said:
I usually upload my equations in a neat screenshot, but I will look at Latex

I don't understand why it is infinity.

(c)^2/(c)^2 = 1
1 - (c)^2/(c)^2 = 0
squareroot of 0 = 0

so how is that infinity?
The zero is in the denominator...

Yes, I understand that, but how does dividing by zero make it infinity? Doesn't that make a number undefined?

I was always taught that means it's unsolvable

fatcats said:
Yes, I understand that, but how does dividing by zero make it infinity? Doesn't that make a number undefined?

I was always taught that means it's unsolvable
Well that may be true (I'm not a math expert), but think about what the value of that fraction does as v --> c from below. The KE rises without bound, right? That's the point, that the KE increases without bound as you try to get closer and closer to c (at least when you approach it from speeds v < c).

Okay, I think I understand
If you look at it as an equation, it's harder to see, but if you visualize it at as a graph that makes sense to me

berkeman

## What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion.

## How is kinetic energy calculated?

The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

## Why is kinetic energy needed for proton acceleration?

Kinetic energy is needed for proton acceleration because it allows the protons to overcome the repulsive forces between them and the positively charged nucleus in order to reach high speeds.

## What factors affect the amount of kinetic energy needed for proton acceleration?

The mass and velocity of the protons are the main factors that affect the amount of kinetic energy needed for their acceleration. Other factors include the strength of the electric fields, the distance the protons need to travel, and any external forces acting on the protons.

## How is kinetic energy used in proton acceleration experiments?

In proton acceleration experiments, kinetic energy is used to accelerate the protons to high speeds before they collide with a target material, producing high-energy particles and radiation that can be studied by scientists to gain insights into the structure and properties of matter.

• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
54
Views
8K
• Introductory Physics Homework Help
Replies
21
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
8K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
362
• Introductory Physics Homework Help
Replies
20
Views
4K
• Introductory Physics Homework Help
Replies
44
Views
849
• Introductory Physics Homework Help
Replies
2
Views
326