Where Will a Proton Stop Before Turning Around in a Collision?

In summary: I have never seen any physics problem where it would be used instead of the elementary charge. I hope you understand the physics of this problem as well as you understand your calculator.In summary, the problem involves finding the position where a proton with an initial speed of 1.23x10^4 m/s stops momentarily before turning around while moving towards another fixed proton. Using the conservation of energy and the equations for initial kinetic energy and electric potential energy, the distance between the two protons can be calculated to be 1.821x10^-9 meters.
  • #1
loganblacke
48
0

Homework Statement


A proton with a speed of 1.23x104 m/s is moving from ∞ directly towards another proton. Assuming that the second proton is fixed in place, find the position where the moving proton stops momentarily before turning around.


Homework Equations


Potential Energy - U/q = V


The Attempt at a Solution


Clueless.
 
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  • #2
The proton is moving initially so it has kinetic energy.
You must think about what energy change takes place as it is repelled by the other proton. The calculation itself will be easy.
 
  • #3
Assuming that the second proton is fixed in place

...in a little proton vice??
 
  • #4
Delphi51 said:
The proton is moving initially so it has kinetic energy.
You must think about what energy change takes place as it is repelled by the other proton. The calculation itself will be easy.

Okay so conservation of energy, I get that.

Ek = .5(1.67x10-27kg)(1.23x104m/s)2

When the proton stops, Ek = 0, I'm not understanding how this relates to the position of the proton when it stops.
 
  • #5
It lost kinetic energy. But energy is conserved; it must have changed into some other kind of energy. It is the energy due to electric charges being near each other and is called electric potential energy. It depends on the distance between charges.

Write "initial kinetic energy = electric potential energy when stopped"
Put in the detailed formulas for both types of energy. Wikipedia has the potential energy one here: http://en.wikipedia.org/wiki/Electric_potential_energy
Then you can solve for the distance between charges.
 
  • #6
Delphi51 said:
It lost kinetic energy. But energy is conserved; it must have changed into some other kind of energy. It is the energy due to electric charges being near each other and is called electric potential energy. It depends on the distance between charges.

Write "initial kinetic energy = electric potential energy when stopped"
Put in the detailed formulas for both types of energy. Wikipedia has the potential energy one here: http://en.wikipedia.org/wiki/Electric_potential_energy
Then you can solve for the distance between charges.

I really really appreciate your help with this but it appears as if there are 50 equations for electric potential energy. I'm assuming the initial kinetic energy would be the 1/2mv2, however it isn't obvious to me which equation to use for the electric potential energy..
 
  • #7
I tried using 1/2mv2 = q*(1/4∏ε0)(Q/r)... and then solving for r but the answer I ended up with was wrong.
 
  • #8
It is the first equation in the Wikipedia article (link in previous post).
Your choice of the form with k or the form with 1/(4∏ε0).
 
  • #9
I tried using both equations and I'm still getting the wrong answer. Here is what I got..

(1/(4∏*(8.854*10-12)))*((Q1*Q2)/r)=.5(1.67*10-27)(1.23*104)2

Q1 and Q2 would be the charge of each proton which I think is 1.6*10-19

If i solve for r, I get 4.92*1018.

Another website said that the charge of a proton is e, so I tired that as well and got 1.89*10-30.

Both answers are not correct.
 
  • #10
Your calc looks okay but I don't get the same answer.
The elementary charge IS 1.6 x 10^-19, so you should not have got a different answer that way!
You know, it is much easier to solve the equation before putting the numbers in - just easier to manipulate letters than lengthy numbers.
k*e²/R = ½m⋅v²
R = 2k*e²/(m⋅v²)
Putting in the numbers at this stage, I get an answer in nanometers.
 
  • #11
Delphi51 said:
Your calc looks okay but I don't get the same answer.
The elementary charge IS 1.6 x 10^-19, so you should not have got a different answer that way!
You know, it is much easier to solve the equation before putting the numbers in - just easier to manipulate letters than lengthy numbers.
k*e²/R = ½m⋅v²
R = 2k*e²/(m⋅v²)
Putting in the numbers at this stage, I get an answer in nanometers.

Finally got it right, 1.821 x 10-9

Apparently e and e on the graphing calculator are two very different things!
 
  • #12
Looks good! Yes, the e on the calculator is 2.718, the base of the natural logarithm.
 

What is the "Proton collision problem"?

The "Proton collision problem" refers to the issue of understanding and predicting the behavior of protons when they collide with each other at high energies, such as in particle accelerators. This is a fundamental problem in the field of particle physics.

Why is the "Proton collision problem" important?

The study of proton collisions allows us to gain insight into the fundamental building blocks of matter and the fundamental forces that govern their interactions. This knowledge can help us understand the origins and evolution of the universe.

What are the challenges of studying the "Proton collision problem"?

The main challenge of studying the "Proton collision problem" is the complexity of the interactions between protons at high energies. These collisions produce a wide range of particles and phenomena, making it difficult to accurately measure and interpret the data.

How do scientists study the "Proton collision problem"?

Scientists study the "Proton collision problem" by conducting experiments at large particle accelerators, such as the Large Hadron Collider (LHC) at CERN. These experiments involve accelerating protons to high energies and observing the particles and energy produced when they collide.

What are some potential applications of understanding the "Proton collision problem"?

Understanding the "Proton collision problem" has many potential applications, including advancements in medical imaging, development of new technologies, and a deeper understanding of the universe. Additionally, it can also contribute to the development of new theories and models in particle physics.

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