Principle of Energy Conservation for Two Protons?

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SUMMARY

The discussion centers on calculating the minimum kinetic energy and initial speed required for two protons to collide using the principle of energy conservation. The solution involves applying the equations for conservation of momentum and kinetic energy, leading to the conclusion that the minimum speed for each proton is 8.3E6 m/s and the minimum kinetic energy is 5.8E-14 J. The relevant equations include the potential energy formula U = kq² / 2R and the relationship between kinetic and potential energy.

PREREQUISITES
  • Understanding of elastic collisions in physics
  • Familiarity with conservation of momentum and kinetic energy equations
  • Knowledge of electric potential energy formula: U = kqQ / r
  • Basic concepts of protons and their properties, including radius
NEXT STEPS
  • Study the principles of elastic collisions in detail
  • Learn about the derivation and application of conservation of momentum equations
  • Explore the concept of electric potential energy in electromagnetism
  • Practice problems involving kinetic and potential energy calculations
USEFUL FOR

Students in electromagnetism or mechanics courses, physics tutors, and anyone interested in understanding energy conservation principles in particle collisions.

musiliu
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Homework Statement



Two protons (radius 10^-15 m) are headed directly toward each other with
equal speeds. Use the principle of energy conservation to compute the minimum
kinetic energy and initial speed each proton must have if they are to
collide.

Homework Equations


W = delta K
W = - delta U
U1 + K1 = U2 + K2

The Attempt at a Solution


I've tried drawing diagram and listing known equations but for some reason, i just can't do this..need some good tips or hints...
 
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musiliu said:

Homework Statement



Two protons (radius 10^-15 m) are headed directly toward each other with
equal speeds. Use the principle of energy conservation to compute the minimum
kinetic energy and initial speed each proton must have if they are to
collide.

Homework Equations


W = delta K
W = - delta U
U1 + K1 = U2 + K2

The Attempt at a Solution


I've tried drawing diagram and listing known equations but for some reason, i just can't do this..need some good tips or hints...

Well firstly, you would need to consider the collision as elastic, so conservation of momentum would play a role as well.

Conservation of KE will give one equation and conservation of momentum will give another equation. Form these two equations.
 
ok, for conservation of momentum i got:

mvi1 + mvi2 = mvf1 + mvf2 (vi is initial and vf is final)

and i am not sure exactly what the conservation of KE is.. is it:

(1/2 mvi1)^2 + (1/2 mvi2)^2 = (1/2 mvf1)^2 + (1/2 mvf2)^2

I am not sure what the values of initial and final velocities are.. the question does not say they start out at rest.

Also, this question is from an Electromagnetism Physics class right before we learned about electric potential.. so I am not sure why the teacher gave us this problem, but I always suffered in Mechanics and was not good at it at all...
 
Do you know the formula for electrical potential energy? (not potential, but potential energy)
 
yes, the electric potential energy is kqQ / r

also, for the 2 equations i wrote, i came up with vi = vf.. and I have no idea what to do with this result..since the mass and and initial velocity of each proton is the same, i just kept canceling out variables..

but I still don't understand how to solve this problem, what do i do with vi = vf?

i don't understand what i am trying to find: what is "minimum kinetic energy" and initial speed the proton must have to collide?
 
OK, i got help from a tutor, and this is how i solved the problem:

i used the potential energy at collision : U = kq^2 / 2R , where R is the radius of the proton
and then i set potential energy = kinetic energy of system

mv^2 = kq^2 / 2R

and found minimum v = 8.3E6 m/s and minimum KE for each proton = 1/2 mv^2 = 5.8E-14 J

is this answer and strategy correct?
 
Yep, that's the right way to do it.
 

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