Asked by maddie
Two protons are moving directly toward one another. When they are very far apart, their initial speeds are 1.5e6 m/s. What is the distance of closest approach?
PLEASE HELP! im very confused.
PLEASE HELP! im very confused.
Answers
Answered by
drwls
It will be at a value of the separation distance R such that the initial kinetic energy, 2*(1/2)M V^2, is equal to the potential energy k e^2/R.
Note that you have to add the kinetic energies of the two particles, but the potential energy only has to be counted for bringing one particle within a distance R of the other.
R = ke^2/[M V^2]
k is the Coulomb constant (which you may need to look up) and e is the proton charge (same as the electron charge, except for sign). M is the proton mass, and V = 1.5*10^6 m/s
Note that you have to add the kinetic energies of the two particles, but the potential energy only has to be counted for bringing one particle within a distance R of the other.
R = ke^2/[M V^2]
k is the Coulomb constant (which you may need to look up) and e is the proton charge (same as the electron charge, except for sign). M is the proton mass, and V = 1.5*10^6 m/s
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.