A problem with electric potential and conservation of energy

AI Thread Summary
The discussion centers on calculating the work needed to move two protons from an atomic distance to a nuclear distance and determining their speed when released. The user initially misapplied energy conservation principles by treating one proton as stationary, leading to an incorrect velocity calculation. After realizing both protons share kinetic energy upon release, the user correctly calculated the work done as 7.6729*10-14 J. The importance of conserving both energy and momentum when dealing with particles of different masses was also highlighted. Ultimately, the user successfully solved the problem after clarifying their approach.
MSZShadow
Messages
9
Reaction score
0

Homework Statement


A) How much work would it take to push two protons very slowly from a separation of 2.00*10-10 m (a typical atomic distance) to 3.00*10-15 m (a typical nuclear distance)?

B) If the protons are both released from rest at the closer distance in part A, how fast are they moving when they reach their original separation?

Homework Equations


Eq 1: Ka + Ua = Kb + Ub
Eq 2: Wa->b = Ua - Ub

The Attempt at a Solution


Using Eq 1: to solve for vb:
0 + Ua - Ub = (1/2)mvb2
sqrt( 2(Ua - Ub)/m ) = vb

Where m = 1.67*10-27 kg, the mass of a proton.

From here, I plug in values and get vb = 9.58598*106 m/s.
Naturally, this is wrong.

I think the error is in that I worked the problem as though one of the protons is held still (which I'm sure is fine for part A, but apparently not B). If this is the case, how should I be handling this problem?

Edit: Whoops. Forgot to re-add this in when my first thread submission attempt got fried...
I've already calculated Wa->b = 7.6729*10-14 J. This, according to the homework problem, is correct.

Edit2: Epic fail. Accidentally posted in the wrong section. This should go under Introductory Physics, as it's just introducing calculus-based physics. If someone could move this thread to there, that would be appreciated.
 
Last edited:
Physics news on Phys.org
When you release the protons in part B, both protons have kinetic energy. Therefore the change in potential energy is equally divided between the two protons. This is true because the masses of the two particles are the same. If they weren't, i.e. if you had a proton and a deuteron or alpha particle, you would have to conserve both energy and momentum to determine how the potential energy change is shared between the particles.
 
Thanks. I had the feeling it was something like that, but I'm always so insecure with physics, so I doubt my thoughts on it too frequently...

I've successfully solved the problem now. :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top