Time it takes for an electron and proton to collide

AI Thread Summary
The discussion centers on calculating the time it takes for an electron and positron, initially at rest and separated by distance R, to collide. The primary challenge is that the force between them is not constant due to their changing separation. Participants suggest using integration to find the position function, but note that the varying force complicates this approach. The concept of reduced mass is introduced to address the symmetry of the problem, clarifying that both particles will meet at the midpoint. Ultimately, the correct method involves adjusting the equations to account for the dynamics of the system.
gsingh2011
Messages
115
Reaction score
1

Homework Statement


An electron and positron have mass me and separation R and are initially at rest. Find the time T it takes for them to collide.


Homework Equations


F=kq1q2/R2


The Attempt at a Solution


My problem is that the force is not constant. Do I integrate the force then? And from what bounds?
 
Physics news on Phys.org
What do you get when you use F=ma?
 
vela said:
What do you get when you use F=ma?

I don't know, acceleration is changing. Let say we have F=kq1q2/r2=ma. Since this problem is symmetrical, I was thinking the two particles would meet in the middle, so each would travel a distance of R/2. So I could divide by m and integrate twice to find the position function, and set that equal to R/2. The only problem with that idea is the force is changing with respect to r, so I can't integrate with respect to t. So I guess that method won't work.
 
gsingh2011 said:
I don't know, acceleration is changing. Let say we have F=kq1q2/r2=ma. Since this problem is symmetrical, I was thinking the two particles would meet in the middle, so each would travel a distance of R/2.
The difference in masses breaks the symmetry. The proton won't move as far as the electron because it's so massive in comparison.
So I could divide by m and integrate twice to find the position function, and set that equal to R/2. The only problem with that idea is the force is changing with respect to r, so I can't integrate with respect to t. So I guess that method won't work.
The standard trick is to multiply by \dot{r}. You'll find it makes both sides of the equation integrable, but first you need to find the correct equation.

Are you familiar with the concept of the reduced mass?
 
Last edited:
The proton won't move as far as the electron because it's so massive in comparison.
It's a positron not a proton, so the mass is the same.
 
Oops, not sure why I thought it was a proton. Yes, you're right, they'll meet in the middle.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top