Proton is moving at an alpha particle, distance when it stops

AI Thread Summary
A proton is shot towards a fixed alpha particle from a distance of 2.00 m at a speed of 3.46x10^6 m/s, and the goal is to determine how close it will come before stopping. The relevant equations for this problem involve kinetic energy (Ek) and electric potential energy (Ee), suggesting that Ek equals Ee at the point of closest approach. The kinetic energy can be expressed as 1/2mv^2, while the electric potential energy can be represented by kq/r^2. The conservation of energy principle is crucial in solving this problem. Understanding these concepts will help in determining the proton's closest distance to the alpha particle.
niveda
Messages
3
Reaction score
0

Homework Statement


an alpha particle with a charge of 2e is fixed at the origin. a proton is aimed at the alpha particle and shot from a distance of 2.00 m at a speed of 3.46x10^6 m/s. How close will the proton come to the alpha particle before coming to a stop?

Homework Equations


Please help I need to know this for my test!

The Attempt at a Solution


Ek=Ee (maybe ?) [/B]
 
Physics news on Phys.org
A plea for help is not a relevant equation. Follow through on your suggested solution approach: What equations are relevant here?
 
So is it correct that Ek=Ee?
If that's the case then 1/2mv^2 and kq/r^2 would be relevant
 
niveda said:
So is it correct that Ek=Ee?
The basic idea is correct. You'll have to expand on it though; What conservation rule is involved?
If that's the case then 1/2mv^2 and kq/r^2 would be relevant
Yes.
 
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