Energy and charged particle collisions

AI Thread Summary
The discussion revolves around calculating the initial velocity of an alpha particle colliding with a stationary gold nucleus, emphasizing energy transformations during charged particle collisions. The participant initially calculated the velocity using kinetic energy equal to electric energy but questioned its accuracy. Clarification was provided that the initial velocity refers to the particle's speed when far from the nucleus, where electric potential energy (EPE) is negligible. The final energy at the closest approach is considered entirely EPE due to the stationary nature of the gold nucleus. Understanding these energy dynamics is crucial for accurately determining the initial velocity.
krbs
Messages
51
Reaction score
3

Homework Statement


Find the initial velocity of an alpha particle with a mass of ##6.64 \times 10^{-27} kg## and a charge of ##+3.2 \times 10^{-19} C##, if it undergoes a head on "collision" with a gold nucleus. You may assume the gold nucleus does not move at all during the interaction. The charge on the gold nucleus is ##+2.53 \times 10^{-17} C## and the distance of closest approach between the two is ##4.7 \times 10^{-15} m##.

Homework Equations


##K_E\ =\ E_E##

or

##K_E\ +\ E_E\ =\ E_E##?

The Attempt at a Solution


I've already calculated initial velocity using ##K_E\ =\ E_E## and got ##6.8 \times 10^{7} m/s## but I suspect this is incorrect.

I'm having trouble understanding energy transformations during collisions of charged particles. I think that by "initial velocity" they mean just before the particles collide? In which case I should take into account EPE for the initial total energy. I think the final energy at the instant of closest approach must be entirely EPE because the gold nucleus is stationary and so the particles' combined velocity must be zero.
 
Physics news on Phys.org
krbs said:
I'm having trouble understanding energy transformations during collisions of charged particles. I think that by "initial velocity" they mean just before the particles collide? In which case I should take into account EPE for the initial total energy. I think the final energy at the instant of closest approach must be entirely EPE because the gold nucleus is stationary and so the particles' combined velocity must be zero.

You're correct on the second point, but not the first. If two particles repel each other, then (unless one has an initial velocity towards the other), they will simply move further apart. Unlike gravity, however, where the velocity gets larger until the two objects actually physically collide, the velocity during repulsion will decrease until the smaller particle stops and then goes back the way it came.
 
Last edited:
So initial velocity refers to the velocity of the particle when it's far away from the other, and EPE is negligible?
 
krbs said:
So initial velocity refers to the velocity of the particle when it's far away from the other, and EPE is negligible?
Yes, you have to assume EPE is negligible initially; otherwise, where it starts will affect the minimum distance.
 
Okay, thanks for your help
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top