Two charged particles fired at each other.

AI Thread Summary
A proton and an alpha particle, each with an initial speed of 0.172c, are analyzed for their distance of closest approach when fired toward each other. The initial kinetic energy is calculated using the combined mass and speed of the particles, while the potential energy is derived from their charges and distance. The solution initially assumed that all kinetic energy converts to potential energy, leading to an incorrect calculation of the closest distance. The realization that momentum is conserved indicates that the final kinetic energy is not zero, prompting a reevaluation of the approach. This highlights the importance of considering both energy and momentum conservation in particle interactions.
mellotron
Messages
2
Reaction score
0
[SOLVED] Two charged particles fired at each other.

Homework Statement


A proton and an alpha particle (q = +2.00e, m = 4.00u ) are fired directly toward each other from far away, each with an initial speed of 0.172c. What is their distance of closest approach, as measured between their centers? (Hint: There are two conserved quantities. Make use of both.)

Homework Equations



E_{i}=\frac{1}{2}(m_{p}+m_{\alpha})v^{2} + k_{e}(\frac{q1q2}{\infty})E_{f}=\frac{1}{2}(m_{p}+m_{\alpha})0^{2} + k_{e}(\frac{q1q2}{r})

The Attempt at a Solution



As I've shown with the equations above, from my understanding, initially the particles only have kinetic energies since they are an infinite distance away. As they get closer their energies will be converted to potential energy and their velocities will momentarily reach 0 when all the energy is potential, which would give us r.

Subbing in values I get 1.11E-11 J for Ei
So I set Ei = Ef do some algebra and get
r = k_{e}(\frac{q1q2}{E_{i}})

When I do this I get 4.16E-17 m, but this is wrong (according to CAPA). What did I miss?
 
Last edited:
Physics news on Phys.org
Figured out that momentum is conserved so I was wrong about my assumption that final kinetic energy would be 0, which in hindsight makes lots of sense.
 
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