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I think it's a conservation of energy problem

by Seraph404
Tags: conservation, energy
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Seraph404
#1
Feb15-09, 03:06 PM
P: 64
1. The problem statement, all variables and given/known data


An alpha particle with kinetic energy 10.5 MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L = p_0 b, where p_0 is the magnitude of the initial momentum of the alpha particle and b=1.0010−12 m. (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

a) find the distance of closest approach

2. Relevant equations

conservation of momentum and energy, I guess; formula for electric potential energy

3. The attempt at a solution


[Edit] Scroll to last post to see my attempt

I definately need some hints on this one. I don't think we've said anything about angular moment of particles during the physics course I'm in. I did a google search to refresh my memory on angular moment, and I recall that it's basically rotation.

Well, in this problem, I can't figure what's rotating around what. Or how this brings on a collision. I appreciate any help. However, I have solved a problem kind of like this that involved conservation of linear momentum and energy to find the closest distance two particles of the same charge got before jutting off in the opposite directions due to their opposing force on one another. But how can I apply the concept of angular momentum?
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tiny-tim
#2
Feb15-09, 03:49 PM
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Quote Quote by Seraph404 View Post
Well, in this problem, I can't figure what's rotating around what.
Hi Seraph404!

Nothing's rotating

any moving object has angular momentum about any point not on its line of travel

and its angular momentum is conserved

angular momentum = speed times perpendicular distance.
Seraph404
#3
Feb15-09, 05:36 PM
P: 64
Wait... does the alpha particle cause the lead nucleus to rotate?

tiny-tim
#4
Feb15-09, 05:38 PM
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I think it's a conservation of energy problem

Quote Quote by Seraph404 View Post
Wait... does the alpha particle cause the lead nucleus to rotate?
You can assume not
Quote Quote by Seraph404 View Post
Assume that the lead nucleus remains stationary and that it may be treated as a point charge.
Seraph404
#5
Feb15-09, 05:40 PM
P: 64
So then it's just approaching at an angle then? That's the only difference between this and what I described?
tiny-tim
#6
Feb15-09, 05:53 PM
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They're both charged, so the path will curve, but at closest approach, it will be moving perpendicular to the line joining them, which is all you need to know about the path.
Seraph404
#7
Feb15-09, 07:50 PM
P: 64
So far is this the correct setup?

L1 = L2
P0b = p(Rmin) [Rmin is the minimum distance the two particles get to each other]
v0b=vf(Rmin) [mass doesn't change so I canceled it]

K0+U0 = Kf+Uf

K0 = .5mV0^2 = 1 mil(1.602E-19) J [ I can use this to find initial velocity, assuming that I can use atomic mass given from the periodic table and divide it by avogadro's number to find the mass of a single atom (one source on the Internet said I could ignore the mass of the electrons - that all the mass is really contained in the nucleus)]
U0 = kq1q2/b [would the charge on on the alpha particle be the atomic # times the charge of one proton? and similar for the second particle?]
Kf = .5mVf^2 [ I think I can change my expression for angular momentum and substitute in for this Vf so that I only have one unknown variable]
Uf = kq1q2/ Rmin

Then I think I can plug in "v0b/Rmin" for Vf. And then I get my terms containing Rmin on one side and everything else on the other. The plug in numbers and take the reciprocal of each term on each side of the equation so that I have a quadratic allowing me to solve for Rmin.
Will this work? I'm working on it now to see, but if this set up is wrong, it sure would save a lot of time to know beforehand.

[Edit] Well, I tried it and I ended up with a negative distance of much greater magnitude than what I started out with. So now that I've expended my best guess, any other ideas on how to approach this problem?
tiny-tim
#8
Feb16-09, 02:39 AM
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Just got up …

Everything looks ok , except …
Quote Quote by Seraph404 View Post
K0 = .5mV0^2 = 1 mil(1.602E-19) J

U0 = kq1q2/b
where does 1 mil(1.602E-19) come from?

and U0 = 0 (you can take it that the alpha particle starts so far away that it's effectively at infinity … the physical significance of b is only that it's the distance by which it would miss the nucleus if there was no repulsion).
Seraph404
#9
Feb16-09, 01:40 PM
P: 64
I looked up on google what MeV stood for and I saw that it was 1 electron volt times a million. Maybe my source was wrong *shrug*.

So is b significant at all in finding the closest distance that the alpha particle comes to the lead nucleus?
tiny-tim
#10
Feb16-09, 01:44 PM
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Quote Quote by Seraph404 View Post
So is b significant at all in finding the closest distance that the alpha particle comes to the lead nucleus?
b is essential to the angular momentum equation, but you can ignore it for the energy equation.


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