Recent content by davidpotts

Yes, I see now that I shouldn't have expressed a general formula that way. Another mistake is that "we need to solve for the initial Ti of the single, moving proton that will produce the same Ef as we just found in the colliding beam case" should read "...that will produce an RE of the same...

For the solution to this problem, see the extended conversation in this thread: https://www.physicsforums.com/showthread.php?p=3488768

Ahh! Much quicker technique. And it gets the same answer as I was getting with my more laborious technique (after correcting the boo-boo you found, and thanks for that). The Science article doesn't provide much illumination beyond explaining why one would be interested in this sort of...

Sorry, but I'm not getting you. Your formula was: "E2=p2c2+m02c4 (m0 is the rest mass)". And my formulas all use "m" for rest mass, multiplying by γ where needed. What am I doing wrong? I got that Science article by the way. I'll read through it tomorrow and see if I can see what's up.

So is the idea the following? Compute the initial kinetic energy from T = mc2(γ - 1). So 3 GeV = 938(γ - 1) MeV (.001 GeV/MeV) 2.062 = .938γ γ = 2.1983 So 1/(1 - V2/c2).5 = 2.1983 1 - V2/c2 = .2069 V = .89c Now use conservation of momentum. Pi = Pf mγV = Pf...

Yeah, I figured this one out. (Sorry, I suppose I should've posted something.) The trick is that since the total energy before the collision, Ei, which is the kinetic energy of the incoming particle, should be just enough to cause the ionization of the helium atom, we can assume it is as low as...

Homework Statement The energy required to ionize a helium atom is 24.6 eV. If helium gas is bombarded with charged particles, at what incident energy will ionization of helium just be observed if the projectiles are (a) alpha particles (b) protons (c) electrons? Homework Equations Most...

Homework Statement What fraction of the energy of a rapidly moving proton is not available for inelastic interactions in proton-proton collisions when the target proton is at rest in the laboratory and the energy of the accelerator is (a) 3 GeV (b) 7 GeV (c) 25 GeV (d) 200 GeV (e) 1000 GeV...

Yes, I did look these up in one of my math books and read the proof of the formula for cos(α+β). Thanks for the tip!

No, I didn't know that formula. Though I might have looked it up and spared myself a flight of logic! So from the formula and the fact that cosθ1cosθ2-sinθ1sinθ2=0, we have cos(θ1+θ2)=0, so θ1+θ2=pi/2, which is our goal. Nice!

Excellent. Thanks for this. (And I like the coding of your notation too!) So keeping your symbols, it goes like this: U2 = v12cos2θ1 + 2v1v2cosθ1cosθ2 + v22cos2θ2 0 = v12sin2θ1 - 2v1v2sinθ1sinθ2 + v22sin2θ2 So U2 = v12cos2θ1 + 2v1v2cosθ1cosθ2 + v22cos2θ2 + v12sin2θ1 -...

I don't know. It would certainly be nice. v1 = v1cos(theta1) - v1sin(theta1) v2 = v2cos(theta2) + v2sin(theta2) So v1 + v2 = v1cos(theta1) + v2cos(theta2) - v1sin(theta1) + v2sin(theta2) Because we have v2sin(theta2) - v1sin(theta1) = 0, the last two terms sum to 0. And because we...

Thanks for this reply -- it certainly helped to reduce the clutter in what I wrote! Obviously, from V^2 = v1^2 + v2^2, we can see that IF theta1 + theta2 = pi/2, then the magnitude V is the length of the hypotenuse of the right triangle formed from sides v1 and v2. And the converse is also...

Homework Statement Prove that in an elastic collision between two identical bodies, one of which is initially at rest, the angle between their velocities after collision is pi/2, except for the case of central impact. Homework Equations Assume particle 1 travels in the x direction with...

Is everybody just supposed to know that m(neutrino) = 0? Maybe so, I have very little formal physics training... Anyway, so using the relativistic equations, we can get: T(muon) = 4.1 = E(muon) - m(muon)c^2 = 4.1 = E(muon) - (105.7 / c^2)(c^2) So E(muon) = 4.1 + 105.7 = 109.8 MeV...