Recent content by AsadaShino92

(E0-E1)2=E02-2E0E1+E12

My mistake. Then squaring both sides should be E22=(E0-E1)2

You mean instead I should use E22=E02-E12?

Using the notation, I have the conservation of energy written as E0=E1+E2 Then using the energy equation for each particle For the K meson E02=P02C2+M2C4 For the Pi stopped E12=m2C4 since P1=0 For Pi moving E22=P22C2+m2C4 Momentum conservation P0=P1+P2 Since P1=0 P0=P2 Do I just plug all...

So am I using this equation to solve for momentum?

P=E/C ?

Then I would use p=γmv but I don't think I have the value for those. I just know that momentum would also be conserved.

Homework Statement A K meson (an elementary particle with approximately 500 Mev rest mass) traveling through the laboratory breaks up into two Pi mesons (elementary particles with 140 Mev rest energies). One of the Pi mesons is left at rest. What is the total energy of the remaining Pi meson...

Since V(R) is given as the potential energy function in the problem, I can find my force function by using F(x)=-du/dx. Is that the right idea?

So then the equilibrium value of R is at R=(B/A)? I found this by leaving A and B as variables and setting V(R)=0.

Thanks for your explanation. I wasn't used to this method so I couldn't see the relationship before. For part B where it asks to find the minimum, taking A=B=1, I found an equilibrium point at R=1. Would it be correct for me to use the first derivative test?

Sorry, I'm not sure how I would plot the function in those terms such as A^2/B and B/A. Do you think I can fix the values like PeroK suggested? My only concern is that the function slightly changes based on what A and B are. But I see that the function runs off to infinity as R approaches 0.

Homework Statement The potential energy V(R) of a two particle system exhibiting oscillatory behavior near a local minimum at the equilibrium separation Ro. V(R)= -(A/R)+(B/R^2) , where R is the interparticle separation. A) Sketch V(R), what happens to V(R) as R→0 B) At what value of R is...

So then I can use the fact that ΔU= Uf-Ui= ½kxf^2-½kxi^2. Where f is final and i is initial? If this is correct, then xi=0 and that term drops out. Then I would be left with ½kxf^2 = -k1½x0^2+k2⅓x0^3

Homework Statement A spring of negligible mass exerts a restoring force on a point mass M given by F(x)= (-k1x)+(k2x^2) where k1 and k2 >0. Calculate the potential energy U(x) stored in the spring for a displacement x. Take U=0 at x=0. Homework Equations ΔU=∫F(x)dx U=½kx^2 The Attempt at a...