Recent content by HarleyM

thats how I interpreted it as well, but I believe I got the question right. I guess its all about how you explain the answer.

Would impulse in this case be equal to ΔPx=(Fnetx)(Δt) ΔPy=(Fnety)(Δt) SO I have an impulse in the x and y dimensions? Can someone please shed some light for me!

BUMP I didn't get full marks for this question, Where exactly do I calculate impulse.. Is it FnetΔt=M(V1-V2) FnetΔt=answer for impulse ? and net force is equal to Fnet= answer for net force?

I don't think R is squared in electric potential energy , only in electric force and field? So are you guys calculating electric force? Its unneeded for the answer ( I just need distance of closest approach) I am just curious thanks everyone for the help!

Did you see this?

OK I found out how to get 8.6 I was just confusing myself, now how are you getting 10 N? I used (9x10^9)(3x10^-4)2/8.6 EE= 94.18 N

so I get 10.67 m/s as velocity at minimum separation M1v1+M2V2=(M1+M2)V v=10.67 m/s EK1+EK2+EE=EE+(M1+M2)(10.67) 1/2m1v12+1/2m2v22 + Kq2/20= kq2/r+1/2(M1+M2)V ( I didn't square V here) R= 7.75 m What am I doing wrong?

Should it look like this? KE (@20 m) + EE(@ 20 m) = KE(@ distance of closest approach) + EE (solve for r) and KE= (M1+M2)V2 ( on both sides of the equation)?

Ok thanks!

How did you get to 8.6 m separation?this equation? EE+EΔK=EE+EK Did you also calculate the mass and momentum as one singular object i.e (M1+M2)V2 for the right side of the equation ?I really want to understand this question so I appreciate any help you can offer a lot!

40.5 n?

I am also wondering is there electric potential energy at 20 m away so it should be EE + EΔK=EE This gives me an answer of 7.3 m, a little better but that still seems huge.. I also get 11.5 m sometimes as well. Im really confused here.

Im still working at this question, 12.7 M seems like way to big of a distance to be the minmum distance these pucks come between each other. Am I doing something wrong in my calculations/equations?

Homework Statement Examine the diagram provided, Sphere 1: q=3.0x10-4C Sphere 2: q=-3.0x10-4C Sphere 3: q= 3x10-4C a) Find the total electric potential energy of the charge distribution b) Find the total electric potential at point Z Homework Equations EE=kq1q2/r...

Uh, how is vertical velocity found, its driving me insane. I know you sort of explained it but I've tried to find it, and tried to get the same number as you but I can't. If you can explain it a little further that would be great ! thanks