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Ahhh back in the linear accelerator days. simpler times 😂

I said for electrons: Davisson germer, Millikan, Thomson. and for Higgs Boson: 2012 CERN synchrotron

Id say high seventies to very low eighties, most likely high seventies in my opinion though.

Also don't worry. looking at the general consensus, everyone was baffled by the paper. You've got this! :)

If it'll make you feel better I had b44's '90's song "get down" stuck in my head. So whenever I encountered a problem I just imagined the physics paper getting 'down on me'... can't get worse than that. DO NOT LISTEN to '90's music before an exam, please DOONNNT!!

I was expecting two 9 markers, maybe that’s why. The actual nine marker provided also had a calculation element to it so it withdrew on the theory a bit. I just feel it was easier, because I hate waffling about theory, I prefer actual mathematical analysis.

Also, did anyone feel the sample questions were a lot harder than the actual questions provided?

Don't worry man, happens to the best of us. Also, you probably still are getting a band 6 anyway. And I doubt you’ll lose more than 1 mark.

I was going to mention tension, however I didn't because I felt it was unecessary to the net forces, because a component of the centripetal force really is a manifestation of the tension in the string. Although, the oblique linear force (net force) is due to tension.

yes lol, I corrected myself in the above post. I was too quick and doubted myself.

Yep. realised that, I didn't realise you commented before I posted

Also @Qiaochu Chen you have to look at F_c from the point of the mass

Hang on... that can't be right actually @Qiaochu Chen, I saw the dot point book had a pretty much identical question, and they obtained the same result as using \tan\theta=\dfrac{v^2}{rg}. Did they do it wrong as well? I'm confused now.

Hmm. Seems like Jimmy got the right answer though, we didn't consider the individual masses.

Fictitious force as in inertial force or centrifugal? Because I saw the Dot Point book use inertia for a similar question.

Oh that’s right! Damn, I didn't consider that.

@Qiaochu Chen this is how I did it. By resolving forces, we see that the centripetal force points inwards, and the bead experiences a downward acceleration. \tan\theta=\dfrac{F_c}{F_g} \tan\theta=\dfrac{\left(\dfrac{mv^2}{r}\right)}{mg} \tan\theta=\dfrac{v^2}{rg}...

Yep, a lot more! there was a good balance between theory, skill and calculations

No

Maybe if I’m lucky 😂 but idek at this point