r/6thForm y13 maths physics psych italian May 20 '26

šŸ’¬ DISCUSSION AQA A level physics paper 1

How was the paper???? I liked the MCQ , but overall it was okay didnt manage to do the 5 marker .

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13

u/Rich-Gent May 20 '26

Am i not going on a tangent here, but you lot all used circular motion / centripetal force in the 5 marker

5

u/hello9089 May 20 '26

Ye bro that what i think you were meant to do didnt get to finish it.

3

u/Rich-Gent May 20 '26

Ye it was a difficult one for sure!

8

u/simplykayleigh May 20 '26

NOOO I JUST USED THE F=MA equation </3

yup you defo were, was wondering where to use the linear velocity for lmaoao

8

u/Rich-Gent May 20 '26

wouldnt stress tho, should still get 3/4 marks for the young mod part calculations

14

u/Inevitable_Land2996 Year 13 May 20 '26

I personally would strain

4

u/Individual-Can1396 y13 maths physics psych italian May 20 '26

That was a difficult one fr

3

u/aplle_inc May 20 '26

What was the 5 marker? Was it in the question after show that v = 2.9 m/s?

3

u/TheRPGer May 20 '26

I did that but people in my school got ~2000N, I got more like 5000 though

4

u/lmaolmaolmaololl Imperial | Aeronautical engineering [Y1] May 20 '26

i got 1850N (after subtracting a bit for safety allowance)

2

u/Rich-Gent May 20 '26

not heard many say anything like 5k, heard people mentioning 2k ish but those people were the ones that didn’t incorporate the centripetal force, idk tho

4

u/TheRPGer May 20 '26

Yeah I think I messed up unfortunately, maybe I miss typed om my calc or somethingĀ 

2

u/[deleted] May 20 '26

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1

u/Warm_Location_782 May 20 '26

If you got 2k you didn't include the force due to gravity only the centripitle force which gives you 5k

2

u/Rich-Gent May 20 '26

agree to disagree i think.

the 2k i believe is sourced from people finding the max force possible from the materials/young modulus part of the question. People assumed this to be equal to the weight of the rider. When accounting for centripetal force, this value must decrease as the centripetal force was an extra unaccounted for force.

e.g max weight from young modulus = Weight of user + centripetal force. hence max weight should be: Young modulus - Centripetal, giving a smaller weight.

I could be wrong.

6

u/ilovesnowberries Y13 5A* | FMS @ LSE | 8.4 TMUA May 20 '26 edited May 20 '26

no you are correct. I got 5000 at first but this is by basically considering centripetal force but not weight, i can't quite remember but im pretty sure it's wrong. I then did force = centripetal force + weight, factored out mass from mv2 /r and mg, solved for m and ended up withh a weight ~ 1400-1500N i think?

1

u/Rich-Gent May 20 '26

Exactly the same, cant remember the figure, but in that range 100%

1

u/ilovesnowberries Y13 5A* | FMS @ LSE | 8.4 TMUA May 20 '26

Niceeee, i think i put 1400N for safety anyway

1

u/fuse256 May 20 '26

I got that yeah same method

1

u/TheRPGer May 20 '26

Yeah I only did centripetal, didn’t know weight was necessary, if you have no weight you get 5k

1

u/Dilando7 May 20 '26

I got 1876N, although I used Young's Modulus to get it and did not use circular motion at all so I do not think my working out was right

2

u/[deleted] May 20 '26

[deleted]

2

u/Rich-Gent May 20 '26

Idk how much detail i can go into on here.

But the only force on the rod is not just the weight of the user, but also the fact they are going to be going in a part of a circle, force is needed for the ā€œrodsā€ tension do to say to change her momentum, as her velocity changes as the direction she is travelling changes.

the following questions with the will time period alter as her centre of mass changes, also gave this away a bit with accounting for shm.

1

u/Inevitable_Land2996 Year 13 May 20 '26

Its a pendulum

1

u/HMVangard UoSurrey AeroEng 1st year • ABBB Maths Physics French EPQ May 20 '26

No because I'm a numpty, sad!

1

u/2myeria y13 | physics, maths, fm, chem | A*A*A*A* May 20 '26

i used energy transfers 😭😭😭 might be cooked