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AQA AS Physics Paper 1 7407 - 14th May 2019

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Reply 40

Nkswaly

Original post by saintboy183

the bungee question and the one with the ruler was really hard.....

I had no idea what to do :frown:

, what did you do

Reply 41

kdemirolcen

I think the grade boundary will be around %55-56. We can’t know what will come up on paper 2 but I feel like 39-40/70 will correspond to an A for paper 1

Reply 42

sapah

@Pangol has made an unofficial mark scheme, which I've just cut and paste here. Thanks again...

1.
Specific charge = 4.79 x 10^7 C kg^-1 (last digit will be different if you use better or worse mass than 1.67 x...)
Exchange particle = pion
Discussion of attractive and repulsive nature of strong force with some distance values, say how it acts between nucleons, therefore keeps them at equilibrium distance. Worth a mention of not having electrostatic repulsion to overcome.
Decay cannot be alpha (not enough particles to make an alpha) or electron capture (would leave no protons!), so must be beta-.

2.
Energy dissipated = 2.1 x 10^5 J (P = VI, P = E/t)
Do a resistance calculation, get 3.5 ohms, which is close enough to 3.7 ohms to be suitable.
pd across R = 7.4 - 2.2. Current in R = 15 mA from graph. So R = 350 ohms (2 sf)

3.
F + mg = 2Tcos(20) leads to 5.78 x 10^4 N, which is 6 x 10^4 N to 1 sf.
F = ma using 2Tcos(20) - mg for F leads to a = 48 m s^-2 (2 sf)
Bit of trig needed to find length + extension, giving an extension of 13.25 m. Then us EPE = .5 x F x extension, double for both cables, leading to 4.90 x 10^5 J = 5 x 10^5 N (1 sf)
If all EPE becomes GPE, h = 41.6 m, so not justified. (Just realised it would be much quicker to work out GPE for h = 50 and show that we haven't got that much.)
KE at that speed is 3.75 x 10^5 J, so we do have enough. Probably all you need for one mark? But what about needing some GPE, and losses to air resistance? I think this is a but ambiguous.

4.
Water is more optically dense than air, so TIR could occur. Angle of incidence needs to be > critical angle, so water will need to bend "gently".
n = c/c(water) leads to c(water) = 2.26 x 10^8 m s^-1.
sin (critical angle) = 1/n leads to critical angle = 48.8.
Can't be bothered with the rest of 4 now!

5.
Resolving vertically and using symmetry, F(A) = 0.56 N.
A couple is two forces equal in magnitude (which we do have) and opposite in direction (which we don't), so not a couple.
Resolving vertically, W = 0.32 N.
Moments about A (other points are available!) leads to d = 0.1425 m.
Need to think about the bit with the non-horizontal ruler, probably something to do with moments now needing sin(angle).

6.
Antiparticle - all the same properties as the original particle except opposite electrical charge.
Proton - antiproton.
Electron - positron
Work our combined masses of proton and electron, use E = mc^2. Electron can be neglected. But do we need to double this? We only need one of each, but the question says they are made using pair production. Without doing this, I get 1.5 x 10^-10 J.
Discussion of energy levels - excitation of orbital electrons/positrons, de-excitation leads to emitted photons, some discussion of fixed energy levels.

Reply 43

saintboy183

any predictions what will come up on paper 2?

Reply 44

darrensandhu07

I’m thinking something to do with collision time for cars - crumple zones etc. might come up as that hasn’t appeared yet in any of the past papers

Original post by saintboy183

any predictions what will come up on paper 2?

Reply 45

Brain-Mind

Original post by LivanC21

I got 0.32N for W
And 0.13m for d
Probably wrong though

wasn't it cm? idk bro... that test was ****...

Reply 46

Brain-Mind

Original post by saintboy183

any predictions what will come up on paper 2?

bro! i reckon is gonna be mechanics added with electricity. similar to mgh of water falling onto generator of some sort of area and what kind of power would it produces or something

Reply 47

Humdan Khalid

what did you guys get for the antihydrogen question?

Reply 48

sapah

bruhh, that's a really specific guess!!

Original post by Brain-Mind

bro! i reckon is gonna be mechanics added with electricity. similar to mgh of water falling onto generator of some sort of area and what kind of power would it produces or something

Reply 49

r.place2001

i got 3x10^-10 J as if you have pair production you are going to need a photon of 2x the energy required to create your particle-antiparticle pair, so:

[br](2p+2e^-)\times 1.6\times 10^{-19}=3\times 10^{-10}J[br]

Original post by Humdan Khalid

what did you guys get for the antihydrogen question?

Guyssssss

Just walked out of the exam room for paper 2, is there a different post for paper 2?

Reply 52

sapah

No idea but how did you find it

Reply 53

aqaboy

Original post by sapah

No idea but how did you find it

Alright I guess, tho I can barely finish the paper in time

What did you put for the internal resistance of the battery in the second last MC?

Reply 54

Vicky Tran

How did you find it? I think section 1 and 2 which are about experimental and practical skill is hard to understand the question.

Original post by aqaboy

Just walked out of the exam room for paper 2, is there a different post for paper 2?

Reply 55

aqaboy

Original post by Vicky Tran

How did you find it? I think section 1 and 2 which are about experimental and practical skill is hard to understand the question.

I would say the questions are pretty straight forward but I’m not really sure what will be the marking points

Yeah defo

Original post by aqaboy

Just walked out of the exam room for paper 2, is there a different post for paper 2?

Yes there is! Please add to so we can create an unofficial mark scheme
https://www.thestudentroom.co.uk/showthread.php?t=5951838

Reply 58

Vicky Tran

Original post by aqaboy

I would say the questions are pretty straight forward but I’m not really sure what will be the marking points

yah, most of the essay questions are lost voltage topics. They will publish the mark scheme on August before the test results day

Reply 59

Jack Hardman

Do u have the paper and markscheme for as physics paper 1

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AQA AS Physics Paper 1 7407 - 14th May 2019

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