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<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

AQA M1 June 2014 Unofficial Mark Scheme

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I have no recollection of 20i+20j. But I'm sure i answered everything hah :')
Thanks again. Did well but not well enough to make up for the disaster that was C4 so oh well, retakes for me.

As for grade boundaries I think this was slightly tougher than the previous 3 exams (didn't do any beyond that) so maybe a grade lower than last year's if anything, probably around the same though. Might have just been me though.
Reply 62
Avatar for Flauta
Flauta
9
Got 3(f) wrong for certain and I got 1200/root3 for the last question which is about 692.8N so I maybe made a rounding error, but other than that it was fine.
Very easy paper. The unofficial "mark scheme" isn't right though. The last question was 4003≈692.8400\sqrt{3} \approx 692.8

Which you get from solving the following equation:

Xcos⁡(30)−tan⁡(30)(40gcos⁡(30)+X2)−40gsin⁡(30)=(0.2)(40)X \cos (30) - \tan (30) (40 g \cos(30) + \frac X 2) - 40 g \sin (30) = (0.2)(40)

Note μ=tan⁡(30)\mu = \tan (30)
(edited 9 years ago)
I'm not so sure for the very last question. Surely you had to use the value of ɥ that you calculated before - 0.577 if I remember correctly - with R in terms of X, then factorise out? If so - I got 353.

Someone post the question paper please?

Posted from TSR Mobile
Original post by tyler.jackoliver
I'm not so sure for the very last question. Surely you had to use the value of ɥ that you calculated before - 0.577 if I remember correctly - with R in terms of X, then factorise out? If so - I got 353.



You forgot to consider the gravitational force acting on the particle which also resists motion.
Reply 67
Avatar for CD223
CD223
10
What was the answer to both assumptions?
For the van I said no air resistance and the projectiles I said model as a particle. Pretty sure I got the first assumption wrong...

Any guess on the cap mark? Last year it was 70/75 for the June and Jan papers so I'm hoping lower as this was ten times harder.


Posted from TSR Mobile
Damn it, I got it all right apart from the final one because I put tan40 as the coefficient of friction because the of the weight being in my head. Put all of the correct working down though.. Massive face palm I checked it so many times aswell.
This paper was a killer :s-smilie: does anyone know what is 34/75 in ums normally?
Original post by yasmine 123
This paper was a killer :s-smilie: does anyone know what is 34/75 in ums normally?


Was 48 last year
Original post by yasmine 123
This paper was a killer :s-smilie: does anyone know what is 34/75 in ums normally?


It's usually 40-45%.
Reply 72
Avatar for EvieH96
EvieH96
Original post by wilson1098
For finding the forward driving force of the van, did you consider the tension and friction from the tip. So would it be P - T - F = ma?

I did this!
Well I completely screwed up the last question, I got 5.33 which is completely wrong, but looking at the mark scheme I think I did pretty good in the rest of the paper. Hopefully the grade boundaries will be lowered :smile:
Reply 74
Avatar for EvieH96
EvieH96
Original post by ramo55
for last question did anyone do: ma=xcos30-mu(mgcos(theta) + xcos60) ? and then subbed in values for m,g, a, and theta to find x?

Yes I did!
Original post by EvieH96
I did this!


I'm not sure that's right, if I understand you correctly. The friction of the tip is acting on the tip, not the van.
Reply 76
Avatar for ubisoft
ubisoft
3
Original post by Tarquin Digby
I'm not sure that's right, if I understand you correctly. The friction of the tip is acting on the tip, not the van.


Yeah, the friction only acted on the skip, the van had no resistive forces.
Reply 77
Avatar for EvieH96
EvieH96
Original post by Tarquin Digby
I'm not sure that's right, if I understand you correctly. The friction of the tip is acting on the tip, not the van.

No I don't think it's right either, I don't understand Mechanics I just blundered through...
Reply 78
Avatar for ubisoft
ubisoft
3
Original post by CD223
What was the answer to both assumptions?
For the van I said no air resistance and the projectiles I said model as a particle. Pretty sure I got the first assumption wrong...

Any guess on the cap mark? Last year it was 70/75 for the June and Jan papers so I'm hoping lower as this was ten times harder.


Posted from TSR Mobile


I'd say 67
Reply 79
Avatar for ramo55
ramo55
3
Original post by Tarquin Digby
Very easy paper. The unofficial "mark scheme" isn't right though. The last question was 4003≈692.8400\sqrt{3} \approx 692.8

Which you get from solving the following equation:

Xcos⁡(30)−tan⁡(30)(40gcos⁡(30)+X2)−40gsin⁡(30)=(0.2)(40)X \cos (30) - \tan (30) (40 g \cos(30) + \frac X 2) - 40 g \sin (30) = (0.2)(40)

Note μ=tan⁡(30)\mu = \tan (30)


how many marks do you think for forgetting the -40gsin30 bit?

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