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2016 Official AQA New Spec AS Level Physics Paper 2 - 9th June 2016

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Original post by 11234
I just put A down


good lad, i tended to go for D when i didnt have an answer
2.7 mm without a shadow of doubt.it was 2mm on the large increment and it matched up on the 7th small increment on the vernier scale.

Screen Shot 2016-06-09 at 18.26.37.png
Original post by optimus9000
2.7 mm without a shadow of doubt.it was 2mm on the large increment and it matched up on the 7th small increment on the vernier scale.

Screen Shot 2016-06-09 at 18.26.37.png


tbh i will 2 u on that, i put it as 2.4 because i am cancer
Original post by optimus9000
2.7 mm without a shadow of doubt.it was 2mm on the large increment and it matched up on the 7th small increment on the vernier scale.

Screen Shot 2016-06-09 at 18.26.37.png


lol it was 1,7
Original post by alhrona
its a mixed opinion , we need to ask a physics professional


I'm not a physics professional but,

If you use 4 resistors: 1/45 + 1/45 + 1/45 + 1/45 = 0.0889
1/0.0889 = 11.25 Ohms resistance.

The Voltage I think was 32V. Power = voltage^2/resistance.
so 32^2/11.25 = 91W

91W is less than 100W so is not quite using the maximum power, 5 resistors would be 114W which goes above the maximum power.

Therefore 4 components maximum before maximum power reached.
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Original post by 83457
For the mass of a fuel decreasing, did people put A- less friction. I simply thought M1 mechanics here.

I put C- Acceleration and deceleration is bigger- because F=ma, F from the engine stays constant, m mass decreases so acceleration increases (newtons law). I watch formula one and I thought that friction is good because the tyres work better when in the working window of a high temperature.
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Original post by Sam Webb
I'm not a physics professional but,

If you use 4 resistors: 1/45 + 1/45 + 1/45 + 1/45 = 0.0889
1/0.0889 = 11.25 Ohms resistance.

The Voltage I think was 32V. Power = voltage^2/resistance.
so 32^2/11.25 = 91W

91W is less than 100W so is not quite using the maximum power, 5 resistors would be 114W which goes above the maximum power.

Therefore 4 components maximum before maximum power reached.


I got 4 for my answer, but I did completely different way(completely random), would I still got full marks?
What was the vernier scale? Was it 2.07mm??
For the mqc question about a girl jogging did anyone else get velocity = 0 and speed = 2.4 ??
(edited 7 years ago)
Original post by uud
I got 4 for my answer, but I did completely different way(completely random), would I still got full marks?


Depends, how did you do it?
Original post by Qwerty4655
For a the mqc question about a girl jogging did anyone else get velocity = 0 and speed = 2.4 ??


Yes i got that too im sure its correct
Original post by Xenon17
lol it was 1,7


you might want to check around page 10/11.
the debate is around the 2.7 - 2.9 mark, unless of course everyone else is blind :wink:
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Original post by Sam Webb
Depends, how did you do it?

it was random , like I used v=ir and then like I divides 100 by something to get 4.4.. ish and then I rounded it down and got 4. lol was just a guess wasn't expecting it to be right. would u just get full marks for the right answer or do u get method marks(fml)
(edited 7 years ago)
Girl runs as 2m/s for 30 seconds, turns around and runs for 20 seconds until she reaches original spot.

2m/s x 30s = 60m, so she ran 60m forward.
60m/20s = 3m/s so she was running at a speed of 3m/s on the way back.

Average Velocity: [(2 x 30) + (-3 x 20)] / 50 = 0m/s
Average speed : [(2 x 30) + (3 x 20)] / 50 = 2.4m/s

So the answer for this question was "A".
Original post by uud
it was random , like I used v=ir and then like I divides 100 by something to get 4.4.. ish and then I rounded it down and got 4. lol was just a guess wasn't expecting it to be right.


XD I think you'll probably get a mark for the correct answer, and another for using a formula, usually examiners just look for the answer. But I think you'll drop like 1 mark max.
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Original post by Sam Webb
Girl runs as 2m/s for 30 seconds, turns around and runs for 20 seconds until she reaches original spot.

2m/s x 30s = 60m, so she ran 60m forward.
60m/20s = 3m/s so she was running at a speed of 3m/s on the way back.

Average Velocity: [(2 x 30) + (-3 x 20)] / 50 = 0m/s
Average speed : [(2 x 30) + (3 x 20)] / 50 = 2.4m/s

So the answer for this question was "A".


Not sure where u have got 50 from and why u are dividing by it. But the average speed is 2.5m/s. because (u+v)/ 2 is average so (3+2)/2 = 2.5 m/s

Velocity is zero but average speed is 2.5 m/s
Original post by katiekat0208
I got around 277 but to be completely honest i had no idea what i was doing haha



I did the same thing but I think it's wrong tbh
This is wrong because theres no unifrom acceleration. 2.4 was correcr.

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Original post by uud
Not sure where u have got 50 from and why u are dividing by it. But the average speed is 2.5m/s. because (u+v)/ 2 is average so (3+2)/2 = 2.5 m/s

Velocity is zero but average speed is 2.5 m/s


that's what i thought. can someone confirm this?
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Original post by Sam Webb
XD I think you'll probably get a mark for the correct answer, and another for using a formula, usually examiners just look for the answer. But I think you'll drop like 1 mark max.

Thanks, I believe that as long as you get the right answer that all that matters. hope the examiner thinks like me

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