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

1. (Original post by 11234)
I just put A down
good lad, i tended to go for D when i didnt have an answer
2. 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.

3. (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.

tbh i will 2 u on that, i put it as 2.4 because i am cancer
4. (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.

lol it was 1,7
5. (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.
6. (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.
7. (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?
8. What was the vernier scale? Was it 2.07mm??
9. For the mqc question about a girl jogging did anyone else get velocity = 0 and speed = 2.4 ??
10. (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?
11. (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
12. (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
13. (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)
14. 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".
15. (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.
16. (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
17. (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
18. This is wrong because theres no unifrom acceleration. 2.4 was correcr.

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19. (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?
20. (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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