2016 Official AQA New Spec AS Level Physics Paper 1 - 24th of May 2016

Original post by SaltandSugar

Why zero? wasnt there 4 resistors in total or something and they were difference along each parallel branch?

I dunno, i said zero cos there were no components in parallel with the voltmeter to begin with at that point

Reply 662

Chickenslayer69

7

Original post by LivToBe

It probably would be significantly higher- compare it to the grade boundaries when the old spec was brought in

Uhhh... except the old spec is 100x easier.

Reply 663

Original post by BainesyA

Hey guys, first post in this thread, quick question...

Did anyone else find the whole paper relatively forgiving except for the electricity questions?

(Also, ticking a box for a final question made me smile)

Same. It was easier than the specimen papers, at least

Reply 664

maximo17000

Does anyone know when an unofficail mark scheme is on its way?

Reply 665

Original post by Pra99

Is the paper out of 60?

70

Reply 666

Original post by studentabcde

Same. It was easier than the specimen papers, at least

I just can't seem to get my head around electricity. Or is that because I revised the whole of the electricity unit in 3 hour last night between 9pm and midnight? :congrats:

But I think it went okay, the worst part is when you come out of the exam and then people are screaming their answers left, right and centre. "Did you get this?" "What did you get for question 1,7,99, 3million and 29?" They just lower my confidence :smile:

Reply 667

edothero

15

Original post by o.s.s15

Did anyone get 2.6 for deceleration

It was 2.7ms^-2, using

F=ma

And the question after it told us the speed when it touched the runway was 68ms^-1, you know a and you know it goes until its at rest
So

s = s[br]u = 68ms^{-1}[br]v = 0ms^{-1}[br]a = -2.71ms^{-2}[br]t = t [br]

You can ignore

t

here, it's irrelevant

Using

v^{2} = u^{2} + 2as

0 = 68^{2} + 2(-2.71)s

\therefore 5.42s = 68^{2}

s = \dfrac{68^{2}}{5.42} = 853.13 = 853m

(edited 7 years ago)

Reply 668

SaltandSugar

Original post by Gurkin

I dunno, i said zero cos there were no components in parallel with the voltmeter to begin with at that point

Was the set up something like this? http://i.imgur.com/bE1Yv5V.png

Wouldnt there be a potential difference as the resistors were different?

I'm not sure tbh I guessed too

Reply 669

Original post by BainesyA

I just can't seem to get my head around electricity. Or is that because I revised the whole of the electricity unit in 3 hour last night between 9pm and midnight? :congrats:

But I think it went okay, the worst part is when you come out of the exam and then people are screaming their answers left, right and centre. "Did you get this?" "What did you get for question 1,7,99, 3million and 29?" They just lower my confidence :smile:

Yeah same. probably shouldn't be going on TSR right now lol

Reply 670

Gurkin

Original post by SaltandSugar

Was the set up something like this? http://i.imgur.com/bE1Yv5V.png

Wouldnt there be a potential difference as the resistors were different?

I'm not sure tbh I guessed too

Yes, the only reason i think the answer may not be as simple as saying the pd is zero is because it mentioned the resistance of one of the components in the question

Reply 671

Original post by studentabcde

Yeah same. probably shouldn't be going on TSR right now lol

I must admit, the majority of TSR answers ring a bell in my head and are similar to what I put, my mates where trying to tell me that the chocolate melted at the nodes :biggrin:

I thought, that surely can't be right, and now TSR have confirmed that it was indeed the antinodes where the chocolate melts.

Reply 672

11234

8

Original post by edothero

It was 2.7ms^-2, using

F=ma

And the question after it told us the speed when it touched the runway was 68ms^-1, you know a and you know it goes until its at rest
So

s = s[br]u = 68ms^{-1}[br]v = 0ms^{-1}[br]a = -2.71ms^{-2}[br]t = t [br]

You can ignore

t

here, it's irrelevant

Using

v^{2} = u^{2} + 2as

0 = 68^{2} + 2(-2.71)s

\therefore 5.42s = 68^{2}

s = \dfrac{68^{2}}{5.42} = 853.13 = 853m

I got a rounding error how many marks would I lose. I got 856

Reply 673

Original post by BainesyA

I must admit, the majority of TSR answers ring a bell in my head and are similar to what I put, my mates where trying to tell me that the chocolate melted at the nodes :biggrin:

I thought, that surely can't be right, and now TSR have confirmed that it was indeed the antinodes where the chocolate melts.

I crossed out nodes and changed it to antinodes when i was checking through. so glad now!

Reply 674

edothero

15

Original post by 11234

I got a rounding error how many marks would I lose. I got 856

Not sure bro
Probably 1?

Reply 675

Original post by edothero

It was 2.7ms^-2, using

F=ma

And the question after it told us the speed when it touched the runway was 68ms^-1, you know a and you know it goes until its at rest
So

s = s[br]u = 68ms^{-1}[br]v = 0ms^{-1}[br]a = -2.71ms^{-2}[br]t = t [br]

You can ignore

t

here, it's irrelevant

Using

v^{2} = u^{2} + 2as

0 = 68^{2} + 2(-2.71)s

\therefore 5.42s = 68^{2}

s = \dfrac{68^{2}}{5.42} = 853.13 = 853m

Can I ask why you haven't rounded the answer to 850, given that the value of 'u' was to two significant figures?

Reply 676

Gurkin

The max sig fig in that question was 2 so i left my answer as 850

Reply 677