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# AQA Physics AS exam 24th MAY and 9th JUNE 2016 (Thread) watch

1. I got 0.06A for a current on one questions and 0.15 ohms for the value of a resistor, anyone get these?
2. (Original post by Akashi)
v^2 = u^2 + 2as

v = 0
u = 68
a = -2.7

-(68^2)/2x-2.7 = s

s = 856.296
= 8.6 x10^2 (2 s.f.)
Yeah I got this too iirc.
3. I used exact values and got 851m
4. (Original post by jcouchman)
I used exact values and got 851m
Yeah I messed up on using exact values for that question as well as the resistance of the probe question so both answers are off by a bit :/
5. (Original post by Akashi)
v^2 = u^2 + 2as

v = 0
u = 68
a = -2.7

-(68^2)/2x-2.7 = s

s = 856.296
= 8.6 x10^2 (2 s.f.)
use the horizontal motion equations.........
not suvat
6. For question part two of the momentum question when they asked about using he laws, i wrote the sum of momentum before = the sum of momentum after providing no external forces acts on it hence the system will be in equilibrium. Air resistance is acting on the system, hence the whole system is not in equilibrium and the momentum of the air changed.

Is this right?
7. (Original post by money-for-all)
use the horizontal motion equations.........
not suvat
That implies that the deceleration was 0 horizontally, but it was (-)2.7...

(Original post by icecubeinferno)
For question part two of the momentum question when they asked about using he laws, i wrote the sum of momentum before = the sum of momentum after providing no external forces acts on it hence the system will be in equilibrium. Air resistance is acting on the system, hence the whole system is not in equilibrium and the momentum of the air changed.

Is this right?
Should be Newton's 3rd law, when object A exerts a force on B, object B exerts and equal and opposite force on A.
8. (Original post by Akashi)
Yeah I messed up on using exact values for that question as well as the resistance of the probe question so both answers are off by a bit :/
If you you used exact values you should be fine, on the circuit questions did you get current ad 0.06A for one part and 0.15 ohms for another part?
9. How do you work out the acceleration? I got it wrong because I got 1.0 ms-2
10. Guyz I accidently thought that the spots for the microwave question were the unheated parts but i did everything else correctly. how many marks am i likely to lose?
11. (Original post by Liam2020)
How do you work out the acceleration? I got it wrong because I got 1.0 ms-2
F=ma rearrange to a=F/m but you had to convert the kN to N for force.
12. Does anyone know how many marks it was out of and the average mark needed for an A?
13. (Original post by jcouchman)
If you you used exact values you should be fine, on the circuit questions did you get current ad 0.06A for one part and 0.15 ohms for another part?
Mhm, that's the thing I used the rounded values I gave in the previous answers :/

Yep I got 0.064 ohms for the current through R3 was it? and I got 0.15 ohms for the resistance of the probe^^
14. Anyone got an unofficial markscheme yet?
15. (Original post by Akashi)
Mhm, that's the thing I used the rounded values I gave in the previous answers :/

Yep I got 0.064 ohms for the current through R3 was it? and I got 0.15 ohms for the resistance of the probe^^
there is usually a bit of leniance with rounded values so you should be fine, I hope there is more mechanics on paper 2 as I'm currently revising mechanics for further maths
16. (Original post by JoshFarrell)
Does anyone know how many marks it was out of and the average mark needed for an A?
out of 70 and i would guess around 50ish as it is a new spec and it seems like many people found it difficult
17. Was it as hard as the specimen papers?
18. (Original post by Akashi)
Didn't the question only ask as to whether the air exerts a force in the direction of motion?
I interpreted it just as 'see which laws of motion can be applied here' but I am sure either way will be fine
19. For the chocolate, it was 14cm so 0.14m and it was therefore 3x10^8 / 0.14 ?
20. (Original post by babyetu)
F=ma rearrange to a=F/m but you had to convert the kN to N for force.
yeah i know now. its because i did 119000-190000 = ma but you were supposed to do 190000 = ma

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