Aqa a-level 24th May 2019 paper 2 physics unofficial mark schemeWatch

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18.24%
#1
Wzup guys I'll add questions/answers on here soon as I leave the exam! I'll update as you guys comment also so... commment the question(and q# if possible) and your answers
Evil Homer

1.1) Thermal physics - find spec.heat capacity [5 marks]
Work out the total energy, 12x890 I think = 10680J,
Work out energy to change state 2x10^5*0.05=10000J
Energy to increase temp = 10680-10000=680J
Work out SHC using Q=mc(deltatheta)
c=680/0.05*(77-70)=1900J/kgK (2sf)

1.2) is x(work done expanding) or y(work done to vaporise) bigger [4 marks]
Is X or Y greater? Calculating work done
X = difference in volume, then W = PV, (0.05/3.8-0.05/810)*1x10^6=1310J
Y = Q = ml = 0.05*2x105=10000J
so Y is greater

2.1) internal energy definition [2 marks]
sum of random distribution of potential and kinetic energies of molecules

2.2) explain absolute zero in terms of gas laws and 'kinetic law theory'(i think it meant KE theory) [2 marks]
At absolute zero, the kinetic energy of gas molecules is zero. And that gas laws show that at absolute zero, gases have no volume or pressure.

2.3) Find the rms speed [3 marks]
3/2kT = 310k i think?
then 1/2m(Crms)^2 = 3/2kT
4x10^-2/6.02x10^23=m
crms = 440 or smth

2.4) describe difference between kinetic energy of argon and helium at the same temp 
ke is same because temperature is the same

2.5) Explain why pressure exists in compressed gas [3 marks]
molecules collide
elastic collisions(KE conserved)
wall exerts force on molecules
molecules exert force on wall
f=delta mv/t so force applied on wall,
p=f/a so pressure exists

2.6) describe how you'd reduce pressure of same mixture of gases by changing 2 independent factors [2 marks]
reduce temperature -

lower ke as ke proportional to temperature
less frequent collisions/collisions per second decrease
less force
lower pressure

increase the volume -

molecules more spaced
collisons per second decrease
less force
hence lower pressure
3.1) calculate the potential of one sphere [3 marks]
23000V (i gott 12000 but apparently you use the capacitance)
3.2) suggest a solution to ONE problem when measuring d [2 marks] use set square to reduce parallax

3.3) show that the electrostatic force between the spheres is 4x10^-3 [3 marks]
F=kQq/r^2 gave you 3.79x10^-3 roughly equal to 4x10^-3

3.4) angle was 7 degrees, explain whether this is consistent [2 marks]
Using components Tcostheta was the weight so you found T(0.03N i think)
Tsintheta was equal to the electrostatic force and finding theta from 0.03sintheta = 3.79x10^-3 gave you roughly 7 degrees

3.6) Deduce whether the gravitational force would have an effect on theta [2 marks]
got something like 1.6x10^-13, this is negligible compared to the electrostatic force hence no effect on angle

4.1) Define gravitational potential [2 marks]
the work done per unit mass, for a small point mass, to move that mass from infinity to that point

4.2) explain how the potential lines show the field is NOT uniform [1 mark]
distance between each line was different despite equal potential spacing

4.3) escape velocity [3 marks]
v=sqrt(2GM/R)
find M by subbing in values from diagram
gave smth like 2373 = 2400 ms-1 to 2sf

EMF coil question

5.1) find number of turns [3 marks]
number of turns was 120

5.2) find flux after rotating [2 marks]
1.3x10^-3

5.3) find max emf [1 mark]
using E=banw or whatever got some small value

5.3) Draw the flux linkage graph for the same time interval as the emf graph shown 
i got it as a cos wave? starting at max because at t=0 emf was 0

Nuclear question

6.1)Explain why the moderator needs to reduce KE of neutrons [1 mark]
To reduced the speed of them so they’re slow enough for fission as KE of neutrons is proportional to their speed

6.2) calculate the amount of energy in MeV left after 5 collisions [3 marks?]
2x10^6 x 0.37^5 = 1.39x10^4

6.3) Explain why the number of collision depends on the nucleon number of the moderator atoms [2 marks]
small nucleon number = less mass so speed of atom will be faster on collision so less collisions needed to lower KE, whereas a bigger mass means the atom would move more slowly so more collisions needed (smth along the lines of that)

6.4) calculation of energy released in MeV [3 marks?]
i think it was smth like 168 MeV not sure

6.5) 3 benefits of nuclear power [3 marks]
1. very large energy yield compared to other sources
2. doesnt produce toxic gases
3. easy to transport them to the reactor
any other valid reasons ngl

MCQ

8) brownian motion C
9)
10) Earth angular speed C
11)
12)
13)
14)
15)
16)
17)
18)
19)
20)
21)
22)
23)
24)
25)
26)
27)
28)
29)
30)
31)
32) what happens to KE and number of neutrons in the control rod C

CBBCDCABDCDDACCBBCCCABDBC but i think theres one missing there [quote badders_ ]

Last edited by mahmed69; 3 weeks ago
8
3 weeks ago
#2
1a) nuetron is absorbed by the coolant rod
1
#3
nice one lool
1a) nuetron is absorbed by the coolant rod
5
3 weeks ago
#4
9 mate
(Original post by xRealistTricks)
what time does your exam start
0
3 weeks ago
#5
tbf i was expecting you to go omg how did you get the paper please send it to me i will do anything
(Original post by mahmed69)
nice one lool
0
#6
one doesnt need the paper when one has the brains to actually do it
tbf i was expecting you to go omg how did you get the paper please send it to me i will do anything
1
3 weeks ago
#7
2
3 weeks ago
#8
Just collating answers I've seen on the main thread, not sure what question numbers these were.

SHC of liquid nitrogen:
Work out the total energy, 12x890 I think = 10680J,
Work out energy to change state 2x10^5*0.05=10000J
Energy to increase temp = 10680-10000=680J
Work out SHC using Q=mc(deltatheta)
c=680/0.05*(77-70)=1900J/kgK (2sf)

Is X or Y greater? Calculating work done
X = difference in volume, then W = PV, (0.05/3.8-0.05/810)*1x10^6=1310J
Y = Q = ml = 0.05*2x105=10000J
so Y is greater

Voltage on sphere:
Using V = Q/C, C = 4piE0r (as given in question) = 2.22x10^-12
V = 52x10^-9/2.22x10^-12=23379V = 23000V
3
#9
i got 12000 somehow not sure if ppl agree?
Just collating answers I've seen on the main thread, not sure what question numbers these were.

SHC of liquid nitrogen:
Work out the total energy, 12x890 I think = 10680J,
Work out energy to change state 2x10^5*0.05=10000J
Energy to increase temp = 10680-10000=680J
Work out SHC using Q=mc(deltatheta)
c=680/0.05*(77-70)=1900J/kgK (2sf)

Is X or Y greater? Calculating work done
X = difference in volume, then W = PV, (0.05/3.8-0.05/810)*1x10^6=1310J
Y = Q = ml = 0.05*2x105=10000J
so Y is greater

Voltage on sphere:
Using V = Q/C, C = 4piE0r (as given in question) = 2.22x10^-12
V = 52x10^-9/2.22x10^-12=23379V = 23000V
2
3 weeks ago
#10
Just collating answers I've seen on the main thread, not sure what question numbers these were.

SHC of liquid nitrogen:
Work out the total energy, 12x890 I think = 10680J,
Work out energy to change state 2x10^5*0.05=10000J
Energy to increase temp = 10680-10000=680J
Work out SHC using Q=mc(deltatheta)
c=680/0.05*(77-70)=1900J/kgK (2sf)

Is X or Y greater? Calculating work done
X = difference in volume, then W = PV, (0.05/3.8-0.05/810)*1x10^6=1310J
Y = Q = ml = 0.05*2x105=10000J
so Y is greater

Voltage on sphere:
Using V = Q/C, C = 4piE0r (as given in question) = 2.22x10^-12
V = 52x10^-9/2.22x10^-12=23379V = 23000V
0
3 weeks ago
#11
0
3 weeks ago
#12
0
3 weeks ago
#13
I would have got 12000V but is r the distance between the surface of both spheres or the distance between the centre of masses of both spheres
(Original post by mahmed69)
i got 12000 somehow not sure if ppl agree?
0
3 weeks ago
#14
Hate to say it but I disagree with 23000V. I think you used r as the radius of the sphere which is incorrect I believe
Just collating answers I've seen on the main thread, not sure what question numbers these were.

SHC of liquid nitrogen:
Work out the total energy, 12x890 I think = 10680J,
Work out energy to change state 2x10^5*0.05=10000J
Energy to increase temp = 10680-10000=680J
Work out SHC using Q=mc(deltatheta)
c=680/0.05*(77-70)=1900J/kgK (2sf)

Is X or Y greater? Calculating work done
X = difference in volume, then W = PV, (0.05/3.8-0.05/810)*1x10^6=1310J
Y = Q = ml = 0.05*2x105=10000J
so Y is greater

Voltage on sphere:
Using V = Q/C, C = 4piE0r (as given in question) = 2.22x10^-12
V = 52x10^-9/2.22x10^-12=23379V = 23000V
0
3 weeks ago
#15
Question 6c advantages to nuclear power
Answer: can generate a large amount of electricity and doesn’t produce greenhouse gases ... (3 marks)
0
3 weeks ago
#16
(Original post by mahmed69)
Wzup guys I'll add questions/answers on here soon as I leave the exam! I'll update as you guys comment also so... commment the question(and q# if possible) and your answers

1.1) Thermal physics - find spec.heat capacity [5 marks]
Work out the total energy, 12x890 I think = 10680J,
Work out energy to change state 2x10^5*0.05=10000J
Energy to increase temp = 10680-10000=680J
Work out SHC using Q=mc(deltatheta)
c=680/0.05*(77-70)=1900J/kgK (2sf)

1.2) is x(work done expanding) or y(work done to vaporise) bigger [4 marks]
Is X or Y greater? Calculating work done
X = difference in volume, then W = PV, (0.05/3.8-0.05/810)*1x10^6=1310J
Y = Q = ml = 0.05*2x105=10000J
so Y is greater

2.1) internal energy definition [2 marks]
sum of random distribution of potential and kinetic energies of molecules

2.2) explain absolute zero in terms of gas laws and 'kinetic law theory'(i think it meant KE theory) [2 marks]
cant remember what i wrote

3.1) calculate the potential of one sphere [3 marks]
12000V i think

3.2) suggest a solution to ONE problem when measuring d [2 marks]

3.3) show that the electrostatic force between the spheres is 4x10^-3 [3 marks]
F=kQq/r^2 gave you 3.79x10^-3 roughly equal to 4x10^-3

3.4) angle was 7 degrees, explain whether this is consistent [2 marks]
Using components Tcostheta was the weight so you found T(0.03N i think)
Tsintheta was equal to the electrostatic force and finding theta from 0.03sintheta = 3.79x10^-3 gave you roughly 7 degrees

3.5) Deduce whether the gravitational force would have an effect on theta [2 marks]
got something like 1.6x10^-13, this is negligible compared to the electrostatic force hence no effect on angle

EMF coil question

.1) find number of turns [3 marks]
number of turns was 120

.2) find flux after rotating [2 marks]
1.3x10^-3

.3) find max emf [1 mark]
using E=banw or whatever got some small value
max emf was 0.038V
Potential of sphere was 23000V
Solution for measuring D:
something to do with making D larger to reduce uncertainty?
1
3 weeks ago
#17
I’m not sure about ‘greenhouse gases’ as it does produce water vapour which is a greenhouse gas. Better to say no CO2.

Not sure if they’ll penalise though.
(Original post by grace frost)
Question 6c advantages to nuclear power
Answer: can generate a large amount of electricity and doesn’t produce greenhouse gases ... (3 marks)
0
3 weeks ago
#18
I also wrote that transport costs are cheap as only a small amount of fuel is required, and it preserves reserves of other fuel such as oil and coal.
(Original post by grace frost)
Question 6c advantages to nuclear power
Answer: can generate a large amount of electricity and doesn’t produce greenhouse gases ... (3 marks)
0
3 weeks ago
#19
Were you high taking this exam or something? SMH
1
3 weeks ago
#20
For multiple choice I got:
3
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