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

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34.95%
Easier than last year (506)
26.91%
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18.24%
mahmed69
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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 [1]
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]

Name:  aqa.png
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Size:  11.4 KBuse 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 [1]
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_ ]

ABDCDBCABDCBDACDEADBBBADC [DivockOrigi27 ]
Last edited by mahmed69; 3 weeks ago
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homemadeclock
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1a) nuetron is absorbed by the coolant rod
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mahmed69
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nice one lool
(Original post by homemadeclock)
1a) nuetron is absorbed by the coolant rod
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homemadeclock
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9 mate
(Original post by xRealistTricks)
what time does your exam start
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homemadeclock
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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
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mahmed69
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one doesnt need the paper when one has the brains to actually do it
(Original post by homemadeclock)
tbf i was expecting you to go omg how did you get the paper please send it to me i will do anything
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The_Governor200
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Whenever you're ready guys.
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ADESONOM
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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
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mahmed69
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i got 12000 somehow not sure if ppl agree?
(Original post by ADESONOM)
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
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meel4d
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do you have a link for the main thread?
(Original post by ADESONOM)
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
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Izz_xox
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Can you link the main thread
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ADESONOM
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https://www.thestudentroom.co.uk/sho....php?t=5835994
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Xopic
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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?
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Xopic
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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
(Original post by ADESONOM)
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
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grace frost
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Question 6c advantages to nuclear power
Answer: can generate a large amount of electricity and doesn’t produce greenhouse gases ... (3 marks)
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Tygasga
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(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?
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tjto
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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)
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The_Governor200
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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)
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DivockOrigi27
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Were you high taking this exam or something? SMH
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GeorgeWakely00
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For multiple choice I got:
CDBCCACBDCCBACCCBCBAADBBC
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