AQA PHYSICS A Level UNOFFICIAL MARK SCHEME - 15th June 2017

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here probably all wrong correct me...

Q1

A) 40K19 + 0E-1 >>>> 40Ar18 + antineutrino
b) weak interaction because proton has converted into a neutron
c)calcualtion...
d) 40K19 >>>> 40Ca20 + BETA PLUS + NEUTRINO
If neutrino is released then the Ca decay happens
If antineutrino is released then the Ar decay happens

Q2))
refracitve

a) i got 39.0

b) n=1.3 something

1.6sin58=nsin 90
c) proabbly wrong but:

red has lower refractive index so its critical angle will be higher than 58 so it does under TIR
but blue has a higher refractive index so it will undergo TIR as its critical angle will belower than 58

nto sure anyone wann say something....

Q3

a)EMF IS y intercept
****ed the internal resitance didn't say to take negative..
b) i used the 75mA and said that the battery doesn't provide enough PD would that be right for the first circuit and for the second circuit i did nothing -.-.
c) it said intensity and using the units Intensity = required Power/surface area.

required power= I"R = (75mA)^2 * 6=0.03375W

as efiieciency = 0.04

the power would be= 0.03375/0.04 = 0.84375W

so the Intensity = 0.84375/surface area forgot what the surface area was and dont remmeber the asnwer.

Q4)

ladder beetch question...

a) there is no friction arrow pointing east on the ground so ffriction is not presnt??? not sure about this.
b)Straight arrow pinting from ground up near bottom of laddder???? not sure about this.
c)taking moments from the bottom.

4(390cos60) - (8)(Fcos30) = 0

F=65 root 3 = 112.58...=113 N .

d) as the person moves about clockwise moment from the bottom increases so the resultant force at the top has to increase to keep the ladder in position.
not sure about this...

did anyone get similar or same???

Q5)

a) just to lines one with T and one with (M+m)gsin35 i think i forgot the g OH NOOOOOO fook.
b) proving i used some long ass method:

for B: T-Mgsin35=Ma
for A: (M+2m)gsin25-T=(M+2m)a

solve this to equations and then i got the required acceleration... THANK YOU M1

c) both A and B have the same acceleration so they will have the same speed but as A has more mass and Momentum=mass*velocity
A has more momentum... not sure..

d) i worked out a using the calues M=95 and m=30
and then did 0.75a and used suvat s=ut + ......
and i got t as 4.2

e)4.2 + 12=16.2
30*60/16.2=111 ....not sure about this..

Q6)
A) divide f by root T and then work out propriotnality constant k=50 so they are proprtional..
B)MEW=A*Density so L=0.67m
C) after some large tension the string would break and there would be 0 frequency so you cannot keep using that formula.

is that right?

Q7)

a) so well we knew f Amax Kinetic Max

so using Vmax=2PIE*f*A
and using EK=1/2 m Vmax " you get m=1.0.... something
and then T=2PIE square root ( m/k) you get k=172.something

so m=1.0,,,
k=170

b) yeh well fook this question....

c) i said the peak amplitude will be lower as ther eis more resistance in oil than in air....
and thta the curve would be wider as it moves more slowly in oil ... not sure about this...

I got 225n for ladder question

Reply 21

AlonsoF

260 for the Intensity anyone? (To 2 sig figs)

Reply 22

RBP_98

Original post by AlonsoF

260 for the Intensity anyone? (To 2 sig figs)

think it was something like 264, but yes 260 to 2sf

Reply 23

Callum_Newton

Q1) d) Surely that is beta minus decay as it gains a proton?

Reply 24

Callum_Newton

Original post by bendrake123

I got 225n for ladder question

Yeah same here but afraid it's wrong, it was a moments question

Reply 25

Becky.s1928

Can anyone remember the marks for each question?

Reply 26

qwerty246

i beg sum1 make a poll

Reply 27

wferverv

Safe g big bois think i got about 11% innait fam

Reply 28

thelegend99

What angle did people put for 'G' on the ladder question? I did 60 degrees

Reply 29

wferverv

Original post by thelegend99

What angle did people put for 'G' on the ladder question? I did 60 degrees

45 degrees

Reply 30

thelegend99

Original post by wferverv

45 degrees

Damn, some others got that.

Reply 31

wferverv

Original post by thelegend99

Damn, some others got that.

Chin up, hard paper I'll be over it in an hour or so

Reply 32

thelegend99

For the unloading/loading question I got 186 but i'm doubting myself. Would you need to x2 the time for loading/unloading at either end? ah

Reply 33

thelegend99

Original post by wferverv

Chin up, hard paper I'll be over it in an hour or so

Fair enough, cheers mate

Reply 34

heshog540

Original post by LightAtTheEnd

I think the surface area was 32cm^2 for the solar cell which gives like 3.2 x 10^-4 m^2 and hence a final answer of 263Wm^-2??

7.3s for blocks to come down.
12s to reload
19.3s total.
Trolleys with block sent down 93 times (1800/19.3)
186 blocks in total (2 blocks per trolley).

The mistake a lot of people made was that they did 0.75 x actual acceleration. Question said in practice a break decreases it TO 25% of maximum acceleration, not BY 25% of maximum acceleration.

Got most of it correct. Did a few mistakes in the forced vibration questions and in the ladder question. I predict bout 20/25 for MC and perhaps 45/60 for rest. About 65/85 is my prediction for that paper. Hopefully, that's an A. Physics doesn't count for me/uni offer. Decent/standard paper maybe slightly harder. I'd say 60-65% for an A would be standard.

I think your original time is wrong. Remember the question gave you the mass of one concrete block, but there were two blocks added. Really sneaky from AQA...

Reply 35

Lokikado

Original post by revisionphysics

here probably all wrong correct me...

Q1

A) 40K19 + 0E-1 >>>> 40Ar18 + neutrino
b) weak interaction because proton has converted into a neutron
c)calcualtion...
d) 40K19 >>>> 40Ca20 + BETA PLUS + NEUTRINO

not sure about reasoning my reason was wrong.

Q2))
refracitve

a) i got 39.0

b) n=1.3 something

1.6sin58=nsin 90
c) proabbly wrong but:

red has lower refractive index so its critical angle will be higher than 58 so it does under TIR
but blue has a higher refractive index so it will undergo TIR as its critical angle will belower than 58

nto sure anyone wann say something....

Q3

a)EMF IS y intercept internal is negative gradient
****ed the internal resitance didn't say to take negative..
b) i used the 75mA and said that the battery doesn't provide enough PD would that be right for the first circuit and for the second circuit i did nothing -.-.
c) it said intensity and using the units Intensity = required Power/surface area.

required power= I"R = (75mA)^2 * 6=0.03375W

as efiieciency = 0.04

the power would be= 0.03375/0.04 = 0.84375W

so the Intensity = 0.84375/surface area where area = 0.32cm2 so Intensit= 264Wm^-2

Q4)

ladder beetch question...

a) there is no friction arrow pointing east on the ground so ffriction is not presnt??? not sure about this.
b)NOT SURE ARROW
c)taking moments from the bottom.

4(390cos60) - (8)(Fcos30) = 0

F=65 root 3 = 112.58...=113 N .

d) as the person moves about clockwise moment from the bottom increases so the resultant force at the top has to increase to keep the ladder in position.
not sure about this...

did anyone get similar or same???

Q5)

a) just to lines one with T and one with (M+m)gsin35 i think i forgot the g OH NOOOOOO fook.
b) proving i used some long ass method:

for B: T-Mgsin35=Ma
for A: (M+2m)gsin25-T=(M+2m)a

solve this to equations and then i got the required acceleration... THANK YOU M1

c) I think the question asked about the loaded mass of both A and B, which would be the same mass and velocity therefore the same momentum. (not sure)
THIS IS NOT GUARANTEED

d) i worked out a using the calues M=95 and m=30
and then did 0.75a and used suvat s=ut + ......
and i got t as 4.2

e)4.2 + 12=16.2
30*60/16.2=111 ....not sure about this..
OR
I got 186 blocks

Q6)
A) divide f by root T and then work out propriotnality constant k=50 so they are proprtional..
B)MEW=A*Density so L=0.67m

C) after some large tension the string would break and there would be 0 frequency so you cannot keep using that formula. PROBABLY WRONG

is that right?

Q7)

a) so well we knew f Amax Kinetic Max

so using Vmax=2PIE*f*A
and using EK=1/2 m Vmax " you get m=1.0.... something
and then T=2PIE square root ( m/k) you get k=172.something

so m=1.0,,,
k=170

b) yeh well fook this question....

c) i said the peak amplitude will be lower as ther eis more resistance in oil than in air....
and thta the curve would be wider as it moves more slowly in oil ... not sure about this...

spring: m=1.094kg, k= 172.758
time taken to reach end of ramp: 7.1 seconds(?)
no times reloaded: I put 75 :/
red and blue light: red = refracted with angle of refraction smaller than 90degrees because of something
blue= TIR because incident angle is larger than critical angle because something.....

guessed quite a few mc because was running out of time and panicking. Could not get the proof and didn't understand the proportionality question so i just bs my way through those.

Resonance question: 1) peak is lower
2) peak is at a greater frequency (weird question imo, didn't know what to put with the oil thing)

IMO, it was a hard paper and i hated it and everyone i spoke to hated it and i'm not gonna be able to get into uni :biggrin:

Reply 36

thelegend99

Original post by heshog540

I think your original time is wrong. Remember the question gave you the mass of one concrete block, but there were two blocks added. Really sneaky from AQA...

No I think the 2 cancels and that's the formula they use.

Reply 37

LightAtTheEnd

Original post by heshog540

I think your original time is wrong. Remember the question gave you the mass of one concrete block, but there were two blocks added. Really sneaky from AQA...

nah in previous question he had to prove that
a = mgsin35/(M+m)
So we only used the mass of 1 block not 2. We already account for two blocks being added when we were deriving formula for acceleration.