# 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 + neutrino

b) weak interaction because proton has converted into a neutron

c)calcualtion... wavelength=8.09*10^-13m

d) 40K19 >>>> 40Ca20 + BETA minus + ANTINEUTRINO

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...

Q1

A) 40K19 + 0E-1 >>>> 40Ar18 + neutrino

b) weak interaction because proton has converted into a neutron

c)calcualtion... wavelength=8.09*10^-13m

d) 40K19 >>>> 40Ca20 + BETA minus + ANTINEUTRINO

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...

1

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Remember they might be wrong correct me and tell me how many marks you think i got out of 60 for section a

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(Original post by

1A wasn't an antineutrino but a neutrino. It was electron capture.

**orangegolf**)1A wasn't an antineutrino but a neutrino. It was electron capture.

what do you think i got out of 3 for the last part?

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#8

In 2(c), I think the refractive index for both the glass and liquid change depending on the light

1.6sin58=nsin 90 seems familiar

4(a) the arrow showed the resultant, which is the same direction as the reaction only if there is no friction (which would act upwards, changing the direction of the resultant)

5(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)

I got 186 blocks and intensity was 264Wm^-2

1.6sin58=nsin 90 seems familiar

4(a) the arrow showed the resultant, which is the same direction as the reaction only if there is no friction (which would act upwards, changing the direction of the resultant)

5(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)

I got 186 blocks and intensity was 264Wm^-2

0

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(Original post by

In 2(c), I think the refractive index for both the glass and liquid change depending on the light

1.6sin58=nsin 90 seems familiar

4(a) the arrow showed the resultant, which is the same direction as the reaction only if there is no friction (which would act upwards, changing the direction of the resultant)

5(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)

I got 186 blocks and intensity was 264Wm^-2

**sername**)In 2(c), I think the refractive index for both the glass and liquid change depending on the light

1.6sin58=nsin 90 seems familiar

4(a) the arrow showed the resultant, which is the same direction as the reaction only if there is no friction (which would act upwards, changing the direction of the resultant)

5(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)

I got 186 blocks and intensity was 264Wm^-2

or maybe i am till on 60 and i gained marks elsewhere such as for ecf... you never knowww

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looking at the 2016 AS paper this paper seems harder as the 2016 one had the most easiest final question.. what is an isotope?....

2016 also had a 6 marker which is normally easier then no 6 marker...

and the 2016 had like 4-5 tricky questions the 3 markers..

whereas in this one there are more tricky question with almost each question having that annoying 3 markers or 2 markers...

so i would say that A=55-60%

and similar to last year 10% is a drop in grade so

but i doubt A is gonna be 55 for an A and max 60

A*=60%+

A=55-60%

B=45-55%

C=35-45%

D=25-35%

2016 also had a 6 marker which is normally easier then no 6 marker...

and the 2016 had like 4-5 tricky questions the 3 markers..

whereas in this one there are more tricky question with almost each question having that annoying 3 markers or 2 markers...

so i would say that A=55-60%

and similar to last year 10% is a drop in grade so

but i doubt A is gonna be 55 for an A and max 60

A*=60%+

A=55-60%

B=45-55%

C=35-45%

D=25-35%

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#12

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.

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.

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#13

**sername**)

In 2(c), I think the refractive index for both the glass and liquid change depending on the light

1.6sin58=nsin 90 seems familiar

4(a) the arrow showed the resultant, which is the same direction as the reaction only if there is no friction (which would act upwards, changing the direction of the resultant)

5(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)

I got 186 blocks and intensity was 264Wm^-2

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#18

What do you guys think was harder: last year's AS paper 1 or this year's A2 paper 1?

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#19

(Original post by

AQA

**Richard971**)AQA

(Original post by

AQA I'm 99% sure

**bpritchard10**)AQA I'm 99% sure

I've added it to the unofficial markscheme list

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(Original post by

What do you guys think was harder: last year's AS paper 1 or this year's A2 paper 1?

**Richard971**)What do you guys think was harder: last year's AS paper 1 or this year's A2 paper 1?

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