AQA PHYSICS A Level UNOFFICIAL MARK SCHEME - 15th June 2017
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also did anyone else put for the question where it says he makes a prediction using his model on the amplitude when the frequency is largely increased that the driving frequency would be larger than the natural frequency of the wire so it would be lower than he expected
Original post by Radial174115
why divide by 0.04 and not times by 0.04 for the intensity question?
The panel is 4% efficient, therefor 4% of the suns energy actually gets turned into electricity. to work out 4% of the suns energy (the electrical energy needed) you times the suns energy by 0.04. You already know the electrical energy needed. To get 100% of the suns energy you divide the electrical energy needed by 0.04.
For the blue and red light refraction question, surely the refractive index in both substances will be greater or less as it said in the question the refractive index is higher/lower in any material, would that not mean the ratio of n1 to n2 would be the same and hence they would both go at 90 degrees as well
Original post by Yepadee
The panel is 4% efficient, therefor 4% of the suns energy actually gets turned into electricity. to work out 4% of the suns energy (the electrical energy needed) you times the suns energy by 0.04. You already know the electrical energy needed. To get 100% of the suns energy you divide the electrical energy needed by 0.04.
i must have read the question wrong, thanks for clearing it up though!
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...
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...
7 Part b is 3.46 Hz
Original post by 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
No 186 is correct
G is 60 Degrees
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...
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...
Hi mate
I'm lost heart attacking😓
Please tell me marks of questions
Thanks
Doing past papers is honestly just a waste of time, like these papers are so much harder. U1 and 2 were easy on the old spec
Original post by Amirali_Mr
G is 60 Degrees
no it would have been 74 degrees. vertical component was 390, horizontal was 113, using trig angle is 74
Original post by Benstorr
no it would have been 74 degrees. vertical component was 390, horizontal was 113, using trig angle is 74
are you talking about drawing the arrow??? Cause we werent given horizontal. and it was a 1 marker so i think your wrong
Original post by roham22
Hi mate
I'm lost heart attacking😓
Please tell me marks of questions
Thanks
I'm lost heart attacking😓
Please tell me marks of questions
Thanks
Particles-1,2,3,3
Refraction-1,2,3
Electricity-2,4,3
String-2,4,2
Ladder-1,2,4,2
Trolleys-1,2,2,3,2
SHM-3,2,4
Original post by roham22
Anybody know about marks of questions?
Thanks
And calculate the wavelength from 1.4MeV was question one or 2?
Thanks
And calculate the wavelength from 1.4MeV was question one or 2?
I think it was 1.46Mev
which will give wavelength=8.09*10^-13m
Particles-1,2,3,3
Refraction-1,2,3
Electricity-2,4,3
String-2,4,2
Ladder-1,2,4,2
Trolleys-1,2,2,3,2
SHM-3,2,4
Original post by Amirali_Mr
are you talking about drawing the arrow??? Cause we werent given horizontal. and it was a 1 marker so i think your wrong
Yeah the drawing arrow one. This angle is correct if you worked out the horizontal component in the next part of the question. However as it was only 1 mark I don't think they would have been looking for an angle at all, just as long as it was diagonal. I didn't put an angle on but if I did it would be 74
Original post by Benstorr
Yeah the drawing arrow one. This angle is correct if you worked out the horizontal component in the next part of the question. However as it was only 1 mark I don't think they would have been looking for an angle at all, just as long as it was diagonal. I didn't put an angle on but if I did it would be 74
alright then we are good. I agree, think that as long as you draw it diagonally you are good.
Original post by Amirali_Mr
are you talking about drawing the arrow??? Cause we werent given horizontal. and it was a 1 marker so i think your wrong
im more than sure this question was a 2 marker because i was wondering where the second mark came from as i didnt realise how to work the angle out for ages stupidly. and i got 225N not 113N dont know which is correct though
Original post by radial174115
im more than sure this question was a 2 marker because i was wondering where the second mark came from as i didnt realise how to work the angle out for ages stupidly. And i got 225n not 113n dont know which is correct though
i am 100% sure it was 1 mark to draw the arrow
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