# AQA PHYA4 ~ 13th June 2013 ~ A2 PhysicsWatch

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6 years ago
#961
(Original post by MisterE1)

14)rotating coil was A q14 from http://papers.xtremepapers.com/AQA/P...W-QP-JAN03.pdf
That link doesn't work for me, try these if they don't for you either:

http://papers.xtremepapers.com/AQA/P...W-QP-JAN03.pdf
http://papers.xtremepapers.com/AQA/P...W-MS-JAN03.pdf

0
6 years ago
#962
(Original post by MisterE1)
1)t root2,
2)460 and current its greater that 0.26,
3)path of electron was A the arrow pointing upwards,
4)potential at mars was -13,
5)maximum value of tension was 24
6) K.E: P>Q, G.P.E: Q<P
7) capacitance graph question was D,
8) fe>fm>fs
9） ratio of y/d=1/3 – I got 2/3 :/
10) R earth/Rmoon=14 ----- >As I remember the question:

Mass of the earth is 81 x mass of the moon
Gravitational field strength on the earth = 9.8
Gravitational field strength on the moon = 1.8

g = GM/r^2.

9.8 = 81GM/re^2

1.8 = GM/rm^2

Re-arrange and you get re^2/rm^2 = ~14.88
11)which one is incorrect: when spring is compressed &suspended,it has minimum PE
12) which one is incorrect: change of momentum=zero
13)moving horizontally,Blv …
14)rotating coil was A q14 from http://papers.xtremepapers.com/AQA/P...W-QP-JAN03.pdf
15) Capacitance value of 0.02F for one question forgot which was, answer choice was B IIRC
16) units of impulse = kgms-1
17) momentum question when the 2 trolleys collided: 12000Ns

8 MORE MULTIPLE CHOICE QUESTIONS THEN DONE!

SECTION B IS COMPLETE
1a)define shm:
acceleration is proportinal to displacement, and acceleration acts toward equilibrium (2)
acceleration was 0.4
frequency of pendulum = 0.503 3s.f (3)
1500 revs per minute (2)
time for oscilations to be in phase again = 38 (3) -- if you think about it,
after 1 oscilation - difference on 0.1s
after 2 oscillations - difference of 0.2s....
after 19 oscillations difference of 1.9 seconds
so on the 19th osciation of the 1.9s pendulum, the 2s pendulum will arrive 1.9s after it, so on the 20th oscillation of the 1.9s pendulum they are in phase. again, on the 19th oscillation of the 2s pendulum they will be in phase.
19 x 2 = 38s
20 x 1.9 = 38s
TOTAL = 12

2) voltage = 30,000 (1)
b) (3) i think this was the capacity was increase from X to Y. find the energy stored? wasnt this the time taken for the spark to be produced?
yes, i think it was similar:

b) i think this was the capacity was increase from X to Y.time taken for the spark to be produced after the last discharge?
then Q=CV, then Q/Current = time was ~ 3.46s

you had to work out the charge stored and then use the current they gave you to find the time??
c) time taken for discharge will increase because time constant increases (2)
d) flash is brighter because more energy stored (2)
TOTAL = 7

3) 6 marker on fields
1. gravity always attractive
2. electric attractive/repulsive
3. both follow inverse square for Force
4. both follow inverse for potential
5. gravity uses mass
6. electric uses charge
7. only electric fields can be shielded
8. only electric fields depend on medium between charges
9. gradient of potential vs distance graph gives field strength
10. potential definition for both revolves around bringing a mass/charge from infinity to a certain point in a field and is zero at infinity
TOTAL = 6

4) 2 conditions when no force is exerted on the partcle (2)
field is parallel to velocity
particle is stationary
b) out of the plane of the paper
b) prove momentum is proportional to radius= mv^2/r = BQv , mv = BQr --> mv = kr
the speed of particle=8.7x107
c)time taken for partile to go through 1 dee = ~6.8x10^-8
d)show the time taken is independent of radius = 2pim/bq (3)
kinetic energy im MEV = 2.44 (3)
TOTAL = 15

5)
angle for max emf = 60 degrees -- this is at 90 degrees to the plane as emf = BANw sin(wt) therefore max emf is when sin(wt) =1, there for wt (which is the angle) = 90, 90-30 = 60 degrees
angle for max flux linkage = 5pi/6 = 2.62 radians -- this is at 0 degrees = 180 degrees, and N(PHI) = BAN cos(THETA), so max is when cos(THETA) = 1 so theta = 180, 180-39 = 150 degrees = 5pi/6
gradient of graph = emf (1)

drawing a graph (2) Positive Sine graph
peak voltage (3) I got 0.55 for peak to peak voltage then divided by 2 to get 27.5V but there is no set agreement on what the value is for this answer yet.. (MSI_10)

calculate flux density = 0.26 (2)( got 0.25 here
TOTAL= 10

Okay i'm going now... this is the best mark scheme ive done so far, ive tried to incorporate the others! please feel free to edit it!
nicely done !! but in section B question 5 for drawing the graph, i am pretty sure it was a negative sin graph, because the flux linkage graph started from maximum NEGATIVE .... and in the book it shows that when linkage is maximum POSITIVE, you get a positive sin graph.
0
6 years ago
#963
(Original post by Robmarkey)
For q17 on section a I think 3 of the options was 8000,12000 and 20000, I'm pretty sure I went with 20000 not 100% sure though.
it was 12000, wish i can remember what i done, but im pretty sure im right, because after exam i checked it with my teacher and it turned out to be 12000
0
6 years ago
#964
(Original post by MisterE1)
1)t root2,
2)460 and current its greater that 0.26,
3)path of electron was A the arrow pointing upwards,
4)potential at mars was -13,
5)maximum value of tension was 24
6) K.E: P>Q, G.P.E: Q<P
7) capacitance graph question was D,
8) fe>fm>fs
9） ratio of y/d=1/3 – I got 2/3 :/
10) R earth/Rmoon=14 ----- >As I remember the question:

Mass of the earth is 81 x mass of the moon
Gravitational field strength on the earth = 9.8
Gravitational field strength on the moon = 1.8

g = GM/r^2.

9.8 = 81GM/re^2

1.8 = GM/rm^2

Re-arrange and you get re^2/rm^2 = ~14.88
11)which one is incorrect: when spring is compressed &suspended,it has minimum PE
12) which one is incorrect: change of momentum=zero
13)moving horizontally,Blv …
14)rotating coil was A q14 from http://papers.xtremepapers.com/AQA/P...W-QP-JAN03.pdf
15) Capacitance value of 0.02F for one question forgot which was, answer choice was B IIRC
16) units of impulse = kgms-1
17) momentum question when the 2 trolleys collided: 12000Ns

8 MORE MULTIPLE CHOICE QUESTIONS THEN DONE!

SECTION B IS COMPLETE
1a)define shm:
acceleration is proportinal to displacement, and acceleration acts toward equilibrium (2)
acceleration was 0.4
frequency of pendulum = 0.503 3s.f (3)
1500 revs per minute (2)
time for oscilations to be in phase again = 38 (3) -- if you think about it,
after 1 oscilation - difference on 0.1s
after 2 oscillations - difference of 0.2s....
after 19 oscillations difference of 1.9 seconds
so on the 19th osciation of the 1.9s pendulum, the 2s pendulum will arrive 1.9s after it, so on the 20th oscillation of the 1.9s pendulum they are in phase. again, on the 19th oscillation of the 2s pendulum they will be in phase.
19 x 2 = 38s
20 x 1.9 = 38s
TOTAL = 12

2) voltage = 30,000 (1)
b) (3) i think this was the capacity was increase from X to Y. find the energy stored? wasnt this the time taken for the spark to be produced?
yes, i think it was similar:

b) i think this was the capacity was increase from X to Y.time taken for the spark to be produced after the last discharge?
then Q=CV, then Q/Current = time was ~ 3.46s

you had to work out the charge stored and then use the current they gave you to find the time??
c) time taken for discharge will increase because time constant increases (2)
d) flash is brighter because more energy stored (2)
TOTAL = 7

3) 6 marker on fields
1. gravity always attractive
2. electric attractive/repulsive
3. both follow inverse square for Force
4. both follow inverse for potential
5. gravity uses mass
6. electric uses charge
7. only electric fields can be shielded
8. only electric fields depend on medium between charges
9. gradient of potential vs distance graph gives field strength
10. potential definition for both revolves around bringing a mass/charge from infinity to a certain point in a field and is zero at infinity
TOTAL = 6

4) 2 conditions when no force is exerted on the partcle (2)
field is parallel to velocity
particle is stationary
b) out of the plane of the paper
b) prove momentum is proportional to radius= mv^2/r = BQv , mv = BQr --> mv = kr
the speed of particle=8.7x107
c)time taken for partile to go through 1 dee = ~6.8x10^-8
d)show the time taken is independent of radius = 2pim/bq (3)
kinetic energy im MEV = 2.44 (3)
TOTAL = 15

5)
angle for max emf = 60 degrees -- this is at 90 degrees to the plane as emf = BANw sin(wt) therefore max emf is when sin(wt) =1, there for wt (which is the angle) = 90, 90-30 = 60 degrees
angle for max flux linkage = 5pi/6 = 2.62 radians -- this is at 0 degrees = 180 degrees, and N(PHI) = BAN cos(THETA), so max is when cos(THETA) = 1 so theta = 180, 180-39 = 150 degrees = 5pi/6
gradient of graph = emf (1)

drawing a graph (2) Positive Sine graph
peak voltage (3) I got 0.55 for peak to peak voltage then divided by 2 to get 27.5V but there is no set agreement on what the value is for this answer yet.. (MSI_10)

calculate flux density = 0.26 (2)( got 0.25 here
TOTAL= 10

Okay i'm going now... this is the best mark scheme ive done so far, ive tried to incorporate the others! please feel free to edit it!
the 6 marker was question 4 was it not?
1
6 years ago
#965
pretty sure it was Earth sun then moon, i worked out the forces
1
6 years ago
#966
(Original post by zbubbly)
pretty sure it was Earth sun then moon, i worked out the forces
How can this be?

The sun is much more massive yes, but its huge distance from earth, coupled with the inverse sqaure law, means its gravitational energy on the satellite must be negligible.

Of course its in orbit so must experience the strongest force from the earth, so I put.

Fe>Fm>Fs
1
6 years ago
#967
(Original post by zbubbly)
pretty sure it was Earth sun then moon, i worked out the forces
I got the same.
0
6 years ago
#968
How can this be?

The sun is much more massive yes, but its huge distance from earth, coupled with the inverse sqaure law, means its gravitational energy on the satellite must be negligible.

Of course its in orbit so must experience the strongest force from the earth, so I put.

Fe>Fm>Fs
just saying the figures i got...
0
6 years ago
#969
(Original post by MisterE1)
1)t root2,
2)460 and current its greater that 0.26,
3)path of electron was A the arrow pointing upwards,
4)potential at mars was -13,
5)maximum value of tension was 24
6) K.E: P>Q, G.P.E: Q<P
7) capacitance graph question was D,
8) fe>fm>fs
9） ratio of y/d=1/3 – I got 2/3 :/
10) R earth/Rmoon=14 ----- >As I remember the question:

Mass of the earth is 81 x mass of the moon
Gravitational field strength on the earth = 9.8
Gravitational field strength on the moon = 1.8

g = GM/r^2.

9.8 = 81GM/re^2

1.8 = GM/rm^2

Re-arrange and you get re^2/rm^2 = ~14.88
11)which one is incorrect: when spring is compressed &suspended,it has minimum PE
12) which one is incorrect: change of momentum=zero
13)moving horizontally,Blv …
14)rotating coil was A q14 from http://papers.xtremepapers.com/AQA/P...W-QP-JAN03.pdf
15) Capacitance value of 0.02F for one question forgot which was, answer choice was B IIRC
16) units of impulse = kgms-1
17) momentum question when the 2 trolleys collided: 12000Ns
18) negative ion moving through 2 plates, the lower earthed the upper at + 50V, the ion moved downwards towards the earthed plate
19. Something about max angular speed when the friction on a turntable of radius r is = mg/2
6 MORE MULTIPLE CHOICE QUESTIONS THEN DONE!
SECTION B IS COMPLETE
1a)define shm:
acceleration is proportinal to displacement, and acceleration acts toward equilibrium (2)
acceleration was 0.4
frequency of pendulum = 0.503 3s.f (3)
1500 revs per minute (2)
time for oscilations to be in phase again = 38 (3) -- if you think about it,
after 1 oscilation - difference on 0.1s
after 2 oscillations - difference of 0.2s....
after 19 oscillations difference of 1.9 seconds
so on the 19th osciation of the 1.9s pendulum, the 2s pendulum will arrive 1.9s after it, so on the 20th oscillation of the 1.9s pendulum they are in phase. again, on the 19th oscillation of the 2s pendulum they will be in phase.
19 x 2 = 38s
20 x 1.9 = 38s
TOTAL = 12

2) voltage = 30,000 (1)
b) (3) i think this was the capacity was increase from X to Y. find the energy stored? wasnt this the time taken for the spark to be produced?
yes, i think it was similar:

b) i think this was the capacity was increase from X to Y.time taken for the spark to be produced after the last discharge?
then Q=CV, then Q/Current = time was ~ 3.46s

you had to work out the charge stored and then use the current they gave you to find the time??
c) time taken for discharge will increase because time constant increases (2)
d) flash is brighter because more energy stored (2)
TOTAL = 7

3) 6 marker on fields
1. gravity always attractive
2. electric attractive/repulsive
3. both follow inverse square for Force
4. both follow inverse for potential
5. gravity uses mass
6. electric uses charge
7. only electric fields can be shielded
8. only electric fields depend on medium between charges
9. gradient of potential vs distance graph gives field strength
10. potential definition for both revolves around bringing a mass/charge from infinity to a certain point in a field and is zero at infinity
TOTAL = 6

4) 2 conditions when no force is exerted on the partcle (2)
field is parallel to velocity
particle is stationary
b) out of the plane of the paper
b) prove momentum is proportional to radius= mv^2/r = BQv , mv = BQr --> mv = kr
the speed of particle=8.7x107
c)time taken for partile to go through 1 dee = ~6.8x10^-8
d)show the time taken is independent of radius = 2pim/bq (3)
kinetic energy im MEV = 2.44 (3)
TOTAL = 15

5)
angle for max emf = 60 degrees -- this is at 90 degrees to the plane as emf = BANw sin(wt) therefore max emf is when sin(wt) =1, there for wt (which is the angle) = 90, 90-30 = 60 degrees
angle for max flux linkage = 5pi/6 = 2.62 radians -- this is at 0 degrees = 180 degrees, and N(PHI) = BAN cos(THETA), so max is when cos(THETA) = 1 so theta = 180, 180-39 = 150 degrees = 5pi/6
gradient of graph = emf (1)

drawing a graph (2) Positive Sine graph
peak voltage (3) I got 0.55 for peak to peak voltage then divided by 2 to get 27.5V but there is no set agreement on what the value is for this answer yet.. (MSI_10)

calculate flux density = 0.26 (2)( got 0.25 here
TOTAL= 10

Okay i'm going now... this is the best mark scheme ive done so far, ive tried to incorporate the others! please feel free to edit it! MOST OF THIS WAS SABA146 not me! but I like to feel I helped

whoops I thought I was editing...
0
6 years ago
#970
(Original post by ankou2612)
I got the same.
Same

Posted from TSR Mobile
0
6 years ago
#971
ratio of y/d=1/3 – I got 2/3...
ermm so what was the answer to this question i got a 1/3..
physics is soo streessfull.
0
6 years ago
#972
Guys regarding the maximum emf question, as someone has already mentioned it can be calculated using BANw

We could find BAN by finding the peak value of flux linkage from the graph which was 0.55, and w (angular speed) could also be calculated using the graph: 2pi/40ms

So

This method to me makes much more sense than attempting to find the gradient of a tangent when the flux graph cuts the x-axis, but who knows maybe they'll have 'alternative method' marks

I myself used the 1.1/40ms method which I've realised now is wrong

Posted from TSR Mobile
0
6 years ago
#973
Can someone explain how to do the time taken for the things to be in phase again really can't figure out how to do it!!! Time periods were 1.9 and 2.0 seconds
0
6 years ago
#974
(Original post by JD1234)
Can someone explain how to do the time taken for the things to be in phase again really can't figure out how to do it!!! Time periods were 1.9 and 2.0 seconds
I did (2pi x 10) / 2
Phase diff = (2pi x t) / T

Posted from TSR Mobile
0
6 years ago
#975
(Original post by Lay-Z)
I did (2pi x 10) / 2
Phase diff = (2pi x t) / T

Posted from TSR Mobile
How do you work out the time before they are in phase again though ??
0
6 years ago
#976
(Original post by JD1234)
Can someone explain how to do the time taken for the things to be in phase again really can't figure out how to do it!!! Time periods were 1.9 and 2.0 seconds
If you think it's just the lowest common multiple...
you could think that after one oscillation they are 0.1 seconds apart
2 oscillations= 0.2 and then keep going until they meet back up that's what most people have done but yea it's just the lowest common multiple.
0
6 years ago
#977
I also agree about that earth,sun,moon question. I worked out the forces and that's the order I got.
Also, I have some doubts about the graph of exs 5. Was it NΦ=f(t)? if it was then the gradient was giving us the emf. And since the gradient was constant for every 20 secs?,doesn't this mean that the emf would be constant for every 20 secs and then change polarity every time the gradient of the graph becomes positive or negative?
0
6 years ago
#978
(Original post by TheGuy117)
I think the 2 Qs which I attached were in the multiple choice, the answers are both D (100%)

Posted from TSR Mobile
0
6 years ago
#979
I think it was a negative sine lads.
Kgm/s not kgm/s2
e>s>m
0
6 years ago
#980
yeah, I just called my teacher and told me so too''/. I also agree about the 2 D answers.what about the angles at exs 5? For the first question: The conductor was moving anticlockwise so if it moved 60 degrees more it would have been parallel to the magnetic field lines and it wouldnt cut any of them so no emf?
0
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