M/C in no particular order:

1)t root2, <--- it was something to do with the time period of a mass on two springs (if I remember right..)
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>fs>fm I just thought that would mean something would be more attracted to the sun over the moon so it would leave the moons orbit? many of us also got:fe>fm>fs
9） ratio of y/d=1/3 – I got 2/3 :/
10) R earth^2/Rmoon^2=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/Physics/2003%20Jan/AQA-PA04-W-QP-JAN03.pdf MS - http://papers.xtremepapers.com/AQA/Physics/2003%20Jan/AQA-PA04-W-MS-JAN03.pdf
15) Capacitance value of 0.02F for one question forgot which was, answer choice was B
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 upwards towards the positive 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)
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) 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
c) time taken for discharge will increase because time constant increases (2)
d) flash is brighter because more energy stored (2)
TOTAL = 7

4) 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

3) 2 conditions when no force is exerted on the particle (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)
1500 revs per minute (2) [I'm sorry MSI_10 :'( ]

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 = size of emf (1)

drawing a graph (2) Negative Sine graph the gradient is the -emf induced, inverting the graph. :'(
peak voltage (3) 86.4V http://www.thestudentroom.co.uk/showthread.php?t=2230354&page=51&p=43151006#post43151006 THANK YOU SO MUCH QWERTISH!
calculate flux density = 0.26 (2)
TOTAL= 10

Reply 1025

aedixon1

Original post by MisterE1

I don't think it was much harder than usual, I think the M/C was a bit harder than normal, but the 6 mark question was nice so it balanced out.

And I wouldn't call you an idiot for finding it hard! it wasn't easy..

Yeh the 6 marker was a god send! But some parts of each question i struggled with, luckily the grade boundaries are usually low and usually 68% for an A

Reply 1026

Jack93o

just checking, you know that question which asked for the angle to be in radians

did they already have the word 'radians' in the answer box?

Reply 1027

MisterE1

Original post by Jack93o

just checking, you know that question which asked for the angle to be in radians

did they already have the word 'radians' in the answer box?

yep

Reply 1028

Jack93o

Original post by MisterE1

yep

and, you're allowed to leave it in decimal places right? (I didn't leave it as a fraction with pi)

Reply 1029

MSI_10

Isn't radius of earth by radius of moon 3.7

http://www.google.co.uk/webhp?source=search_app#sclient=psy-ab&q=radius+of+earth+divided+by+radius+of+moon&oq=radius+of+earth+divided+by+radius+of+moon&gs_l=hp.3..33i29i30.7309.7815.1.7964.4.4.0.0.0.0.88.349.4.4.0.cqrwrth.1.0.0.0..1.1.17.psy-ab.bReAMcN57BA&pbx=1&bav=on.2,or.r_qf.&bvm=bv.47883778,d.d2k&fp=f85c2b3b14f6a49c&biw=1920&bih=955

Reply 1030

fizzbizz

If the flux graph was -cosine surely the emf graph is the derivative of -cosine which is +sine?

Reply 1031

MSI_10

Original post by Jack93o

and, you're allowed to leave it in decimal places right? (I didn't leave it as a fraction with pi)

2.6 radians?

Reply 1032

Alivere

Is there an unofficial markscheme for the multiple choice yet? :smile:

Reply 1033

Jack93o

Original post by MSI_10

2.6 radians?

can't remember the exact number I got, all I know was that I pressed the '='' sign on my calculator after calculation involving pi, and wrote down a decimal place number

Reply 1034

Jack93o

Original post by fizzbizz

If the flux graph was -cosine surely the emf graph is the derivative of -cosine which is +sine?

emf goes the other way apparently (due to lenz law), so its negative sine (negative of the gradient of the flux)

I know, it hurts right? I put positive sine as well :frown:

Reply 1035

MSI_10

Original post by Jack93o

can't remember the exact number I got, all I know was that I pressed the '='' sign on my calculator after calculation involving pi, and wrote down a decimal place number

http://www.mathinary.com/calculation.jsp?calcname=Grader&grader=&pi=Pi&radianer=2.6&

Reply 1036

saba146

Original post by Jack93o

emf goes the other way apparently (due to lenz law), so its negative sine (negative of the gradient of the flux)

I know, it hurts right? I put positive sine as well :frown:

the y-axis said change in flux linkage, not just the flux linkage, so the gradient gave emf - faradys law rate of change of flus linkage is proportianl to emf. the graph wee were given started at a negative maximum
so it was a positive sine graph
look at page 127 of the nelson thornes book