Newtonian World 11/6/2015 Unofficial mark scheme

I'm predicting G485 will be hardest paper yet. They've done the same in Chem and Bio this year......

yeah it probably will be the bastards f******* us over.

Yeah I'm pretty sure it did, and that's how I solved it too. I actually really liked that question, even though in the exam I thought I'd messed up somewhere because it was a small value. Exactly the ame for the titanium and laser question

I'm predicting G485 will be hardest paper yet. They've done the same in Chem and Bio this year......

I haven't been working on G485, I sacrificed G485 for G484, and at the and i don't think I got an A or a B, this is so unfair, no brownian motion, no experiments, I learn all of them, if there were questions about kinetic theory, brownian motion, or specific heat capacity, I'd have got an A easily, why are they doin this, only people with a few exams can spend time and get A*, I had so many exams....i'm pissed

G484 Newtonian World 11/6/15 Unofficial mark scheme

Usual disclaimer: these are just my answers and are in no sense official. I don't mark this paper and I have no access to the mark schemes.
The answers may contain errors and typos. other correct answers may also be possible.

First impression: pretty straight down the middle paper. Heavy on maths and not much heat and gas stuff compared to some papers.

Q1 a) i) nice to see this - its something I spend quite a while teaching
cant be 3rd law pairs because
act on same object
different kinds of force
don't have to be equal if object is accelerating (2)
ii) W = grav force of moon on object down
so 3rd law pair is
grav force of object on moon up
magnitude =W and it acts at centre of moon (1)
b) standard proof in M2
resolve into horiz (u cos theta) and vert (u sin theta) components
calc time of flight as 2 time to reach highest point
v=u+at so 0 = u sintheta - gt
so time of flight = u sin theta /g
so range = time of flight x horiz component which is x = 2u^2 sin theta x cos theta / g
which is prop to v^2 (3)
Could do this by proportionalities all way rather then algebra Total (6)

Q2 a) i) m = mass of 1 mole / no of ions in a mole = 0.131/6.0E23 = 2.18E-25 kg (1)
ii) Force on ions = rate of change of momentum of ions
= no per sec x change in momentum of 1 ion
= 2.18E-25 x 3.2E4 x 9.5E18 = 0.066 N
Force on spacecraft = force on ions
acc = force / mass = 0.066 / 5.2E4 = 1.27E-5 ms-2 (3)
iii) Seems to me you need both Newton 2 and newton 3 so either will do.
N2 : force - rate of change of momentum etc (1)
iv) Mass decreasing as fuel ejected
a = F/m so acceleration increase (3 - really? 3 marks for that??)
b) i) Area under graph = impulse = change in momentum
Triangles + rectangles -> 11.1 (did you spot ms on x axis?)
Change in vel = impulse / mass = 11.1 / 180 (did you use the right mass?) = 0.062 ms-1 (4)
ii) Acc - increase uniformly for first 3ms, then increases non uniformly until reaches a max at 6.5 ms
then decreases no uniformly until 10ms (2)
Total (14)

Q3 a) i) straight line through origin with a negative gradient \ (2)
ii) grad = (2 pi f )^2 so f = sqr(gradient) / 2 pi (2)
b) i) vmax = 2 pi f A = 0.09 ms-1
so A = 0.09 /( 2 pi x 8) = .179E-3m (2)
ii) amax = (2 pi f )^2 A or vmax x 2 pi f = 4.52 ms-2 (2)
c) same period; amplitude decreases exponentially (2)
d) resonance curve.
max amplitude when driving freq = natural frequency at 8 Hz
small amplitude away from peak (3)
Total (13)

Q4 a) grav force between 2 masses = grav constant x mass1 x mass2 / separation squared (1)
b) standard proof which I make my lot learn and write out lots!
F = ma
GMm/r^2 = mv^2/r
so v^2 = GM/r v = 2 pi r / T
4 pi^2 r^2 = GM/r
rearrange
GM T^2 = 4 pi^2 r^3 so T^2 is prop to r^3 (4)
c) i) straight line through origin so T^2 is prop to r^3 (1)
ii) r^3 = (GM/4 pi^2) x T^2 so grad = GM / 4 pi^2
so M = 4 pi^2 x Grad / G = 1.97E30 kg (ie same as our sun!) (3)
Total (9)

Q5 a) E = hf = hc / lambda = 6.6E-34 x 3.0E8/1.1E-6 = 1.8E-19J (1)
b) P = 6.3E19 x 1.8E-19 = 11.34W
W=Pt so t = W/P m = rho x V and W = mcdT
so t = 8.1E-12 x 4.5E3 x 520 x (1700-20) / 11.34 = 2.8E-3s (wow seems a bit quick but numbers are right) (3)
c) some heat will conduct through titanium into rest of metal / some heat lost to surroundings (by convection and radiation)
(not sure about second bit - I dont think heat loss will happen in ms) (2)
d) temp will be constant while melting (1)
Total (7)

Q6 a) Volume is zero at absolute zero - draw line on graph (2)
b) i) internal energy = sum of random distribution of kinetic and potential energies for all the molecules in the system (1)
ii) Need to put energy to convert liquid to a gas so gas has
more potential energy since intermolecular forces are less
more kinetic energy if temp is higher (2)
c) i) PV=nRT so P = 45 / 8.31 x (273+20) / 1.2E-2 = 9.13E6 Pa (2)
ii) no of moles in cylinder = PV/RT = 41.1 moles.
Total no of moles stays same = 45+41.1 = 86.1
Use total volume n=PV/RT 86 = P x (1.2E-2 +2.0E-3) / 8.31 x 293
so P = 1.5E7 Pa (3)
Tricky but we've seen this before.
iii) If n and V are constant, P is prop to T
so if T decreases (293 -> 277) P decreases (1)
Total (11)

Not too bad.
Last year looked Ok but the Mark scheme was very picky. This paper is more clear cut. Expecting very tight grade boundaries...

A 43
B 40
C 37
D 34
E 31

Good Luck
Col

43 for an A? Are you ****ing kidding me? That was a horrendous paper.

Everyone complains, but it was a great paper imo. Not that I did very well on it. They got rid of the questions with mark schemes that change every year and the memorisation bull ****, it was actually about understanding physics this year.

But I genuinely feel sorry for the competent A Level physicists who would be able to solve all those problems were it not under the stress of exam conditions. You feel guilty for every second you spend thinking about how to answer a question when you've got such little time.

G484 Newtonian World 11/6/15 Unofficial mark scheme

Usual disclaimer: these are just my answers and are in no sense official. I don't mark this paper and I have no access to the mark schemes.
The answers may contain errors and typos. other correct answers may also be possible.

First impression: pretty straight down the middle paper. Heavy on maths and not much heat and gas stuff compared to some papers.

Q1 a) i) nice to see this - its something I spend quite a while teaching
cant be 3rd law pairs because
act on same object
different kinds of force
don't have to be equal if object is accelerating (2)
ii) W = grav force of moon on object down
so 3rd law pair is
grav force of object on moon up
magnitude =W and it acts at centre of moon (1)
b) standard proof in M2
resolve into horiz (u cos theta) and vert (u sin theta) components
calc time of flight as 2 time to reach highest point
v=u+at so 0 = u sintheta - gt
so time of flight = u sin theta /g
so range = time of flight x horiz component which is x = 2u^2 sin theta x cos theta / g
which is prop to v^2 (3)
Could do this by proportionalities all way rather then algebra Total (6)

Q2 a) i) m = mass of 1 mole / no of ions in a mole = 0.131/6.0E23 = 2.18E-25 kg (1)
ii) Force on ions = rate of change of momentum of ions
= no per sec x change in momentum of 1 ion
= 2.18E-25 x 3.2E4 x 9.5E18 = 0.066 N
Force on spacecraft = force on ions
acc = force / mass = 0.066 / 5.2E4 = 1.27E-5 ms-2 (3)
iii) Seems to me you need both Newton 2 and newton 3 so either will do.
N2 : force - rate of change of momentum etc (1)
iv) Mass decreasing as fuel ejected
a = F/m so acceleration increase (3 - really? 3 marks for that??)
b) i) Area under graph = impulse = change in momentum
Triangles + rectangles -> 11.1 (did you spot ms on x axis?)
Change in vel = impulse / mass = 11.1 / 180 (did you use the right mass?) = 0.062 ms-1 (4)
ii) Acc - increase uniformly for first 3ms, then increases non uniformly until reaches a max at 6.5 ms
then decreases no uniformly until 10ms (2)
Total (14)

Q3 a) i) straight line through origin with a negative gradient \ (2)
ii) grad = (2 pi f )^2 so f = sqr(gradient) / 2 pi (2)
b) i) vmax = 2 pi f A = 0.09 ms-1
so A = 0.09 /( 2 pi x 8) = .179E-3m (2)
ii) amax = (2 pi f )^2 A or vmax x 2 pi f = 4.52 ms-2 (2)
c) same period; amplitude decreases exponentially (2)
d) resonance curve.
max amplitude when driving freq = natural frequency at 8 Hz
small amplitude away from peak (3)
Total (13)

Q4 a) grav force between 2 masses = grav constant x mass1 x mass2 / separation squared (1)
b) standard proof which I make my lot learn and write out lots!
F = ma
GMm/r^2 = mv^2/r
so v^2 = GM/r v = 2 pi r / T
4 pi^2 r^2 = GM/r
rearrange
GM T^2 = 4 pi^2 r^3 so T^2 is prop to r^3 (4)
c) i) straight line through origin so T^2 is prop to r^3 (1)
ii) r^3 = (GM/4 pi^2) x T^2 so grad = GM / 4 pi^2
so M = 4 pi^2 x Grad / G = 1.97E30 kg (ie same as our sun!) (3)
Total (9)

Q5 a) E = hf = hc / lambda = 6.6E-34 x 3.0E8/1.1E-6 = 1.8E-19J (1)
b) P = 6.3E19 x 1.8E-19 = 11.34W
W=Pt so t = W/P m = rho x V and W = mcdT
so t = 8.1E-12 x 4.5E3 x 520 x (1700-20) / 11.34 = 2.8E-3s (wow seems a bit quick but numbers are right) (3)
c) some heat will conduct through titanium into rest of metal / some heat lost to surroundings (by convection and radiation)
(not sure about second bit - I dont think heat loss will happen in ms) (2)
d) temp will be constant while melting (1)
Total (7)

Q6 a) Volume is zero at absolute zero - draw line on graph (2)
b) i) internal energy = sum of random distribution of kinetic and potential energies for all the molecules in the system (1)
ii) Need to put energy to convert liquid to a gas so gas has
more potential energy since intermolecular forces are less
more kinetic energy if temp is higher (2)
c) i) PV=nRT so P = 45 / 8.31 x (273+20) / 1.2E-2 = 9.13E6 Pa (2)
ii) no of moles in cylinder = PV/RT = 41.1 moles.
Total no of moles stays same = 45+41.1 = 86.1
Use total volume n=PV/RT 86 = P x (1.2E-2 +2.0E-3) / 8.31 x 293
so P = 1.5E7 Pa (3)
Tricky but we've seen this before.
iii) If n and V are constant, P is prop to T
so if T decreases (293 -> 277) P decreases (1)
Total (11)

Not too bad.
Last year looked Ok but the Mark scheme was very picky. This paper is more clear cut. Expecting very tight grade boundaries...

A 43
B 40
C 37
D 34
E 31

Good Luck
Col

The general consensus at my sixth form was that this paper was a lot harder than last year. We were never really taught the information needed for question 1, and so it sent a lot of us into panic. It had also never come up in any past papers.

I haven't been working on G485, I sacrificed G485 for G484, and at the and i don't think I got an A or a B, this is so unfair, no brownian motion, no experiments, I learn all of them, if there were questions about kinetic theory, brownian motion, or specific heat capacity, I'd have got an A easily, why are they doin this, only people with a few exams can spend time and get A*, I had so many exams....i'm pissed

I had ten exams, all physics modules and maths modules. And I've got between 48-52. 😉

Everyone complains, but it was a great paper imo. Not that I did very well on it. They got rid of the questions with mark schemes that change every year and the memorisation bull ****, it was actually about understanding physics this year.

But I genuinely feel sorry for the competent A Level physicists who would be able to solve all those problems were it not under the stress of exam conditions. You feel guilty for every second you spend thinking about how to answer a question when you've got such little time.

I hope not. I have/had 18 exams and think I got around 53.



What'd you get in the practicals?
An A* would be much easier if I only had 10 exams.

it will not be more than 40 for an A. 43 is too high.

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