Newtonian World 11/6/2015 Unofficial mark scheme

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

Reply 1

Elcor

14

Thanks a lot!

For Q.a.i), do you think a valid reason for W and N not being equal is 'the spacecraft is traveling in a circle due to the Moon's rotation, and hence there must be a centripetal force so W would be greater than N'?

Reply 2

MrChemKid

4

can i ask for the area of force and time

does it have to be a specific value or could you say have 10Ns instead of 11?

Reply 3

verello12

14

Those grade boundaries are way off. A grade B was 35marks last year and most people are saying this exam was harder than last years...

Reply 4

teachercol

OP

9

I don't think this paper was harder than last years - and neither did my students. In 2013 45 was the A grade, in 2012 it was 42. I think this paper was on a par with those.
Just my opinion.

6c was hard. The rest?

Reply 5

ecsefem

2

31.....game over...inabit university

Reply 6

teachercol

OP

9

Original post by MrChemKid

can i ask for the area of force and time

does it have to be a specific value or could you say have 10Ns instead of 11?

They will allow a range = probably something like 10 to 12

Reply 7

L'Evil Fish

18

Original post by teachercol

I don't think this paper was harder than last years - and neither did my students. In 2013 45 was the A grade, in 2012 it was 42. I think this paper was on a par with those.
Just my opinion.

Would I get any marks for plotting

Vmax against f (proportional)

Reply 8

teachercol

OP

9

Original post by ecsefem

31.....game over...inabit university

A good FoP paper always makes up for a poor NW paper

Reply 9

teachercol

OP

9

Original post by L'Evil Fish

Would I get any marks for plotting

Vmax against f (proportional)

. Doubt it - its a resonance question!

Reply 10

teachercol

OP

9

Original post by Elcor

Thanks a lot!

For Q.a.i), do you think a valid reason for W and N not being equal is 'the spacecraft is traveling in a circle due to the Moon's rotation, and hence there must be a centripetal force so W would be greater than N'?

Yes - that's the same as saying its accelerating.

Reply 11

TadiwaM

Sorry what was 5 d about? I forgot the context. Thanks for this btw, very grateful.

Reply 12

Elcor

14

Think I've got over 50, which hopefully puts me at full UMS!!

For Q5.c) I think the second point to make would be that not all of the photons' energy would be adsorbed by the titanium, i.e. some would be reflected.

Reply 13

L'Evil Fish

18

Original post by teachercol

. Doubt it - its a resonance question!

It said describe the motion of the metal plate, without any axes given.

I talked about how displacement graph would be transformed, the velocity graph, and drew the Vmax against F.

Bit unfair, oh well. Still did shockingly

Reply 14

ibanezmatt13

2

For 1a)ii Did we have to draw lines on the diagram and things like that. I simply said "Equipment exerts gravitational force on moon. Force is of same magnitude as W and acts at centre of moon"

Just wondering because it's only 1 mark and unofficial ms has quite a bit in there

Reply 15

BrokenS0ulz

3

If you got the area wrong for the F/t graph, but all your subsequent working was correct, would you get 3/4 or 2/4?
Also thanks for doing this.

(edited 8 years ago)

Reply 16

L'Evil Fish

18

Original post by teachercol

I don't think this paper was harder than last years - and neither did my students. In 2013 45 was the A grade, in 2012 it was 42. I think this paper was on a par with those.
Just my opinion.

6c was hard. The rest?

I thought 6 was the easiest question :colonhash:

That resonance screwed me over, along with first question. And ashamed to say I didn't do the acceleration, assumption, or change in acceleration. Ooops

Reply 17

gothmog827

2

Hi, for the damping graph, my graph decreased in amplitude and period increased very slightly? Do you think this will be allowed?

Reply 18

Flux_Dubstep

3

Original post by verello12

Those grade boundaries are way off. A grade B was 35marks last year and most people are saying this exam was harder than last years...

tsr is a small vocal minority

Reply 19

fatart123

15

46-48 then. Was so concerned about not being able to answer any of q1 I didn't even realise that there was a resonance question and just drew an x/t graph instead. Rest of the paper was fairly easy tho at least. Did you have to draw a line on the graph for 6a? I just said that the gradient continues and then it intercepts the axis at temp = -273 with a volume of 0.

Newtonian World 11/6/2015 Unofficial mark scheme

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