# Newtonian World 11/6/2015 Unofficial mark scheme

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#1
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)

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
5
6 years ago
#2
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'?
0
6 years ago
#3
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?
0
6 years ago
#4
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...
1
#5
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?
0
6 years ago
#6
31.....game over...inabit university
0
#7
(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
0
6 years ago
#8
(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)
0
#9
(Original post by ecsefem)
31.....game over...inabit university
A good FoP paper always makes up for a poor NW paper
0
#10
(Original post by L'Evil Fish)
Would I get any marks for plotting

Vmax against f (proportional)
. Doubt it - its a resonance question!
0
#11
(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.
0
6 years ago
#12
Sorry what was 5 d about? I forgot the context. Thanks for this btw, very grateful.
0
6 years ago
#13
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.
0
6 years ago
#14
(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
0
6 years ago
#15
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
0
6 years ago
#16
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.
0
6 years ago
#17
(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 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
0
6 years ago
#18
Hi, for the damping graph, my graph decreased in amplitude and period increased very slightly? Do you think this will be allowed?
0
6 years ago
#19
(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
0
6 years ago
#20
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.
1
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