# Newtonian World 11/6/2015 Unofficial mark scheme

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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

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

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#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'?

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'?

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#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?

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

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#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...

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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?

Just my opinion.

6c was hard. The rest?

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(Original post by

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?

**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?

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#8

(Original post by

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.

**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.

Vmax against f (proportional)

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(Original post by

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

**ecsefem**)31.....game over...inabit university

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(Original post by

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'?

**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'?

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#12

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

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#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.

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.

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#14

(Original post by

. Doubt it - its a resonance question!

**teachercol**). Doubt it - its a resonance question!

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

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#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

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

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#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.

Also thanks for doing this.

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#17

(Original post by

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?

**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?

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

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#18

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

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#19

(Original post by

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...

**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...

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#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.

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