# G484 Newtonian World 20/6/13

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OCR G484 Newtonian World 20 June 2013

Usual disclaimers. These are just my answers and mark scheme. These are in no sense official.

I dont know what the grade boundaries will be.

I dont know what you will score.

My lot came out smiling. I'd covered over 2/3 the paper the previous afternoon in a revision session.

They thought it was pretty easy - which seems to be the general impression here too..

One or two tricky bits though - and in a 60 mark paper, you cant afford to drop too many.

Every 3 marks is a grade usually!

Q1 a) i) rate of change of momentum is proportional to the resultant force

and in the same direction (1)

ii) F is prop to d(mv)/dt. If m is constant then F is prop to mdv/dt = ma

If using N m kg and s , this becomes an equality. (2)

b) i) Change of momentum = impulse = area under graph

= area of triangle = 1/2 x 4 x 20 = 40 Ns

so change in vel = Impulse / mass = 40/2.5 = 16 ms-1 (3)

ii) Mean acc = change in vel / time = 16/4 = 4.0 ms-2 (1)

iii) acc increases uniformly from 0 to 8ms-2 at t=2.0s then decreases uniformly to zero at t=4.0s (2)

Should have been a nice easy opener. F/t graph was widely predicted.

Total 9.

Q2 a) i) Lines drawn radially inwards touching surface (at right angles) (1)

ii) Lines are straight parallel (Spelling) and evenly spaced and towards surface (2)

b) i) g = GM/r^2 = 6.67E-11 x 5.7E26 / (6.0E7)^2 = 10.6ms-2 (2)

ii) mv^2/r = GMm/r^2 so v= sqrt(GM/r) (Mass of SATURN not satellite!)

so v = 8470 ms-1

KE = 1/2 mv^2 (mass of RHEA) = 8.25E28 J (1)

Total 9.

Seems easy but some will mix up the masses.

Q3 a) Still arguing about this one.

1) is true

2) is definitely false

4) is definitely false

3) It depends what you mean by increase and decrease! Does acc decrease from -2 to -3 or decrease?

a gets more negative as x gets more positive. I'd say 3 is wrong. (2)

b) i) T=1.2s so f=1/T = 0.83 Hz (1)

ii) A = max velocity from graph = 0.080m

so A = max v / 2 pi f = 0.015m

iii) max acc = (2 pi f)^2 A = 0.417 ms-2 (2)

c) i) Usual resonance curve y axis = amplitude / x-axis = driving freq

When driving freq = natural freq , get large amplitude

natural freq = freq of free oscillations

when f <> fo then amp is small. (4)

ii) eg Microwave: when EM radiation at same freq as natural freq of (scissoring) oscillation of water molecule

amplitude of oscillation of water molecule become large and heats food (2)

some did MRI since they'd revised it and it didnt come up last week.

everyday example ? maybe , maybe not.

Mostly routine apart from part a. Total 13

Q4 a) Vol of room = 43.2 m^3 so mass of air = vol x density of air 56.2kg

E = mc dT = 5.0E5J (3)

b) i) t = E/P = 5.0E5/2.3E3 = 218s (2)

ii) Vol needed = 0.50 MJ / 39 = 0.013 m^3

so mass needed = vol x density of heating gas 9.2E-3 kg (2)

c) Some heat lost via walls and ceiling

Also heating other things in room + heater itself etc (2)

Total 9

Part bii is cute. Trying to drown you in data.

Rest is easy.

Q5a) Momentum is a vector so when reverse direction, changes from + to - ( or vice versa)

so change = mv - (-mv) = 2mv (2)

b) P / F/A

F = no of collisions per unit time x change in momentum per collision

Higher temp -> higher velocity

so more collisions per unit time AND greater change of momentum per collision

so force is bigger

so pressure is bigger (3)

c) i) P/T is constant so P = 2.2E5 x (273+54) / (273 + 18) = 2.47E5 Pa (2)

ii) P = F/A F = weight of car

Initial area = 1200 x 9.81 / 2.2E3 = 0.0535 m^2

Final area = 1200 x9,91 / 2.47E5 = 0.0477 m^2

so change = 5.8E-3 m^2 (3)

Thats a bit tricky. Cant just use change in pressure.

Total 10

Q6 a) i) For a fixed mass of gas at constant temp

pressure is inversely prop to volume (2)

ii) P against 1/V (or V against 1/P) (1)

b) i) 1 PV=nRT

n = PV/RT = 1.2E7x 0,050 / 8.31 x (273+21) = 246

2 m = n x mm = 246 x 0.029 = 71.kg (3)

ii) Finally the tough one.

We are supposed to know that gas cylinders are made of metal rather than balloons so their volume doesnt change. You just put more moles in and increase the pressure.

For the extra gas n = PV/RT = 1.0E5 x 1.5/8.31x294 = 61.4 mols

so new total = 246+61 = 307 mol

Can now use P/n = constant - or recalculate P = nRT/V

P = 1.5E7Pa either way. (4)

Total 10

Pretty easy on the whole

Tougher bits were 3a, 4b 5cii and 6bii.

The A grade boundary isnt going to be less than 48.

Edited for mitake in Q3b and typos.

Usual disclaimers. These are just my answers and mark scheme. These are in no sense official.

I dont know what the grade boundaries will be.

I dont know what you will score.

My lot came out smiling. I'd covered over 2/3 the paper the previous afternoon in a revision session.

They thought it was pretty easy - which seems to be the general impression here too..

One or two tricky bits though - and in a 60 mark paper, you cant afford to drop too many.

Every 3 marks is a grade usually!

Q1 a) i) rate of change of momentum is proportional to the resultant force

and in the same direction (1)

ii) F is prop to d(mv)/dt. If m is constant then F is prop to mdv/dt = ma

If using N m kg and s , this becomes an equality. (2)

b) i) Change of momentum = impulse = area under graph

= area of triangle = 1/2 x 4 x 20 = 40 Ns

so change in vel = Impulse / mass = 40/2.5 = 16 ms-1 (3)

ii) Mean acc = change in vel / time = 16/4 = 4.0 ms-2 (1)

iii) acc increases uniformly from 0 to 8ms-2 at t=2.0s then decreases uniformly to zero at t=4.0s (2)

Should have been a nice easy opener. F/t graph was widely predicted.

Total 9.

Q2 a) i) Lines drawn radially inwards touching surface (at right angles) (1)

ii) Lines are straight parallel (Spelling) and evenly spaced and towards surface (2)

b) i) g = GM/r^2 = 6.67E-11 x 5.7E26 / (6.0E7)^2 = 10.6ms-2 (2)

ii) mv^2/r = GMm/r^2 so v= sqrt(GM/r) (Mass of SATURN not satellite!)

so v = 8470 ms-1

KE = 1/2 mv^2 (mass of RHEA) = 8.25E28 J (1)

Total 9.

Seems easy but some will mix up the masses.

Q3 a) Still arguing about this one.

1) is true

2) is definitely false

4) is definitely false

3) It depends what you mean by increase and decrease! Does acc decrease from -2 to -3 or decrease?

a gets more negative as x gets more positive. I'd say 3 is wrong. (2)

b) i) T=1.2s so f=1/T = 0.83 Hz (1)

ii) A = max velocity from graph = 0.080m

so A = max v / 2 pi f = 0.015m

iii) max acc = (2 pi f)^2 A = 0.417 ms-2 (2)

c) i) Usual resonance curve y axis = amplitude / x-axis = driving freq

When driving freq = natural freq , get large amplitude

natural freq = freq of free oscillations

when f <> fo then amp is small. (4)

ii) eg Microwave: when EM radiation at same freq as natural freq of (scissoring) oscillation of water molecule

amplitude of oscillation of water molecule become large and heats food (2)

some did MRI since they'd revised it and it didnt come up last week.

everyday example ? maybe , maybe not.

Mostly routine apart from part a. Total 13

Q4 a) Vol of room = 43.2 m^3 so mass of air = vol x density of air 56.2kg

E = mc dT = 5.0E5J (3)

b) i) t = E/P = 5.0E5/2.3E3 = 218s (2)

ii) Vol needed = 0.50 MJ / 39 = 0.013 m^3

so mass needed = vol x density of heating gas 9.2E-3 kg (2)

c) Some heat lost via walls and ceiling

Also heating other things in room + heater itself etc (2)

Total 9

Part bii is cute. Trying to drown you in data.

Rest is easy.

Q5a) Momentum is a vector so when reverse direction, changes from + to - ( or vice versa)

so change = mv - (-mv) = 2mv (2)

b) P / F/A

F = no of collisions per unit time x change in momentum per collision

Higher temp -> higher velocity

so more collisions per unit time AND greater change of momentum per collision

so force is bigger

so pressure is bigger (3)

c) i) P/T is constant so P = 2.2E5 x (273+54) / (273 + 18) = 2.47E5 Pa (2)

ii) P = F/A F = weight of car

Initial area = 1200 x 9.81 / 2.2E3 = 0.0535 m^2

Final area = 1200 x9,91 / 2.47E5 = 0.0477 m^2

so change = 5.8E-3 m^2 (3)

Thats a bit tricky. Cant just use change in pressure.

Total 10

Q6 a) i) For a fixed mass of gas at constant temp

pressure is inversely prop to volume (2)

ii) P against 1/V (or V against 1/P) (1)

b) i) 1 PV=nRT

n = PV/RT = 1.2E7x 0,050 / 8.31 x (273+21) = 246

2 m = n x mm = 246 x 0.029 = 71.kg (3)

ii) Finally the tough one.

We are supposed to know that gas cylinders are made of metal rather than balloons so their volume doesnt change. You just put more moles in and increase the pressure.

For the extra gas n = PV/RT = 1.0E5 x 1.5/8.31x294 = 61.4 mols

so new total = 246+61 = 307 mol

Can now use P/n = constant - or recalculate P = nRT/V

P = 1.5E7Pa either way. (4)

Total 10

Pretty easy on the whole

Tougher bits were 3a, 4b 5cii and 6bii.

The A grade boundary isnt going to be less than 48.

Edited for mitake in Q3b and typos.

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#2

Question 3a)

I put:

1.True

2.False

3.False

4.True.

4 was as acceleration increases speed decreases. This is true, as at max speed you have 0 accelertion, and at max acceleration you have 0 speed.

Edit: I think you got mixed up you should switch your 3 for 4, i think that is what you meant. But it is definitely TRUE.

You got question 6 right on second thoughts. I got it wrong, I thought pressure was x10 to the 5, but it was x10 to the 7. Hopefully that isn't too many marks lost though.

Also damn, everyone in my school thought total volume was 1.55, but it remained at 0.05. FFS. Personally I don't think the grade bounaries will be ridiculously high tbh.

I put:

1.True

2.False

3.False

4.True.

4 was as acceleration increases speed decreases. This is true, as at max speed you have 0 accelertion, and at max acceleration you have 0 speed.

Edit: I think you got mixed up you should switch your 3 for 4, i think that is what you meant. But it is definitely TRUE.

You got question 6 right on second thoughts. I got it wrong, I thought pressure was x10 to the 5, but it was x10 to the 7. Hopefully that isn't too many marks lost though.

Also damn, everyone in my school thought total volume was 1.55, but it remained at 0.05. FFS. Personally I don't think the grade bounaries will be ridiculously high tbh.

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#4

Thanks for the unofficial mark scheme! But for 3bii, wasn't it a velocity-time graph? I believe 0.080 was the maximum velocity of the oscillation.

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#5

(Original post by

Thanks for the unofficial mark scheme! But for 3bii, wasn't it a velocity-time graph? I believe 0.080 was the maximum velocity of the oscillation.

**Cosectant**)Thanks for the unofficial mark scheme! But for 3bii, wasn't it a velocity-time graph? I believe 0.080 was the maximum velocity of the oscillation.

Well if it shows people made these silly mistakes, then maybe the grade boundaries won't be too high. I think the only question that really caught me out was that tyre question, which I figured out how to do at the end, but didn't have time to finish!

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#6

**Cosectant**)

Thanks for the unofficial mark scheme! But for 3bii, wasn't it a velocity-time graph? I believe 0.080 was the maximum velocity of the oscillation.

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#7

Q1)ai. i didn't put net or resultant force, I just wrote force, does the mark scheme normally accept that?

biii. Do you need to include "uniformly"?

biii. Do you need to include "uniformly"?

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#8

(Original post by

Yeah, you're right, I was thinking that too.

Well if it shows people made these silly mistakes, then maybe the grade boundaries won't be too high. I think the only question that really caught me out was that tyre question, which I figured out how to do at the end, but didn't have time to finish!

**SyntaxZogre**)Yeah, you're right, I was thinking that too.

Well if it shows people made these silly mistakes, then maybe the grade boundaries won't be too high. I think the only question that really caught me out was that tyre question, which I figured out how to do at the end, but didn't have time to finish!

_{max}= (2pif)A and put the max velocity in and rearrange to give A?

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#9

(Original post by

Did yous use V

**martynsteel**)Did yous use V

_{max}= (2pif)A and put the max velocity in and rearrange to give A?
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#10

(Original post by

Yeah. Also what did you put for question 3a) Did you say for part 4, that it was false or true?

**SyntaxZogre**)Yeah. Also what did you put for question 3a) Did you say for part 4, that it was false or true?

a) True

b) False

c) False

d) True

D has to be true as when the velocity decreases and it heads towards the equilibrium point its' acceleration increases.

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#11

(Original post by

I went

a) True

b) False

c) False

d) True

D has to be true as when the velocity decreases and it heads towards the equilibrium point its' acceleration increases.

**martynsteel**)I went

a) True

b) False

c) False

d) True

D has to be true as when the velocity decreases and it heads towards the equilibrium point its' acceleration increases.

I did that too (:

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#12

(Original post by

I did that too (:

**eggfriedrice**)I did that too (:

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#13

(Original post by

Question 3a)

I put:

1.True

2.False

3.False

4.True.

4 was as acceleration increases speed decreases. This is true, as at max speed you have 0 accelertion, and at max acceleration you have 0 speed.

Edit: I think you got mixed up you should switch your 3 for 4, i think that is what you meant. But it is definitely TRUE.

Also you got question 6 wrong, I made the same error initially then changed it afterwards. It was 2.46 moles, 246 was a ridiculous value, so I typed it into my calculator again, and also all my friend got the same as me for that. So for part 2 of that question total moles = 63.56 or something like that.

EDIT: I typed it in my calculator again and I got 245 moles again, aaah I don't know! Oh wait I know what happened, it was 1.2 x 10to the 5 not 10 to the 7? Hopefully? I think it was, I know that all my friends got the total value for n number of moles as 63.56 or similar.

And it couldn't have been 245 moles, as it was a minute volume of gas compared to the second volume, which had 61 moles! So it was 2.45.

I got 100600 Pa (to 3sf) which makes sense, as it is a pressure in between the values of 100,000 Pa and 120,000 Pa.

**SyntaxZogre**)Question 3a)

I put:

1.True

2.False

3.False

4.True.

4 was as acceleration increases speed decreases. This is true, as at max speed you have 0 accelertion, and at max acceleration you have 0 speed.

Edit: I think you got mixed up you should switch your 3 for 4, i think that is what you meant. But it is definitely TRUE.

Also you got question 6 wrong, I made the same error initially then changed it afterwards. It was 2.46 moles, 246 was a ridiculous value, so I typed it into my calculator again, and also all my friend got the same as me for that. So for part 2 of that question total moles = 63.56 or something like that.

EDIT: I typed it in my calculator again and I got 245 moles again, aaah I don't know! Oh wait I know what happened, it was 1.2 x 10to the 5 not 10 to the 7? Hopefully? I think it was, I know that all my friends got the total value for n number of moles as 63.56 or similar.

And it couldn't have been 245 moles, as it was a minute volume of gas compared to the second volume, which had 61 moles! So it was 2.45.

I got 100600 Pa (to 3sf) which makes sense, as it is a pressure in between the values of 100,000 Pa and 120,000 Pa.

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#14

(Original post by

I was reasonably confident for this exam I can see a couple of mistakes I've done ie. the mass of Rhea not the mass of Saturn and the final question but I think I'll pick up method marks along the way as it was a relatively straightforward paper

**martynsteel**)I was reasonably confident for this exam I can see a couple of mistakes I've done ie. the mass of Rhea not the mass of Saturn and the final question but I think I'll pick up method marks along the way as it was a relatively straightforward paper

The question I couldn't do was Q4 bii, but 2 marks lost isn't the end of the world.

I have a feeling the grade boundaries will be quite high, a lot of people found it straight forward plus there were hardly any long answer questions.

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#15

For the last question I think I got the same answer but I did it a different way?! Will I still get the marks?

Posted from TSR Mobile

Posted from TSR Mobile

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#16

(Original post by

I can't remember if I made the mistake of the masses :s I'll have to see the paper again. What mistake did you make on the final question?

The question I couldn't do was Q4 bii, but 2 marks lost isn't the end of the world.

I have a feeling the grade boundaries will be quite high, a lot of people found it straight forward plus there were hardly any long answer questions.

**eggfriedrice**)I can't remember if I made the mistake of the masses :s I'll have to see the paper again. What mistake did you make on the final question?

The question I couldn't do was Q4 bii, but 2 marks lost isn't the end of the world.

I have a feeling the grade boundaries will be quite high, a lot of people found it straight forward plus there were hardly any long answer questions.

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#17

(Original post by

For the last question I think I got the same answer but I did it a different way?! Will I still get the marks?

Posted from TSR Mobile

**jmt95**)For the last question I think I got the same answer but I did it a different way?! Will I still get the marks?

Posted from TSR Mobile

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#18

(Original post by

The pressure was much higher in the cylinder originally, wasn't it? so it could fit a lot of moles? I got 245

**combinekidd**)The pressure was much higher in the cylinder originally, wasn't it? so it could fit a lot of moles? I got 245

Ah, we don't get error carried forward marks. If I get it wrong, then at least I get some working marks.

All of my firends said they got about 63 moles for total moles though, and they all said 2.45, so maybe all of us got it wrong? Wait one of them got 300 something, so yeah we did all get it wrong lol, you're right.

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#19

(Original post by

For the last question I didn't even consider moles I equated PV = constant for the gas before and gas added and did the sum of the constants over the sum of the volumes and got 4.8x10^5

**martynsteel**)For the last question I didn't even consider moles I equated PV = constant for the gas before and gas added and did the sum of the constants over the sum of the volumes and got 4.8x10^5

EDIT no it isn't. OP is completely right for everything in question 6.

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#20

(Original post by

That's the correct value.

**SyntaxZogre**)That's the correct value.

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