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1. OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
OCR Physics A G484 NW 24 Jan 2012

Usual disclaimers - these are purely my answers and are in no sense official.
They are just to give you an idea of how well you did.
They may contain errors and typos.
They are not intended to be a complete set of every possible answer.

Generally I give answers to 3sf - but 2sf is always Ok.

Q1 a) i) Linear momentum = mass x velocity (1)
ii) Momentum is a vector because - velocity is a vector and a scalar (mass) x a vector is always a vector
or because it has a direction or because you need to use vector addition when adding . (2)
b) i) 1 a=dv/dt = -7.5/0.28 = (-) 26.8 ms-2. (1)
( I wish they would ask for acceleration when a negative sign would be clear = I'm sure it will be ignored)
2 F=ma = 850 x 26.8 = (-) 2.28E4 N. (2)
ii) 1/2 mvmax^2 = 0.45E6 so v = 32.5 ms-1 (2)
c) m1v1 + 0 = (m1 + m2) v v = 3.11 ms-1 (2)
[Total 10]

Q2 a) i) Amplitude = 0.40m; Period = 5.0s (1)
ii) f = 1/T = 0.2Hz; w = 2 pi f = 1.26 rads-1 (2)
b) All crosses are on x=0 . (3) v=max here; acc=zero here; PE is min where KE is max ie here.
c) i) amax = (2 pi f)^2 A = 4.09E5 ms-2 (3) convert kHz and mm ...
ii) Cute question. Stretch and challenge.
1st Approack WD = force x distance. For one cycle distance = 4 x amplitude. Power = WD / T = (F x 4A)/T
2nd Approach Power = F x average speed = F x (4A/T) = same answer
P = 0.25 x4 x 1.8E-3 x 2.4E3 = 4.32W (3)
[Total 12]

Q3 a) i) geostationAry (spelling) (1)
ii) so that stays above sma eplace on earths surface / doesnt move in sky (1)
iii) so can use fixed dishes for transmitting and receiving from satellite. (1) WTTE
iv) Use GMT^2 = 4 pi^2 r^3 T = 24x 60 x 60s
gives r = 4.2E7 m (3)
b) i) T^2 is prop to r^3 (1) - had a discussion with my students last week about whether this is on the syllabus!
ii) t1^2/t2^2 = r1^3/r2^3 so ratio is 27.3 ^2/3 = 9.1 (2)
[Total 9]

Q4 a) Latent heat of fusion (spelling) (1)
b)i) Internal energy = the sum of the random distribution of KE and PE for all the particles in the system (2)
ii) Energy is supplied to break bonds / PE of particles increases. (2)
c) i) Pt = m c dT 12 x 60 x 60 = 15 x 1.2 x 99 x dT so dT = 2.4K (3)
ii) Heat lost to surroundings / heat is needed to heat greenhouse frame too /
heat to heat contents of greenhouse /
rate of energy transfer is likely to fall over the hour - all seem Ok to me. (3)
Looking forward to the mark scheme
[Total 10]

Q5 a) i) Perfectly elastic = total KE stays constant (1)
ii) molecules bouce off walls perfectly eleastically.
molecules momentum changes ( = 2mv )
force on molecule is equal and opp to force on wall
force on moecule = change in momentum oer collision x no of collisions per unti time
pressure = force / area
lots of molecules colliding - add up ll forces. (4)
iii) when temp increases, velocity increases
larger change in momentum in each collision
more collisions per unti time
so bigger force so higher pressure. (2)
b) i) PV/T is constant either work out n and sub or use p1V1/T1 = p2V2/T2
Did you spot answer had to be in kPa? p2 = 36 kPa. (3)
ii) Balloon will expand. If gets too big, it may burst. Limit to how far envelope can stretch. (1)
[Total 11]

Q6 a) m=Mm/NA = 2.02E-3/6.02E23 = 3.36E-27 or you could just add masses of two protons together (2)
b) E=3/2 k T = 2.28E-20 J (2)
c) E = 1/2 mv^2 so v=2.63E3 ms-1 (2)
d) E is the mean Ke per molecule - some have much more than the mean / Maxwell- Boltzman distribution sketch.
so the highest energy molecules can escape. (2)
[Total 8]

Overall I thought the paper was pretty similar to June 2011. Yes there were one or two tricky bits but there were a
lot of easy marks too.
The grade boudaries were pretty low in June 44/40/36/33/30 . I'd expect about the same.

Most of my students reckon they dropped about 10 marks.
Of course the ones that messed it up proably wouldnt admit it to me!

Hope it went well for you. On to Frontiers of Physics.
Thats a monster paper.

Col

Edited answer for 2 c i
Last edited by teachercol; 24-01-2012 at 19:56.
2. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
Thanks teachercol again
3. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
Thank you!
4. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
Based on that, and being harsh on myself as i know ocr aren't the kindest markers, id predict i got 50/60.
5. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
(Original post by teachercol)
OCR Physics A G484 NW 24 Jan 2012

ii) Cute question. Stretch and challenge.
1st Approack WD = force x distance. For one cycle distance = 4 x amplitude. Power = WD / T = (F x 4A)/T
2nd Approach Power = F x average speed = F x (4A/T) = same answer
P = 0.080W (3)
If P= F x (4A/T), doesn't that mean that P= A/T in this question(as F=0.25N). Since A= 1.8 x 10^-3 and T= 1/(2.4 x 10^3), shouldn't P=4.32W?
Last edited by soulcrasher; 24-01-2012 at 19:54.
6. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
I agree with soulcrasher, P=4.32 W.
7. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
I cannot believe it 46... im stunned.. finally 3rd time doing this piece of **** exam and i was moderately harsh with this mark scheme. I only need a minimum B and 46 is an A yay! (hopefully this will reflect on results day, dont want to be too cocky/pleased though)
8. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
(Checks working) - I used the wrong amplitude and frequency!

It should be 0.25 x 4 x 1.8E-3 x 2.4E3 = 4.32W as you said.
9. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
(Original post by teachercol)
OCR Physics A G484 NW 24 Jan 2012

Usual disclaimers - these are purely my answers and are in no sense official.
They are just to give you an idea of how well you did.
They may contain errors and typos.
They are not intended to be a complete set of every possible answer.

Generally I give answers to 3sf - but 2sf is always Ok.

Q1 a) i) Linear momentum = mass x velocity (1)
ii) Momentum is a vector because - velocity is a vector and a scalar (mass) x a vector is always a vector
or because it has a direction or because you need to use vector addition when adding . (2)
b) i) 1 a=dv/dt = -7.5/0.28 = (-) 26.8 ms-2. (1)
( I wish they would ask for acceleration when a negative sign would be clear = I'm sure it will be ignored)
2 F=ma = 850 x 26.8 = (-) 2.28E4 N. (2)
ii) 1/2 mvmax^2 = 0.45E6 so v = 32.5 ms-1 (2)
c) m1v1 + 0 = (m1 + m2) v v = 3.11 ms-1 (2)
[Total 10]

Q2 a) i) Amplitude = 0.40m; Period = 5.0s (1)
ii) f = 1/T = 0.2Hz; w = 2 pi f = 1.26 rads-1 (2)
b) All crosses are on x=0 . (3) v=max here; acc=zero here; PE is min where KE is max ie here.
c) i) amax = (2 pi f)^2 A = 4.09E5 ms-2 (3) convert kHz and mm ...
ii) Cute question. Stretch and challenge.
1st Approack WD = force x distance. For one cycle distance = 4 x amplitude. Power = WD / T = (F x 4A)/T
2nd Approach Power = F x average speed = F x (4A/T) = same answer
P = 0.080W (3)
[Total 12]

Q3 a) i) geostationAry (spelling) (1)
ii) so that stays above sma eplace on earths surface / doesnt move in sky (1)
iii) so can use fixed dishes for transmitting and receiving from satellite. (1) WTTE
iv) Use GMT^2 = 4 pi^2 r^3 T = 24x 60 x 60s
gives r = 4.2E7 m (3)
b) i) T^2 is prop to r^3 (1) - had a discussion with my students last week about whether this is on the syllabus!
ii) t1^2/t2^2 = r1^3/r2^3 so ratio is 27.3 ^2/3 = 9.1 (2)
[Total 9]

Q4 a) Latent heat of fusion (spelling) (1)
b)i) Internal energy = the sum of the random distribution of KE and PE for all the particles in the system (2)
ii) Energy is supplied to break bonds / PE of particles increases. (2)
c) i) Pt = m c dT 12 x 60 x 60 = 15 x 1.2 x 99 x dT so dT = 2.4K (3)
ii) Heat lost to surroundings / heat is needed to heat greenhouse frame too /
heat to heat contents of greenhouse /
rate of energy transfer is likely to fall over the hour - all seem Ok to me. (3)
Looking forward to the mark scheme
[Total 10]

Q5 a) i) Perfectly elastic = total KE stays constant (1)
ii) molecules bouce off walls perfectly eleastically.
molecules momentum changes ( = 2mv )
force on molecule is equal and opp to force on wall
force on moecule = change in momentum oer collision x no of collisions per unti time
pressure = force / area
lots of molecules colliding - add up ll forces. (4)
iii) when temp increases, velocity increases
larger change in momentum in each collision
more collisions per unti time
so bigger force so higher pressure. (2)
b) i) PV/T is constant either work out n and sub or use p1V1/T1 = p2V2/T2
Did you spot answer had to be in kPa? p2 = 36 kPa. (3)
ii) Balloon will expand. If gets too big, it may burst. Limit to how far envelope can stretch. (1)
[Total 11]

Q6 a) m=Mm/NA = 2.02E-3/6.02E23 = 3.36E-27 or you could just add masses of two protons together (2)
b) E=3/2 k T = 2.28E-20 J (2)
c) E = 1/2 mv^2 so v=2.63E3 ms-1 (2)
d) E is the mean Ke per molecule - some have much more than the mean / Maxwell- Boltzman distribution sketch.
so the highest energy molecules can escape. (2)
[Total 8]

Overall I thought the paper was pretty similar to June 2011. Yes there were one or two tricky bits but there were a
lot of easy marks too.
The grade boudaries were pretty low in June 44/40/36/33/30 . I'd expect about the same.

Most of my students reckon they dropped about 10 marks.
Of course the ones that messed it up proably wouldnt admit it to me!

Hope it went well for you. On to Frontiers of Physics.
Thats a monster paper.

Col
Thank You, this has reassured me that I didn't completely mess up the exam, I tend to assume the worst :P
10. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
11. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
Think I dropped anywhere between 9-13 ish marks based on that, praying it's enough for an A. Can't seem to shake the horrible feeling that it didn't go well! It's probably down to making very silly mistakes that i'm kicking myself for!
12. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
(Original post by teachercol)
(Checks working) - I used the wrong amplitude and frequency!

It should be 0.25 x 4 x 1.8E-3 x 2.4E3 = 4.32W as you said.
Why do you 4x the amplitude?

I would have thought it would be 2x the amplitude as energy is used to bring the cone up to the amplitude, but no energy is required to bring it back down to the equilibrium position, then again to the negative amplitude and no energy required again to bring it back to equilibrium position, hence 2.16W. I think I am misunderstanding how the damping force actually effects it.

Thank you for the mark scheme!
13. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
I forget to times the amplitude by 4.......other then that my method was correct in da damping qustion.....how many marks would I lose?
14. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
(Original post by teachercol)
OCR Physics A G484 NW 24 Jan 2012

Spoiler:
Show
Usual disclaimers - these are purely my answers and are in no sense official.
They are just to give you an idea of how well you did.
They may contain errors and typos.
They are not intended to be a complete set of every possible answer.

Generally I give answers to 3sf - but 2sf is always Ok.

Q1 a) i) Linear momentum = mass x velocity (1)
ii) Momentum is a vector because - velocity is a vector and a scalar (mass) x a vector is always a vector
or because it has a direction or because you need to use vector addition when adding . (2)
b) i) 1 a=dv/dt = -7.5/0.28 = (-) 26.8 ms-2. (1)
( I wish they would ask for acceleration when a negative sign would be clear = I'm sure it will be ignored)
2 F=ma = 850 x 26.8 = (-) 2.28E4 N. (2)
ii) 1/2 mvmax^2 = 0.45E6 so v = 32.5 ms-1 (2)
c) m1v1 + 0 = (m1 + m2) v v = 3.11 ms-1 (2)
[Total 10]

Q2 a) i) Amplitude = 0.40m; Period = 5.0s (1)
ii) f = 1/T = 0.2Hz; w = 2 pi f = 1.26 rads-1 (2)
b) All crosses are on x=0 . (3) v=max here; acc=zero here; PE is min where KE is max ie here.
c) i) amax = (2 pi f)^2 A = 4.09E5 ms-2 (3) convert kHz and mm ...
ii) Cute question. Stretch and challenge.
1st Approack WD = force x distance. For one cycle distance = 4 x amplitude. Power = WD / T = (F x 4A)/T
2nd Approach Power = F x average speed = F x (4A/T) = same answer
P = 0.25 x4 x 1.8E-3 x 2.4E3 = 4.32W (3)
[Total 12]

Q3 a) i) geostationAry (spelling) (1)
ii) so that stays above sma eplace on earths surface / doesnt move in sky (1)
iii) so can use fixed dishes for transmitting and receiving from satellite. (1) WTTE
iv) Use GMT^2 = 4 pi^2 r^3 T = 24x 60 x 60s
gives r = 4.2E7 m (3)
b) i) T^2 is prop to r^3 (1) - had a discussion with my students last week about whether this is on the syllabus!
ii) t1^2/t2^2 = r1^3/r2^3 so ratio is 27.3 ^2/3 = 9.1 (2)
[Total 9]

Q4 a) Latent heat of fusion (spelling) (1)
b)i) Internal energy = the sum of the random distribution of KE and PE for all the particles in the system (2)
ii) Energy is supplied to break bonds / PE of particles increases. (2)
c) i) Pt = m c dT 12 x 60 x 60 = 15 x 1.2 x 99 x dT so dT = 2.4K (3)
ii) Heat lost to surroundings / heat is needed to heat greenhouse frame too /
heat to heat contents of greenhouse /
rate of energy transfer is likely to fall over the hour - all seem Ok to me. (3)
Looking forward to the mark scheme
[Total 10]

Q5 a) i) Perfectly elastic = total KE stays constant (1)
ii) molecules bouce off walls perfectly eleastically.
molecules momentum changes ( = 2mv )
force on molecule is equal and opp to force on wall
force on moecule = change in momentum oer collision x no of collisions per unti time
pressure = force / area
lots of molecules colliding - add up ll forces. (4)
iii) when temp increases, velocity increases
larger change in momentum in each collision
more collisions per unti time
so bigger force so higher pressure. (2)
b) i) PV/T is constant either work out n and sub or use p1V1/T1 = p2V2/T2
Did you spot answer had to be in kPa? p2 = 36 kPa. (3)
ii) Balloon will expand. If gets too big, it may burst. Limit to how far envelope can stretch. (1)
[Total 11]

Q6 a) m=Mm/NA = 2.02E-3/6.02E23 = 3.36E-27 or you could just add masses of two protons together (2)
b) E=3/2 k T = 2.28E-20 J (2)
c) E = 1/2 mv^2 so v=2.63E3 ms-1 (2)
d) E is the mean Ke per molecule - some have much more than the mean / Maxwell- Boltzman distribution sketch.
so the highest energy molecules can escape. (2)
[Total 8]

Overall I thought the paper was pretty similar to June 2011. Yes there were one or two tricky bits but there were a
lot of easy marks too.
The grade boudaries were pretty low in June 44/40/36/33/30 . I'd expect about the same.

Most of my students reckon they dropped about 10 marks.
Of course the ones that messed it up proably wouldnt admit it to me!

Hope it went well for you. On to Frontiers of Physics.
Thats a monster paper.

Col

Edited answer for 2 c i
Thanks once again.

Looking at your answers has once again actually made me more confident... I was expecting to not recognize anything. I feel similar to your students I think, dropped 10-12 marks perhaps.

I suck at stretch and challenge questions though If I don't recognize the connection I panic. I know this is pretty much how they identify A grade students but I'm sure I could do it if I realized that I'm supposed to interpret not automatically know it... any advice on how one could improve in these situations?
Last edited by shorty.loves.angels; 24-01-2012 at 20:24.
15. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
What was everyone discussing about 'oh no I found Vmax instead of amax'... confused.com. On the bright side, pretty sure I did what teachercol put for that.
16. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
(Original post by teachercol)
Q4 a) Latent heat of fusion (spelling) (1)
b)i) Internal energy = the sum of the random distribution of KE and PE for all the particles in the system (2)
ii) Energy is supplied to break bonds / PE of particles increases. (2)
c) i) Pt = m c dT 12 x 60 x 60 = 15 x 1.2 x 99 x dT so dT = 2.4K (3)
ii) Heat lost to surroundings / heat is needed to heat greenhouse frame too /
heat to heat contents of greenhouse /
rate of energy transfer is likely to fall over the hour - all seem Ok to me. (3)
Looking forward to the mark scheme
[Total 10]
I think it said that the 12W was the 'average' energy transfer over the time? so that answer wouldnt be valid.. dont hold me to it though

thanks a lot anyway - much appreciated!
17. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
(Original post by shorty.loves.angels)
What was everyone discussing about 'oh no I found Vmax instead of amax'... confused.com. On the bright side, pretty sure I did what teachercol put for that.
The Vmax thing came from the loudspeaker cone question. Me and many others used Force x Vmax to work out the power.

I only dropped about 7 marks from this mark scheme, so all is not too bad!
18. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
No - it says the average energy transfer rate at the start is 12W
19. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
(Original post by Europe Skies)
The Vmax thing came from the loudspeaker cone question. Me and many others used Force x Vmax to work out the power.

I only dropped about 7 marks from this mark scheme, so all is not too bad!
Oh right, I totally flunked that question anyway so I won't worry any more about that... I thought it was implying the question in which we WERE asked for a(max), if I'm even remembering it right.
20. Re: OCR Physics A G484 Newtonian World - unofficial answers and mark scheme
(Original post by teachercol)
No - it says the average energy transfer rate at the start is 12W
oh, fair enough then i just said 1. some of the thermal energy transfers to the surroundings e.g. the glass, and 2. the greenhouse isnt perfectly sealed so some of the heated up air will escape the greenhouse/be replaced by cooler air from outside..
do you think the 2nd is a valid answer?
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