# Physics A G484 NW - my answers and mark scheme Watch

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OCR Physics A G484 Newtonian World 16/1/13

Usual disclaimer. These are just my answers and are in no sense 'official'.

They may contain errrors and typos.

They are not intended to be a complete set of everything that will score marks.

I DONT know the grade boundaries. I dont know what grade you will get.

So here we go.

1 a Rate of change of momentum is proportional to the resultant force and in the same direction 92)

b i Conservation of momentum

(3.0 x 5.0) + (7.0 x -2.0) = (3.0 +7.0) x v

so v = 0.10 ms-1 ro right (+ve direction) (2+1)

Had a discussion with my guys as to whether 0.1 would lose a mark for sig figs. Answer is 'maybe'

ii Impulse = change of momentum of A = momentum after - momentum before

= (3.0 x 0.10) - (3.0 x 5.0) = - 14.7 Ns to left (2)

iii Impulse on B is equal and opposite to A ie 14.7 Ns to right

because Impulse = Ft. T is same and F is equal and opposite (2)

Total 9

2 a i mg = mv^2/r so v = sqrt (gr) = sqrt( 7.7 x 7.2E6) = 7450 ms-1 (3)

ii T = 2 pi r / v = 6070s (2)

b i No of orbits = 24 x 60 x 60 / 6070 = 14.2 (about 14) (1)

ii Circumference = 2 x pi x 6400 km

Each orbit crosses equator twice so no of orbits needed = (2 pi x 6400) / (2 x 3000) = 6.7 so yes (3)

c Surveillance / Weather / Surveying - NOT GPS /TV / radio / communications (1)

Total 10

3 a Every object attracts every other object with a force that is proportional to the product of the masses

and inversely proportional to the square of the distance between them (AW) (1)

b i GMT^2 = 4 pi^2 r^3

6.67E-11 x M x (7.2x60x60)^2 = 4 pi^2 x (1.3E8)^3

M = 1.98E27 kg (3)

ii g = GM/r^2 = 6.67E-11 x 1.98E27 / (2.4e10)^2 = 2.3E-4 Nkg-1 (2) ecf possible here

iii T^2 is prop to r3 so (TM/TA)^2 = (rM/rA)^3

so TM = TA x sqrt((RM/RA)^3) = 18060 h (3)

Total 9

4 a At last an experiment all my students have actually done!

Measure time for ten oscillations twice and average

divide by ten to get average period

Repeat for a range of masses (known)

Fiducial mark. Eye level. Small ampplitude so obeys Hooke's law.

Calculate f = 1/T and 1/m and 1/sqrt(m)

Plot grpahs f v 1/m and f v 1/sqrt(m) see if straight line through origin

(its the second one) 94)

b i f = 1/T = 1/1/2 = 0.83s

vmax = 2 pi f A = 2 pi 0.83 x 36E-3 = 0.188ms-1

KEmax = 1/2 m vmax^2 = 7.11E-3J (3)

ii Amax = (2 pi f)^2 x A = 0.99 ms-2 (2)

c KE of mass / grav PE of mass / elastic PE of spring

At bottom grav PE is min / KE = 0 / elastic PE = max

In middle grav PE is more / KE is max / elastic PE is less

At top grav PE is max / KE is zero / elastic PE is min (4)

Total energy syats constant.

Total 13

5 a i n = number of moles / N = number of molecules (1)

ii PV = nRT if mass and Temp are constant then n and T are constants

(n=m/mm) so PV is constant

so P is inversely prop to volume (Boyles law0 (2)

iii PV = Nm-2 x m^3 = Nm (1)

b Several approaches here. I'm going to show graph goes through origin.

Grad = (0.9-0.6)/(12-8) = 0.075

Graph goes through (8,0.6) so y int = 0.6 - 0.075x 8 =0

so its a straight line through origin so obeys Boyles law so T is constant. (3)

c PV=nRT (And yes Oxygen WOULD be diatomic at this temp so molar mass is confusing)

0.6E5 x 1/8 = (0.050/0.016) x 8.31 xT so T - 288K = 15 C (3)

Total 9

6 a i Atoms oscillate about equilibrium point (SHM) (1)

ii Amplitude increases. (Period will stay same). max vel increases. (1)

iii When melts, PE increases so internal energy increases. T stays constant (so KE stays same) (2)

b i Heat supplied by heater = heat gained by block + heat gained by insulation

48 x 720 = 0.98 x c x (54-18) + 0.027 x 850 x (38-18)

34560 = 35.28c + 459

c = 967 J kg-1K-1

ii If no insulating material, more heat will be lost to surroundings

so our figure for heat to block will be too big (we'll use 34560 insetad of something smaller)

so our figure for c will be too big (c = heat / m dT)

so it will be higher than previously. (2)

Total 10

My reaction to the paper was one of relief. Nothing too unusual there.

Generally my students liked it. That doesnt seem to be the general reaction here.

Col

Usual disclaimer. These are just my answers and are in no sense 'official'.

They may contain errrors and typos.

They are not intended to be a complete set of everything that will score marks.

I DONT know the grade boundaries. I dont know what grade you will get.

So here we go.

1 a Rate of change of momentum is proportional to the resultant force and in the same direction 92)

b i Conservation of momentum

(3.0 x 5.0) + (7.0 x -2.0) = (3.0 +7.0) x v

so v = 0.10 ms-1 ro right (+ve direction) (2+1)

Had a discussion with my guys as to whether 0.1 would lose a mark for sig figs. Answer is 'maybe'

ii Impulse = change of momentum of A = momentum after - momentum before

= (3.0 x 0.10) - (3.0 x 5.0) = - 14.7 Ns to left (2)

iii Impulse on B is equal and opposite to A ie 14.7 Ns to right

because Impulse = Ft. T is same and F is equal and opposite (2)

Total 9

2 a i mg = mv^2/r so v = sqrt (gr) = sqrt( 7.7 x 7.2E6) = 7450 ms-1 (3)

ii T = 2 pi r / v = 6070s (2)

b i No of orbits = 24 x 60 x 60 / 6070 = 14.2 (about 14) (1)

ii Circumference = 2 x pi x 6400 km

Each orbit crosses equator twice so no of orbits needed = (2 pi x 6400) / (2 x 3000) = 6.7 so yes (3)

c Surveillance / Weather / Surveying - NOT GPS /TV / radio / communications (1)

Total 10

3 a Every object attracts every other object with a force that is proportional to the product of the masses

and inversely proportional to the square of the distance between them (AW) (1)

b i GMT^2 = 4 pi^2 r^3

6.67E-11 x M x (7.2x60x60)^2 = 4 pi^2 x (1.3E8)^3

M = 1.98E27 kg (3)

ii g = GM/r^2 = 6.67E-11 x 1.98E27 / (2.4e10)^2 = 2.3E-4 Nkg-1 (2) ecf possible here

iii T^2 is prop to r3 so (TM/TA)^2 = (rM/rA)^3

so TM = TA x sqrt((RM/RA)^3) = 18060 h (3)

Total 9

4 a At last an experiment all my students have actually done!

Measure time for ten oscillations twice and average

divide by ten to get average period

Repeat for a range of masses (known)

Fiducial mark. Eye level. Small ampplitude so obeys Hooke's law.

Calculate f = 1/T and 1/m and 1/sqrt(m)

Plot grpahs f v 1/m and f v 1/sqrt(m) see if straight line through origin

(its the second one) 94)

b i f = 1/T = 1/1/2 = 0.83s

vmax = 2 pi f A = 2 pi 0.83 x 36E-3 = 0.188ms-1

KEmax = 1/2 m vmax^2 = 7.11E-3J (3)

ii Amax = (2 pi f)^2 x A = 0.99 ms-2 (2)

c KE of mass / grav PE of mass / elastic PE of spring

At bottom grav PE is min / KE = 0 / elastic PE = max

In middle grav PE is more / KE is max / elastic PE is less

At top grav PE is max / KE is zero / elastic PE is min (4)

Total energy syats constant.

Total 13

5 a i n = number of moles / N = number of molecules (1)

ii PV = nRT if mass and Temp are constant then n and T are constants

(n=m/mm) so PV is constant

so P is inversely prop to volume (Boyles law0 (2)

iii PV = Nm-2 x m^3 = Nm (1)

b Several approaches here. I'm going to show graph goes through origin.

Grad = (0.9-0.6)/(12-8) = 0.075

Graph goes through (8,0.6) so y int = 0.6 - 0.075x 8 =0

so its a straight line through origin so obeys Boyles law so T is constant. (3)

c PV=nRT (And yes Oxygen WOULD be diatomic at this temp so molar mass is confusing)

0.6E5 x 1/8 = (0.050/0.016) x 8.31 xT so T - 288K = 15 C (3)

Total 9

6 a i Atoms oscillate about equilibrium point (SHM) (1)

ii Amplitude increases. (Period will stay same). max vel increases. (1)

iii When melts, PE increases so internal energy increases. T stays constant (so KE stays same) (2)

b i Heat supplied by heater = heat gained by block + heat gained by insulation

48 x 720 = 0.98 x c x (54-18) + 0.027 x 850 x (38-18)

34560 = 35.28c + 459

c = 967 J kg-1K-1

ii If no insulating material, more heat will be lost to surroundings

so our figure for heat to block will be too big (we'll use 34560 insetad of something smaller)

so our figure for c will be too big (c = heat / m dT)

so it will be higher than previously. (2)

Total 10

My reaction to the paper was one of relief. Nothing too unusual there.

Generally my students liked it. That doesnt seem to be the general reaction here.

Col

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#2

What was the mass of block in shc ?i cnt remember it!

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#4

In the 5th question,i wrote gradient is equal to nRT. So T=gradient/nR and then after finding n i got answer of T. Is that method right? Cause i think my answer is wrong.

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#5

i got similar answers like yours but a small difference. e.g for the velocity i got something like 7446 instead of 7450....do u recon they'd still accept that?

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#6

(Original post by

i got similar answers like yours but a small difference. e.g for the velocity i got something like 7446 instead of 7450....do u recon they'd still accept that?

**Sinkim**)i got similar answers like yours but a small difference. e.g for the velocity i got something like 7446 instead of 7450....do u recon they'd still accept that?

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If you write more sf than required, you wont lose marks - provided your answer is correct to the number of sf you gave

so the actual answer here was 7445.804188!

7400

7450

7446

7445.8

7445.80 etc

are all acceptable.

But 7445 or 7447 would lose a mark for incorrect rounding

so the actual answer here was 7445.804188!

7400

7450

7446

7445.8

7445.80 etc

are all acceptable.

But 7445 or 7447 would lose a mark for incorrect rounding

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#8

hi teachercol

um, for the satellite orbital period question I did get the answer as 18060 hours, but I didnt write it on the dotted answer line. Although it is clearly written for the examiner to see, will I still get the mark?

um, for the satellite orbital period question I did get the answer as 18060 hours, but I didnt write it on the dotted answer line. Although it is clearly written for the examiner to see, will I still get the mark?

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(Original post by

hi teachercol

um, for the satellite orbital period question I did get the answer as 18060 hours, but I didnt write it on the dotted answer line. Although it is clearly written for the examiner to see, will I still get the mark?

**omar1995**)hi teachercol

um, for the satellite orbital period question I did get the answer as 18060 hours, but I didnt write it on the dotted answer line. Although it is clearly written for the examiner to see, will I still get the mark?

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#11

do I get error carried forward mark, for 5(c) I didn't add 0.6E^5 instead i used 0.6.

Rest of my process was the same.

Rest of my process was the same.

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(Original post by

What would you guess at the grade boundaries being?

**Rough Silk**)What would you guess at the grade boundaries being?

(Original post by

OCR Physics A G484 Newtonian World 16/1/13

I DONT know the grade boundaries. I dont know what grade you will get.

Col

**teachercol**)OCR Physics A G484 Newtonian World 16/1/13

I DONT know the grade boundaries. I dont know what grade you will get.

Col

Best guess 45/6 for an A 30 for an E.

A 46

B 42

C 38

D 34

E 30

That would put the 90% level at 50 and the 100% level at 54/60.

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(Original post by

do I get error carried forward mark, for 5(c) I didn't add 0.6E^5 instead i used 0.6.

Rest of my process was the same.

**singh224**)do I get error carried forward mark, for 5(c) I didn't add 0.6E^5 instead i used 0.6.

Rest of my process was the same.

General rule for power of ten errors (POT) is lose one mark then ecf.

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#14

(Original post by

Tricky - given there isnt a part c

General rule for power of ten errors (POT) is lose one mark then ecf.

**teachercol**)Tricky - given there isnt a part c

General rule for power of ten errors (POT) is lose one mark then ecf.

Thanks for the mark scheme

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(Original post by

lol, I meant the question where we had to calculate the temperature.

Thanks for the mark scheme

**singh224**)lol, I meant the question where we had to calculate the temperature.

Thanks for the mark scheme

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#16

Lost 5 marks of physics and say 2 marks for spelling mistakes, at worst 9 marks lost. Thank you teacher col. Made my day. Now I'm gonna start revising for tomorrow's exam.

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#17

(Original post by

I guees you get a very silly low temp (below oxygen liquifying point) if you miss out 1E5. The examiners may take the view that you should have spotted that.

**teachercol**)I guees you get a very silly low temp (below oxygen liquifying point) if you miss out 1E5. The examiners may take the view that you should have spotted that.

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#19

Can anyone have a guess as to what raw mark I will need for 82/90 UMS cos I need 82 to get an A overall (retaking)

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#20

Dont want to bombard you with questions but for energies on the spring would I gain any credit for mentioning potential energy, not gravitational potential energy??

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