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# Official Thread For OCR Physics A G484 Jan 2011 watch

1. I'll post my answers and mark scheme later . I've just sat down and worked through the whole paper. Harder than Jan or June 2010 - but maybe not as hard as my first reaction suggested.

I'd expect the A grade to be around 43/44 out of 60. The usually grades go in steps of 4 marks or so.
2. (Original post by teachercol)
I'll post my answers and mark scheme later . I've just sat down and worked through the whole paper. Harder than Jan or June 2010 - but maybe not as hard as my first reaction suggested.

I'd expect the A grade to be around 43/44 out of 60. The usually grades go in steps of 4 marks or so.
How much later?
3. My first impression was its quite hard with a few tricky bits - but on reflection and wroking through it, maybe not so bad.

Usual disclaimer -this are my answers and may and probably will contain mistakes and typos. They are in no way connected to the exam's boars enswers or mark scheme.

1)a)i) For an isolated system, total momentum is constant WTTE (2)
ii) Kinetic energy is not conserved (1)
iii) 1) Total momentum before = total momentum after
( 2.4 x 3.0 ) + ( 1.2 x -2.0 ) = ( 3.6 x v )
v = 1.33 ms-1 (2)
2) KE = 1.2 mv^2
Total Ke before = 1/2 x 2.4 x 3.0^2 + 1/2 x 1.2 x 2.0^2 = 13.2J
Total KE after = 1/2 x 3.6 x 1.33^2 = 3.2 J
so not conserved so inelastic (2)

b)i) Volume of air = pi r^2 H
= pi x 5.0^2 x ( 12x 5) = 4712 m^3

Mass = vol x density = 4712 x 1.3 = 6130 kg (about 6000 kg) (2)
ii) 1) momentum = mv = 6130 x 12 = 7.36E4 kg ms-1 (1)
2) F = dp/dt = 7.36E4 / 5 = 1.47E4N (2)
3) F = mg so m = 1.47E4/9,81 = 1500 kg (1)
[ Total 13 ]

2a) i) Constant force perpendicular to velocity (1)
ii) Velocity is changing because direction is changing / v is a vector
Acceleration is towards centre of circle (2)
b) F = ma
GMm/r^2 = mv^2 /r
v^2 = GM/r r = 6.67E-11 x 6.0E24 / (3700)^2 = 2.92E7m (4)
c) i) Use rockets / jets of gas. (1)
ii) v = sqrt(GM/r) = sqrt(6.67E-11x6.0E24/2.0E7) = 4473 ms-1 (2)
[ Total 10 ]

3 a) i) 1kW.h = energy converted by a devise of power 1kW operating for a time of 1h. (1)
ii) cost = 12/100 x 70/1000 x (7 x 24) = £1.41 (2)
b) i) E = mctheta = 2.0 x 3800 x (18-3) = 1.14E5 J (2)
ii) rate = E/t = 1.14E5/(100 x 60) = 10 Js-1 (1)
c) Straight line from 18 to 0; stays at o for a time; steeper straight line down (3)
[ Total 9 ]

4a) i) Displacment is distance from equilibrium point in a specified direction
Amplitude is max displacement (2)
ii) frequency is no of oscillations per unit time; measured in s-1 or Hz
Angular freq is 2pi x freq = angle per unit time ; measured in radians s-1 (2)
b) i) 1) Amplitude = 2.5m (equilibrium point is 15.5m ) (1)
2) f = 1/T = 1/(12.5 x 3600) = 2.31E-6 Hz (2)
ii) vmax = 2 pi f A = 3.64E-4 ms-1 (2)
iii) x = 15.5+ 2.5 sin (1.45E-5t) (2 pi f = 1.45e-5) ; could use cos. (2)
[ Total 11 ]

5a) i) Random motion (spell random ..) in different directions with different speeds (1)
ii) Smoke particles change motion when collide with 'air' molecules
so molecules are moving in random directions with a range of speeds (high)
molecules are small since cant see them (3)
b) Same temp so same Ke so mv^2 is constant
m is 16x bigger for Oxygen so v is 1/4x ie 1800/4 = 450 ms-1 (3)
[Total 7 ]

6a) i) For a fixed mass of gas at constant temp, pressure is inversely proportional to volume (2)
ii) 1) 1/x graph / rectangular hyperbola
2) straight line through origin (3)
b) i) PV=nRT so n = (5.0E5 x 0.040)/(8.31 x 288) = 8.36 mlol (2)
ii) pV = nRT , V and T are constant so p/n = constant
n2/n1 = p2/p1 so n2 = 8.36 x 4.5E5/5.0E5 = 7.52 mol
Loss of gas = 8.36 - 7.52 = 0.84 mol
Loss of mass = 0.84 x 0.028 = 0.0235 kg (3)
[ Total 10 ]
4. Great apart from a few minor errors here and there, I seem to have done pretty well in that paper. I'm hoping for an A in this and an A in the G482 unit, as I had to retake after getting a C in June. I am comfortable seeing as everyone is predicting low grade boundaries for this paper, but even if they are high, I don't think I got any less than 50/60 anyway. Good luck everyone
5. Better than two of my best students who lost the last 5 marks because they didnt see the questions on the back page
6. (Original post by teachercol)
Better than two of my best students who lost the last 5 marks because they didnt see the questions on the back page
Hello Teachercol, do you think they will be carrying errors forward, particularly for the amplitude question? I got 5m for the amplitude and had used this throughout. Also, for the last part of that the question I had obtained all of that which you had got (although slight different values as I had used amplitude=5), however I did not include the 15.5m. Do you think I may get a mark?

Thank you!
7. They will usualy ecf forward in the amplitude result. You should get 1/2 for the Asin (2 pi f t) part if you left out the 15.5 (or used a different value)
8. How did everyone do?
9. (Original post by Right Guard 3.D)
How did everyone do?
90/90 you?

(thought i'd dropped a few points but obviously not many )
10. horrible
11. (Original post by Right Guard 3.D)
How did everyone do?
12. (Original post by Right Guard 3.D)
How did everyone do?
100%

Yourself?
13. (Original post by anshul95)
oh shut up. who are you anyway? I swear you got an E, i looked myself...
14. 74/90 I'm happy with that
15. (Original post by teachercol)
b) i) 1) Amplitude = 2.5m (equilibrium point is 15.5m ) (1)
2) f = 1/T = 1/(12.5 x 3600) = 2.31E-6 Hz (2)
ii) vmax = 2 pi f A = 3.64E-4 ms-1 (2)
iii) x = 15.5+ 2.5 sin (1.45E-5t) (2 pi f = 1.45e-5) ; could use cos. (2)
[ Total 11 ]
Your answer to this question confused me slightly. You have 2.31E-6 as your answer to b) i) 2) with a working of 1/(12.5 X 3600), but while I got exactly the same working I got a value of 2.2E-5.

I was wondering if I'd done anything wrong or if you made an error yourself. It's relatively important as this obviously affects the rest of the question.

EDIT: Also, on the last question I used 273.15 to convert to K to get 288.15 for T, is this a problem? It gave me a value of 8.35 for b) i) and for b) ii) 0.0234.
16. No prob using 273.15 but its normal to work to 3sigfigs.

Just checked calc and yes it should be 2.22E-5. My bad. I hope you'll ecf my answers
17. Can anyone explain why in question 2)c)ii)

It says the satellite is decelerated in order to reduce the radius of orbit, but when you work out the new velocity it is GREATER then the original velocity of the larger orbit?

Initial velocity = 3700ms^-1

New Velocity = 4500ms^-1

That hasn't decelerated!

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