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.