OCR Physics A - G484 The Newtonian World Watch

Katermerang
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(Original post by hoddy)
The force on the wall/ball was 8.4 N

12--9 = 21 m/s (Not 3 m/s as velocity is a vector and it rebounds at 9 m/s not slowed to 9 m/s)

I = Ft ---> F = I/t

F = (0.06 x 21) / 0.15 = 8.4 N

Can anyone clarify?

Also for the last question, I missed the point about increasing the specific heat capacity. Any chance I could blag a mark saying putting holes in the discs (like motorcycle brakes) could help ventilate heat to the surroundings as the holes are put in for heat dissipation. Doubt I'll get this mark though.
Yes, that calculation is correct. I do remember putting down 8.4N.

I have no idea if that explanation is any good.
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edris
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(Original post by hoddy)
The force on the wall/ball was 8.4 N

12--9 = 21 m/s (Not 3 m/s as velocity is a vector and it rebounds at 9 m/s not slowed to 9 m/s)

I = Ft ---> F = I/t

F = (0.06 x 21) / 0.15 = 8.4 N

Can anyone clarify?

Also for the last question, I missed the point about increasing the specific heat capacity. Any chance I could blag a mark saying putting holes in the discs (like motorcycle brakes) could help ventilate heat to the surroundings as the holes are put in for heat dissipation. Doubt I'll get this mark though.
Yes the force was 8.4N, I did that mark scheme in like 10 mins so didnt look back on the minor mistakes
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edris
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(Original post by Katermerang)
Yes, that calculation is correct. I do remember putting down 8.4N.

I have no idea if that explanation is any good.
Yes the force was 8.4N, I did that mark scheme in like 10 mins so didnt look back on the minor mistakes
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Im_a_cyborg
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(Original post by hoddy)
Also for the last question, I missed the point about increasing the specific heat capacity. Any chance I could blag a mark saying putting holes in the discs (like motorcycle brakes) could help ventilate heat to the surroundings as the holes are put in for heat dissipation. Doubt I'll get this mark though.
You are right, it increases service area and therefore aids heat loss. Therefore you should get the mark, i guess it depends on the examiner. However holed disc brakes are not normally used in cars due to a high risk of cracking, I don't think this will matter though to be honest.
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zoolander0110
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edris, do you have a copy of the paper?
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teachercol
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Usual disclaimers. These are in no sense official and my contain typos and errors

OCR A G484 Newtonian World 29 June 2010

I thought this was a pretty easy paper - and easy to predict if you looked at what wasn't on the Jan 2010 paper

Q1 a) 1 and 3. Impulse is same / Total energy is conserved [2]

b) i) Loss in KE = Final KE - Initial KE = 1/2 x 0.06 x (12^2 - 9^2) = 1.89J [2]
ii) Force = change in momentum / time = 0.06 x (12 - (-9) ) / 0.15s = 8.4 N (watch for signs) [2]
iii) Newton 3rd law says same as ii) = 8,4N [1]
c) i) particles are in constant random motion with (MB distribution of velocities)
no intermolecular forces
vol of molecules is negligible compared to vol of gas
duration of collision is negligible compared to time between collisions
can apply newtonian mechanics
etc [3]
ii) Molecules collide elastically with wall and bounce off
Wall exerts force on molcules changing their momentum
so molcules exert force on wall (=change in momentum per unit time)
Pressure = total force / area
(pretty much a repeat of steps in b)
(p = 1/3 rho csquared bar) [3]
Total 13

Q2 a) i) Vertical component of L = weight
Horizontal component of L = centripetal force [2]
ii) F = mv^2/r 1.8E6 = 1.2E5 x V^2 / 2E3 so v = 1.73E2 ms-1 [2]
b) F = ma
so GMm/r^2 = mv^2/r cancel and sub v = 2 pi r / T then rearrange [3]
c) i) Period is 24h + orbit is over equator [2]
ii) r^3 = GMT^2/4pi^2 sub T = (24 x 60 x 60) gives r = 4.2E7m [3]
Total 12
Q3 a)1) acc is proportional to displacment
2) acc is directed towards a fixed point / is in opposite direction to displacment [2]
b) i) KE graph is zero at end s and max at x = 0 / reflection of PE graoph in line E = 9mJ [2]
ii) Total Energy is horizontal line through E=18mJ [1]
c) i) Amplitude = max displacment = 0.04m [1]
ii) KEmax = 18E-3J so use 1/2 mv^2 to get v=0.548ms-1 [2]
iii) vmax = 2 pi f A so f = 2.18Hz [2]
d) i) Useful eg MRI - RF pulse at Larmor frequency makes presessing proton move to higher energy state [2]
ii) problem : bridge oscillating with large amplitude when freq of
gusting wind / steps of walkers = natural freq of bridge [2]
Total 14

Q4 a) i) Brownian motion [1]
ii) Air molecules move randomly in all directions at high speeds [2]
b) i) PV = nRT so n = 0.602 mol so m= 1.68E-2 kg [4]
ii) 1) Thermal equilibrium means gains heat at same rate as loses it / constant temp [1]
2) m and V are constant so p/T is constant p = 2.6E5 * 273/290 = 2.45E5 Pa [2]
Total 10

Q5 a) i) KE = 1/2 mv^2 (again!) = 3.54E5 J [1]
ii) Work done = change in energy = Fd
so F = 3.54E5/40 = 8.84E3 N (or use uvast and F=ma)
assume : braking force is only force acting ie ignore other resistive forces like drag [3]
b) i) 1/4 energy loss at each brake = 8.85E4J
= mcdT so dT = 141K = 141 C (trap a change of temp is same in Kelvin or Celsius) [2]
ii) Front brakes do more work so will be higher temp at front and lower at back
Discs will lose heat due to forced convection so wont reach as high a temp
Drag also act so less force needed so lower temps. [4}
iii) Use larger disc / bigger mass
Use material with higher shc
Use more discs / twin discs at each wheel. [1]
Total 11

So there we have it. Most of my students doing resits seem to have found it pretty straightforward
(apart from one who didnt spot the last question! )

Col

(edited typos)
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DanPok
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Thank you for this!

hopefully i'll get an okayish grade, what would you expect a B to be?
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teachercol
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Hard to say as ever - it was an easy paper so I'm expecting the grade boudaries to be quite steep.

so the B grade might be anything from 42-45 raw marks on the paper.
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03wgrego
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For the last part of question 4, would the pressure go up? cos if Volume is proportional to Temperature (Charles's Law), then if the temperature decreases, the Volume would decrease as well?
Then also Volume is inversely proportional to pressure (Boyle's law) so if the Volume decreases wouldnt the pressure increase?
So i got something like 2.8x10^5 Pa or summin
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dave2k
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I need to slap myself....HARD

for the gas molecules 1 marker, I wrote chaotic movement....:/
I am SO dumb, it was soo obvious
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DanPok
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(Original post by 03wgrego)
For the last part of question 4, would the pressure go up? cos if Volume is proportional to Temperature (Charles's Law), then if the temperature decreases, the Volume would decrease as well?
Then also Volume is inversely proportional to pressure (Boyle's law) so if the Volume decreases wouldnt the pressure increase?
So i got something like 2.8x10^5 Pa or summin
the volume doesn't decrease because the radius stays the same.. what your saying only applies to the gas inside, the volume of the ball stays the same.
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03wgrego
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ahh no way, never mind
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edris
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(Original post by teachercol)
Usual disclaimers. These are in no sense official and my contain typos and errors

OCR A G484 Newtonian World 29 June 2010

I thought this was a pretty easy paper - and easy to predict if you looked at what wasn't on the Jan 2010 paper

Q1 a) 1 and 3. Impulse is same / Total energy is conserved [2]

b) i) Loss in KE = Final KE - Initial KE = 1/2 x 0.06 x (12^2 - 9^2) = 1.89J [2]
ii) Force = change in momentum / time = 0.06 x (12 - (-9) ) / 0.15s = 8.4 N (watch for signs) [2]
iii) Newton 3rd law says same as ii) = 8,4N [1]
c) i) particles are in constant random motion with (MB distribution of velocities)
no intermolecular forces
vol of molecules is negligible compared to vol of gas
duration of collision is negligible compared to time between collisions
can apply netonian mechanics
etc [3]
ii) Molecules collide elastically with wall and bounce off
Wall exerts force on molcules changing their momentum
so molcules exert force on wall (=change in momentum per unit time)
Pressure = total force / area
(pretty much a repeat of steps in b)
(p = 1/3 rho csquared bar) [3]
Total 13

Q2 a) i) Vertical component of L = weight
Horizontal component of L = centripetal force [2]
ii) F = mv^2/r 1.8E6 = 1.2E5 x V^2 / 2E3 so v = 1.73E2 ms-1 [2]
b) F = ma
so GMm/r^2 = mv^2/r cancel and sub v = 2 pi r / T then rearrange [3]
c) i) Period is 24h + orbit is over equator [2]
ii) r^3 = GMT^2/4pi^2 sub T = (24 x 60 x 60) gives r = 4.2E7m [3]
Total 12
Q3 a)1) acc is proportional to displacment
2) acc is directed towards a fixed point / is in opposite direction to displacment [2]
b) i) KE graph is zero at end s and max at x = 0 / reflection of PE graoph in line E = 9mJ [2]
ii) Total Energy is horizontal line through E=18mJ [1]
c) i) Amplitude = max displacment = 0.04m [1]
ii) KEmax = 18E-3J so use 1/2 mv^2 to get v=0.548ms-1 [2]
iii) vmax = 2 pi f A so f = 2.18Hz [2]
d) i) Useful eg MRI - RF pulse at Larmor frequency makes presessing proton move to higher energy state [2]
ii) problem : bridge oscillating with large amplitude when freq of
gusting wind / steps of walkers = natural freq of bridge [2]
Total 14

Q4 a) i) Brownian motion [1]
ii) Air molecules move randomly in all directions at high speeds [2]
b) i) PV = nRT so n = 0.602 mol so m= 1.68E-2 kg [4]
ii) 1) Thermal equilibrium means gains heat at same rate as loses it / constant temp [1]
2) m and V are constant so p/T is constant p = 2.6E5 * 273/290 = 2.45E5 Pa [2]
Total 10

Q5 a) i) KE = 1/2 mv^2 (again!) = 3.54E5 J [1]
ii) Work done = change in energy = Fd
so F = 3.54E5/40 = 8.84E3 N (or use uvast and F=ma)
assume : braking force is only force acting ie ignore other resistive forces like drag [3]
b) i) 1/4 energy loss at each brake = 8.85E4J
= mx dT so dT = 141K = 141 C (trap a change of temp is same in Kelvin or Celsius) [2]
ii) Front brakes do more work so will be higher temp at front and lower at back
Discs will lose heat due to forced convection so wont reach as high a temp
Drag also act so less force needed so lower temps. [4}
iii) Use larger disc / bigger mass
Use material with higher shc
Use more discs / twin discs at each wheel. [1]
Total 11

So there we have it. Most of my students doing resits seem to have found it pretty straightforward
(apart from one who didnt spot the last question! )

Col

(edited typos)
I forgot to divide the Brake temperature by 4, I left it as 564 degrees, would i still get a mark? And the 1st question, I ticked the last 2 answer. I though momentum would be the same for each object because their masses would add together so both travel with same momentum.

But I have to admit, I found this paper very easy! probz lost about 7maks! hopefuly its still enough for an A!
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shepstar1
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How strict are OCR with sig figs and rounding? Their leniency varies from paper to paper and I'm not sure what's tolerable and what isn't.
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hackman®
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anyone got a copy of the paper?
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skipp
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42 with harsh marking (ie not giving myself a mark if I don't remember the question) so hopefully got a B...lets hope its 42 for the B boundary and not anything higher
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shepstar1
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46 being realistic. Hope I get my B overall, I'm on a complete knife edge!
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Katermerang
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57/60 being realistic. :woo: :woo:

Please please please A* :eek3:
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ViolinGirl
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(Original post by Katermerang)
57/60 being realistic. :woo: :woo:

Please please please A* :eek3:
Same

Well, guessing that is. No idea how actual markscheme will be.
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Katermerang
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(Original post by ViolinGirl)
Same

Well, guessing that is. No idea how actual markscheme will be.
I know, I took those 3 marks away for the wordy questions because If they are pick they will easily take those marks off Hopefully it's enough for Cambridge but at least I can end exams on a high.
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