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OCR (not MEI) M2 - 13/01/2012

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Reply 40
Avatar for rogersnm
rogersnm
4
I got 233.9m for the power question, which I'm pretty sure is right. However, if you think about it, the car can never reach 20 m/s - equating forces and calculating acceleration means it's top speed up the slope is 12.3m/s.

Regarding the ball bouncing, the equation is t=0.8+1.6e+1.6e2+1.6e3...+1.6e∞=5t = 0.8 + 1.6e + 1.6e^2 + 1.6e^3 ... + 1.6e^\infty = 5
The sum to infinity of a geometric progression sums the first term, through to the last term, the first term is u1=1.6e0=1.6u_1 = 1.6e^0 = 1.6, therefore you have to take 1.6 away, which means the equation is:

t=0.8+S∞−1.6=5t = 0.8 + S_\infty - 1.6 = 5

t=S∞−0.8=5t = S_\infty - 0.8 = 5

S∞=5+0.8=5.8S_\infty = 5+0.8 = 5.8

a1−r=5.8=1.61−e\dfrac{a}{1-r} = 5.8 = \dfrac{1.6}{1-e}

5.8(1−e)=1.65.8(1-e) = 1.6

1.6−5.8=−5.8e1.6 - 5.8 = -5.8e

e=1.6−5.8−5.8=2129≈0.724e = \dfrac{1.6-5.8}{-5.8} = \frac{21}{29} \approx 0.724
(edited 12 years ago)
Reply 41
Avatar for Stainton93
Stainton93
you sure the 12.6 wasnt an average speed?
Reply 42
Avatar for rogersnm
rogersnm
4
Original post by Steve0606
why was the max speed 12.6?



m=1500[br]θ=5∘[br]R=750[br]P=25000[br][br]F=mam = 1500[br]\theta = 5^\circ[br]R = 750[br]P = 25000[br][br]F = ma
Pv−750−1500gsin⁡(5∘)=1500a\frac{P}{v} - 750 - 1500g \sin(5^\circ) = 1500a

Find the speed when a = 0:
25000v−750−1500gsin(5∘)=0\frac{25000}{v} - 750 - 1500g sin(5^\circ) = 0
25000v=750+1500gsin(5∘)\frac{25000}{v} = 750 + 1500g sin(5^\circ)
v=25000750+1500gsin⁡(5∘)=12.31m/sv = \dfrac{25000}{750 + 1500g \sin(5^\circ)} = 12.31 m/s
Reply 43
Avatar for ahmedm0209
ahmedm0209
Original post by Steve0606
why was the max speed 12.6?


P = 25000W = Dv
Resistive forces + Weight down slope = 750 + 1500gsin5 = 2031N (0 d.p)
At max speed, D = R + W => P/v = R + W => 25000/v = 2031 And so Vmax = 25000/2031 = 12.3 m/s
Reply 44
Avatar for Coulson72
Coulson72
Original post by Steve0606
why was the max speed 12.6?


from power = force times velocity
resolve down the slope you get something like 2030N
so max velocity is Power divided by force needed to overcome resistances
or about 12.6ish :smile:
Reply 45
Avatar for rogersnm
rogersnm
4
My actual working for the power question (using energies):

ΔGPE=mgΔh=1500g×Dsin(5∘)\Delta \mathrm{GPE} = mg \Delta h = 1500g \times Dsin(5^\circ)

ΔKE=12m202−12m102=121500(202−102)=225000J\Delta \mathrm{KE} = \frac{1}{2}m20^2 - \frac{1}{2}m10^2 = \frac{1}{2}1500(20^2 - 10^2) = 225000\mathrm{J}

Work done against friction=Fs=750×D=750D\mathrm{Work\ done\ against\ friction} = Fs = 750 \times D = 750D


Power=Et=25000=E28Power = \frac{E}{t} = 25000 = \frac{E}{28}
Therefore: E=25000×28=700000JE = 25000 \times 28 = 700000\mathrm{J}

Equate the energy produced to the energy changes:
E=ΔGPE+ΔKE+Work done against frictionE = \Delta \mathrm{GPE} + \Delta \mathrm{KE} + \mathrm{Work\ done\ against\ friction}

700000=1500g×Dsin(5∘)+225000+750D=D(1500g×sin(5∘)+750)+225000700000 = 1500g \times Dsin(5^\circ) + 225000 + 750D = D(1500g \times sin(5^\circ) + 750) + 225000

700000−225000=D(1500g×sin(5∘)+750)700000 - 225000 = D(1500g \times sin(5^\circ) + 750)


D=700000−2250001500g×sin(5∘)+750=233.9mD = \dfrac{700000 - 225000}{1500g \times sin(5^\circ) + 750} = 233.9\mathrm{m}
(edited 12 years ago)
Reply 46
Avatar for Sazhns2
Sazhns2
Well, looking at it now, I ballsed up the entire thing :/ Good thing i can count this as a AS Further maths unit, which I don't need a grade in :P
Original post by rogersnm
My actual working for the power question (using energies):

ΔGPE=mgΔh=1500g×Dsin(5∘)\Delta \mathrm{GPE} = mg \Delta h = 1500g \times Dsin(5^\circ)

ΔKE=12m202−12m102=121500(202−102)=225000J\Delta \mathrm{KE} = \frac{1}{2}m20^2 - \frac{1}{2}m10^2 = \frac{1}{2}1500(20^2 - 10^2) = 225000\mathrm{J}

Work done against friction=Fs=750×D=750D\mathrm{Work\ done\ against\ friction} = Fs = 750 \times D = 750D


Power=Et=25000=E28Power = \frac{E}{t} = 25000 = \frac{E}{28}
Therefore: E=25000×28=700000JE = 25000 \times 28 = 700000\mathrm{J}

Equate the energy produced to the energy changes:
E=ΔGPE+ΔKE+Work done against frictionE = \Delta \mathrm{GPE} + \Delta \mathrm{KE} + \mathrm{Work\ done\ against\ friction}

700000=1500g×Dsin(5∘)+225000+750D=D(1500g×sin(5∘)+750)+225000700000 = 1500g \times Dsin(5^\circ) + 225000 + 750D = D(1500g \times sin(5^\circ) + 750) + 225000

700000−225000=D(1500g×sin(5∘)+750)700000 - 225000 = D(1500g \times sin(5^\circ) + 750)


D=700000−2250001500g×sin(5∘)+750=233.9mD = \dfrac{700000 - 225000}{1500g \times sin(5^\circ) + 750} = 233.9\mathrm{m}


Assuming your numbers are correct, I can't see an error :s-smilie:
Reply 48
Avatar for rogersnm
rogersnm
4
Original post by Xero Xenith
Assuming your numbers are correct, I can't see an error :s-smilie:


Yeah but the issue is it's sort of meaningless, if you do s=vts=vt and find the average velocity if it travels 233.9m in 28 seconds then it comes out as 8.4m/s. Which is impressive seeing as it starts at 10 m/s and ends up at 20 m/s.
Reply 49
Avatar for Sazhns2
Sazhns2
Original post by rogersnm
Yeah but the issue is it's sort of meaningless, if you do s=vts=vt and find the average velocity if it travels 233.9m in 28 seconds then it comes out as 8.4m/s. Which is impressive seeing as it starts at 10 m/s and ends up at 20 m/s.


So the maximum distance was 560m assuming it instantly accelerates to 20ms-1 and the minimum distance was 280m assuming it goes all the way to B and then instantly accelerates? I got 233 as well, can't see a problem with my working either :confused:
Reply 50
Avatar for rogersnm
rogersnm
4
It looks like yet another error in an OCR exam paper to me.
Reply 51
Avatar for Steve0606
Steve0606
1
I must have got 233.85 then! Don't know where I got 127 from but I worked it out using the values given above (assuming they're correct of course) and got 233.85!
That max speed = 12.6 is confusing. I didn't realise that in the exam - I can't see how it's wrong.
It would be interesting if this is an actual error - Might mean low grade boundaries!
Reply 52
Avatar for CJAW
CJAW
9
Even if it is an error, OCR will be unlikely to admit to it because it was still possible to work out the distance. I think the grade boundaries will be low regardless, but more due to the difficulty of the paper. The question still seems to have created a lot of confusion though, so this will contribute to the low grade boundaries.
Reply 53
Avatar for Steve0606
Steve0606
1
To be honest, I don't think the paper was difficult enough to warrant low grade boundaries. Only one or two question caused problems but theres always one or two problematic questions with every paper. I reckon at least 56 or 57 for an A.
God, I messed up those last two questions. There goes my A*, or an A, or a B or a C or a D.... :facepalm:

And the collision question - I'd ran out of time before I even got there. And I couldn't do the progression question either.
Original post by Steve0606
To be honest, I don't think the paper was difficult enough to warrant low grade boundaries. Only one or two question caused problems but theres always one or two problematic questions with every paper. I reckon at least 56 or 57 for an A.


Judging by the amount of :facepalm: in this thread, I think they'll be comparable to last June's.

(Which were, starting at 100UMS and counting down to 40UMS: 64 57 50 43 36 29 22)

Original post by Contrad!ction.
God, I messed up those last two questions. There goes my A*, or an A, or a B or a C or a D.... :facepalm:

And the collision question - I'd ran out of time before I even got there. And I couldn't do the progression question either.


Yep, feel this way too! But check the above grade boundaries, last June you could lose 8 marks and still get 100UMS :smile:
Reply 56
Avatar for wibletg
wibletg
3
Original post by Xero Xenith
Judging by the amount of :facepalm: in this thread, I think they'll be comparable to last June's.

(Which were, starting at 100UMS and counting down to 40UMS: 64 57 50 43 36 29 22)



Yep, feel this way too! But check the above grade boundaries, last June you could lose 8 marks and still get 100UMS :smile:


I agree with you. June was a difficult paper, and this one was pretty ridiculous.

The centre of mass question, for example - wasn't too difficult but I can see it swallowing up a lot of people's time :tongue:
Original post by wibletg
I agree with you. June was a difficult paper, and this one was pretty ridiculous.

The centre of mass question, for example - wasn't too difficult but I can see it swallowing up a lot of people's time :tongue:


That was another panic point for me! The distance just had to be negative, right?

EDIT: I'm going to stop posting here. M2 went badly :facepalm:
Reply 58
Avatar for wibletg
wibletg
3
Original post by Xero Xenith
That was another panic point for me! The distance just had to be negative, right?

EDIT: I'm going to stop posting here. M2 went badly :facepalm:


I didn't actually do the paper but I've seen a copy and to be honest it was pretty nasty. Took my further maths teacher a while to do the CoM question himself :tongue:

And at least you have a chance to resit it in the summer :smile: not all bad :tongue:
Original post by Xero Xenith
Judging by the amount of :facepalm: in this thread, I think they'll be comparable to last June's.

(Which were, starting at 100UMS and counting down to 40UMS: 64 57 50 43 36 29 22)



Yep, feel this way too! But check the above grade boundaries, last June you could lose 8 marks and still get 100UMS :smile:

Those boundaries made me feel a bit better, thanks. I knew it was 50 for an A last year, I'm hoping for similar this time. Thing is, I did Jan 11 the day before the exam and only dropped 2 marks, with about half an hour to spare. This time, I had huge timing issues and couldn't do some questions at all.

Let's hope I can drop 20 and still get an A - I think I may have overreacted yesterday. I couldn't do the progression, my mind just went blank, and then I attempted the last question. I did the first part of the last question up until I had sin theta being greater than one and then I realised that I'd misread the beginning of the question and just sat there crying with ten minutes to spare. Didn't even get time to start the last nine-marker.

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