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    (Original post by oli_G)
    Damn I see my mistake now, if I did all the other parts of the method correctly (with a correct a) ) how many marks do you think I'll lose?
    Generally it is 2 for that sort of mistake, because you lose a mark for the actual mistake and then you get the answer wrong so that's the other mark. But possibly 3
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    (Original post by MAD Phil)
    I might be wrong, but I don't think so. The little diagram on the last page is just a unit-circle diagram to establish which value of theta I was looking for. (You look for intersections with x = a to solve cos(theta) = a and with y = b to solve sin(theta) = b.) Because these unit circle diagrams conventionally measure theta anticlockwise from the rightwards horizontal, and the question measures theta clockwise from the downward vertical, the little diagram is rotated 90 degrees compared to the main diagram.

    In fact, to avoid this confusion, (compounded by the fact that according to their diagram, the critical value of theta is negative), I almost decided to do the Newton and CofE again, instead of using the result proved in (a).
    Oh ok I get it now, thanks - damn though that means I seriously mucked up the last question, and I already know another one I screwed up on, any ideas what the grade boundaries might be like?
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    (Original post by oli_G)
    Damn I see my mistake now, if I did all the other parts of the method correctly (with a correct a) ) how many marks do you think I'll lose?

    I'd guess two lost I think. Any more than that would be incredibly harsh.
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    (Original post by mashmammad)
    About the boundaries, I've noticed that m3 in the recent years has only had two types of boundaries, one is the one that you need 63 for 90% and the other is the one that you need 70 for 90%. I'm not an expert obviously, but I think sadly this one will be one of those that you need 70 for 90%. Obviously it's just a guess. No offence intended.
    There' s a bit more range than that... there were a couple of 69s...

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    (Original post by kettlechips)
    There' s a bit more range than that... there were a couple of 69s...

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    Eh eh... 69's? I think you want the relationship forum
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    I badly messed up the centre of mass question, no idea how done so many practice questions never got this type wrong.

    I somehow expanded (x+1)^4 wildy wrong through either nerves or rushing, I used the correct methods but got different answers obviously (thought not different enough that they where feasible ), will I get error carried forward to part b) if I use my wrongly calculated mass and distance. And will I get some method marks for part a)? How many marks was the total question out of, 13?

    My other mistake was in question 7 I took away a an extra 1/2gt^2 through misusing the basic SUVAT equations (used s=(v+u)2*t + 1/2at^2), how many marks do you think I will lose there (everything else perfect + well explained methods)

    So disappointed with this paper, losing all those marks from basic errors that might well cost me my further maths *.

    Think I have any chance of 84+ UMS taking those mistakes into account?
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    Yeah, but tbh 69 and 70 aren't that different since we are only guessing.
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    (Original post by Boy_wonder_95)
    How was M3?
    the paper was fair, M3 is just damned hard, ran out of time for the last part of the last question, resisting looking at the unofficial mark scheme thanks for asking
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    (Original post by MAD Phil)
    These papers contain some quite awkward questions, and seem quite long. (But partly that's just me trying to get exact answers when any sane person in an exam would be using a calculator.)

    I haven't checked the international one at all, so there will probably be even more mistakes in that. (See below.) I'll check out the errors found in the first one now, and answer other points after that.
    Thanks. But I think in question 5 you have made a mistake. The correct answer to a is (1094/726) which is equal to (547/363). And consequently the part b is wrong with a small difference as well. Double check please. Thanks again for sharing.
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    (Original post by MAD Phil)
    These papers contain some quite awkward questions, and seem quite long. (But partly that's just me trying to get exact answers when any sane person in an exam would be using a calculator.)

    I haven't checked the international one at all, so there will probably be even more mistakes in that. (See below.) I'll check out the errors found in the first one now, and answer other points after that.
    Sorry. I just read your other post saying you've realised. My bad. Sorry.
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    (Original post by YDQ)
    For Question 5-

    The answer should be 547/363? Just done it on my calculator...
    that's also my answer for 5a. Unfortunately in 5b i found the volume of a whole sphere instead of the hemisphere how stupid is that
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    i did everything right but did not take away from 2 for 5B, how many marks will i lose?
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    Grade boundaries?
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    (Original post by YDQ)
    For Question 5-

    The answer should be 547/363? Just done it on my calculator...
    Is this international or UK paper and if UK what was the question about.
    If it is centre of mass:

    http://i40.tinypic.com/2z6eezt.jpg
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    (Original post by Sujman)
    i did everything right but did not take away from 2 for 5B, how many marks will i lose?
    I would say 1 or possibly 2
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    How many marks do you guys reckon I will lose for the centre of mass question - i forgot to square the yterm so i just integrated x((x+1)^2) divided by the integral of (x+1)^2
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    (Original post by tornzy)
    How many marks do you guys reckon I will lose for the centre of mass question - i forgot to square the yterm so i just integrated x((x+1)^2) divided by the integral of (x+1)^2
    What about mass. Did you do the same for finding the mass i.e. M = ρπ∫y² dx.
    If so, I would say you would lose all the answer marks but will get a few method mark, especially if your method of proving the thing was correct. You would probably get about 5-6-7-8 / 13(might be 14 not sure).
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    the radius is 1, before it rotates the y coordinate is 1 when x is 0, so when the graph rotate 360 degree the smaller end is a circle of radius one which is the same as the radius of the hemisphere.
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    (Original post by cant_think_of_name)
    Eh eh... 69's? I think you want the relationship forum
    Lol I knew someone would say that

    (Original post by mashmammad)
    Yeah, but tbh 69 and 70 aren't that different since we are only guessing.
    yeah, just the principle of the thing... and it would make a difference if you got 69


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    My thoughts on the two papers:-

    M3 UK Version
    Q1) The simplest sort of horizontal circles question. Very reasonable first question to get people started.
    Q2) Acceleration depending on t. Needs two simple integrations and solving a quadratic; I had to re-read the question to check it was OK to ignore the negative solution for t. I'm pretty sure it was, but it would have been nice of them to make it more obvious by saying the motion started at t = 0. Otherwise O.K.
    Q3) A slightly unusual conical pendulum question in that instead of being given the tension, you had to find it by considering Q. Otherwise quite easy if you have practiced conical pendulums.
    Q4) Reasonable energy question. Important to read carefully; the second answer would be different if the string had been replaced by a spring.
    Q5) I think this one was rather nasty. Originally I did all the integration by hand, and it takes ages (if you can stay awake through the boredom). It would be even worse if you multiplied out the 4th power of the bracket instead of (implicitly or explicitly) changing variable to u = x + 1. I also made a sign mistake (pointed out by The Burgeoning - thanks). The question only gives a reasonable rate of return on your effort if you use a calculator to do the definite integrations. I've now changed my answers to do it that way. However, I've just noticed that it says "use algebraic integration", so now I'm worried that they actually might want you to do it the long way. (I've put that version at the end of my answers.)
    Q6) SHM with two (unequal) horizontal strings. The first two parts fall nicely to the standard method that I teach my students, and the last part to the reference-circle method, or to just substituting into x = a cos(wt).
    Q7) A rather awkward type 2 vertical circles/projectile question. The awkwardness is mainly in trying to use the working done in part (a) to get the initial conditions for the projectile motion in part (c). The critical value of theta is negative and of size greater than 90 degrees. I was tempted to do the Newton and CofE afresh for point B instead of reusing the earlier working. Also, getting x and y in terms of t is going to take a while if you use uvast (or suvat, as some weird people call it ) or if you integrate, find the constant of integration and then substitute back, 4 times! This is where "iauic" (integrate and use initial conditions) comes in useful.

    M3 International version.
    Q1) Horizontal circles inside a smooth cone. Quite quick, especially if you use Newton perpendicular to the normal reaction so that you don't have to find it.
    Q2) Acceleration depending on t, one integration, followed by finding the work done. This would be horrendous if you tried to find the work done by integrating power with respect to time (which was my first, stupid, idea). Instead use WEP (I made 2 careless errors in one line! Thanks, Goodkwong.)
    Q3) Newton, Hooke for (a), CofE for (b). The quadratic factorises over decimals, but it might be quicker to use the formula or a calculator. Reasonable.
    Q4) Rather unusual; I had to think about this one. For (a), write v as dx/dt, then svauiafc (separate variables and use initial and final conditions). Part (b) is the second time recently that they have asked you to find an acceleration using vdv/dx, (rather to than use that formula to set up a differential equation). This might throw people who hadn't seen the previous example.
    Q5) Another vertical circles/projectile question. A bit easier than the UK version for two reasons - you didn't need to solve simultaneous equations to find out where the particle left the circle, and you didn't have the same awkwardness in getting the angle of elevation for the projectile motion. Also, uvast could be used conveniently. (Nevertheless, I got this one wrong, leaving out the 1/2 when substituting into the no-v formula. Thanks again, Goodkwong. By the way, how's Badkwong?) As with Q5 of the UK paper, making sensible use of the more advanced facilities of modern calculators (in this case, solving quadratic equations), shortened the answer very significantly.
    Q6) The first part was quite reasonable so long as you decided to get y-bar by integration and get x-bar from it by symmetry. The integration would be much more awkward if you did it the other way around. Second part falls to the table method as usual. Third part is simple moments.
    Q7) Similar to Q6 of the UK paper overall, but a bit more awkward because you want the time until a particular speed was achieved, instead of a particular position.

    I found both papers fairly challenging time-wise, because of the way that they now seem to ask questions that take a long time to do using traditional methods, but are quick using the more advanced calculator facilities. It would have been nice to be given prior warning of this change in style; I don't remember seeing any. Certainly both papers seemed on the hard side.

    Sorry to have taken a long time to reply to people - I've had internet problems at home, and at work there is still teaching going on. (Also we've had a lot of leaving parties - more people have left this term than we usually have leaving in a period of many years, for some reason.) I may not be able to reply to people's individual queries till tomorrow. Many apologies.
 
 
 
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