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    (Original post by Will1692)
    It was certainly challenging in each m1 area. I messed up the overtaking business and the pulley system. I even had coefficent of friction as 1.2 ish so basically it was screwed up!
    I got that, it's OK, friction can be greater than one, it just isn't usual. Also, think how great friction would have had to be to stop that heavier ball sliding down a slope that steep.
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    (Original post by mcp2)
    The 9 mark question, what time did you get? I got 9.50s
    :five:
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    (Original post by MaccyG)
    Looks like I may have got full marks. When you had to integrate I added a constant but then thought there cannot have been one on question 6.
    the constant was zero as it had not travelled anywhere when t = 0 s
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    (Original post by NCIShippo)
    Will you be paying a visit to the D1 OCR thread today? :ninja: Thanks!
    nope! sorry
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    (Original post by Undersc0re)
    Regarding the coefficient of friction question. Every time I attempted it I got a value above 1. This was however, taking the tension of the pulley into account, as I was under the impression that it always has to be between 0 and 1 I returned to the question and took the wording to mean that T was unimportant as it was addressed in the last part.
    This resulted in obtaining a ~0.47 coefficient.
    Well, if the string had not been present, the object would be at limiting equilibrium when the coefficient of friction = the tangent of the inclination of the slope = tan 60 = sqrt 3 = same answer so I'm not sure where your 0.47 came from.
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    Ok, here's what i've found re. coefficient of friction: when a slope is inclined at <45 degrees it cannot be above 1, so those who got 1.7 something will be right because at >45 it can go above 1.
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    (Original post by Mr M)
    ...
    Could you explain to me how the R parallel to the slope was 0.849 (question 4 I think). I thought R had no component parallel to the slope. The wording of that question really confused me.
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    (Original post by MathsHamster)
    Could you explain to me how the R parallel to the slope was 0.849 (question 4 I think). I thought R had no component parallel to the slope. The wording of that question really confused me.
    It was referring to the frictional force between particle P and the slope that is acting up the plane. This force comes from the objects being in contact with each other.
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    (Original post by mrdude)
    Does anybody have today's paper please?
    Somebody uploaded it a few days back, PM them. Why doesn't anyone take the paper with them?!
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    coeff of friction can be greater than 1.

    For instance, if you have 2 aluminium surfaces in contact, it is about 1.4
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    (Original post by Mr M)
    It was referring to the frictional force between particle P and the slope that is acting up the plane. This force comes from the objects being in contact with each other.
    Aaah why couldn't it just say that? :mad: I actually worked out the friction in the next bit (correctly), but I guess my statement that the component of R parallel to the plane is 0 will have lost me the marks.
    Oh well, everything else seems to have gone well.

    Thanks very much, I can see why you have so much rep!
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    (Original post by Undersc0re)
    Ok, here's what i've found re. coefficient of friction: when a slope is inclined at <45 degrees it cannot be above 1, so those who got 1.7 something will be right because at >45 it can go above 1.
    Clearly not true. If the slope in this question was inclined at 25 degrees, its coefficient of friction would not suddenly change. It is a property of the two surfaces. I think I know what you are thinking of though ...

    http://library.thinkquest.org/C00557...leFriction.htm
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    (Original post by mcp2)
    Somebody uploaded it a few days back, PM them. Why doesn't anyone take the paper with them?!
    Most centres retain them.
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    Argh, hate the paper. Did loadsa past papers and none of them were that hard...

    Q7, 9 marker was solid, forgot that friction turns the other way once the log starts moving downwards /facepalm.

    Also, can anyone remember what the momentum question was? the answers given don't seem to be familiar...
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    (Original post by mcp2)
    You think? Jan boundaries are usually lower than June and this paper was harder than past ones. OK, Maybe not 53, but not 58 either! 55, it could be.
    Well looking at the grade boundaries from past papers on the OCR website:

    Jan 07 - 61/72
    June 07 doesn't give it for some reason
    Jan 08 - 59/72
    June 08 - 56/72
    Jan 09 - 62/72
    June 09 - 62/72.

    Like you say, 55 might be more like it. But we could both be completely wrong!
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    I'll defer to your knowledge, mine was just a quick trawl through Goggle to see what came up, in response to how I obtained ~0.47:
    Somehow I came up with 0.4gsin60/0.4gsin60= "sqrt3"/4 which I realised in the exam didn't make any sense but it came up twice, so despairingly I used it.
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    June 08, I remember doing it and it being nothing compared to this.
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    (Original post by mrdude)
    Well looking at the grade boundaries from past papers on the OCR website:

    Jan 07 - 61/72
    June 07 doesn't give it for some reason
    Jan 08 - 59/72
    June 08 - 56/72
    Jan 09 - 62/72
    June 09 - 62/72.

    Like you say, 55 might be more like it. But we could both be completely wrong!
    June 2007 was 57. 55 is too low in my opinion.
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    55 in Jan 2006 though so it is possible!
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    We shall just have to wait and see I s'pose. Roll on March 11th
 
 
 
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