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    (Original post by andyd_y02)
    Cheers mate - il prob need about full marks - i havnt got a "TSR grade" for p3 but im pretty that was C or below! so even with good mark on M2 its going to be a close run thing...
    you should be fine. a 100 is always a nice thing.
    i think i might have blown mine :mad: - for the proving of 5b to show the balls dont collide again, i accidently took the mass of B as 3m (it was 2m) :eek: should still get method marks. hopefully 5/7 for that question.
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    Not a bad paper.
    70/75 raw for 100UMS I reckon

    Hopefully I've got 95+
    Fingers crossed.
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    Yea youl only lose the A1 marks which in a proof probably wasnt that many. BTW for that proof of no vertical component, i didnt actually find out what the angle of the new pole to the wall was...just put it as theta for moments about A, and as theta for resolving vertically and the Tcos(theta)'s cancel each other out if you see what i mean, so you dont need to know the actual angle thats why there was so few marks for it i guess - hard paper though i thought, because they havnt done wire laminas and stuff for ages

    My written proof for that as far as i can remember was

    Res vert: Tcos(theta) + X = 30g (where X is vertical component at A)
    Mom about A: 1.5 x Tcos(theta) = 1.5 x 30g
    Therefore divide through by 1.5: Tcos(theta) = 30g
    Then sub that into the res vert and you get X=0
    So no VERTICAL (sorry, i typoed and put horizontal before) component

    Hope that clears it up for people/or identifies that iv actually got it wrong as well?!
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    (Original post by Widowmaker)
    Not a bad paper.
    70/75 raw for 100UMS I reckon

    Hopefully I've got 95+
    Fingers crossed.
    you think its that low for 100UMS?
    i've been told the formula for 100UMS = 3a-2b
    and usually a = b+7
    so, b+21=70
    so b=49 and a=56??? (out of 75)
    a bit low if you ask me
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    (Original post by Zuber)
    you think its that low for 100UMS?
    i've been told the formula for 100UMS = 3a-2b
    and usually a = b+7
    so, b+21=70
    so b=49 and a=56??? (out of 75)
    a bit low if you ask me
    yes, sounds reasonable to me.
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    (Original post by andyd_y02)
    Yea youl only lose the A1 marks which in a proof probably wasnt that many. BTW for that proof of no vertical component, i didnt actually find out what the angle of the new pole to the wall was...just put it as theta for moments about A, and as theta for resolving vertically and the Tcos(theta)'s cancel each other out if you see what i mean, so you dont need to know the actual angle thats why there was so few marks for it i guess - hard paper though i thought, because they havnt done wire laminas and stuff for ages

    My written proof for that as far as i can remember was

    Res vert: Tcos(theta) + X = 30g (where X is vertical component at A)
    Mom about A: 1.5 x Tcos(theta) = 1.5 x 30g
    Therefore divide through by 1.5: Tcos(theta) = 30g
    Then sub that into the res vert and you get X=0
    So no horizontal component

    Hope that clears it up for people/or identifies that iv actually got it wrong as well?!
    very good point. you didnt HAVE to calculate the angle. it doesnt matter either way - we both showed what the question asked for.
    although if we spotted taking moments around midpoint of AB, it would have been much much easier!!!
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    (Original post by andyd_y02)
    Hi there, i noticed an answer thread for M1 and looked like a good idea cos i dont like waiting til August for my results, but i couldnt actually remember what Id put for each one - so today I thought Id write all my answers down in my calculator lid... Id much appreciate it if someone could have a quick look through and give me a mark/tell me which ones are wrong...

    1a) 35m/s
    b) 14.6m/s

    2a) 3cm
    b) k=1/7

    3a) 4
    b) 36.9m/s/s

    4a) 1.8m
    b) 6.75

    5a) e=2/3
    b) Some proof that the speed of B after its collision with C is less than that of A after its collision with B

    6a) 1018N
    b) 778N
    c) Some proof that the vertical component is zero because Tcos(angle)= 30g

    7a) 117.6J
    b) 10.0N
    c) coeff fric = 0.393
    d) 5.39m/s

    Thanks everyone

    that's a risky business. I thought about doing that but decided against it as knowing my luck a teacher would see it and think I took the answers into the exam.
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    there isnt a formular as such for converting to UMS - the examiners actually have a meeting and decide what raw mark should translate to what grade - or so it says on that central exam website Q...somethingorother (sorry i cant be more specific). But i think its fair to say that was a harder exam than usual, 100UMS is still pretty hard though, hazard a guess at 71/75 would do it...but the marks below that will have to be spread so about 60/75 raw is 80UMS so i doubt full marks could b alot lower? anyway who knows? evidently not me lol....
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    (Original post by Widowmaker)
    yes, sounds reasonable to me.
    the M2 jan 2005 paper had these boundaries
    A-59
    B-52
    C-45
    D-39
    E-33
    (from edexcel website)
    in this case 100UMS = 73/75
    i think Jun 05 will not differ too much (maybe 72 for 100UMS)
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    yea i see what you mean - i suppose i was over confident knowing that justice would be done as in fact i wasnt actually cheating - guess they werent to know that though. having said that, our invigilators are totally clueless - i had a history module clash so i actually had to start M2 at 10.30 after my history, so they tried to put me in isolation between exams, but totally f***ed it up and ended up putting me in "isolation" with someone who had just sat M2 and was waiting to do a biology paper!! lucky i genuinely didnt feel the need/wish to cheat, because it wouldnt exactly have been hard to
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    (Original post by andyd_y02)
    there isnt a formular as such for converting to UMS - the examiners actually have a meeting and decide what raw mark should translate to what grade - or so it says on that central exam website Q...somethingorother (sorry i cant be more specific). But i think its fair to say that was a harder exam than usual, 100UMS is still pretty hard though, hazard a guess at 71/75 would do it...but the marks below that will have to be spread so about 60/75 raw is 80UMS so i doubt full marks could b alot lower? anyway who knows? evidently not me lol....
    what i meant was that yes the examiners do have meetings to calculate the boundaries for As Bs Cs and so on.
    BUT, i think the formula for 100UMS depends on the grade boundaries that the examiners have put into place (i.e. 3a-2b).
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    ah i get ya - that sounds right yea well it'll be pretty close then - you might well still get that 600/600
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    (Original post by andyd_y02)
    ah i get ya - that sounds right yea well it'll be pretty close then - you might well still get that 600/600
    i still have C4 to go, and i havent yet learnt it (i.e. i havent got passed chapter 2 in my text book yet!!!)
    damn my maths teachers for letting me pace myself
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    (Original post by andyd_y02)
    Yea youl only lose the A1 marks which in a proof probably wasnt that many. BTW for that proof of no vertical component, i didnt actually find out what the angle of the new pole to the wall was...just put it as theta for moments about A, and as theta for resolving vertically and the Tcos(theta)'s cancel each other out if you see what i mean, so you dont need to know the actual angle thats why there was so few marks for it i guess - hard paper though i thought, because they havnt done wire laminas and stuff for ages

    My written proof for that as far as i can remember was

    Res vert: Tcos(theta) + X = 30g (where X is vertical component at A)
    Mom about A: 1.5 x Tcos(theta) = 1.5 x 30g
    Therefore divide through by 1.5: Tcos(theta) = 30g
    Then sub that into the res vert and you get X=0
    So no horizontal component

    Hope that clears it up for people/or identifies that iv actually got it wrong as well?!
    Should be vertical~ :rolleyes: Just make sure those who visit this thread later won't get confused~
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    Who wants Kaiser_mole's answers lol, I too thought question 2 was about a lamina, rest ought to have been alright though....on question 2 would I get full marks for part (b) with error carried forward?

    There is going to be an AWFUL LOT of people that thought question 2 was about a lamina, so I predict the mark scheme will be scrutinised...overall I found it a middle difficulty paper, not actually stuggling with anything, but instead not paying close enough attention ...and my motto is READ THE BLOODY QUESTION!!!! as I always, and I mean always, lose the vast majority of my marks by not reading the question explicitly

    edit: raw mark wise I think I will get about 65-72, depending on the aforementioned scrutiny
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    I am quite pleased in some ways... For example getting all that crap about thrust right... I am annoyed that my obvious mistake was resolving wrongly to find R in the last question, thus bodging my friction coefficient!! Lol... But seeing most of my answers crop up in this thread, I have some faith that I may JUST achieve 100% UMS, or if not a strong 90.

    It was a fairly unusual paper, however... Last 2 questions were not easy, and I imagine many were caught out and didnt change the projectile displacement (dart) to metres.. I myself almost treated the thingy as a lamina, didnt even READ it. Then had to lose 10 mins for that error!

    Good luck though everyone.
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    Could someone tell me what question 4 was? Someone said the answers were:

    4a) 1.8m
    b) 6.75

    I remember writing them but don't remember the actual question :rolleyes:
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    hash thats correct (it was the darts question)
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    (Original post by Zuber)
    hash thats correct (it was the darts question)
    Ah yes. Thanks!
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    hey zuber...

    where did u get tha formula from (3a-2b) and do they use that for other exams aswell....chem phy etc??
 
 
 
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