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Edexcel AS/A2 Mathematics M1 - 8th June 2016 - Official Thread watch

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    Okay so I am finally adding up my marks, and I just have one question about how many marks I would get.

    For question 7b you had to find the magnitude of the velocity at t=3 so V = (whatever) +(fsdfds)3

    i did that however, i found the magnitude before adding and multiplying so i got the wrong answer and i was wondering how many marks you guys think I would get.

    i tried to compare to previous similar questions however this one was unique, the closest i could find are

    question 7 part a:
    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

    question 6 part c:
    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

    question 6 part a:
    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

    thanks.
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    but if you get the final answer you get all the marks dont you?
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    For q4 for time at Y i got 32.5 instead of 40 because i didn't divide 30 by 2 and ultimately got 16.25, the method was correct though, could anyone tell me how many marks i've lost out 8
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    Has anyone done working for the magnitude of impulse question? I definitely got it wrong but was wondering how many marks i would get for my workings? Also for question 7 I had no idea where to start so i worked out the angles relative to N for F1 F2 and RF will i get any marks for that? Out of 7?
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    (Original post by KloppOClock)
    ok question 7 part b, what am i doing wrong. i seem to be getting a different answer to everyone else so lets hear it!

    ok ive done this twice now (once in the exam, once again now) and ive got the same answer which is different to what I have been seeing.

    V= U + AT
    V = ? U = (3i-22j) A=(3i+9j) T=3
    |U| = root(3^2 + (-22)^2) = root(493)
    |A| = root(3^2 + 9^2) = root(90)

    V = root(493) + 3*root(90)


    V= 50.66410225
    V = 50.7


    anyone?

    was i not meant to find the magnitude until the end or something?
    Ye you find the magnitude at the end. This should explain why to you:

    V = U + AT
    |V| = |U + AT|
    What you did was:
    |V| = |U| + |AT|.

    This is incorrect as |a+b| is not the same as |a| + |b|
    You will learn more about modulus when you do c3 and c4. I just happen to be an early entry but I can assure you of the above statement.
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    (Original post by LightAtTheEnd)
    Ye you find the magnitude at the end. This should explain why to you:

    V = U + AT
    |V| = |U + AT|
    What you did was:
    |V| = |U| + |AT|.

    This is incorrect as |a+b| is not the same as |a| + |b|
    You will learn more about modulus when you do c3 and c4. I just happen to be an early entry but I can assure you of the above statement.
    yeah that makes sense now that I think about it! thank you for confirming, Im kinda annoyed with myself that I didn't come across this during my revision, anyway im gonna copy + paste this reply from before, could you give me your opinion on it?


    Okay so I am finally adding up my marks, and I just have one question about how many marks I would get.

    For question 7b you had to find the magnitude of the velocity at t=3 so V = (whatever) +(fsdfds)3

    i did that however, i found the magnitude before adding and multiplying so i got the wrong answer and i was wondering how many marks you guys think I would get.

    i tried to compare to previous similar questions however this one was unique, the closest i could find are

    question 7 part a:
    https://3b0a7b1bc87f5381e60f8f717510...% 20Edexcel.pdf

    question 6 part c:
    https://3b0a7b1bc87f5381e60f8f717510...% 20Edexcel.pdf

    question 6 part a:
    https://3b0a7b1bc87f5381e60f8f717510...% 20Edexcel.pdf

    thanks.
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    (Original post by Nizza98)
    but if you get the final answer you get all the marks dont you?
    Not necessarily - the mark scheme can specify "correct solution only" or "no errors seen" etc. You can also have dependent marks i.e. marks that you can only get if you got the previous mark(s)
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    Predicted UMS for 68/75?
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    How many method marks would I get for the moments my method was spot on but I messed up the numbers somewhere?
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    (Original post by SirRaza97)
    I put AS = d but I meant A to centre of mass = d




    Here you go. This is how I did it.

    Hi for this questions, I did a really stupid mistake, I assumed that A to the centre of the mass equals to D, which means that I used a different distance from the point T. I did rest of the questions right, with wrong values, how many marks did I drop? And is it out of 10 marks for that question? This is how I did it30g(4-x) I got 1m for distance and 90 for mass
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    (Original post by KloppOClock)
    yeah that makes sense now that I think about it! thank you for confirming, Im kinda annoyed with myself that I didn't come across this during my revision, anyway im gonna copy + paste this reply from before, could you give me your opinion on it?


    Okay so I am finally adding up my marks, and I just have one question about how many marks I would get.

    For question 7b you had to find the magnitude of the velocity at t=3 so V = (whatever) +(fsdfds)3

    i did that however, i found the magnitude before adding and multiplying so i got the wrong answer and i was wondering how many marks you guys think I would get.

    i tried to compare to previous similar questions however this one was unique, the closest i could find are

    question 7 part a:
    https://3b0a7b1bc87f5381e60f8f717510...% 20Edexcel.pdf

    question 6 part c:
    https://3b0a7b1bc87f5381e60f8f717510...% 20Edexcel.pdf

    question 6 part a:
    https://3b0a7b1bc87f5381e60f8f717510...% 20Edexcel.pdf

    thanks.
    You would receive no more than 1 mark, if any.
    This 1 mark depends on the mark scheme. If it gives a mark for correct identification of which SUVAT equation to use and attempting to use it then you may receive 1 mark. Unfortunately you dropped both method and answer marks and since this question was independent of the first part there are also no ECF marks to be obtainted.
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    (Original post by SirRaza97)
    I put AS = d but I meant A to centre of mass = d




    Here you go. This is how I did it.
    That's a much better way than what I did, I can't believe I didn't do it that way.

    Instead, I took moments about A and B, and resolved vertically for both scenarios in order to cancel out the reaction forces of S and T.

    I then solved simultaneously again for d and M.
    I got the same answer and did it a ridiculously long way, so it's likely correct.
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    (Original post by LightAtTheEnd)
    You would receive no more than 1 mark, if any.
    This 1 mark depends on the mark scheme. If it gives a mark for correct identification of which SUVAT equation to use and attempting to use it then you may receive 1 mark. Unfortunately you dropped both method and answer marks and since this question was independent of the first part there are also no ECF marks to be obtainted.
    in the other questions, its been 1 mark for the suvat, 1 mark for answer
    1 mark for magnitude 1 mark for answer

    i mean I used the right suvat, and i did calculate magnitude (altho before I was meant to) maybe if there was follow through marks i could maybe get 3 marks (maybe). but yeah, its not lookin good, thanks for taking a look at that for me.
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    I would like to address a misconception that some people are having.
    It is regarding the question where you had to find the resultant force of the string on the pulley and the direction of it.
    Most people got the magnitude correct via Pythagoras' Theorem. However the issue comes with the direction part.
    Most people managed to figure out the angle was 45 degrees but decided to express this as a bearing and used 225 degrees.

    Now here is where you go wrong. Do NOT use bearings for these sorts of questions. Bearings give the direction relative to the North Line. There is no specified north line in this question.
    Why can we use bearings in vectors? Well at the beginning of vector questions it clearly states that vectors i and j represent east and NORTH respectively.
    In this scenario we aren't told that the vertical represents the north line. In fact, if we used common sense then a vertical to a table (given it is upright) would point upwards to the sky, and not to North.
    The direction would be: inclined downwards at an angle of 45 degrees to the horizontal/table.
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    I'm thinking I've got around 61, what are my chances for an A? Any predicted grade boundaries??😁😁
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    (Original post by LightAtTheEnd)
    I would like to address a misconception that some people are having.
    It is regarding the question where you had to find the resultant force of the string on the pulley and the direction of it.
    Most people got the magnitude correct via Pythagoras' Theorem. However the issue comes with the direction part.
    Most people managed to figure out the angle was 45 degrees but decided to express this as a bearing and used 225 degrees.

    Now here is where you go wrong. Do NOT use bearings for these sorts of questions. Bearings give the direction relative to the North Line. There is no specified north line in this question.
    Why can we use bearings in vectors? Well at the beginning of vector questions it clearly states that vectors i and j represent east and NORTH respectively.
    In this scenario we aren't told that the vertical represents the north line. In fact, if we used common sense then a vertical to a table (given it is upright) would point upwards to the sky, and not to North.
    The direction would be: inclined downwards at an angle of 45 degrees to the horizontal/table.
    Would drawing it in on the diagram and quoting that this was done provide sufficient for the mark?
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    you know what? i'm happy they didnt put any vertical motion questions in there.
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    (Original post by LightAtTheEnd)
    I would like to address a misconception that some people are having.
    It is regarding the question where you had to find the resultant force of the string on the pulley and the direction of it.
    Most people got the magnitude correct via Pythagoras' Theorem. However the issue comes with the direction part.
    Most people managed to figure out the angle was 45 degrees but decided to express this as a bearing and used 225 degrees.

    Now here is where you go wrong. Do NOT use bearings for these sorts of questions. Bearings give the direction relative to the North Line. There is no specified north line in this question.
    Why can we use bearings in vectors? Well at the beginning of vector questions it clearly states that vectors i and j represent east and NORTH respectively.
    In this scenario we aren't told that the vertical represents the north line. In fact, if we used common sense then a vertical to a table (given it is upright) would point upwards to the sky, and not to North.
    The direction would be: inclined downwards at an angle of 45 degrees to the horizontal/table.
    I didnt put my answer as a bearing because they never have in mark schemes, but i never understood why.... UNTIL NOW thanks a lot man this has legit made my life
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    (Original post by LightAtTheEnd)
    I would like to address a misconception that some people are having.
    It is regarding the question where you had to find the resultant force of the string on the pulley and the direction of it.
    Most people got the magnitude correct via Pythagoras' Theorem. However the issue comes with the direction part.
    Most people managed to figure out the angle was 45 degrees but decided to express this as a bearing and used 225 degrees.

    Now here is where you go wrong. Do NOT use bearings for these sorts of questions. Bearings give the direction relative to the North Line. There is no specified north line in this question.
    Why can we use bearings in vectors? Well at the beginning of vector questions it clearly states that vectors i and j represent east and NORTH respectively.
    In this scenario we aren't told that the vertical represents the north line. In fact, if we used common sense then a vertical to a table (given it is upright) would point upwards to the sky, and not to North.
    The direction would be: inclined downwards at an angle of 45 degrees to the horizontal/table.
    I agree with this. However, if they have stated that the angle is measured clockwise from the upwards vertical then it should be fine. But a bearing alone is a bit risky.
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    (Original post by IAROX15)
    you know what? i'm happy they didnt put any vertical motion questions in there.
    I find it a bit odd they put in two vector questions tho lmao. I think that is the first paper I've done with no vertical motion maybs they forgot
 
 
 
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