AQA Mechanics 1B 24 May 2012 Unofficial Markscheme

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  1. 1platinum's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by karchun)
    +1

    This was posted from The Student Room's Android App on my GT-N7000
    +1?
  2. Miyata's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by 1platinum)
    Can't you do v=u+at?
    you do you use v = u+at, and then you equate i = -j values from the equation to work out that time is 20. Then you substitute back into equation to get v = i -j and then to find speed you do root(1^2 + 1^2) which gives you root2 or 1.41
  3. poyyo's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by darkshadow1111)
    No, this is not trial and error. You dont just put as many answers as you want and then get the marks.
    oh noooo.... ewbsfbgssjk im an idiot

    edit: oh wait no i remembered i didnt put several answers but i put different working outs so are my method marks invalidated too>??
    Last edited by poyyo; 24-05-2012 at 11:47.
  4. darkshadow1111's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by louisjevans)
    Use v=u +at, sub in u and a

    i component equals negative j component because its SE.

    So t = 20
    Therefore v= i - j

    Speed = root (1^2 + 1^2)
    = 1.41





    This was posted from The Student Room's iPhone/iPad App
    I got t=40 for some reason.
  5. Dirtybit's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    Whats the grade boundaries usually for mechanics?
  6. tom472's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by louisjevans)
    Use v=u +at, sub in u and a

    i component equals negative j component because its SE.

    So t = 20

    Therefore v= i - j

    Speed = root (1^2 + 1^2)
    = 1.41





    This was posted from The Student Room's iPhone/iPad App
    I equated i = -j, then i factored out the t and i'm pretty sure i got t = 40 seconds. Then speed as 5.86 or something....
  7. Miyata's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    can someone remember question 6? like values or exact question, I jsut want to make sure I got c) right, because I forgot what i wrote for it. Would appreciate if someone would upload the question paper.
  8. Americaniamh's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by tom472)
    I equated i = -j, then i factored out the t and i'm pretty sure i got t = 40 seconds. Then speed as 5.86 or something....
    I think my friend got that too, but we think he used the position equation rather than velocity?
  9. darkshadow1111's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by tom472)
    I equated i = -j, then i factored out the t and i'm pretty sure i got t = 40 seconds. Then speed as 5.86 or something....
    Exactly what i done and got. Hopefully were right. My teacher is going to do the exam in 2 hours so i will post the right answers.
  10. 1platinum's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by Miyata)
    you do you use v = u+at, and then you equate i = -j values from the equation to work out that time is 20. Then you substitute back into equation to get v = i -j and then to find speed you do root(1^2 + 1^2) which gives you root2 or 1.41
    I got 5.88 though?
  11. Miyata's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by tom472)
    I equated i = -j, then i factored out the t and i'm pretty sure i got t = 40 seconds. Then speed as 5.86 or something....
    Did you make new equation v = u + at? You might have used equation from the previous question which was for r.
  12. Dirtybit's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    I didn't know how to find t But i knew how to do the next 6marker but i needed t in order to do it FML
  13. louisjevans's Avatar
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    (Original post by darkshadow1111)
    I got t=40 for some reason.
    -1-0.1t = -(3-0.2t)
    0.1t = 2
    T = 20




    This was posted from The Student Room's iPhone/iPad App
  14. Dirtybit's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by louisjevans)
    -1-0.1t = -(3-0.2t)
    0.1t = 2
    T = 20




    This was posted from The Student Room's iPhone/iPad App
    OMG i Can't believe i didn't realise that!
  15. tom472's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    I think there was a question before it where you used v=u+at to find a time when its travelling east of the origin. I used s=ut+1/2at^2 for the last one though, i may be wrong gah. i thought it was an ok paper overall.
  16. louisjevans's Avatar
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    (Original post by darkshadow1111)
    Exactly what i done and got. Hopefully were right. My teacher is going to do the exam in 2 hours so i will post the right answers.
    That would be delightful


    This was posted from The Student Room's iPhone/iPad App
  17. Miyata's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by tom472)
    I think there was a question before it where you used v=u+at to find a time when its travelling east of the origin. I used s=ut+1/2at^2 for the last one though, i may be wrong gah. i thought it was an ok paper overall.
    Last one asked you to find Speed. To find speed you find v, and then square root i and j component and root it. The first part asked for s though. and more precisely it is r, not s but w/e.
    Last edited by Miyata; 24-05-2012 at 11:59.
  18. 1platinum's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    For part 1(b) isn't the bearing 022 degrees rather than 338 degrees?
  19. samkanwoods's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by darkshadow1111)
    Werent those two minuses.
    Plus 7c i got 5.38
    Same i got 5.38 for the speed when it is travelling south-east
  20. louisjevans's Avatar
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    Re: AQA Mechanics 1B 24 May 2012 Unofficial Markscheme
    (Original post by 1platinum)
    For part 1(b) isn't the bearing 022 degrees rather than 338 degrees?
    22 degrees anticlockwise from north
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