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    (Original post by oShahpo)
    Yea you're right. It was -1/6 I think.
    Nope


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    (Original post by physicsmaths)
    Na i had a pi there. Im paranoid i switcced g and L. I doubt i did since I looked at my answer and then looked at the show that answer. And turned the page thinking yep. Its just i recall 2g/l but i think im just getting confused lol.
    All right. What do you think of what I said above?


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    (Original post by Insight314)
    Nope


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    I am fairly certain there was 1/6 in the question.
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    Guys for question 2, like some people here I also flipped the triangle and got y as 2x/3 and ended up getting 6 cm as my answer but then I thought I remembered the equation wrong and just added a 1/2 in front of the integral of xy dx and ended up getting the correct answer which is 3. Im wondering how many marks I would lose for that :/
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    (Original post by Tgyhujikol59)
    Everyone seems to be saying the first question is 1/3N, but i swear the acceleration was 1/3 and the mass was 0.5, so the force was F = ma = 1/6N?
    I got acceleration magnitude as 2/3 making the resultant force 1/3N

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    Are we allowed for error carry forward in M3? Because in the question 7, I got root(20) instead of 5 for omega, and i used that omega to calculate the rest for the question.
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    (Original post by nthyy)
    yea me too and i couldn't find a thread for M3 IAL

    i remember there's a question about showing that OB<l ? sigh idk how to prove that but i think the answer for part b) is 15l/16?

    oh and there's a question about SHM (the water level and ladder one)
    i'm so confused about the last part of the question (i think i've calculated 11 hours? idk) i think i've ****ked up my paper
    Yup its 15l/16. For part A showing OB<l, i showed that the work done against friction when it moves 1/4l is less than the EPE stored inside the string initially. (i.e. 0.1mgl<0.125mgl), so that the particle P is still moving, but idk if this is the most appropriate method.

    The ladder one is so confusing(i mean part b)!! I dun even understand what the hell it is asking, i had to skip that part and do it later. my answer is 9.xx hour. (provided that the question means water level rises from "bottom of ladder" to top and falls back to "bottom of ladder", 9.xx hour would be the time for 1 cycle for this) Sigh.
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    (Original post by Tgyhujikol59)
    Everyone seems to be saying the first question is 1/3N, but i swear the acceleration was 1/3 and the mass was 0.5, so the force was F = ma = 1/6N?
    (Original post by Insight314)
    Nope


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    (Original post by oShahpo)
    I am fairly certain there was 1/6 in the question.
    nope

    v = 12/(x+3)

    m = 0.5

    work out F at x = 3

    F = ma = mv(dv/dx)

    v = 12(x+3)^-1
    dv/dx = -12(x+3)^-2

    so at x = 3, dx/dx = -12(6)^-2 = 1/3

    v = 12(3+3) = 2

    m = 1/2

    mvdv/dx = 2 * 1/2 * 1/3 = 1/3N
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    Unofficial mark scheme anyone?
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    (Original post by Euclidean)
    I got acceleration magnitude as 2/3 making the resultant force 1/3N

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    Do you remember what the value of x was?
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    (Original post by oShahpo)
    I am fairly certain there was 1/6 in the question.
    I think velocity was 12/x+3, making the acceleration -144/(x+3)^3. So Force was -72/(x+3)^3..... when x=3, then magnitude of force was 1/3??.. I think??
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    (Original post by Tgyhujikol59)
    Everyone seems to be saying the first question is 1/3N, but i swear the acceleration was 1/3 and the mass was 0.5, so the force was F = ma = 1/6N?
    no 12x12x0.5/(x+3)^3 was force magnitude.
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    (Original post by Student403)
    nope

    v = 12/(x+3)

    m = 0.5

    work out F at x = 3

    F = ma = mv(dv/dx)

    v = 12(x+3)^-1
    dv/dx = -12(x+3)^-2

    so at x = 3, dx/dx = -12(6)^-2 = 1/3

    v = 12(3+3) = 2

    m = 1/2

    mvdv/dx = 2 * 1/2 * 1/3 = 1/3N
    Do you remember what the second bit of the question was? Was there even a second bit?
    Do you remember how many marks this bit was for? I might have not done it correctly, can't exactly remember though.
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    (Original post by Insight314)
    Well, I do need to get A in Physics for my offer, and unit 1 is the easiest module so I should aim to get 100 UMS on it. When saying that I will be revising M4/M5 during the holiday, and not that much (still a few papers) in the next two weeks, I didn't mean that I will be doing only M4 and M5 during the holiday. My plan is to do a lot of papers and revision for Unit 1 Physics this weekend and get ready for the exam on Tuesday, then revise for Chemistry unit 1 next Friday (2 day gap). Whilst doing so, I will be trying to do some M4/M5 papers and DEFINITELY continue doing 3 STEP papers a week. During the holiday, I will revise the other exams I got during June which includes M4 and M5, and also do 1 STEP paper a day.


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    sounds good
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    (Original post by oShahpo)
    Do you remember what the second bit of the question was? Was there even a second bit?
    Do you remember how many marks this bit was for? I might have not done it correctly, can't exactly remember though.
    not sure on how many marks. It was like find t when x = 10
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    (Original post by Student403)
    not sure on how many marks. It was like find t when x = 10
    Yea I got that bit right, 7 seconds.
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    (Original post by oShahpo)
    Yea I got that bit right, 7 seconds.
    Yep nice
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    What do people think the 90 ums boundary will be??
    also is there anyone whos done fp2 and fp3, how do they compare to m3 (made stupid mistakes in m3 so hoping to get 270+ with m2/fp2 and fp3)
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    (Original post by Student403)
    nope

    v = 12/(x+3)

    m = 0.5

    work out F at x = 3

    F = ma = mv(dv/dx)

    v = 12(x+3)^-1
    dv/dx = -12(x+3)^-2

    so at x = 3, dx/dx = -12(6)^-2 = 1/3

    v = 12(3+3) = 2

    m = 1/2

    mvdv/dx = 2 * 1/2 * 1/3 = 1/3N
    Lol, I'm such an idiot. I got the expression correct but then substituted x=5 instead of x=3. Would I only lose 1 mark for that?


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    q1 was overall 9. 4 for first part 5 for second part.
 
 
 
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