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16May 2018 Edexcel M3 Unofficial Mark Scheme watch

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    2a. 11.3
    2b. 2.5
    3. 2R
    4. k=1.86
    5b. Centre of Mass = 9a(16-k)/4(k-8)
    5c. k=16
    6b. u>=sqrt(3ag)
    6c. u=sqrt(5ag)
    7b. a=-49x therefore SHM
    7c. v=2.42
    7d. Time taken = 0.547

    Please tell if there are any mistakes
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    2. 11.3
    B. 2.5
    3. R/5
    4. 1.7
    5. 9a(k-16)/4(8+k)
    b. 35.2
    6. Sqrt 3ag
    b sqrt 5ag
    7. -49x
    b V=2.4
    c t=0.55
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    Same as amber but 3R/4 for Q3
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    I got 1.86 for 4, 3sf
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    How did you get Q5b and what was Q4?
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    Q4 was when the particle went down a slope with friction
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    2a. λ=(98 root 3) divided by 15
    5b k=16
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    (Original post by Hasad)
    I got 1.86 for 4, 3sf
    same coz it was to 3sf
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    These were my answers:

    1. Show that...

    2.(a) 2g/sqrt(3) = 11.3 (3 s.f.)
    2.(b) omega = 2.51 rad s^-1 (3 s.f.)

    3. Distance = R/5

    4. k = 1.72 (3 s.f.)
    EDIT: I made a silly mistake - the answer was actually k = 1.86.

    5.(a) Prove centre of mass = 3r/8 from O (I couldn't remember how to do it, FFS!)
    5.(b) Distance from O = Modulus of {3a(16-k)}/{16(8+k)}
    5.(c) k = 16

    6.(a) Show that...
    6.(b) Minimum u = sqrt(3ag)
    6.(c) u = sqrt(5ag)

    7.(a) Show that l = 1.2
    7.(b) Prove SHM (acceleration = -49x)
    7.(c) v = 2.42 ms^-1 (3 s.f.)
    7.(d) Time taken = 0.547 seconds (3 s.f.)
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    (Original post by Pitsillides)
    2a. λ=(98 root 3) divided by 15
    5b k=16
    I got k=16 too!!
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    I swear question 4 k was 1.86
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    (Original post by mupsman2312)
    These were my answers:

    1. Show that...

    2.(a) 2g/sqrt(3) = 11.3 (3 s.f.)
    2.(b) omega = 2.51 rad s^-1 (3 s.f.)

    3. Distance = R/5

    4. k = 1.72 (3 s.f.)

    5.(a) Prove centre of mass = 3r/8 from O (I couldn't remember how to do it, FFS!)
    5.(b) Distance from O = Modulus of {3a(16-k)}/{16(8+k)}
    5.(c) k = 16

    6.(a) Show that...
    6.(b) Minimum u = sqrt(3ag)
    6.(c) u = sqrt(5ag)

    7.(a) Show that l = 1.2
    7.(b) Prove SHM (acceleration = -49x)
    7.(c) v = 2.42 ms^-1 (3 s.f.)
    7.(d) Time taken = 0.547 seconds (3 s.f.)
    I got pretty much the same as you! I couldn't remember 5a too, I tried to do a circle of radius r around centre (r,0) but i'm not sure it worked lol. I did however get that the distance was 0 because my x = R and it was from the surface, seems i may have gotten that one a tad wrong lol.
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    (Original post by mupsman2312)
    These were my answers:

    1. Show that...

    2.(a) 2g/sqrt(3) = 11.3 (3 s.f.)
    2.(b) omega = 2.51 rad s^-1 (3 s.f.)

    3. Distance = R/5

    4. k = 1.72 (3 s.f.)

    5.(a) Prove centre of mass = 3r/8 from O (I couldn't remember how to do it, FFS!)
    5.(b) Distance from O = Modulus of {3a(16-k)}/{16(8+k)}
    5.(c) k = 16

    6.(a) Show that...
    6.(b) Minimum u = sqrt(3ag)
    6.(c) u = sqrt(5ag)

    7.(a) Show that l = 1.2
    7.(b) Prove SHM (acceleration = -49x)
    7.(c) v = 2.42 ms^-1 (3 s.f.)
    7.(d) Time taken = 0.547 seconds (3 s.f.)
    Q 4 is defo wrong! the ans is 1.86 🙃
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    (Original post by boodledoodle123)
    I got k=16 too!!
    got k=16 as well but seems many ppl get other things?
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    How did you get 7)d)?
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    (Original post by gnehcnosna)
    got k=16 as well but seems many ppl get other things?
    Me and my friend both got k=16
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    (Original post by gnehcnosna)
    got k=16 as well but seems many ppl get other things?
    Well, to get to 16 I didn't do much? Like it was 2 marks? So lol I hope it's 16 xD if not oh dearie me
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    2. Lamda = 11.3
    Omega = 2.51
    3. Distance = R/5 (was from surface)
    4. k = 1.86
    5. (C) k=16
    6. (B) Minimum u = sqrt(3ag)
    (C) u = sqrt(5ag)
    7. (B) a = -49x therefore SHM
    (C) v = 2.42 m/s
    (D) t = 0.55 secs
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    (Original post by gnehcnosna)
    got k=16 as well but seems many ppl get other things?
    K=16 is right
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    (Original post by Legend123456)
    How did you get 7)d)?
    I got 0.443 for 7d so haha idk
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