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    Name:  image.jpg
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Size:  443.8 KB Wouldn't this be above the horizontal but still a push
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    (Original post by EconomicsSA)
    Name:  image.jpg
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    nope
    draw the horizontal line from the start of the force, not the end
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    if your force was acting such that it had a vertical component acting in an opposite direction to the reaction force, it's wrong - I don't think ocr will allow two different answers on account of ambiguity because it very clearly states 'above the horizontal'.
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    (Original post by EconomicsSA)
    Name:  image.jpg
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    Yeah I think we all did the right drawings but I did a pull instead of a push
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    Wtf, this pull or push question is too polarised
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    I would just like to thank all of you who have helped me over the past few days. I have got from a U to a somewhat good grade. Wasn't the biggest fan of M1 as our teacher wasn't very good.

    MsFahima
    kawehi
    chloe-jessica
    Lilli1997


    Hope the exam went well for you. Good luck to you if you are doing C3 and C4.
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    Could someone please take me through the whole of question 6? The whole integration and differentiation bit, I think I really screwed that bit up
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    (Original post by Shaagggyy)
    Could someone please take me through the whole of question 6? The whole integration and differentiation bit, I think I really screwed that bit up
    I can go through it if someone has the question, the whole paper was posted before but I think it's been deleted, or at least I can't find it.
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    (Original post by Super199)
    I would just like to thank all of you who have helped me over the past few days. I have got from a U to a somewhat good grade. Wasn't the biggest fan of M1 as our teacher wasn't very good.

    MsFahima
    kawehi
    chloe-jessica
    Lilli1997


    Hope the exam went well for you. Good luck to you if you are doing C3 and C4.
    That's so good to hear, I'm happy for you! I'm doing C3 but not C4, hope your exams all went/go okay!
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    Uh. For the last part of Q6, I added 68 to 54 and got 112 and then divided by 2 to get 61 :/ I didn't think the negative sign on the velocity mattered.
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    Also, can someone please do Q4 because I got 13.3 and 25.1 degrees I think. How did people get like 10 or something?
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    (Original post by chloe-jessica)
    I can go through it if someone has the question, the whole paper was posted before but I think it's been deleted, or at least I can't find it.
    I think it was to do with particle P passing through a point O and the velocity of the particle when it passes O is 2ms^-1 and the acceleration at the time it passes O is (4+12t)ms^-2


    Then the questions were
    Find the velocity of P when t = 3
    Find the displacement of OP when t = 3
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    (Original post by Shaagggyy)
    I think it was to do with particle P passing through a point O and the velocity of the particle when it passes O is 2ms^-1 and the acceleration at the time it passes O is (4+12t)ms^-2


    Then the questions were
    Find the velocity of P when t = 3
    Find the displacement of OP when t = 3
    That sounds familiar
    (i) To get from acceleration to velocity, you integrate (as acceleration is change in velocity over time, a=dv/dt so to get back to v to integrate). So integrating 4+12t gives us 4t+6t2+c. When t=0, v=2 as it states in the question. Therefore 2=4(0)+6(0)2+c => c=2. You end up with v=4t+6t2+2. At t=3, sub in 3 to get v=4(3)+6(3)2+2 = 68m.
    (ii) To get from velocity to displacement, you integrate again. Integrating 4t+6t2+2 gives you s=2t2+2t3+2t+c. When t=0, s=0 so c=0. Sub in t=3 to get s=2(3)2+2(3)3+2(3) = 78m.
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    What kind of grade do you reckon 55 marks would be?
    Also, can someone please go through question 7iv)b. because I got 2.32 and I'm not sure where/why I went wrong
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    (Original post by chloe-jessica)
    That sounds familiar
    (i) To get from acceleration to velocity, you integrate (as acceleration is change in velocity over time, a=dv/dt so to get back to v to integrate). So integrating 4+12t gives us 4t+6t2+c. When t=0, v=2 as it states in the question. Therefore 2=4(0)+6(0)2+c => c=2. You end up with v=4t+6t2+2. At t=3, sub in 3 to get v=4(3)+6(3)2+2 = 68m.
    (ii) To get from velocity to displacement, you integrate again. Integrating 4t+6t2+2 gives you s=2t2+2t3+2t+c. When t=0, s=0 so c=0. Sub in t=3 to get s=2(3)2+2(3)3+2(3) = 78m.

    Ah thank you, it all makes sense now, I forgot the plus C coefficient :///
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    (Original post by artbreaker)
    What kind of grade do you reckon 55 marks would be?
    Also, can someone please go through question 7iv)b. because I got 2.32 and I'm not sure where/why I went wrong
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    Starting from (a) because it's easier. Using deceleration = 0.9 from part (iii).
    (a) F=ma for P, gives T-0.3gsin30=0.3*-0.9 => T=0.3*-0.9+0.3gsin30 => T=1.2N.
    (b) Tension must be constant throughout the string, so using F=ma for Q => 0.4gsin30-T-Fr=0.4*-0.9 => Fr=0.4gsin30-1.2-0.4*-0.9 (using T=1.2N from above) => Fr=1.12N.
    Not sure about the grade sorry!
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    (Original post by Shaagggyy)
    Ah thank you, it all makes sense now, I forgot the plus C coefficient :///
    Sure you'll get method marks for attempting integration and substituting it, then you'll probably get ecf for the second bit
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    (Original post by chloe-jessica)
    Starting from (a) because it's easier. Using deceleration = 0.9 from part (iii).
    (a) F=ma for P, gives T-0.3gsin30=0.3*-0.9 => T=0.3*-0.9+0.3gsin30 => T=1.2N.
    (b) Tension must be constant throughout the string, so using F=ma for Q => 0.4gsin30-T-Fr=0.4*-0.9 => Fr=0.4gsin30-1.2-0.4*-0.9 (using T=1.2N from above) => Fr=1.12N.
    Not sure about the grade sorry!
    Ooooh thank you! What a ridiculous mistake to make damn
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    (Original post by Super199)
    I would just like to thank all of you who have helped me over the past few days. I have got from a U to a somewhat good grade. Wasn't the biggest fan of M1 as our teacher wasn't very good.

    MsFahima
    kawehi
    chloe-jessica
    Lilli1997


    Hope the exam went well for you. Good luck to you if you are doing C3 and C4.
    You're welcome! And good luck to you too! See you at the C3/C4 thread
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    (Original post by LukeBarnett)
    Thank you!
    Could u post the link again please bcos I don't see any link
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    (Original post by Amzii_b)
    Also, can someone please do Q4 because I got 13.3 and 25.1 degrees I think. How did people get like 10 or something?
    4) i. I used the cosine rule:
    R^2 = 10^2 + 6^2 -2 x 10 x 6 x cos(70)
    which gives
    R = 9.74 (3 s.f)
    then I used the sine rule to find the angle:
    sin(x)/6 = sin(70)/9.74
    which gives x = 35.4 (3 s.f)
    4) ii. I did:
    20 - 9.74 = 10.3 (3 s.f)
    because the particle is at rest so I figured the resultant force must be acting directly upward (90 degrees above the horizontal)
    4) iii. I then did:
    90 - 35.4 = 54.6

    Hope that's right and I hope that helps!
 
 
 

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