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    Q1 P= 5.32N Q=10.5N
    Q2 (i) u = 14.7 m/s
    Q2 (ii) s = 11.0 m (3sf)


    Q3 (i) 7120 N
    Q3 (ii) Maximum number of people = 8
    Q3 (iv) No air resistance. Treat lift as particle


    Q4 (i) t = 3/2, 6 seconds
    Q4 (ii) s = (2/3)t^3 - (15/2)t^2 + 18t metres
    Q4 (iii) 31.6m


    Q5 (i) v = 13 m/s
    Q5 (ii) m = 4 or 12 kg


    Q6 (ii) a = (6+8y) (y = coefficient of friction)
    Q6 (iii) To get answer, use v = u + at, rearrange to get u = -(a)t = -(-8y - 6)T = (8y + 6)T


    Q7 (ii) P = 48.3 N (3sf)
    Q7 (iii) Q = 54.4N and 33.4 degrees


    Anyone give feedback?????
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    (Original post by Chriispy)
    Q1 P= 5.32N Q=10.5N
    Q2 (i) u = 14.7 m/s
    Q2 (ii) s = 11.0 m (3sf)


    Q3 (i) 7120 N
    Q3 (ii) Maximum number of people = 8
    Q3 (iv) No air resistance. Treat lift as particle


    Q4 (i) t = 3/2, 6 seconds
    Q4 (ii) s = (2/3)t^3 - (15/2)t^2 + 18t metres
    Q4 (iii) 31.6m


    Q5 (i) v = 13 m/s
    Q5 (ii) m = 4 or 12 kg


    Q6 (ii) a = (6+8y) (y = coefficient of friction)
    Q6 (iii) To get answer, use v = u + at, rearrange to get u = -(a)t = -(-8y - 6)T = (8y + 6)T


    Q7 (ii) P = 48.3 N (3sf)
    Q7 (iii) Q = 54.4N and 33.4 degrees


    Anyone give feedback?????
    Q3 (ii) Maximum number of people = 8

    Just wondering how did you get 8?
    did u round up from 7.52?
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    (Original post by Chriispy)
    Q1 P= 5.32N Q=10.5N
    Q2 (i) u = 14.7 m/s
    Q2 (ii) s = 11.0 m (3sf)


    Q3 (i) 7120 N
    Q3 (ii) Maximum number of people = 8
    Q3 (iv) No air resistance. Treat lift as particle


    Q4 (i) t = 3/2, 6 seconds
    Q4 (ii) s = (2/3)t^3 - (15/2)t^2 + 18t metres
    Q4 (iii) 31.6m


    Q5 (i) v = 13 m/s
    Q5 (ii) m = 4 or 12 kg


    Q6 (ii) a = (6+8y) (y = coefficient of friction)
    Q6 (iii) To get answer, use v = u + at, rearrange to get u = -(a)t = -(-8y - 6)T = (8y + 6)T


    Q7 (ii) P = 48.3 N (3sf)
    Q7 (iii) Q = 54.4N and 33.4 degrees


    Anyone give feedback?????
    For 5 i) v= -5ms^-1 because Impulse was -18Ns not 18Ns

    (Original post by jashfield730)
    Q3 (ii) Maximum number of people = 8

    Just wondering how did you get 8?
    did u round up from 7.52?
    I got 8.3 something then rounded down to 8.
    Then I scribbled this answer out in favour of a wrong one....
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    (Original post by jashfield730)
    Q3 (ii) Maximum number of people = 8

    Just wondering how did you get 8?
    did u round up from 7.52?

    Accel up F= 800 x 0.9 = 720

    800g + 720 = 8560

    15000-8560 =6440N

    6440/80g = 8.21 hence I got 8
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    (Original post by upthegunners)
    how did you even do 5ii)?
    They specified the final speed of B, but didn't give the final direction of motion of B, meaning that you cannot just assume that the particle was travelling to the right, giving a positive speed, but as direction was not given, this means that the particle could have also been travelling 1.5m/s to the left, which in these style of questions, is actually a speed of -1.5m/s. So, in order to do the question, you had to apply the conservation of momentum to both 1.5 and -1.5 and they gave 4kg and 12kg respectfully
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    did anyone get K as 1/16 - C1 EXAM
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    (Original post by JasonH27)
    They specified the final speed of B, but didn't give the final direction of motion of B, meaning that you cannot just assume that the particle was travelling to the right, giving a positive speed, but as direction was not given, this means that the particle could have also been travelling 1.5m/s to the left, which in these style of questions, is actually a speed of -1.5m/s. So, in order to do the question, you had to apply the conservation of momentum to both 1.5 and -1.5 and they gave 4kg and 12kg respectfully
    I can't believe I did this, but instead of dividing 18 by 1.5 and 4.5 to give 12kg and 4kg, I divided 28 by 1.5 and 4.5. I must have copied down the mass wrong and got 20 instead of 10! No!! How could I have done this!?
    Anyway, aside from this sheer stupidity, my method was correct. How many marks do you think I will lose? 2,3, or 4? I'm hoping it's only 3 and not more.
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    (Original post by patterson)
    did anyone get K as 1/16 - C1 EXAM

    Ccea C1 isn't until 24th may?
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    (Original post by Chriispy)
    Accel up F= 800 x 0.9 = 720

    800g + 720 = 8560

    15000-8560 =6440N

    6440/80g = 8.21 hence I got 8
    Heres how I did it look at the attached file
    Attached Images
  1. File Type: pdf M1.pdf (92.8 KB, 169 views)
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    (Original post by jashfield730)
    Heres how I did it look at the attached file
    I had a quick look at that, but I got 7 as well and I rounded down from 7.5 or something like that.

    Also, when you substitute N as 7, the Tension is approx. 14700 and when it's 8, tension is approx. 15200, so that makes me think that 7 must be correct.
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    The test was alright apart from Q7. I'll probably get a high B now even though I wanted that A.
    For the lift i got 7 and a lot of people in my year got 7.
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    (Original post by Chriispy)
    Ccea C1 isn't until 24th may?
    edexcel
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    I don't understand how people are getting 10.5 for Q for q1 defs 11.2
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    (Original post by patterson)
    edexcel
    Just so you know ccea has got nothing to do with edexcel

    ccea is the Northern Ireland exam board
    edexcel is an england based board(mainly)
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    OMG I am the biggest idiot ever, cryyyyyyyyyyyyyy. I did 10 sin70 instead of 10sin60 fmlllllllll
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    and I did 10cos 60, just when I was copying out the 70 degree angle was so close aaaaaaaaaaahhh. feck sake, i've lost enough marks through lack of understanding i dont need to lose more through stupidity wehhh
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    Here is my working for 1 P = 5.32N and Q = 10.5N
    Too bad I got Q to equal 11.2N on the actual exam, fsfsName:  IMG_3971.jpg
Views: 138
Size:  507.2 KB
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    I found the exam was brilliant until question 5 (ii) and onwards. Eventually got part 6 but I doubt the examiner will be able to read my last minute scribble. I feel that question 7 was so difficult due to CCEA's awful wording. I interpreted "Tom exerts a force P newtons, at right angles to the pole, at C and a force Q newtons at D", meaning Q and P were both right angles until part 3. Anyone else have the same problem?
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    (Original post by jashfield730)
    Heres how I did it look at the attached file
    I did the same in that I took the Normal Reaction of the people in the lift as the downwards. Unfortunately, I forgot that I was considering forces acting on the lift only so I did the equation:
    15,000-7840-856x=0.9(800+80x) where x is the number of people.
    Basically, I made the equation equal to mass of both the lift and all the people times the acceleration. 3 or more marks just tossed in the bin...wonderful.
    Combined with the stupid marks lost in the momentum question, I'm not very happy.

    Alternatively, there is a much easier way:
    F=ma
    15000-800g-80g=0.9(800+80x)
    which gives the same answer of 7.5something.
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    Here's my working for 2, which I also got wrong when sitting the actual paper, yaaay
    2(i) u = 14.7 ms^-1
    (ii) s = 11.0m to 3sfName:  IMG_3972.jpg
Views: 140
Size:  520.3 KB
 
 
 
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