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    I'm so screwed for M1 😓 my past paper marks fluctuate so much; goes from 100-73-94-69 % :'(


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    (Original post by Don Joiner)
    I'm so screwed for M1 😓 my past paper marks fluctuate so much; goes from 100-73-94-69 % :'(


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    That's not bad!

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    (Original post by flowerspace)
    That's not bad!

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    Thanks but whenever I do badly in a paper all the confidence I had gained from doing well in others just disappears

    I'm panicking that I won't get an A in maths now :'(


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    M1 Jan 2012 --> 63/75 = 80% = A. (I should have scored 66/75 raw marks) :sigh:

    Mistakes & tips for future past papers:
    - Reread the SUVAT question and try another method to see if you would get the same answer.
    - Understand the last section of the vectors question + draw a diagram & not be lazy.
    - Resolve in the direction of motion.
    - Read the question context and underline any key words I.e acceleration/constant speed etc.

    I'll be recording every past paper score that I do and my overall notes/opinion on that past paper.

    Past papers completed = 1/13. (edexcel regular + edexcel IAL) [excluding x2 MAM pp closer to exam date]
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    (Original post by XxKingSniprxX)
    M1 Jan 2012 --> 63/75 = 80% = A. (I should have scored 66/75 raw marks) :sigh:

    Mistakes & tips for future past papers:
    - Reread the SUVAT question and try another method to see if you would get the same answer.
    - Understand the last section of the vectors question + draw a diagram & not be lazy.
    - Resolve in the direction of motion.
    - Read the question context and underline any key words I.e acceleration/constant speed etc.

    I'll be recording every past paper score that I do and my overall notes/opinion on that past paper.

    Past papers completed = 1/13. (edexcel regular + edexcel IAL) [excluding x2 MAM pp closer to exam date]
    Damn, you've still got lots to do! I'm almost over with my past papers. 66 aint bad you probably will improve as you go onwards, Recent papers are slightly harder though!
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    Are the January IAL Edexcel papers harder than regular Edexcel papers?


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    hi could someone help me with this question please?

    Name:  Screenshot 2016-06-02 at 14.18.27.png
Views: 117
Size:  48.4 KB

    (a) find the tension in the string
    (b) the value of F
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    (Original post by peacefully_)
    hi could someone help me with this question please?

    Name:  Screenshot 2016-06-02 at 14.18.27.png
Views: 117
Size:  48.4 KB

    (a) find the tension in the string
    (b) the value of F
    Just resolve forces.
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    (Original post by Marxist)
    Just resolve forces.
    Yes i know that, im just confused on how to go about answering this question in particular and wanted somebody to explain it step by step.
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    (Original post by peacefully_)
    Yes i know that, im just confused on how to go about answering this question in particular and wanted somebody to explain it step by step.
    I'll do it in a few minutes unless someone does it before me.
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    (Original post by peacefully_)
    hi could someone help me with this question please?

    Name:  Screenshot 2016-06-02 at 14.18.27.png
Views: 117
Size:  48.4 KB

    (a) find the tension in the string
    (b) the value of F
    PART A:

    Okay, so what I always do is put a lil' arrow on the strong (OP) pointing towards the origin to and label it "TN" (T newtons).

    Now, we're given an angle of 30°. If you draw a straight line going up from the particle, you create a set of parallel lines like this:

    Name:  1.png
Views: 115
Size:  12.0 KB

    Using the "Z angles" rule, you'll know that the angle between the string and this new line will be the same as the angle between the string and the vertical, like this:

    Name:  1.png
Views: 101
Size:  16.8 KB

    Now, we can resolve the force T vertically and horizontally, like this:


    Attachment 542211542213542200

    Now we are told in the question that the weight of the particle is 6N, so we should mark this on the diagram also (you'd do this at the start). Since the WEIGHT is given, there is no need for any "g"s.

    You then resolve vertically to find T:
    Using F = ma,

    Tsin(30) - 6 = 0
    Tsin(30) = 6
    T = 6/sin(30)
    T = 12 newtons.

    PART B:
    Now you resolve horizontally to find F:

    F - Tcos(30) = 0 (We can sub in the value of T which we now know).
    F - (12)cos(30) = 0.
    F = 12*cos(30) = 12 * (root(3)/2)
    F = 6root(3) newtons.

    Hope that helps !
    Attached Images
          
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    (Original post by Don Pedro K.)
    PART A:

    Okay, so what I always do is put a lil' arrow on the strong (OP) pointing towards the origin to and label it "TN" (T newtons).

    Now, we're given an angle of 30°. If you draw a straight line going up from the particle, you create a set of parallel lines like this:

    Name:  1.png
Views: 115
Size:  12.0 KB

    Using the "Z angles" rule, you'll know that the angle between the string and this new line will be the same as the angle between the string and the vertical, like this:

    Name:  1.png
Views: 101
Size:  16.8 KB

    Now, we can resolve the force T vertically and horizontally, like this:


    Attachment 542211542213542200

    Now we are told in the question that the weight of the particle is 6N, so we should mark this on the diagram also (you'd do this at the start). Since the WEIGHT is given, there is no need for any "g"s.

    You then resolve vertically to find T:
    Using F = ma,

    Tsin(30) - 6 = 0
    Tsin(30) = 6
    T = 6/sin(30)
    T = 12 newtons.

    PART B:
    Now you resolve horizontally to find F:

    F - Tcos(30) = 0 (We can sub in the value of T which we now know).
    F - (12)cos(30) = 0.
    F = 12*cos(30) = 12 * (root(3)/2)
    F = 6root(3) newtons.

    Hope that helps !
    The attachments aren't in order, but I hope you can see the order they are supposed to be in :s!!
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    For questions where they tell you that "a slope is inclined at angle α, where tan(α) = 3/4", would you get full method marks for doing α = tan-1(4/3), working out the angle and then doing the question normally from there or do you have to find sin(α) = 3/5 and cos(α) = 4/5 ??

    example question:
    Spoiler:
    Show
    I prefer doing it this method but i've noticed mark schemes do it the other method
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    (Original post by GarlicBread01)
    In a question where they tell you that "a slope is inclined at angle α, where tan(α) = 3/4", would you get full method marks to do α = tan-1(4/3), work out the angle and continue the question from there or do you have to do it like the mark schemes do where you have to find sinα = 3/5 and cosα = 4/5 ??

    example question:


    I prefer doing it this method but i've noticed mark schemes do it the other method
    The questions are designed that you use 3/5 and 4/5. If you convert to angles you may not get exactly the same solution (if your calculator loses a bit of accuracy) and you will also waste quite a bit of time that you may not have in M1.
    In summary - a bad idea to use the angles
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    (Original post by GarlicBread01)
    For questions where they tell you that "a slope is inclined at angle α, where tan(α) = 3/4", would you get full method marks for doing α = tan-1(4/3), working out the angle and then doing the question normally from there or do you have to find sin(α) = 3/5 and cos(α) = 4/5 ??

    example question:
    Spoiler:
    Show
    I prefer doing it this method but i've noticed mark schemes do it the other method
    No, just use them in your answers and then substitute values in.
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    (Original post by SaadKaleem)
    Damn, you've still got lots to do! I'm almost over with my past papers. 66 aint bad you probably will improve as you go onwards, Recent papers are slightly harder though!
    Yeah, I know but I still have time. I'm doing x5 pp today, x4 pp tomorrow, x3 pp Saturday & then I'm done with past papers & have sun-Tuesday to do x2 mam past papers during that time & review every example/q's on the topics im dropping marks on.

    I've already done a few M1 pp today. I will update the scores on this thread later on today when I'm free.
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    (Original post by Don Pedro K.)
    PART A:

    Okay, so what I always do is put a lil' arrow on the strong (OP) pointing towards the origin to and label it "TN" (T newtons).

    Now, we're given an angle of 30°. If you draw a straight line going up from the particle, you create a set of parallel lines like this:

    Name:  1.png
Views: 115
Size:  12.0 KB

    Using the "Z angles" rule, you'll know that the angle between the string and this new line will be the same as the angle between the string and the vertical, like this:

    Name:  1.png
Views: 101
Size:  16.8 KB

    Now, we can resolve the force T vertically and horizontally, like this:


    Attachment 542211542213542200

    Now we are told in the question that the weight of the particle is 6N, so we should mark this on the diagram also (you'd do this at the start). Since the WEIGHT is given, there is no need for any "g"s.

    You then resolve vertically to find T:
    Using F = ma,

    Tsin(30) - 6 = 0
    Tsin(30) = 6
    T = 6/sin(30)
    T = 12 newtons.

    PART B:
    Now you resolve horizontally to find F:

    F - Tcos(30) = 0 (We can sub in the value of T which we now know).
    F - (12)cos(30) = 0.
    F = 12*cos(30) = 12 * (root(3)/2)
    F = 6root(3) newtons.

    Hope that helps !
    thank you for this !!
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    (Original post by peacefully_)
    thank you for this !!
    No problemo
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    (Original post by Don Pedro K.)
    PART A:

    Okay, so what I always do is put a lil' arrow on the strong (OP) pointing towards the origin to and label it "TN" (T newtons).

    Now, we're given an angle of 30°. If you draw a straight line going up from the particle, you create a set of parallel lines like this:

    Name:  1.png
Views: 115
Size:  12.0 KB

    Using the "Z angles" rule, you'll know that the angle between the string and this new line will be the same as the angle between the string and the vertical, like this:

    Name:  1.png
Views: 101
Size:  16.8 KB

    Now, we can resolve the force T vertically and horizontally, like this:


    Attachment 542211542213542200

    Now we are told in the question that the weight of the particle is 6N, so we should mark this on the diagram also (you'd do this at the start). Since the WEIGHT is given, there is no need for any "g"s.

    You then resolve vertically to find T:
    Using F = ma,

    Tsin(30) - 6 = 0
    Tsin(30) = 6
    T = 6/sin(30)
    T = 12 newtons.

    PART B:
    Now you resolve horizontally to find F:

    F - Tcos(30) = 0 (We can sub in the value of T which we now know).
    F - (12)cos(30) = 0.
    F = 12*cos(30) = 12 * (root(3)/2)
    F = 6root(3) newtons.

    Hope that helps !
    I got 6.93 N and 3.4N for F though...
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    (Original post by coconut64)
    I got 6.93 N and 3.4N for F though...
    Really? Hmm...Well, I don't see anywhere where I went wrong in my working! Maybe show your working?
 
 
 
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