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AQA A2 Mathematics MM2B Mechanics 2 - Monday 22nd June 2015 [Exam Discussion Thread] Watch

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    I don't really understand (mv^2)/r directions when dealing with balls flying in circles please help
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    (Original post by Stepidermis)
    I don't really understand (mv^2)/r directions when dealing with balls flying in circles please help
    Newton's Second Law of F = ma applies horizontally in vertical motion, where a = v^2/r, hence F = (mv^2)/r

    The F is usually in form of Tsinx - mg.

    Also, why are the grade boundaries so high? Nerds, please, I need to get an A/B on this!
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    (Original post by Artur96)
    Newton's Second Law of F = ma applies horizontally in vertical motion, where a = v^2/r, hence F = (mv^2)/r

    The F is usually in form of Tsinx - mg.

    Also, why are the grade boundaries so high? Nerds, please, I need to get an A/B on this!
    Bolded not true. F depends on the position of the ball around the circle, as it is the resultant force acting towards the centre of the circle - it'll only be -mg in the lower half, I think.

    They're so high because everyone taking M2 is doing their A2 maths and loads of them find it their easiest unit, and also it's very popular in further maths classes which tends to drive grade boundaries up.

    (Original post by Stepidermis)
    I don't really understand (mv^2)/r directions when dealing with balls flying in circles please help
    It might help you to draw the particle on a bit of the circle in the exam and draw the forces acting (weight and tension for a particle on a string, weight and reaction force for a particle on a wire or similar). Then F = resultant force towards the centre = mv2/r
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    Could someone kindly explain Q3(b) on June 2012?


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    (Original post by CD223)
    Could someone kindly explain Q3(b) on June 2012?


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    Hope this helps, it's not very neat though.
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    (Original post by CD223)
    Could someone kindly explain Q3(b) on June 2012?


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    Take moments about the line AD to work out the tension of the string at B. So you have  (1.6g \times 6) +(0.4g \times 12) = TB x 12. . Then to work out the sum of the tensions is equal to the sum of the weights of the laminas. Hence [latex] TA + TB = 2g [latex]. Then just rearrange to work out the tension at A.
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    I find these past papers to be going really well or not too great, been getting either about 70-72 or about 60-62. This is the one exam for maths this year I find some questions where I don't even know where to start, and usually they're 7 markers. Just hope there's not a horrible moments question tomorrow because that's definitely my weakness, I like elasticity questions though
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    please may anyone help with question 9c june 2013. Its hookes law question with conservation of energy.
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    (Original post by porters)
    please may anyone help with question 9c june 2013. Its hookes law question with conservation of energy.
    Most of it is straightforward conservation of energy stuff, but the important thing is to factor in the work done against friction by A during its motion.
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    (Original post by Tiwa)
    Take moments about the line AD to work out the tension of the string at B. So you have  (1.6g \times 6) +(0.4g \times 12) = TB x 12. . Then to work out the sum of the tensions is equal to the sum of the weights of the laminas. Hence [latex] TA + TB = 2g [latex]. Then just rearrange to work out the tension at A.
    (Original post by Tibooster)
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    Hope this helps, it's not very neat though.
    Thank you both good luck tomorrow



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    (Original post by Lau14)
    Bolded not true. F depends on the position of the ball around the circle, as it is the resultant force acting towards the centre of the circle - it'll only be -mg in the lower half, I think.

    They're so high because everyone taking M2 is doing their A2 maths and loads of them find it their easiest unit, and also it's very popular in further maths classes which tends to drive grade boundaries up.



    It might help you to draw the particle on a bit of the circle in the exam and draw the forces acting (weight and tension for a particle on a string, weight and reaction force for a particle on a wire or similar). Then F = resultant force towards the centre = mv2/r
    Always towards the centre for circles?
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    (Original post by Stepidermis)
    Always towards the centre for circles?
    Yes, tension in a string is always towards the centre of the circle.
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    Praying for a paper like June 2014 tbh. Anyone else feel that last year's was pretty good compared to recent years?


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    this is going to be dreadful
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    (Original post by Stepidermis)
    this is going to be dreadful
    Oh dear, How so? What do you struggle with?


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    Which is the hardest paper?


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    (Original post by CD223)
    Praying for a paper like June 2014 tbh. Anyone else feel that last year's was pretty good compared to recent years?


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    I think the first 6 question were relatively straight forward and then the ladder question on moments and the EPE question threw me off. I agree with you.
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    Could someone please explain question 8d on january 2012?
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    (Original post by Pangol)
    Most of it is straightforward conservation of energy stuff, but the important thing is to factor in the work done against friction by A during its motion.
    How can we tell if the string is taut or slack when it comes to rest for the first time? In the mark scheme it has EPE when it comes to rest so i can see this means it is still taut, but why?
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    (Original post by CD223)
    Oh dear, How so? What do you struggle with?


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    This is just the one module that I can't do! I always miss things when doing questions or make stupid mistakes and then I can't deal with the outside of the box questions... I think it just hasn't clicked for me which is fine since I can do every other module, this just gives me brain veins
 
 
 
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