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Edexcel Mathematics: Mechanics M3 6679 01 - 17 May 2017 [Exam Discussion] Watch

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    (Original post by solC)
    Are you still stuck?
    Could you potenitally help me with it bro? Just finished the rest of the exercise and
    found it alright but not grasping that question
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    (Original post by TrueDAN)
    Could you potenitally help me with it bro? Just finished the rest of the exercise and
    found it alright but not grasping that question
    Consider the moment at which the cone is about to topple. Recall from M2 that the weight of the body acts through the point about which toppling occurs, which in this case is going to be the point A shown in the picture. Therefore the reaction R must also pass through A.

    Next we take moments about A to get an equation involving P, M and r.
    Note that Resolving vertically gives  F = P\sin60, then taking moments about A we get  Mgr = Pr\sin60 + 2Pr\cos60 and the rest is just rearranging for P.

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    (Original post by solC)
    Consider the moment at which the cone is about to topple. Recall from M2 that the weight of the body acts through the point about which toppling occurs, which in this case is going to be the point A shown in the picture. Therefore the reaction R must also pass through A.

    Next we take moments about A to get an equation involving P, M and r.
    Note that Resolving vertically gives  F = P\sin60, then taking moments about A we get  Mgr = Pr\sin60 + 2Pr\cos60 and the rest is just rearranging for P.

    Attachment 632034
    Quality stuff man, thanks a lot!!
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    (Original post by TrueDAN)
    Quality stuff man, thanks a lot!!
    No worries, sorry for the late reply
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    Should have all questions done by Tuesday, maybe even tomorrow - really have improved with strings - much more confident so I am glad that I
    persevered with them; my next weakness is some SHM stuff and then the papers grind can begin!
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    Yes plz send the diagrams that would help alot! thankss
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    (Original post by helpme:))
    Yes plz send the diagrams that would help alot! thankss
    Here is an example each for SHM and circular motion.
    Not sure how the Edexcel book does the diagrams for SHM but doing them like this is much more clear IMO.

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    I know it's late in the year but if you can find it cheap I recommend the book 'Applied Mathematics I' by Bostock and Chandler. Very good textbook with very good explanations, cannot recommend their A level textbooks enough.
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    ooh that helped me remember a q... if it's the particle in the second diagram is being released it would go upwards right? Why does x(double dot) act downwards and not upwards? I may try and get that book thx. And another question, how do you determine the amplitude in simple harmonic motion questions? on June 2011 m3 paper q7d i got amplitude as 4, when it was 0.2m, would you be able to explain why?
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    (Original post by helpme:))
    ooh that helped me remember a q... if it's the particle in the second diagram is being released it would go upwards right? Why does x(double dot) act downwards and not upwards? I may try and get that book thx. And another question, how do you determine the amplitude in simple harmonic motion questions? on June 2011 m3 paper q7d i got amplitude as 4, when it was 0.2m, would you be able to explain why?

    As the particle is released it's going to be moving with SHM. Therefore by definition the acceleration acts in the opposite direction to the displacement of the particle, and since the particle moves upwards from C the acceleration will act downwards.

    (I forgot to mention what the letters are - C is the lowest point of the particle i.e. the point at which it is released, P is a general point through which the particle passes, E is the position of equilibrium and AB is the natural length of the string.)

    The amplitude clearly can't be 4 because the distance AB is only 2m
    The centre of oscillation is going to be O and it's released from C meaning C is the maximum displacement from the equilibrium. Hence the amplitude is OC = 0.2m (It's also given in the question)
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    (Original post by helpme:))
    ooh that helped me remember a q... if it's the particle in the second diagram is being released it would go upwards right? Why does x(double dot) act downwards and not upwards?
    x double dot is positive in the direction of increasing x (it's the rate of the rate of the INCREASE of x). x is measured downwards from a fixed point, so increases downwards.

    When the particle is released, it goes upwards as you say, so at that moment x double dot will be negative.
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    Tried 2 papers yesterday. First one went well except for a 15 marker on SHM and I didn't attempt it really which was costly. Next one I got 67/75 with full marks on SHM - Unfortunately I am a bit inconsistent with SHM - theres one scenario which confuses me - I do not understand how to draw a solid diagram for those like the Last Q on JUNE 2011. If anyone gets the chance please can you take a look at the last Q on June 2011, not 100% sure I understand everything that is going on even though it is probably relatively straight forward, just not clicked with me fully.
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    (Original post by TrueDAN)
    Tried 2 papers yesterday. First one went well except for a 15 marker on SHM and I didn't attempt it really which was costly. Next one I got 67/75 with full marks on SHM - Unfortunately I am a bit inconsistent with SHM - theres one scenario which confuses me - I do not understand how to draw a solid diagram for those like the Last Q on JUNE 2011. If anyone gets the chance please can you take a look at the last Q on June 2011, not 100% sure I understand everything that is going on even though it is probably relatively straight forward, just not clicked with me fully.
    The string is lying on a horizontal table. I've attached a diagram. Put the particle at a variable point Q a distance x from O, the equilibrium position.

    BP is now of length 1 - x, so its extension is 1 - x - 0.7 (half the natural length of the string). Use Hooke's law on this to do part a and then on AP to do part b.

    You now need to use Newton's second law to get an SHM equation. This involves the tensions in the two strings and x double dot. Since x increases to the right, take x double dot as positive to the right, so T(BP) - T (AP) = m x double dot.

    Have a go at this.
    Attached Files
  1. File Type: docJune 11 M3 last question SHM.doc (36.0 KB, 22 views)
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    (Original post by tiny hobbit)
    The string is lying on a horizontal table. I've attached a diagram. Put the particle at a variable point Q a distance x from O, the equilibrium position.

    BP is now of length 1 - x, so its extension is 1 - x - 0.7 (half the natural length of the string). Use Hooke's law on this to do part a and then on AP to do part b.

    You now need to use Newton's second law to get an SHM equation. This involves the tensions in the two strings and x double dot. Since x increases to the right, take x double dot as positive to the right, so T(BP) - T (AP) = m x double dot.

    Have a go at this.
    Thank you so so much for taking the time to help, really appreciate it and
    has been extremely useful!! Much appreciated man
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    Every question in the book done!! Except for a few C.O.M integration derivations - think they can be well painful. Finally start papers properly!
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    (Original post by TrueDAN)
    Every question in the book done!! Except for a few C.O.M integration derivations - think they can be well painful. Finally start papers properly!
    The ones working from first principles are the most fun
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    (Original post by solC)
    The ones working from first principles are the most fun
    Think I will have to re do some of them then xD
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    June 2008 - it was 57 for an A and I got 62 marks which is not too bad, but need to push into the higher 60s / 70s. Atleast the boundaries do actually reflect the difficulty of the paper, most of the time.The last part of the last question I thought was insane!
    Think If I can, I will redo some papers twice. Currently needing to use the full 90 minutes too, which is not great. Just over 1 month to bump it up
    10 marks!
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    Hi guys does anyone know why in June 2015 m3 paper q3b why do you make the BP string = 0?? i dont rly get why you do not make the AP=0 as it lifts it so is more crucial to the ball moving around in the string? if anyone knows the answer let me know
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    (Original post by helpme:))
    Hi guys does anyone know why in June 2015 m3 paper q3b why do you make the BP string = 0?? i dont rly get why you do not make the AP=0 as it lifts it so is more crucial to the ball moving around in the string? if anyone knows the answer let me know
    Assuming that you are talking about the tensions in the strings, not the length of the strings, the tension in AP could not equal 0 since the particle is hanging at the bottom of it. If you don't whizz the particle round fast enough however, the string BP could go slack. We are told that both strings are taut, so we need to check that the tension in the bottom string is > = to 0. = 0 means that there is no tension but the string is still stretched out to its full length.
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    Persevered and have polished off weaknesses - nearly done the solomons and have actually found them worthwile - there was a question where you bascially had to calculate the escape velocity on the moon which I thought was quite interesting. With c.o.m derivations from first principles, say if it is a cone for example, the markscheme formulated the equation of the line and rotated about the x axis. I did it by forming a line, with y-intercept h and rearranged y=mx+ h to get an expression for x i.e. x = (y-h)/m. I then used the formula for rotating about the y-axis and got the same answer - presumably both methods are valid, even though my way wasnt stated in the markscheme? That was solomon B for reference
 
 
 
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