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    (Original post by .A.)
    On being difficult. 1 to 10.

    Thats what I hated the most, no hard questions on chap 1. I was so good on it.

    Yea if you can post the answers that would be great.

    I know for the last part it was cos theta = -2/3 (Unstable i think)

    let me think about them.

    .A.
    My memory isn't that great. so here goes

    Question 1: 260 degrees
    Question 2: the velocity i got was either (-i + j) or ( i -j) one of those and ten the speed is root 2.

    Question 3: was this the find tan theta question? if so I remember getting tan theta = 1/2 as my answer.

    Question 4 : Was this the air resistance one with the ball travelling at speed U and find the time it takes to get to 2U ? if so My answer something like  \frac{1}{k} \ln \left( \frac{g - kU}{g-2kU}  \right)

    Question 5: I think this was that funny DE question, They I think they didn't think they allocated marks fairly, as the first part was 6 marks for using Newton second law and the more difficult part of solving the DE was also 6 marks but it was much harder, I dont remember exactly what i got as the solution but i remember it being double roots, but for the final answer i am fairly sure i got  \frac{2}{3w}

    Question 6: The first part you need to draw a velocity triangle, I had 5 ms-1 as he hypotenuse, V as the vertical and x/10 as he horizontal, then use Pythagorus to get their thing. for part b) I differentiated with respect to x to get their equation. and for the final part i got the time as 6.4 seconds.

    Question 7: -2/3 sounds familiar, I forget which one i got an stable/unstable, but i do remember that one was stable and the other wasn't.
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    (Original post by bobx2001)
    My memory isn't that great. so here goes

    Question 1: 260 degrees
    Question 2: the velocity i got was either (-i + j) or ( i -j) one of those and ten the speed is root 2.

    Question 3: was this the find tan theta question? if so I remember getting tan theta = 1/2 as my answer.

    Question 4 : Was this the air resistance one with the ball travelling at speed U and find the time it takes to get to 2U ? if so My answer something like  \frac{1}{k} \ln \left( \frac{g - kU}{g-2kU}  \right)

    Question 5: I think this was that funny DE question, They I think they didn't think they allocated marks fairly, as the first part was 6 marks for using Newton second law and the more difficult part of solving the DE was also 6 marks but it was much harder, I dont remember exactly what i got as the solution but i remember it being double roots, but for the final answer i am fairly sure i got  \frac{2}{3w}

    Question 6: The first part you need to draw a velocity triangle, I had 5 ms-1 as he hypotenuse, V as the vertical and x/10 as he horizontal, then use Pythagorus to get their thing. for part b) I differentiated with respect to x to get their equation. and for the final part i got the time as 6.4 seconds.

    Question 7: -2/3 sounds familiar, I forget which one i got an stable/unstable, but i do remember that one was stable and the other wasn't.
    My answers for Q 1,3,4 and 7 (ofcourse) match yours.
    Q2. Flopped. Hoping to get couple of method marks for momentum and stuff.
    Q5 I think it was something similar to you but from what i remeber it had something else too. W^2=l/g or something like that. But there was something else there too.

    O6. Got part b right. I guess for part c you had to have got part a right so prob 3/9 for part c (as i coudnt figure out what would be the value of dx/dt at t=o).



    .A.
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    can someone explain how they got 260 for q1? wasn't it -2i+11j the relative velocity?

    ...

    Does anybody have the paper btw?
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    Yeh, I got -2i + 11j for the velocity so the bearing was 349.7.
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    yes i think i got something like that too !! i remember using tan theta is 11/2 ..
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    1. 349.7
    2. root 2
    3. tan theta = 0.5
    4. can't remember
    5. 2/3w
    6. 6.44 s
    7. -2/3 unstable
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    yeah your right the bearing was 349.7 , I did 180 + arctan(11/2) instead of the more correct 270 + arctan (11/2)
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    (Original post by bobx2001)
    My memory isn't that great. so here goes

    Question 1: 260 degrees
    Question 2: the velocity i got was either (-i + j) or ( i -j) one of those and ten the speed is root 2.

    Question 3: was this the find tan theta question? if so I remember getting tan theta = 1/2 as my answer.

    Question 4 : Was this the air resistance one with the ball travelling at speed U and find the time it takes to get to 2U ? if so My answer something like  \frac{1}{k} \ln \left( \frac{g - kU}{g-2kU}  \right)

    Question 5: I think this was that funny DE question, They I think they didn't think they allocated marks fairly, as the first part was 6 marks for using Newton second law and the more difficult part of solving the DE was also 6 marks but it was much harder, I dont remember exactly what i got as the solution but i remember it being double roots, but for the final answer i am fairly sure i got  \frac{2}{3w}

    Question 6: The first part you need to draw a velocity triangle, I had 5 ms-1 as he hypotenuse, V as the vertical and x/10 as he horizontal, then use Pythagorus to get their thing. for part b) I differentiated with respect to x to get their equation. and for the final part i got the time as 6.4 seconds.

    Question 7: -2/3 sounds familiar, I forget which one i got an stable/unstable, but i do remember that one was stable and the other wasn't.
    Your Question 1 is wrong I think, I got 350

    Question two is correct

    Question three correct

    Question four is correct,

    Question five is fine,

    Question six was 6.44 seconds for me, a weird relative motion question... they combined it with DE

    Question 7 was cos X = -2/3 Unstable eq. and cos X = 3/4 stable eq.
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    (Original post by MaLvI)
    can someone explain how they got 260 for q1? wasn't it -2i+11j the relative velocity?

    ...

    Does anybody have the paper btw?
    I thought everybody had the paper by now...
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    Have fun with it
    Attached Images
  1. File Type: pdf 6680_01_que_20080612.pdf (500.8 KB, 94 views)
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    thnx for the paper!

    ok can somebody tell me how u prove exercise 6b) since the differentiation of V^2 is 2dxt/dt d^2x/dt^2
    I could not prove it using that so i did
    V^2 is 2d^2x/dt^2

    what's the proper way?
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    (Original post by MaLvI)
    thnx for the paper!

    ok can somebody tell me how u prove exercise 6b) since the differentiation of V^2 is 2dxt/dt d^2x/dt^2
    I could not prove it using that so i did
    V^2 is 2d^2x/dt^2

    what's the proper way?
     100v^2 + x^2 = 2500
     \frac{d}{dx}(100v^2 + x^2 )= 0
     200v \frac{dv}{dx} + 2x = 0
     \frac{d^2x}{dt^2} + \frac{x}{100} = 0
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    ok thats more sensible than what i did..

    I did..

    100v^2=2500-x^2
    200d^2x/dt2=2x

    :PP i guess this will score zero marks out of four?
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    I found the paper hard, I wasn't happy with it. Unfortunately, one of the other (two) people who did it said "it could have been worse", I was kinda hoping for an "argh that was terrible" type thing :p:

    I think I can max get a C... and seeming as I need to get an A to get to Imperial, I need 76 UMS points average on FP1 and FP2, and FP2 has never really been my strong suit...

    Oh well, I'll just have to hope and do loads of past papers :p: unfortunately I've got French on Monday too...

    Pretty hard paper I thought though, overall
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    (Original post by MaLvI)
    ok thats more sensible than what i did..

    I did..

    100v^2=2500-x^2
    200d^2x/dt2=2x

    :PP i guess this will score zero marks out of four?
    I mean, I really can't understand where you're coming from

    v= dx/dt
    hence
    100 v^2 is equivalent to 100(dx/dt)^2 and then what I did was to differentiate implicitly on both sides of the equation with respect to t hence

    100(dx/dt)^2 = 2500 - x^2

    Thus by implicit differentiation

    200(dx/dt)(d^2x/dt^2) = -2x(dx/dt) --> you cancel a dx/dt on both sides and divide by 200 to get

    (d^2x/dt^2) = -(1/100)x

    Which rearranged gives you

    (d^2x/dt^2) + (1/100)x = 0
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    I didnt do 6a. But went on to get 6b right. Will I get 4 marks for 6b as or not as the questions said HENCE show that.

    .A.
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    (Original post by .A.)
    I didnt do 6a. But went on to get 6b right. Will I get 4 marks for 6b as or not as the questions said HENCE show that.

    .A.
    Yeah you'll get the marks, cos the answer was given in part a.
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    (Original post by .A.)
    I didnt do 6a. But went on to get 6b right. Will I get 4 marks for 6b as or not as the questions said HENCE show that.

    .A.
    All you had to do for 6a was to draw a rather easy right angled triangle with sides 5 has the hypotenuse, (1/10)x and v.

    v^2 + 1/100 x^2 = 25 and you're good..

    But you will get the marks for 6b though
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    1 wrong, 2 wrong, 3 wrong, 4 right, 5 wrong, 6 and 7 right

    :mad:

    looking like c. 45/75 pour moi
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    What do you people reckon A would be. I reckon 60/75.

    .A.
 
 
 

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