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POST ALL YOUR: Edexcel M2 Sophisticated Questions ---> Here!!!! watch

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    Hello M2 people, if there is a question on M2 that you think is particularly "skewed away from the norm of M2 questions" (i.e. distinctly harder than usual or there's a different original concept to understand which Edexcel likes to confuse people with), then please post it up here along with completed methods and answers.

    If you cannot do the question then feel free to post it up, and me and other candidates will attempt to complete it and cross-check our answers.

    If you cannot do the question but have the answer, then please post up the answer and me and other candidates will attempt to complete it and check the answer.

    Remember, there are some highly sophisticated questions on M2 papers which require possibly beyond M2 standard knowledge - these are the ones that have to be tracked down and eliminated so that during the exam we will be comfortable with it.
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    A stone is projected from a point O on a cliff with a speed of 20 m/s at an angle of elevation of 30 degrees. T seconds later the angle of depression of the stone from O is 45 degrees. Find the value of T?
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    θ = 30°
    ø = 45°

    Vh = 20cos30°
    Vh = 10√3 m/s
    ===========

    The particle will travel, initially upwards, in a parabolic curve as depicted in Fig 1.
    The particle will then fall below the level of the cliff until its angle of depression is 45°.
    At this point, the horizontal distance travelled will be equal to the vertical distance below the cliff top. See Fig 2.

    Sh = VhT
    Sh = 10√3T
    =========

    Sv = ut - ½gt² - (measured positive upwards)
    Sv = 20sin30°T - 4.9T²
    Sv = 10T - 4.9T²
    ============

    at P, Sh = -Sv

    10√3T = 4.9T² - 10T
    4.9T² - 10(1 + √3)T = 0
    T(4.9T - 10(1 + √3)) = 0
    T = 0, or T = 10(1 + √3)/4.9
    T = 5.576 sec
    ==========
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    thanks/ good math head
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    Qno.1. The velocity V m/s of a particle P at time t seconds is given by
    V=(3t^2-12)i+5tj.
    with respect to a fixed origin O, the position vector of P when t=2 is 4j. Find the distance between OP when t=1.

    Qno.2. A particle P of mass 0.5kg is at at rest on a horizontal table. It recieves a blow of impulse 2.5Ns.
    (a) calculate the speed with which P is moving immediately after the blow.
    The height of the table is 0.9m and the floor is horizontal. In an initial model of the situation the table is assumed to be smooth.
    (b) calculate the horizontal distance from th edge of the table to the point where P hit the ground.
    In a refinement of the model the table is assumed to be rough. The coefficient of friction between P and the table is 0.2.
    (c) calculate the deceleration of P.
    Given that P travels 0.4m to the edge of the table,
    (d) calculate the time which elapses between P recieving the blow to P hitting the floor. ( i could do all parts of Qno2. except the (d) part.
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    (Original post by bodomx)
    Qno.1. The velocity V m/s of a particle P at time t seconds is given by
    V=(3t^2-12)i+5tj.
    with respect to a fixed origin O, the position vector of P when t=2 is 4j. Find the distance between OP when t=1.
    Only time for the first one

    ds/dt = v = (3t²-12)i + 5tj
    s = (t³ - 12t + A)i + (2.5t² + B)j

    at t=2 , s = 4j

    s(t=2) = (8 - 24 + A)i + (10 + B)j

    8 - 24 + A = 0 -> A = 16
    10 + B = 4 -> B = -6

    s = (t³ - 12t + 16)i + (2.5t² - 6)j
    s(t=1) = (1 - 12 + 16)i + (2.5 - 6)j
    s = 5i + 3.5j
    =========
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    (Original post by bodomx)
    Qno.2. A particle P of mass 0.5kg is at at rest on a horizontal table. It recieves a blow of impulse 2.5Ns.
    (a) calculate the speed with which P is moving immediately after the blow.
    The height of the table is 0.9m and the floor is horizontal. In an initial model of the situation the table is assumed to be smooth.
    (b) calculate the horizontal distance from th edge of the table to the point where P hit the ground.
    In a refinement of the model the table is assumed to be rough. The coefficient of friction between P and the table is 0.2.
    (c) calculate the deceleration of P.
    Given that P travels 0.4m to the edge of the table,
    (d) calculate the time which elapses between P recieving the blow to P hitting the floor. ( i could do all parts of Qno2. except the (d) part.
    a)u = 0m/s, m = 0.5kg, I = 2.5Ns
    I = mv - mu , so I/m = v - u
    v - 0 = 2.5/0.5 = 5
    v = 5 m/s

    b)The table is smooth so P travels with constant velocity: v = 5m/s, h = 0.9m.
    Horizontal motion: x = vt
    Vertical motion: y = - gt^2/2
    When P hits the ground, y = -h (I take the upward vertical is positive)
    Gives: -0.9 = -9.8t^2/2
    so t = 0.4286s
    The horizontal distance: x = 5*0.4286 = 2.143m

    c)coefficient of friction = 0.2
    Frictional force = 0.2mg
    F = ma so a = 0.2g = 1.96m/s^2. (just using the modulus)

    d)Total time = time for P travels on the table + time for P reaches the floor (t1 +t2)
    Using V^2 = v^2 + 2as (V is the speed at the edge of the table)
    V^2 = 25 - 2*1.96*0.4 ( a= -1.96m/s^2)
    V = 4.84m/s
    Using V = u + at1
    4.84 = 5 - 1.96t1
    t1 = 0.0813s
    Similarly, you have x = Vt and y = -gt2^2/2
    and t2 = 0.4286s when P hits the ground.
    So T = t1 + t2 = 0.51s.

    But when Im doing part d), i found something strange:
    If i used s = vt1 + at1^2/2, I would get 2 values of t, both positive. How can we eliminate one?
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    (Original post by BCHL85)
    .....

    But when Im doing part d), i found something strange:
    If i used s = vt1 + at1^2/2, I would get 2 values of t, both positive. How can we eliminate one?
    The particle starts at a point, called O say, and travels under a decelerating force.
    After 0.08 s, P reaches a dist 0.4m from O, call it Q. It then continues decelerating until it reaches zero velocity, then it travels backwards. Finally reaching the point Q again about 5 secs after leaving O.
    In other words, the smaller value of t is for when P reaches Q on its outward journey, and the greater value of t is for when it reaches Q on the way back.
    This explanation ignores the fact that there is an edge to the table!! It's what would happen it the particle didn't fall off the table, but was simply subject to that particular equation of motion only.
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    Thank you Fermat. I know why i had 1 value for t if using the 1st way, but had 2 if using the 2nd one, just cuz i forgot when v^2 = u^2 + 2as, v would have 2 values, 1 (+) and 1 (-).
    However in this case, even if there were no edge of the table (it might be long enough for all the journey), P couldn't return because the applied force here is frictional force which resists the motion. When v = 0, there wouldn't have force applying on P.
    Is it right, Fermat?
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    (Original post by BCHL85)
    Thank you Fermat. I know why i had 1 value for t if using the 1st way, but had 2 if using the 2nd one, just cuz i forgot when v^2 = u^2 + 2as, v would have 2 values, 1 (+) and 1 (-).
    However in this case, even if there were no edge of the table (it might be long enough for all the journey), P couldn't return because the applied force here is frictional force which resists the motion. When v = 0, there wouldn't have force applying on P.
    Is it right, Fermat?
    Yes that's right, BCHL85.
    Considering trhe circumstances when the table is long enough, but the deceleration is suppplied by friction only then the particle will stop at v=0 and stay stopped. It won't return.
    The eom (equation of motion) simply describes how the particle moves. It (the eom) doesn't know that there is an edge to the table, or that the deceleration will disappear when zero velocity is reached
    It's a bit like working with eqns that have a fixed domain. For example,

    y = √(x²-9), x >= 3
    f(ø) = sin²ø - 2cos(3ø) , ø ε [π, π/2]

    In our case, the eom is really only valid for that period of time during which it (the eom) is applicable, which is while travelling under that deleration on the table, so when it leaves the table the eom is no longer applicable or valid.
    The eom should could be written more (mathematically) precise as ,

    s = vt1 + at1^2/2, t1 ε [0, 0.08]
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    The figure is attached. Find the centre of mass form the corner A
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    Get a pen and paper to make it easier
    Going anticlockwise label the corners A B C D E (from A already marked)
    Now draw a line from C to the midpoint AE.
    Now we have a 4x4 square, 8x4 rectangle and doubled up triangle.
    Now call E the origin and let 1cm = 1 unit

    shape - ratio of mass - x - y
    8x4 - 32 - 6 - 4
    4x4 - 16 - 2 - 2
    triangle - 16 - 8/3 - 16/3
    overall - 64 - a - b

    (32*6) + (16*2) + [16*(8/3)] = 64a
    a = 25/6

    (32*4) + (16*2) + [16*(16/3)] = 64b
    b = 23/6

    Distance between (25/6 , 23/6) and (8 , 0)
    = sqrt[(8 - 25/6)^2 + (0 - 23/6)^2]
    = 5.42 cm (2dp)
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    I forgot to mention that one corner is folded over as shown on the figure attached. It is a square of side 8 cm with one corner folded. the length folded is shown in the figure. I think it is more difficult now.
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    (Original post by bodomx)
    I forgot to mention that one corner is folded over as shown on the figure attached. It is a square of side 8 cm with one corner folded. the length folded is shown in the figure. I think it is more difficult now.
    Bodomx, SsEe has already answered the question, and got the correct result.
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    (Original post by bodomx)
    I forgot to mention that one corner is folded over as shown on the figure attached. It is a square of side 8 cm with one corner folded. the length folded is shown in the figure. I think it is more difficult now.
    Oh, I dont think it's M2 question :confused: . If the corner is folded, the paper is not uniform. But let's do as normal. Hope it's right.
    Let see my attached file
    Call x, y are the distances from AB and AE to centre of mass. Mass of FCG is doubled
    ABCD DEFG FCG Overall
    Ratio of mass 32 16 8*2(16) 64
    x 2 6 16/3 x
    y 4 2 16/3 y

    Taking the moment, we have:
    64x = 64 + 96 + 256/3 , so x = 23/6 cm
    64y = 376/3 , so y = 23/6 cm.
    Check it for me, maybe i have some mistake
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    (Original post by BCHL85)
    Oh, I dont think it's M2 question :confused: . If the corner is folded, the paper is not uniform. But let's do as normal. Hope it's right.
    Let see my attached file
    Call x, y are the distances from AB and AE to centre of mass. Mass of FCG is doubled
    ABCD DEFG FCG Overall
    Ratio of mass 32 16 8*2(16) 64
    x 2 6 16/3 x
    y 4 2 16/3 y

    Taking the moment, we have:
    64x = 64 + 96 + 256/3 , so x = 23/6 cm
    64y = 376/3 , so y = 23/6 cm.
    Check it for me, maybe i have some mistake
    No mistake!
    I just checked.
    SsEe and BCHL85 both have 23/6 for the y-ordinate.
    SsEe has 23/6 for the x-ordinate and BCHL85 has 25/6, but they are the same since one is from the left of the diagram and the other is from the right

    p.s. BCHL85, when you post info like you did, with lots of space formatting, unfortunately it gets messed up. Have you tried using the "post as Code" optin when writing your post. That's the one that looks like '#'. That might keep the formatting.

    Let's try it.
    Code:
    a        b                     c
    d        e                     f
    g        h                     i
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    I feel very sorry about that I have edited the question without checking. Fermat is right.
 
 
 
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