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M1 Vectors

Discussion in 'Mathematics' started by GoldMaths, Feb 16, 2012.

  1. GoldMaths

    GoldMaths New commenter

  2. DM

    DM New commenter

    Method 1 is correct. You can't just ignore the directions and work with the magnitudes.
  3. Method 2 assumes that the acceleration is acting in the same direction as the initial velocity and clearly from the vectors given this is not the case, hence the incorrect answer.
  4. GoldMaths

    GoldMaths New commenter

    Cheers for your replies it was I told the students the error is to do with signs (directions) but they wernt and to be fair still arent happy with just being told because it is, I think their logic is valid and intially I was happy with their working.
    Least it shows they're thinking. Thanks again.
    N.B. If the acceleration was of the same direction (gradient) then I assume method 2 would be valid?
  5. DM

    DM New commenter

    This worries me a little so I think more exposition might be required.
    On the January 2012 OCR M1 paper three forces acted at a point. They were 8N, 15N and 20N.
    The question asked "State the greatest and least magnitudes of the resultant force".
    The greatest magnitude is when all three forces act in the same direction so 8 + 15 + 20 = 43N.
    The least magnitude is NOT 8 + 15 - 20 = 3N as many candidates thought as it is possible to construct a triangle with sides 8 cm, 15 cm and 20 cm so clearly the answer must be 0 N.
    Using "signs" as a synonym for "directions" leads to this sort of misconception.
    I am equally uncomfortable with your use of the word "gradient" here.
  6. Maths_Mike

    Maths_Mike New commenter

    I agree of course although I am not surprised many students got this wrong!
    Presumably if a trinagle can not be formed (say 20N, 5N and 5N) then the answer would indeed be the largest takeaway the other 2)
    This then gets me thinking about 4 or more forces. Would it be possible to determine the least magnitude of the resultnat force?
  7. DM

    DM New commenter

    My instinct says the same geometrical reasoning would apply. So long as one length is not greater than the sum of the other three you could always form a quadrilateral. Likewise for shapes with more sides.
  8. Piranha

    Piranha Star commenter

    I think that the answer for any number of forces >=3 is the same. If the largest force is smaller than the sum of the rest, then it will always be possible to form a polygon, so the lowest force needed will be 0. If not, it is the largest force minus the sum of the others.

  9. fieldextension

    fieldextension New commenter

    In method 1 the vectors a, u, s, i and j should be underlined. The step from the seventh to the eighth line is invalid since it equates a vector with its magnitude. There are also units missing on the answer. None of that is a problem for us - many of us write quickly and informally here - but I would not want to present your method 1 to students as written without these alterations. I think this ought to be said.

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