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    Im doing this paper:

    http://www.ocr.org.uk/download/pp_10...n_gce_g484.pdf

    And Im stuck on question 1 B i)

    Yes, its early but ive always had problems on newtons laws etc.

    So, the area of the graph is 2.2NS^-1 but how? I dont understand how you can get that answer.
    The mark scheme says:
    area: number of squares correctly counted: 20 - 24 (500 – 600)
    = 2.2 Ns {allow 2.0 to 2.4}

    Can anyone explain the method of counting squares and working out the impulse of a force using this question as an example?

    thanks
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    The impulse is the area under the force-time graph. You have to work out what this area is. In this case, you can only do that by counting the number of small squares under the graph, although you can take out some large rectangles and only have to bother with counting the extra squares round the edge, or simply estimate by approximating the shape to a triangle.

    Once you've got a rough number of small squares, you have to work out what each small square represents, i.e. what is its height (in N) and its width (in s). Note that the horizontal scale is in ms rather than s.

    Does this make it any clearer?
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    (Original post by Pangol)
    The impulse is the area under the force-time graph. You have to work out what this area is. In this case, you can only do that by counting the number of small squares under the graph, although you can take out some large rectangles and only have to bother with counting the extra squares round the edge, or simply estimate by approximating the shape to a triangle.

    Once you've got a rough number of small squares, you have to work out what each small square represents, i.e. what is its height (in N) and its width (in s). Note that the horizontal scale is in ms rather than s.

    Does this make it any clearer?

    Thanks for your time to write that answer

    I've done what you said and got 2.2, but can I jut clarify this to be the method that you would do?

    I counted there to be 550 squares approximately and worked out each square to be worth 0.04x10^-3 (0.2x10^-3/5) then multiply that by 100N and each square is 4x10^-3 NS^-1

    So 550 x 4x10^-3 = 2.2NS^-1

    So that's the right answer, but I cant help but feel that counting all those squares is very time consuming. And using a triangle, I can't get the answer- 0.5 x 2800 x 1.2x10^-3 = 1.68NS^-1

    So would you count squares or use a triangle if you were sitting the exam?

    Thanks so much for your help though!
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    What you have done is broadly correct (although you should be measuring impulse in N s, not N s-1).

    In an exam, you want to count the number of squares as accurately as possible, so I wouldn't treat the curve as a triangle. I would use the fact that it is symmetric, so that you only have to count half of the area and then double it, and would remove as many large rectangles from the area as possible and then count the extra squares. Probably best to divide the area up into convenient areas, work them out, then add them all up. No harm in doing this on the diagram provided.
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    (Original post by Pangol)
    What you have done is broadly correct (although you should be measuring impulse in N s, not N s-1).

    In an exam, you want to count the number of squares as accurately as possible, so I wouldn't treat the curve as a triangle. I would use the fact that it is symmetric, so that you only have to count half of the area and then double it, and would remove as many large rectangles from the area as possible and then count the extra squares. Probably best to divide the area up into convenient areas, work them out, then add them all up. No harm in doing this on the diagram provided.
    Thanks for your help I forgot about symmetry so it can take half the time haha

    I shall give you some rep as its the least I can do for my gratitude
 
 
 
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