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    paper: http://www.freeexampapers.com/get_pa...M2+2004-01.pdf

    mark scheme: http://www.freeexampapers.com/get_pa...2004-01+MS.pdf

    can someone please explain where the I = 9i - 9j comes from for question 2b in the mark scheme? how do i do this question?

    thanks

    jb
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    The impulse is in the direction of i-j, so I = C(i-j) for some constant C. You're told that the magnitude of the impulse is 9\sqrt{2}, and so |C(\mathbf{i}-\mathbf{j})| = 9\sqrt{2}, and so C=\cdots [you do this bit].
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    (Original post by nuodai)
    The impulse is in the direction of i-j, so I = C(i-j) for some constant C. You're told that the magnitude of the impulse is 9\sqrt{2}, and so |C(\mathbf{i}-\mathbf{j})| = 9\sqrt{2}, and so C=\cdots [you do this bit].

    so its just guess work at what i and j components could be? surely there are several possibilities?
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    (Original post by jumblebumble)
    so its just guess work at what i and j components could be? surely there are several possibilities?
    There's no guesswork at all. A vector is defined by its direction and magnitude. It says that the impulse is in the same direction as \mathbf{i}-\mathbf{j} and has magnitude 9\sqrt{2}, so you can work out what the vector is. Saying "it's in the same direction as \mathbf{i}-\mathbf{j}" means "it's a (positive) scalar multiple of \mathbf{i}-\mathbf{j}"; in other words, \mathbf{I} = C(\mathbf{i}-\mathbf{j}) for some positive number C. The magnitude then determines the value of C, which you can find by taking the length of C(\mathbf{i}-\mathbf{j}) in terms of C, and then solving for C.

    If it helps, write \begin{pmatrix} 1 \\ -1 \end{pmatrix} instead of \mathbf{i}-\mathbf{j}. They mean the same thing.
 
 
 
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