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Math Help - Immediate Mechanics Help Required!!!

  1. #1
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    Immediate Mechanics Help Required!!!

    please visit this link:
    http://www.meidistance.co.uk/npapers/m201ja.pdf
    and see question 1(iv), the mark scheme answer is on page 7 but i need someone to explain how 24.6m/s was obtained.

    Also, on page 8, how do you find 3.45 and 1.82 from those vectors on the top of the page? Thanks,
    AAKhan07
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by AAKhan07 View Post
    please visit this link:
    http://www.meidistance.co.uk/npapers/m201ja.pdf
    and see question 1(iv), the mark scheme answer is on page 7 but i need someone to explain how 24.6m/s was obtained.

    Consevation of linear momentum. The initial momentum of the system is 300 \times20 kg m/s the momentum of the component is -20\times 44.5 kg m/s. So the momentum of the remaining two parts is the initial momentum of the system plus 20\times 44.5, so:

    280 \times v_{final} = 300 \times 20 + 20 \times 44.5

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by AAKhan07 View Post
    Also, on page 8, how do you find 3.45 and 1.82 from those vectors on the top of the page? Thanks,
    AAKhan07
    Combine the vectors on the left into a single vector, then equate components on the right and left to get an equation for x and an equation for y. Solve these equations.

    RonL
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