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    Can someone please explain how to do 5c). I am unsure on what the markscheme did, but I am assuming they rotated the velocity vector so that it became parallel with the line of action, if this is the case I understand the reason behind this. But I am unsure on how they did the rotation (I think they used a transformation matrix).

    Markscheme; https://eiewebvip.edexcel.org.uk/Rep...s_20090819.pdf

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    (Original post by mubmoh)
    Can someone please explain how to do 5c). I am unsure on what the markscheme did, but I am assuming they rotated the velocity vector so that it became parallel with the line of action, if this is the case I understand the reason behind this. But I am unsure on how they did the rotation (I think they used a transformation matrix).

    Markscheme; https://eiewebvip.edexcel.org.uk/Rep...s_20090819.pdf

    Name:  M4 2009 Q5.png
Views: 75
Size:  45.1 KB
    There is no rotation.

    Having worked out the line of centres, they then worked out the velocities along the line of centres using the dot product with a unit vector in that direction.
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    (Original post by ghostwalker)
    There is no rotation.

    Having worked out the line of centres, they then worked out the velocities along the line of centres using the dot product with a unit vector in that direction.

    A.B=|A||B|cos(pheta)
    If B is a unit vector |B| is 1. If I want A such that it is along the line of centres (parallel to B) then pheta is equal to 0, therefore cos(pheta)=1, leaving A.B=|A|
    where |A| is the speed along the line of centre. Am I right with this reasoning?

    Also, is using a unit vector for B mandatory as if B had a scaler they would all cancel when calculating e?
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    (Original post by mubmoh)
    A.B=|A||B|cos(pheta)
    If B is a unit vector |B| is 1. If I want A such that it is along the line of centres (parallel to B) then pheta is equal to 0, therefore cos(pheta)=1, leaving A.B=|A|
    where |A| is the speed along the line of centre. Am I right with this reasoning?
    That's a bit confusing, you seem to have used A with two meanings.

    If r is a unit vector, then A.r, will give you the magnitude of the component of A in the r direction.

    A.r = |A| |r| cos(angle between them)

    And if |r| =1, then
    A.r = |A| cos(angle between them)
    i.e. the magnitude of the component of A in the r direction. Effectively resolving A into components; this one being in the direction r.
    Diagram may help clarify.

    Also, is using a unit vector for B mandatory as if B had a scaler they would all cancel when calculating e?
    It's wise to use a unit vector as then you've have the velocity in the the desired direction, otherwise it's a multiple of the velocity. And you may have other uses for the velocity within the question.

    It would still work with a non-unit vector, when calculating e, as you surmised.
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    (Original post by ghostwalker)
    That's a bit confusing, you seem to have used A with two meanings.

    If r is a unit vector, then A.r, will give you the magnitude of the component of A in the r direction.

    A.r = |A| |r| cos(angle between them)

    And if |r| =1, then
    A.r = |A| cos(angle between them)
    i.e. the magnitude of the component of A in the r direction. Effectively resolving A into components; this one being in the direction r.
    Diagram may help clarify.
    Thanks a lot, I understand now , we're just taking the part of A which is parallel/in the direction of r.
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    (Original post by mubmoh)
    Thanks a lot, I understand now , we're just taking the part of A which is parallel/in the direction of r.
    Yep.
 
 
 
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