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    I don't understand questions with a wall that isn't in the x or y-axis.
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    I understand up to "(1/\sqrt2)(-i+j) is a unit vector perpendicular to the wall" but I do not know how to get the next line. Please could someone explain what is going on?

    I know that the unit vector parallel to the wall is (1/\sqrt2)(i+j), and I am familiar with the dot product.
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    (Original post by ombtom)
    ...
    This is FP3 vector stuff. The resolved part of a vector \mathbf{a} in the direction of another vector \mathbf{b} is given by:

    \displaystyle \left(\frac{\mathbf{a} . \mathbf{b}}{|\mathbf{b}|}\right)

    This is given to you in your formula booklet, page 10 under FP3 vectors.
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    (Original post by Zacken)
    This is FP3 vector stuff. The resolved part of a vector \mathbf{a} in the direction of another vector \mathbf{b} is given by:

    \displaystyle \left(\frac{\mathbf{a} . \mathbf{b}}{|\mathbf{b}|}\right) \mathbf{a}

    This is given to you in your formula booklet, page 10 under FP3 vectors.
    (a.b/|b|)a

    Okay, so we know the initial velocity, u = (4i + 2j). We want the 'amount' of this that acts parallel to the wall, and the 'amount' that acts perpendicular to the wall.

    Parallel to the wall, the unit vector is (i+j)/\sqrt2. So a = (4i + 2j), b = (i + j) (the equation of the line representing the wall), |b| = r(1+1) = r2, and so the 'amount' of a acting parallel to the wall equals [(4i + 2j).(i + j)/r2](4i + 2j). But in the solution, it has the unit vector parallel to the wall at the end instead of (4i + 2j), as if the FP3 equation is (a.b/|b|)b. I'm still confused!
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    (Original post by ombtom)
    (a.b/|b|)a
    as if the FP3 equation is (a.b/|b|)b. I'm still confused!
    Oh darn! My apologies, that's the one I meant. If it's in the direction of b and and (a.b)/|b| is a scalar then it makes sense for resolving a in the direction of b as b(a.b/|b|). Sorry again!
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    (Original post by Zacken)
    Oh darn! My apologies, that's the one I meant. If it's in the direction of b and and (a.b)/|b| is a scalar then it makes sense for resolving a in the direction of b as b(a.b/|b|). Sorry again!
    So what I said made sense? :eek:
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    (Original post by ombtom)
    So what I said made sense? :eek:
    Pretty much.
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    (Original post by Zacken)
    Pretty much.
    Great

    So for the 'amount' of (4i + 2j) acting perpendicular to the wall, we use [(4i + 2j).(-i + j)/r2](-i + j)... Which is not what the solution says. It has (i - j) then (-i + j); could it be a mistake?
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    (Original post by ombtom)
    Great

    So for the 'amount' of (4i + 2j) acting perpendicular to the wall, we use [(4i + 2j).(-i + j)/r2](-i + j)... Which is not what the solution says. It has (i - j) then (-i + j); could it be a mistake?
    This is probably a good time to drop the disclaimer that I've never touched an M4 book in my life and that I generally suck at mechanics...

    Buuut, yeah, I'd wager that's a mistake? Hopefully. Don't take my word for it, though.
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    (Original post by Zacken)
    This is probably a good time to drop the disclaimer that I've never touched an M4 book in my life and that I generally suck at mechanics...

    Buuut, yeah, I'd wager that's a mistake? Hopefully. Don't take my word for it, though.
    Probably is; I was finding 4 mistakes per question in the FP2 solutionbank. If it is a mistake, I think I understand it now, so thank you!
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    (Original post by ombtom)
    Great

    So for the 'amount' of (4i + 2j) acting perpendicular to the wall, we use [(4i + 2j).(-i + j)/r2](-i + j)... Which is not what the solution says. It has (i - j) then (-i + j); could it be a mistake?
    Ooor: I don't think it's a mistake, actually. I think it has something to do with the direction of resolving. The perpendicular of the wall thingy is point "away" from the wall, if you get my drift? But if you resolve the vector thingy perpendicular it kind of points perpendicular "into" the wall, get me? So you multiply be a negative sign or something I don't know?
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    (Original post by Zacken)
    Ooor: I don't think it's a mistake, actually. I think it has something to do with the direction of resolving. The perpendicular of the wall thingy is point "away" from the wall, if you get my drift? But if you resolve the vector thingy perpendicular it kind of points perpendicular "into" the wall, get me? So you multiply be a negative sign or something I don't know?
    Oh. I was happy with the mistake.
    I think I'll ask someone from my school about this one.
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    (Original post by ombtom)
    Oh. I was happy with the mistake.
    I think I'll ask someone from my school about this one.
    Haha, sorry for being so unhelpful. let me know what the person from your school says? I'd be interested in knowing.
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    (Original post by Zacken)
    Haha, sorry for being so unhelpful. let me know what the person from your school says? I'd be interested in knowing.
    Will do.
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    (Original post by Zacken)
    Haha, sorry for being so unhelpful. let me know what the person from your school says? I'd be interested in knowing.
    I've just done a similar question and it makes me think the last example had a mistake.
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    (Original post by ombtom)
    I've just done a similar question and it makes me think the last example had a mistake.
    It's probably a mistake, then. Yay!
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    (Original post by ombtom)
    I don't understand questions with a wall that isn't in the x or y-axis.
    Name:  Screen Shot 2016-02-26 at 22.07.41.png
Views: 105
Size:  223.1 KB
    I understand up to "(1/\sqrt2)(-i+j) is a unit vector perpendicular to the wall" but I do not know how to get the next line. Please could someone explain what is going on?

    I know that the unit vector parallel to the wall is (1/\sqrt2)(i+j), and I am familiar with the dot product.
    The coefficient of restitution is only applied to the component of the velocity perpendicular to the wall (not the X and Y axis).

    The dot product of the original vector and the unit vector parallel to the wall will give you the magnitude of the final velocity parallel to the wall. Then the scalar is multiplied with the unit vector parallel to the wall to find the final velocity parallel to the wall.

    The same is done for the perpendicular velocity and it is also multiplied by the coefficient of restitution. Both vectors are added together to find the answer.
 
 
 
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