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    (Original post by mikelbird)
    Perhaps I am not understanding what you mean by cartesian trajectories...the actual trajectories are certainly are not straight lines and so I feel further explanation is needed. I suppose you are making the point about the fact that gravity seems to disappear in the second part when establishing the relationships with regard to theta but then you are not really dealing with a cartesian graph anymore (or the rescaling of the height coordinate needs some explanation).
    He is looking at initial trajectories so the situation is mirrored as after that the forces on them are the same.


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    (Original post by StrangeBanana)
    Isn't the point that the "gravitational part" of their trajectories (so to speak) is the same, so you can just forget about it? It's like having the same term present on two sides of an equation, allowing you to cancel.
    Yu took it banana boy. Me sed it first


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    (Original post by mikelbird)
    Perhaps I am not understanding what you mean by cartesian trajectories...the actual trajectories are certainly are not straight lines and so I feel further explanation is needed. I suppose you are making the point about the fact that gravity seems to disappear in the second part when establishing the relationships with regard to theta but then you are not really dealing with a cartesian graph anymore (or the rescaling of the height coordinate needs some explanation).
    Yes, I had shown that \mathbf{r}^{\text{bullet}} = \mathbf{r}^{\text{target}} \iff \mathbf{r}^A=\mathbf{r}^B i.e. that gravity is irrelevant and that it is necessary (and sufficient, but that's not important here) that A, B collide in order for the bullet to hit the target.

    Furthermore, by "cartesian trajectory", I meant the equation of motion that relates x and y and I presume this is what you meant by "actual trajectory". The trajectories of A and B (that is, of the particles of part (i)) are straight lines - this is immediate from the parametric forms of part (i) by eliminating t (mentally or otherwise) and indeed what one expects if gravity does not influence the motion. So, for the bullet to hit the target, it is therefore necessary that these lines must intersect in the upper right quadrant. My conclusion follows.*

    I must say this is all in my explanation for the last part, and I felt this was the whole point of the question. I'm still unsure where you feel an issue lies.
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    (Original post by Farhan.Hanif93)
    Yes, I had shown that \mathbf{r}^{\text{bullet}} = \mathbf{r}^{\text{target}} \iff \mathbf{r}^A=\mathbf{r}^B i.e. that gravity is irrelevant and that it is necessary (and sufficient, but that's not important here) that A, B collide in order for the bullet to hit the target.

    Furthermore, by "cartesian trajectory", I meant the equation of motion that relates x and y and I presume this is what you meant by "actual trajectory". The trajectories of A and B (that is, of the particles of part (i)) are straight lines - this is immediate from the parametric forms of part (i) by eliminating t (mentally or otherwise) and indeed what one expects if gravity does not influence the motion. So, for the bullet to hit the target, it is therefore necessary that these lines must intersect in the upper right quadrant. My conclusion follows.*

    I must say this is all in my explanation for the last part, and I felt this was the whole point of the question. I'm still unsure where you feel an issue lies.
    I understand what you are saying completely....perhaps I worded what I said too strongly but on first reading I did not feel what you were saying was crystal clear...so probably my response was 'over enthusiastic'!!
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    I Think this solves it!!
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  1. File Type: pdf Step2016Paper3Question7.pdf (84.9 KB, 97 views)
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    (Original post by mikelbird)
    I Think this solves it!!
    The other method is fine aswell. By putting z=1 you aren't dividing by zero if that was your issue since all you are doing is plugging z=1 into two equal polynomials. All you were noting was that those polynomials are equal due to the factored forms.


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    step is bulsht
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    (Original post by Llewellyn)
    Well, actually, I think it could maybe be done.

    https://docs.google.com/document/d/1...it?usp=sharing

    Very tricky to know the exact mark allocation... I'm tempted to reduce marks by line for each part and add an overall "get the gist of the question" or "bonus mark(s) at the end for getting a full solution"

    I'll add in the rest of the pure section later. Anyone else free to edit/ improve.
    Was the rest of this ever made?

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    (Original post by fa991)
    Was the rest of this ever made?

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    Nah
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    Step III Question 7

    [attach]6.135646135666136e+23[/attach]
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    (Original post by Zacken)
    I left my answer was ((7^7 + 1)^3 - sum of 7's)((7^7 + 1)^3 + sum of 7's).
    In your solution how did you go from the power of 6 to 2 as you would surely use the 4th row of the binomial expansion which is 14641
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    (Original post by Salimshady007)
    In your solution how did you go from the power of 6 to 2 as you would surely use the 4th row of the binomial expansion which is 14641
    Not sure what you're asking, but 2*3 = 6. So you can view it as a difference of cubes.
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    (Original post by Zacken)
    Not sure what you're asking, but 2*3 = 6. So you can view it as a difference of cubes.
    sorry just realised how you got that (sigh kmn)
 
 
 
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