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    (Original post by EnglishMuon)
    Insight314 Would you be able help with a probable silly slip on an M4 q? On June 2011 q2, when applying Newton's Law of restitution to the wall I got  - \frac{3}{4} u \sin \theta = v \sin \phi (where  \theta , \phi are the angles between the wall and initial velocity and wall and final velocity respectively, but I shouldnt have a -ve there apparently. Why is this the case?
    It doesn't matter, all they're doing is absorbing the negative into the "u" - i.e: you've made sure that both your u and v are positive, they will have u and v being of differing signs.
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    (Original post by Zacken)
    It doesn't matter, all they're doing is absorbing the negative into the "u" - i.e: you've made sure that both your u and v are positive, they will have u and v being of differing signs.
    hmm thanks, ok thats what I thought but my final answer is incorrect as I have a negative in my workings that shouldnt be there later on. I guess I must have to 'account' for this negative somehow?
    E.g. I have
    Let  x be distance XC.
     u \cos \theta = v \cos \phi .
    Dividing the 1st expression by 2nd,  - \frac{3}{4} \tan \theta = \tan \phi \Rightarrow - \frac{3}{4} \dfrac{4}{5-x} = \frac{7.5}{x} and rearranging I get  x= 25/3 instead of 25/7 or whatever it should be. Im guessing its just a really dumb mistake I missed!?
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    (Original post by EnglishMuon)
    hmm thanks, ok thats what I thought but my final answer is incorrect as I have a negative in my workings that shouldnt be there later on. I guess I must have to 'account' for this negative somehow?
    E.g. I have
    Let  x be distance XC.
     u \cos \theta = v \cos \phi .
    Dividing the 1st expression by 2nd,  - \frac{3}{4} \tan \theta = \tan \phi \Rightarrow - \frac{3}{4} \dfrac{4}{5-x} = \frac{7.5}{x} and rearranging I get  x= 25/3 instead of 25/7 or whatever it should be. Im guessing its just a really dumb mistake I missed!?
    Oh yeah, my bad!

    You are resolving perpendicular to the wall, so what the negative sign is really saying is v\sin \phi - \frac{3}{4} u \sin \theta = 0 - think about it like conservation of momentum.
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    (Original post by Zacken)
    Oh yeah, my bad!

    You are resolving perpendicular to the wall, so what the negative sign is really saying is v\sin \phi - \frac{3}{4} u \sin \theta = 0 - think about it like conservation of momentum.
    ah ok I see, thanks!
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    (Original post by EnglishMuon)
    ah ok I see, thanks!
    I think that's how it's shown in the textbook (if you look at the fist few pages of the chapter)?
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    Insight314 physicsmaths Dont suppose you remember doing that q6 from M4 June 05? So my argument is the standard thing that if interception occurs aVb= k i . By looking at an expression for the rel. velocity in terms of theta I get that theta either equals -arctan4/3 or arctan 4/3 not anywhere in between. Any ideas where they get that range from in the q?
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    (Original post by EnglishMuon)
    Insight314 physicsmaths Dont suppose you remember doing that q6 from M4 June 05? So my argument is the standard thing that if interception occurs aVb= k i . By looking at an expression for the rel. velocity in terms of theta I get that theta either equals -arctan4/3 or arctan 4/3 not anywhere in between. Any ideas where they get that range from in the q?
    will try now
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    (Original post by physicsmaths)
    will try now
    Thanks
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    (Original post by EnglishMuon)
    Insight314 physicsmaths Dont suppose you remember doing that q6 from M4 June 05? So my argument is the standard thing that if interception occurs aVb= k i . By looking at an expression for the rel. velocity in terms of theta I get that theta either equals -arctan4/3 or arctan 4/3 not anywhere in between. Any ideas where they get that range from in the q?
    Haven't done the question, but doesn't it have to do with the fact that A travels at maximum speed so the value of the speed of A would vary so theta would also vary since tan theta is relative speed over speed of A.


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    (Original post by Insight314)
    Haven't done the question, but doesn't it have to do with the fact that A travels at maximum speed so the value of the speed of A would vary so theta would also vary since tan theta is relative speed over speed of A.


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    Ah yes ofcourse! Sorry that was me missing out a rather important word from the question XD Thanks
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    (Original post by EnglishMuon)
    Thanks
    Name:  ImageUploadedByStudent Room1464957713.736710.jpg
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    Very different to the Mark scheme but who cares haha.


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    nice, thanks. I think I did the same but just missed out every single inequality with an equality
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    (Original post by EnglishMuon)
    nice, thanks. I think I did the same but just missed out every single inequality with an equality
    I dnt get what the **** the mark schemes going on about.
    I tend to do most questions different to the MS.



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    (Original post by physicsmaths)
    I dnt get what the **** the mark schemes going on about.
    I tend to do most questions different to the MS.



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    yeah same with me, its really offputting sometimes as the methods shown seem to vary between purely algebraic to purely geometrical for similar problems.
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    (Original post by physicsmaths)
    I dnt get what the **** the mark schemes going on about.
    I tend to do most questions different to the MS.



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    Old spec uses different textbooks so your method is different to them.


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