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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 34usinθ=vsinϕ - \frac{3}{4} u \sin \theta = v \sin \phi (where θ,ϕ \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 uu and vv are positive, they will have uu and vv being of differing signs.
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 uu and vv are positive, they will have uu and vv 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 x be distance XC.
ucosθ=vcosϕ u \cos \theta = v \cos \phi .
Dividing the 1st expression by 2nd, 34tanθ=tanϕ3445x=7.5x - \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 x= 25/3 instead of 25/7 or whatever it should be. Im guessing its just a really dumb mistake I missed!?
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 x be distance XC.
ucosθ=vcosϕ u \cos \theta = v \cos \phi .
Dividing the 1st expression by 2nd, 34tanθ=tanϕ3445x=7.5x - \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 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 vsinϕ34usinθ=0v\sin \phi - \frac{3}{4} u \sin \theta = 0 - think about it like conservation of momentum.
Original post by Zacken
Oh yeah, my bad!

You are resolving perpendicular to the wall, so what the negative sign is really saying is vsinϕ34usinθ=0v\sin \phi - \frac{3}{4} u \sin \theta = 0 - think about it like conservation of momentum.


ah ok I see, thanks!
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)?
@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?
(edited 7 years ago)
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 :smile:
Original post by physicsmaths
will try now :smile:


Thanks :smile:
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 :smile:
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 :tongue:
Original post by EnglishMuon
nice, thanks. I think I did the same but just missed out every single inequality with an equality :tongue:


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.
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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