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Official M4 Edexcel Thread – Wednesday 15th June 2016

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Original post by Euclidean
Is this the best method to show that two objects cannot intercept eachother?


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A proof by contradiction didn't take too much work
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@A Slice of Pi @Euclidean isn't it easiest just to look at the j components only? I mean it's like a 1 line thing which immediately provides a contradiction. Basically Euclideans method but without the need for writing out the full relative velocity expression.
Original post by A Slice of Pi
A proof by contradiction didn't take too much work


This is a nice method :smile:

Original post by EnglishMuon
@A Slice of Pi @Euclidean isn't it easiest just to look at the j components only? I mean it's like a 1 line thing which immediately provides a contradiction. Basically Euclideans method but without the need for writing out the full relative velocity expression.


Probably could have saved myself some time yeah :redface:
Original post by Euclidean
Is this the best method to show that two objects cannot intercept eachother?

I think trying to equate position vectors is a little more complicated, I probably wrote more lines than necessary for this too

Tagging @physicsmaths cause I heard he's got sik geometry skills

Tagging @A Slice of Pi cause mechanics wizard

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Dude. if i was the examiner and ended up marking your paper. i'd give you full marks just for that buffting handwriting. pls explain how you do that.

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Original post by ChuckNorriss
Dude. if i was the examiner and ended up marking your paper. i'd give you full marks just for that buffting handwriting. pls explain how you do that.

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:lol:

It's all intended to distract the examiner from the fact that I don't have a f***in clue what I'm doing :redface:
Wot.
M4 June 2006, Q6a.
Isnt there an impulse horizontal and vertical in this case? And they are different masses so it must be the case that they are bot unchanged.


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Nvm Im retarded.


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Original post by euclidean
:lol:

It's all intended to distract the examiner from the fact that i don't have a f***in clue what i'm doing :redface:


looooooooool
Original post by physicsmaths
Nvm Im retarded.


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I think we can all agree this isn't true :lol: :lol:
On Q6 June 2005, the last A mark of part (a) credits an explanation for the range of values of theta. I don't understand why theta has a range of values rather than just one value of arctan(0.75).

Does anyone know why?

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Original post by Euclidean
On Q6 June 2005, the last A mark of part (a) credits an explanation for the range of values of theta. I don't understand why theta has a range of values rather than just one value of arctan(0.75).

Does anyone know why?

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that's question 13 from rev ex 1. solution bank should explain it properly. i flopped that question myself tbh.



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Original post by Euclidean
On Q6 June 2005, the last A mark of part (a) credits an explanation for the range of values of theta. I don't understand why theta has a range of values rather than just one value of arctan(0.75).

Does anyone know why?

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The question says maximum value is 30km/h. So there is a range of values of Ship A with a corresponding value of theta that will make the two ships intercept. Maybe you have to find an equation of the velocity of A , call it V(A), as a function of theta and find a maximum of theta since we know max velocity of A is 30km/h? I will try and see myself.
Original post by Pyslocke
The question says maximum value is 30km/h. So there is a range of values of Ship A with a corresponding value of theta that will make the two ships intercept. Maybe you have to find an equation of the velocity of A , call it V(A), as a function of theta and find a maximum of theta since we know max velocity of A is 30km/h? I will try and see myself.


Ah my brain skimmed over this when I read the question!

Thanks
Original post by Euclidean
On Q6 June 2005, the last A mark of part (a) credits an explanation for the range of values of theta. I don't understand why theta has a range of values rather than just one value of arctan(0.75).

Does anyone know why?

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I did this question last week when english muon brought it up. It is on his blog thing since i can't upload it here since i can't find he picture haha.
Basically V<=30 and some manipulation gets the result after sin rule. Now two poistions on either side for intersections.


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Here it is
ImageUploadedByStudent Room1465568123.939738.jpg


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Original post by physicsmaths
I did this question last week when english muon brought it up. It is on his blog thing since i can't upload it here since i can't find he picture haha.
Basically V<=30 and some manipulation gets the result after sin rule. Now two poistions on either side for intersections.


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@EnglishMuon


Would you mind sharing a link?

I see now, I didn't see how to formulate the inequality. Nice one m8 :biggrin:
Original post by physicsmaths


PRSOM
Original post by Euclidean
Ah my brain skimmed over this when I read the question!

Thanks


I just realized I'm overthinking things again :s-smilie:. I guess you just just make two triangles and the second triangle will give -arctan(4/3) where theta is 306.89 degrees due north.
Original post by physicsmaths


Or this :biggrin:
Original post by Euclidean
@EnglishMuon


Would you mind sharing a link?

I see now, I didn't see how to formulate the inequality. Nice one m8 :biggrin:


Oh yea basically the thing I missed is that the question says the Maximum speed is .... and I missed that part so I only got the two endpoint values for theta only :smile: So since we know its velocity is less than or equal to the value given, the angle can lie anywhere in that range.

Full convo is at the bottom the page http://www.thestudentroom.co.uk/showthread.php?t=3973271&page=8

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