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Edexcel AS/A2 Mathematics M1 - 8th June 2016 - Official Thread

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For this question, why is it not possible to just say x + 9 = 41 --> X = 41 - 9 = 32N? I thought resultant forces are the sum of the other forces :/?

kik.JPG
Original post by KloppOClock
c2 was pretty sound altho i messed up the discrimination question on c1 :/

you?


I'm in A2 so I've got C3 and 4 this year, along with M1. I just have to do well in M1 tomorrow to make life easier in C4 :biggrin:.

I'm just working through Gold papers today, I don't know what else I can do...
Reply 642
Original post by Don Pedro K.
For this question, why is it not possible to just say x + 9 = 41 --> X = 41 - 9 = 32N? I thought resultant forces are the sum of the other forces :/?

kik.JPG


the forces are not acting in the same direction. You need to draw a force diagram (triangle in this case), showing the forces in series (one after the other). The resultant is the diagonal joining the start point to the end point
Good luck for tomorrow guys :biggrin: I'm shitting it as always!!
Original post by candol
the forces are not acting in the same direction. You need to draw a force diagram (triangle in this case), showing the forces in series (one after the other). The resultant is the diagonal joining the start point to the end point


Ah alright, thanks :smile:

Also, is it just me being stupid (again) or is the equation for solving (-->) in this solution wrong? Surely it should be
F + Rcos(60) = 30cos(30)?

kik.JPG
Reply 645
Original post by Don Pedro K.
Ah alright, thanks :smile:

Also, is it just me being stupid (again) or is the equation for solving (-->:wink: in this solution wrong? Surely it should be
F + Rcos(60) = 30cos(30)?

kik.JPG


that's the same thing
Original post by candol
that's the same thing


But in their equation, F - 30cos(30) = Rcos(60) so F = Rcos(60) + 30cos(30)

but if it was F + Rcos(60) = 30cos(30) to begin with, F would = 30cos(30) - Rcos(60), which isn't the same as above?
Reply 647
Original post by Don Pedro K.
But in their equation, F - 30cos(30) = Rcos(60) so F = Rcos(60) + 30cos(30)

but if it was F + Rcos(60) = 30cos(30) to begin with, F would = 30cos(30) - Rcos(60), which isn't the same as above?


quite right its not - do apologise. wasn't paying full attention. There equation looks right, but why would you draw 2 diagrams for a question?
Original post by candol
quite right its not - do apologise. wasn't paying full attention. There equation looks right, but why would you draw 2 diagrams for a question?


Hmm I'm not sure :/ I have a feeling that it's wrong but I don't know to be honest! @Zacken could you have a look?
Reply 649
Original post by Don Pedro K.
Hmm I'm not sure :/ I have a feeling that it's wrong but I don't know to be honest! @Zacken could you have a look?

Have you got the original question - the diagram is fine, (it is BASIC forces) but you really should have everything on the one diagram. It is not obvious to me what the original question was (cant be statics, it would be moving down??)
Original post by Don Pedro K.
Hmm I'm not sure :/ I have a feeling that it's wrong but I don't know to be honest! @Zacken could you have a look?


It's correct.

If you resolve to the right, you get F - 30 cos 30.

This is the resultant force to the right. The resultant force to the right is also given by R cos 60

So F - 30 \cos 30= R \cos 60.

They've been drawn on different diagrams for a reason, you're treating them as though they are the same diagram.
Screen Shot 2016-06-07 at 13.15.24.png sorry for the crap quality but does anyone know how to do this question? I can't find a mark scheme for it anywhere
Original post by Zacken
It's correct.

If you resolve to the right, you get F - 30 cos 30.

This is the resultant force to the right. The resultant force to the right is also given by R cos 60

So F - 30 \cos 30= R \cos 60.

They've been drawn on different diagrams for a reason, you're treating them as though they are the same diagram.


I see, I see. Why do we use two separate diagrams then?
Original post by Strom
Good luck for tomorrow guys :biggrin: I'm shitting it as always!!


Ayy, optimism ftw

Original post by candol
the forces are not acting in the same direction. You need to draw a force diagram (triangle in this case), showing the forces in series (one after the other). The resultant is the diagonal joining the start point to the end point


Post 666 and you mention triangle :teehee:
Reply 654
Original post by bertolli1
Screen Shot 2016-06-07 at 13.15.24.png sorry for the crap quality but does anyone know how to do this question? I can't find a mark scheme for it anywhere


part a) suvat
s=h
A=-9.8
t=5
u=20

partb)
Find V using suvat for the first ball
You can now work out 3/4 v for the second ball
Do suvat with v=3/4V, u=w, a=9.8, s= height found from part a
Original post by bertolli1
Screen Shot 2016-06-07 at 13.15.24.png sorry for the crap quality but does anyone know how to do this question? I can't find a mark scheme for it anywhere


For (a)

s = -h, u = 20, v = v, a = -9.8, t = 5

Shove into suvat and solve for h.

Original post by Don Pedro K.
I see, I see. Why do we use two separate diagrams then?


I wouldn't use two diagrams. I'd just say resultant to the right is F - 30 cos 30 and to the bottom is 30 sin 30 or w/e and work from there without using R.
Original post by candol
part a) suvat
s=h
A=-9.8
t=5
u=20



Check your s.
Reply 657
Original post by Zacken
Check your s.


true, but as its a distance (not a displacement in this case) it can only have magnitude - but I agree, strictly it should be negative
Original post by Don Pedro K.
What is this from?


Solomon Paper J
Original post by candol
true, but as its a distance (not a displacement in this case) it can only have magnitude - but I agree, strictly it should be negative


No, it's displacement. You will get the wrong answer if you do not use a negative.

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