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Edexcel - M3 - 18th May 2016

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Guys I don't understand how to go about June 2004 4.) b in this link:

https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/June%202004%20QP%20-%20M3%20Edexcel.pdf
ImageUploadedByStudent Room1462539560.650056.jpg

http://qualifications.pearson.com/content/dam/pdf/A%20Level/Mathematics/2013/Exam%20materials/M3_Jan_2007_Paper.pdf

Can someone explain the geometry of 2 to me? I can get that the c.o.m acts at 3a/4 above base but then how do I work out theta maybe I'm being retarded idk
(edited 7 years ago)
Original post by J.M.Keynes


You now need to include the tension in the string in your horizontal equation. If the disc whizzes round too fast, the particle is likely to fly outwards so the friction is acting inwards.

If the disc goes round too slowly, the tension in the elastic string is likely to pull the particle inwards, so the friction will be acting outwards.

You can find the tension in the string using Hooke's law because you know where the particle is and so how long the string is.
Original post by BBeyond
ImageUploadedByStudent Room1462539560.650056.jpg

http://qualifications.pearson.com/content/dam/pdf/A%20Level/Mathematics/2013/Exam%20materials/M3_Jan_2007_Paper.pdf

Can someone explain the geometry of 2 to me? I can get that the c.o.m acts at 3a/4 above base but then how do I work out theta maybe I'm being retarded idk


You'd have helped yourself if your diagram was a bit more precise.

The angle between the vertical line you've drawn from A and the axis of the cone (dotty on the original diagram) is theta. So the angle that you are working with is 90 - theta.
Original post by tiny hobbit
You'd have helped yourself if your diagram was a bit more precise.

The angle between the vertical line you've drawn from A and the axis of the cone (dotty on the original diagram) is theta. So the angle that you are working with is 90 - theta.


Oh I see cheers yh my diagram was pretty awful didn't help at all :tongue:
Original post by BBeyond
ImageUploadedByStudent Room1462539560.650056.jpg

http://qualifications.pearson.com/content/dam/pdf/A%20Level/Mathematics/2013/Exam%20materials/M3_Jan_2007_Paper.pdf

Can someone explain the geometry of 2 to me? I can get that the c.o.m acts at 3a/4 above base but then how do I work out theta maybe I'm being retarded idk


That diagram xD

Original post by tiny hobbit
You'd have helped yourself if your diagram was a bit more precise.

The angle between the vertical line you've drawn from A and the axis of the cone (dotty on the original diagram) is theta. So the angle that you are working with is 90 - theta.

;-; a sec too l8
Original post by Imperion
That diagram xD


;-; a sec too l8


Mate my diagrams are world class shhhh
Original post by BBeyond
Mate my diagrams are world class shhhh


Better than mine tbh
Original post by samb1234
Better than mine tbh


Tbf u don't need a decent diagram for most questions
Original post by BBeyond
Mate my diagrams are world class shhhh


LOL I thought this was a triangular lamina. :lol:
****ing sorcery.
M3 2004 june q2.
Why can't i resolve for tensions since it is at rest?
Is it because it isnt in equilibrium and will spring back up after the instant.


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Original post by physicsmaths
****ing sorcery.
M3 2004 june q2.
Why can't i resolve for tensions since it is at rest?
Is it because it isnt in equilibrium and will spring back up after the instant.


Yeah, it needs to be in equilibrium for you to use something like "mg = 2T". When it reaches its lowest point, the tensions will be greater than the weight (hence why it moves up again) so you'd need to know the acceleration to use F = ma; that's M4 stuff! :goofy:
Original post by ombtom
Yeah, it needs to be in equilibrium for you to use something like "mg = 2T". When it reaches its lowest point, the tensions will be greater than the weight (hence why it moves up again) so you'd need to know the acceleration to use F = ma; that's M4 stuff! :goofy:


Thought so. I always make mistakes in M3 but never M4.


Posted from TSR Mobile
Original post by physicsmaths
Thought so. I always make mistakes in M3 but never M4.


Wish I could say the same... :lol:
Could anyone help me on question 3 June 2005: https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/June%202005%20QP%20-%20M3%20Edexcel.pdf
Surely when one string is taut, the other is slack and hence there is only one elastic energy. However the mark scheme talks about 4 different elastic energies.
Original post by J.M.Keynes
Could anyone help me on question 3 June 2005: https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/June%202005%20QP%20-%20M3%20Edexcel.pdf
Surely when one string is taut, the other is slack and hence there is only one elastic energy. However the mark scheme talks about 4 different elastic energies.


AB is 4l. The natural length of the string is 2l, so when you release the particle, both parts of the string are stretched and will remain so.
Original post by 1asdfghjkl1
can someone explain the method to use for question 2 from the exam style paper in the textbook?


If you send a picnof it I can since i don't have the text book anymore.


Posted from TSR Mobile
Is there a M2 thread?
I'm finding the second half of the centre of mass chapter quite difficult.

Any tips?
Could someone explain second half of example 17 and 19 please. I'm so confused.

EDIT: Centre of mass chapter.
(edited 7 years ago)

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