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Edexcel Mathematics: Mechanics M1 6677 - Wednesday 14 June 2017 [Official Thread]

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Find the unofficial markscheme HERE!

Edexcel M1 Official Thread - June 2017

Here's a thread for the best exam of them all!! Join fellow students preparing for the exam who like you are climbing slopes, raising pulleys, accelerating in lifts and playing on see-saws.





Key Information

Date: Wednesday 14th June 2017
Time: Morning
Duration: 1 hour 30 minutes


Resources



Past papers




And some of my personal M1 tips








Good Luck!!



(edited 6 years ago)

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Thank you for creating this.


Graham is in the process of creating new Bronze Silver and Gold papers, removing the occasional repeated topics and using questions upto 2016. He has done new papers for C1 - C4 so far.
Reply 3
Hi!! I'm really worried for my Alevel Maths, and i was just wondering what we should be doing right now? Textbook questions? Solomon papers? or Practice papers? And how much should we be doing, I'm doing C1 C2 C3 C4 M1 S1 this year.

Thanks
Original post by ndk123
Hi!! I'm really worried for my Alevel Maths, and i was just wondering what we should be doing right now? Textbook questions? Solomon papers? or Practice papers? And how much should we be doing, I'm doing C1 C2 C3 C4 M1 S1 this year.

Thanks


Right now I'm doing past papers and occasionally Solomon papers but I think past papers are more useful, especially the more recent ones! Often the questions in the textbook aren't exam style so if you've finished "learning" the course the textbooks aren't that useful in my opinion. :smile:
Reply 5
Are you guys doing IAL or just AS?


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Original post by Droneon
Are you guys doing IAL or just AS?


Posted from TSR Mobile


The IAL papers are written by the same team as do the ordinary AS and A2 papers, so they are definitely worth doing.
Checking in! Doing M1 as my applied module either as part of AS further maths or the full A level single
Reply 8
ImageUploadedByStudent Room1494155396.425347.jpg

Can someone pls explain part c? I don't know how to do it.
T= 1.2mg , mu=0.69


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Original post by Droneon
ImageUploadedByStudent Room1494155396.425347.jpg

Can someone pls explain part c? I don't know how to do it.
T= 1.2mg , mu=0.69

Posted from TSR Mobile

The two forces on the pulley are tension acting on the pulley. So it's basically the square root of 2T
m1 ..png

In this question, how would you determine the direction of movement of each particle after the collision.
Does particle B having a greater momentum mean that A moves in the opposite direction? but then how do you determines B's direction?
Original post by NotNotBatman
m1 ..png

In this question, how would you determine the direction of movement of each particle after the collision.
Does particle B having a greater momentum mean that A moves in the opposite direction? but then how do you determines B's direction?


You can assume the direction of the particles after the collision. Lets say you assume they both go to the right but then you calculate the velocity of A to be negative, then the maths shows that A was in fact reversed, since velocity is a vector and you assumed the direction to the right was positive.
(edited 6 years ago)
Original post by mary.athaide
The two forces on the pulley are tension acting on the pulley. So it's basically the square root of 2T


But the force diagram is not a right angled triangle.


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Original post by Droneon
ImageUploadedByStudent Room1494155396.425347.jpg

Can someone pls explain part c? I don't know how to do it.
T= 1.2mg , mu=0.69


Posted from TSR Mobile


Can someone pls help?


Posted from TSR Mobile
Original post by hellomynameisr
You can assume the direction of the particles after the collision. Lets say you assume they both go to the right but then you calculate the velocity of A to be negative, then the maths shows that A was in fact reversed, since velocity is a vector and you assumed the direction to the right was positive.


Yes, I've tried the different ways, it all works, thanks.
Original post by Droneon
But the force diagram is not a right angled triangle.


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my bad- I got confused with physics, best thing here is a force vector triangle, then cosine rule
Original post by Droneon
Can someone pls help?


Posted from TSR Mobile


Both sides of the string pull diagonally down on the pulley. Their horizontal components cancel out, with one going to the left and one to the right. So the resultant force on the pulley is their 2 vertical components, i.e. 2Tcos 60 vertically downwards.
(edited 6 years ago)
Original post by tiny hobbit
Both sides of the string pull diagonally down on the pulley. Their horizontal components cancel out, with one going to the left and one to the right. So the resultant force on the pulley is their 2 vertical components, i.e. 2Tcos 60 vertically downwards.


Oh okay. It also says give the magnitude and direction, but T is in terms of mg, so what is the magnitude? And the direction?


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Original post by mary.athaide
my bad- I got confused with physics, best thing here is a force vector triangle, then cosine rule


Yeah I did that but didn't know how to get the magnitude. Do I use 1.2mg and square it while applying the cosine rule or does mg get cancelled somewhere along the way?


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Reply 19
Original post by Droneon
Can someone pls help?


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For resultant force questions e.g. resultant force on a pulley, there's always the option to draw a vector diagram i.e. a triangle of forces. If two forces acting are F1\mathbf{F_1} and F2\mathbf{F_2} then the resultant force is F1+F2\mathbf{F_1} + \mathbf{F_2}. So you can draw the force vectors head to tail in a triangle:



It's not hard to show that this is an equilateral triangle so the magnitude of the resultant must be equal to TT.

I always like this method but students are often not keen on it for some reason :smile:

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