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

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Original post by MJS107
Could you perhaps post your working? I've seen that in the mark schemes a couple of times as well but resolving parallel and perpendicular should still give you the correct answer it may just be a longer method for this particular question.


IMG_3366.jpg

I did this
Original post by muctarrio97
In honesty...the only topic that I think would be a nightmare has to be vectors...though in my case, my teacher really sucked with the topic.


omg same ugh
Original post by Ahmed3651998
Is this thread for GCE M1 and/or IAL M1 ??
doesnt matter the subjects are same though......


arent they like the same lol
Original post by SnowLeopard27
IMG_3366.jpg

I did this


I want to be very honest with you, but that diagram is not the best. You need to be very careful with how you do your diagram, as not doing it properly can cost you a LOT of marks in mechanics. Your lines of action are what I assume is confusing you.
Original post by SnowLeopard27
Can anyone explain 5b on this paper?http://qualifications.pearson.com/content/dam/pdf/A%20Level/Mathematics/2013/Exam%20materials/6677_01_que_20090113.pdf

I have been taught to resolve parallel then perpendicular, which is what I did, but I didn't get the right answers and the mark scheme is for resolving vertical/horizontal. This has happened on another paper too, so I don't know what I'm doing wrong.

Thanks anyone:smile:


The mark scheme shows the best way of doing this problem which is to resolve vertically using angles. But the long method is to resolve using the lines of action perpendicular and parallel to the package.
Original post by Marxist
I want to be very honest with you, but that diagram is not the best. You need to be very careful with how you do your diagram, as not doing it properly can cost you a LOT of marks in mechanics. Your lines of action are what I assume is confusing you.


Thanks for the honesty:wink: so what have I done wrong precisely? Is it the working as opposed to the method? Thanks :smile:
Original post by SnowLeopard27
Thanks for the honesty:wink: so what have I done wrong precisely? Is it the working as opposed to the method? Thanks :smile:


Yes, I just want to point out that you have to do the basics a lot better than that. The basics are what will get you to any answer in dynamics and equilibrium type questions - there's not much else to be said.

This video should help a lot with what I pointed:

https://www.youtube.com/watch?v=VfRc3fDdF28
Original post by Marxist
Yes, I just want to point out that you have to do the basics a lot better than that. The basics are what will get you to any answer in dynamics and equilibrium type questions - there's not much else to be said.

This video should help a lot with what I pointed:

https://www.youtube.com/watch?v=VfRc3fDdF28


Woah ok. I've been trying so hard in mechanics and finally predicted an A in it despite not doing/liking physics, so generally I think my 'basics' aren't that bad:wink:

Just this particular style of question, with the extra force 'P' (or X or whatever) is throwing me recently. The video is quite useful but does anyone know what exactly I have done wrong?! Lol if not I will ask my teacher after half term so no worries:smile:
Original post by SnowLeopard27
Woah ok. I've been trying so hard in mechanics and finally predicted an A in it despite not doing/liking physics, so generally I think my 'basics' aren't that bad:wink:

Just this particular style of question, with the extra force 'P' (or X or whatever) is throwing me recently. The video is quite useful but does anyone know what exactly I have done wrong?! Lol if not I will ask my teacher after half term so no worries:smile:


Sorry if I'm being counter-productive, my initial assessment made me think that you may have just needed a video to provide you with more understanding.

In a previous reply, I explained how you would do it. You can either use simultaneous equations - which is not a good method to eliminate P to find R - then substitute the value of R back into one of your two equations. But the best method for this is to resolve vertically to find R and finding P should be easy from there.
Original post by SnowLeopard27
Woah ok. I've been trying so hard in mechanics and finally predicted an A in it despite not doing/liking physics, so generally I think my 'basics' aren't that bad:wink:

Just this particular style of question, with the extra force 'P' (or X or whatever) is throwing me recently. The video is quite useful but does anyone know what exactly I have done wrong?! Lol if not I will ask my teacher after half term so no worries:smile:


Negative sign of p when you flipped it for your resolving down the plane
Original post by samb1234
Negative sign of p when you flipped it for your resolving down the plane


Oh I see it! Thanks so much:smile: So silly - never see stuff like that when I look back. Thanks again!
Original post by SnowLeopard27
Oh I see it! Thanks so much:smile: So silly - never see stuff like that when I look back. Thanks again!


Do it what ever way suits you, but I am pointing out to you that the easiest way would be resolve vertically. You still get the same answer of course, but it's just a lot easier and less time-consuming to do that.
Original post by Marxist
Do it what ever way suits you, but I am pointing out to you that the easiest way would be resolve vertically. You still get the same answer of course, but it's just a lot easier and less time-consuming to do that.


I'm sure you're right, but unfortunately I haven't been taught that way, so will stick with what I know. Thanks for your help anyways :smile:
Original post by SnowLeopard27
I'm sure you're right, but unfortunately I haven't been taught that way, so will stick with what I know. Thanks for your help anyways :smile:


Well, that's okay.It just saves you a lot of time. And it will come in handy to know your angles well.
Original post by notnek
Both trains start from the same station and end at the same station so the total distance travelled by both trains is the same.

This means that the areas under the graphs for the 2 trains will be the same.

First find the area under the train A graph (the trapezium).

If you get stuck, post all your working / ideas.


Perfect thatnks this will help me a lot I imagine.
Original post by pondsteps
arent they like the same lol


they are 100% same but the EXAMINATION SCRIPTS are 100% different !!
https://534ff63a01f6cb44c9c4e69801769b0628cd58d1.googledrive.com/host/0B1ZiqBksUHNYdG9SOEVLaTVSYUE/CH6.pdf


For page 21. Exd q4

do we say x degress above the vertical similar to how we talk about the horizontal?
A particle A starts at the point with position vector 12i+12j. The initial velocity of A is (-i+j) ms^-1, and it has constant acceleration (2i-4j)ms^-2. Another particle B has initial velocity i ms^-1 and constant acceleration 2j ms^-2. After 3 seconds the two particles collide.

Find
A. The position vector of the point where the two particles collide
B. The position vector of B at the starting point

=========

I know where in the case velocity is constant the position vector is (initial position vector +vt) but in this case there is an acceleration and I dont know how to take it into account

This is supposed to be done without any calculus by the way
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Avatar for candol
candol
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Original post by L'Evil Wolf
A particle A starts at the point with position vector 12i+12j. The initial velocity of A is (-i+j) ms^-1, and it has constant acceleration (2i-4j)ms^-2. Another particle B has initial velocity i ms^-1 and constant acceleration 2j ms^-2. After 3 seconds the two particles collide.

Find
A. The position vector of the point where the two particles collide
B. The position vector of B at the starting point

=========

I know where in the case velocity is constant the position vector is (initial position vector +vt) but in this case there is an acceleration and I dont know how to take it into account

This is supposed to be done without any calculus by the way

Step1. Find the distance travelled by A in terms of i and j using suvat
s=ut +1/2at^2
=3(-i+j) + .5(2i-4j)*3^2
=-3i+3j+9i-18j
=6i+15j
Add this to the original position of 12i+12j to get 18i+33j
Part B is simply position vector eqn using this result as you have v and t
what paper?

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