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

ok question 7 part b, what am i doing wrong. i seem to be getting a different answer to everyone else so lets hear it!

ok ive done this twice now (once in the exam, once again now) and ive got the same answer which is different to what I have been seeing.

V= U + AT
V = ? U = (3i-22j) A=(3i+9j) T=3
|U| = root(3^2 + (-22)^2) = root(493)
|A| = root(3^2 + 9^2) = root(90)

V = root(493) + 3*root(90)


V= 50.66410225
V = 50.7


anyone?

was i not meant to find the magnitude until the end or something?

Ye you find the magnitude at the end. This should explain why to you:

V = U + AT
|V| = |U + AT|
What you did was:
|V| = |U| + |AT|.

This is incorrect as |a+b| is not the same as |a| + |b|
You will learn more about modulus when you do c3 and c4. I just happen to be an early entry but I can assure you of the above statement.

but if you get the final answer you get all the marks dont you?

I put AS = d but I meant A to centre of mass = d




Here you go. This is how I did it.

yeah that makes sense now that I think about it! thank you for confirming, Im kinda annoyed with myself that I didn't come across this during my revision, anyway im gonna copy + paste this reply from before, could you give me your opinion on it?


Okay so I am finally adding up my marks, and I just have one question about how many marks I would get.

For question 7b you had to find the magnitude of the velocity at t=3 so V = (whatever) +(fsdfds)3

i did that however, i found the magnitude before adding and multiplying so i got the wrong answer and i was wondering how many marks you guys think I would get.

i tried to compare to previous similar questions however this one was unique, the closest i could find are

question 7 part a:
https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

question 6 part c:
https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

question 6 part a:
https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

thanks.

I put AS = d but I meant A to centre of mass = d




Here you go. This is how I did it.

You would receive no more than 1 mark, if any.
This 1 mark depends on the mark scheme. If it gives a mark for correct identification of which SUVAT equation to use and attempting to use it then you may receive 1 mark. Unfortunately you dropped both method and answer marks and since this question was independent of the first part there are also no ECF marks to be obtainted.

I would like to address a misconception that some people are having.
It is regarding the question where you had to find the resultant force of the string on the pulley and the direction of it.
Most people got the magnitude correct via Pythagoras' Theorem. However the issue comes with the direction part.
Most people managed to figure out the angle was 45 degrees but decided to express this as a bearing and used 225 degrees.

Now here is where you go wrong. Do NOT use bearings for these sorts of questions. Bearings give the direction relative to the North Line. There is no specified north line in this question.
Why can we use bearings in vectors? Well at the beginning of vector questions it clearly states that vectors i and j represent east and NORTH respectively.
In this scenario we aren't told that the vertical represents the north line. In fact, if we used common sense then a vertical to a table (given it is upright) would point upwards to the sky, and not to North.
The direction would be: inclined downwards at an angle of 45 degrees to the horizontal/table.

I would like to address a misconception that some people are having.
It is regarding the question where you had to find the resultant force of the string on the pulley and the direction of it.
Most people got the magnitude correct via Pythagoras' Theorem. However the issue comes with the direction part.
Most people managed to figure out the angle was 45 degrees but decided to express this as a bearing and used 225 degrees.

Now here is where you go wrong. Do NOT use bearings for these sorts of questions. Bearings give the direction relative to the North Line. There is no specified north line in this question.
Why can we use bearings in vectors? Well at the beginning of vector questions it clearly states that vectors i and j represent east and NORTH respectively.
In this scenario we aren't told that the vertical represents the north line. In fact, if we used common sense then a vertical to a table (given it is upright) would point upwards to the sky, and not to North.
The direction would be: inclined downwards at an angle of 45 degrees to the horizontal/table.

I would like to address a misconception that some people are having.
It is regarding the question where you had to find the resultant force of the string on the pulley and the direction of it.
Most people got the magnitude correct via Pythagoras' Theorem. However the issue comes with the direction part.
Most people managed to figure out the angle was 45 degrees but decided to express this as a bearing and used 225 degrees.

Now here is where you go wrong. Do NOT use bearings for these sorts of questions. Bearings give the direction relative to the North Line. There is no specified north line in this question.
Why can we use bearings in vectors? Well at the beginning of vector questions it clearly states that vectors i and j represent east and NORTH respectively.
In this scenario we aren't told that the vertical represents the north line. In fact, if we used common sense then a vertical to a table (given it is upright) would point upwards to the sky, and not to North.
The direction would be: inclined downwards at an angle of 45 degrees to the horizontal/table.

you know what? i'm happy they didnt put any vertical motion questions in there.

I agree with this. However, if they have stated that the angle is measured clockwise from the upwards vertical then it should be fine. But a bearing alone is a bit risky.

I find it a bit odd they put in two vector questions tho lmao. I think that is the first paper I've done with no vertical motion maybs they forgot