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Edexcel AS/A2 Mathematics M2 - 17th June 2016 - Official Thread

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Original post by callum9977
how many marks lost if I integrated between 3 and 0 instead of 3 and 2?


Same i think 2 maybe 3 marks
which question did you have to do by using the work energy principle? I dont remember seeing one but apparently there was? if so - got the correct answer using a different method - how many would i have lost?
Thank you, but still really don't understand what's going on tbh I don't even understand how come position vector can be minus(-)

Could you please go through more details? Or if you are going to solve in on paper and post the picture I will wait for it

Thank you very much in advance
Original post by SHJBHB
Has to be 3sf in mechanics doesn't it, or can you get away with 4sf or greater for all questions?


it's whenever you use g=9.8

If you use g=9.8 then you have to have the answer as 2 or 3 s.f., if you don't you can have it to more.
Original post by callum9977
how many marks lost if I integrated between 3 and 0 instead of 3 and 2?


I did the same thing, that question was so annoyingly worded!

I reckon that I lost 3 marks to be honest, I would have got a couple for integrating I guess. Maybe I lost 4 marks since I didn't have much working.
Poll for the UK paper: http://www.thestudentroom.co.uk/showthread.php?t=4171568&p=65871248#post65871248
Original post by mani119
Same i think 2 maybe 3 marks


probably 10 marks mate
Reply 467
Avatar for Nerrad
Nerrad
7
Original post by Cong
How did you got e in q7


Centre of mass angle was 114 degrees
Anyone know how many marks each question was?
Original post by ombtom
They are, but looking at June 2012, the questions are quite similar. That year was 57 for 80 UMS. :eek:


That paper is far easier than the one we just sat! Famous last words maybe but I'm predicted a sub 58 80 UMS boundary and a sub 67 90 UMS boundary.
Reply 470
Avatar for LLk1
LLk1
1
am i the only one who made the impulse negative? to get e=4/7 as it stated in the question it was the magnitude of the impulse.
Original post by saarah_soldier
which question did you have to do by using the work energy principle? I dont remember seeing one but apparently there was? if so - got the correct answer using a different method - how many would i have lost?


It was the one with the car on the slope - I used suvat first and then double checked and it said use the work energy principle! I don't think you'd get any marks because you didn't do what the question asked :frown:
Reply 472
Avatar for RuariJp
RuariJp
How many marks was the question when you had to find e and say is the collided again ?
Original post by angelitajin
Thank you, but still really don't understand what's going on tbh I don't even understand how come position vector can be minus(-)

Could you please go through more details? Or if you are going to solve in on paper and post the picture I will wait for it

Thank you very much in advance


Because it's in terms of vectors you start at the origin (0i+0j). There's no ground so when the particle goes up and back down it'll pass through the i vector (x axis). This means that'll it'll go into the -j region, and as A was at lambda(i-j), the position vector of A is (lambda i, -lambda j). I don't really have time to go through the question again right now but that's how the position vectors work. You then have to work through using displacement and time in terms of lambda (it goes to (lambda i, -lambda j) so the i displacement will be lambda and j will be -lambda).
Reply 474
Avatar for linaxq
linaxq
2
Original post by ombtom
They are, but looking at June 2012, the questions are quite similar. That year was 57 for 80 UMS. :eek:

would love for that to be the gb xD
Original post by Harriet3006
Anyone know how many marks each question was?


1
a) 8
b) 5
2
a) 6
b) 4
3
a) 6
4
a) 5
b) 4
5
a) 8
b) 4
6
a) 6
b) 7
7
a) 7
b) 3
c) 2
Reply 476
Avatar for linaxq
linaxq
2
http://www.strawpoll.me/10508393/r Please be honest so we can have a clear indication of what the grade boundaries may be :smile:
Original post by LLk1
am i the only one who made the impulse negative? to get e=4/7 as it stated in the question it was the magnitude of the impulse.


You could do it either way. I used the impulse as negative because I took the velocity afterwards as negative. This still comes out as e=1/14.
Seemed alright, looking at this thread does inspire confidence. The wording in 1(b) threw me off though and I integrated between t=0 and t=3. Probably a loss of 2-3 marks if they give marks for the method and integration. In this question I found the determinant just to be sure that it didn't cross the t-axis, so maybe that's a method mark too?
Original post by zandneger
lambda(i-j) = (lambda)i+(-lambda)j
Since you know horizontal speed is constant, using speed = distance/time you get
distance(horizontal) = 3t
Hence, lambda = 3t and -lamba = -3t
So you want the value for t when displacement(vertical) = -3t
s=-3t u=4 v=/ a=-9.8 t=t
s=ut+1/2at^2
-3t=4t-4.9t^2
7t-4.9t^2=0
t(7-4.9t)=0
s=-3t when t=0 or when t=7/4.9
7/4.9 = 10/7
hence lambda = 3t = 30/7


I write lambda equals 42/g would I lose any marks for writing it in that form.

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