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

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Reply 440
Avatar for USB111
USB111
3
I think i put wrong fprces for the the moments question what were the right forces
Original post by TKA1998
For question 1b) it said in the THIRD second therefore would the boundaries for the integral not be t=2 and t=3 meaning you would get 34/3 or 11.66 as the distance


The distance was 11/3 though.
Reply 442
Avatar for Cong
Cong
Original post by Mattematics
My answers:
Q1) Acceleration: 4 ms^-2 (p = 2, q = -8)
Distance = 11/3 m

Q2) R = 784
P = 4900
d = 58.1 m

Q3) c = 10/3

Q4) Distance below AD = 0.830
Couldn't work out the angle

Q5) R = 5g - something cos^2theta
F = something * dg*costheta*sintheta

d = ~1.94

Q6) lambda = 30/7
v = 10.4 ms^-1
73.3 degrees below horizontal

Q7) e = 1/14
No collision because they're moving the same speed in the same direction


How did you got e in q7
Original post by TKA1998
For question 1b) it said in the THIRD second therefore would the boundaries for the integral not be t=2 and t=3 meaning you would get 34/3 or 11.66 as the distance


The first part of what you've said is true but the answer was 11/3 not 34/3
Original post by cjlh
Began at 0, projected with velocity (3i-4j) and ended up at A which had position vector lambda(i-j)


Gutted, thought it said i+j, that's why. Oh well, cheers
Original post by TKA1998
For question 1b) it said in the THIRD second therefore would the boundaries for the integral not be t=2 and t=3 meaning you would get 34/3 or 11.66 as the distance


Integrating 2t^2 - 8t + 11 and then using the limits 2 and 3 gives you 11/3m (This is what I did in the exam).
Reply 446
Avatar for USB111
USB111
3
I also was baffled on the lambda question
Reply 447
Avatar for linaxq
linaxq
2
Original post by edothero
Grade boundaries:
I reckon

66 - 90 UMS
58 - 80 UMS
51 - 70 UMS


those are pretty low no??
Original post by angelitajin
I still don't understand how to do question 6 got completely wrong the lambda question
Since the exam ended I don't really care how to solve it (good bye m2 Let's never meet again ) but then again I want to know how many marks I can pick up hopefully few method marks
Can anybody please explain me how to do that question?


I said a few pages back:

Original post by cjlh
suvat

horizontal: s=lambda
vertixcal: s=-lamda

Fill in the other knowns, find t in terms of lambda using the horizontal (t=1/3 lambda) and then solve substituting this into s=ut+1/2 at^2 with the vertical values

If you need it cleared up I can go into more detail

Original post by Jakester11
Gutted, thought it said i+j, that's why. Oh well, cheers


No problem, sucks but hopefully you'll score the Ms.
(edited 7 years ago)
Original post by Jakester11
Ahh was the position vector lambda(i+j) or lambda (i-j)


Lambda(I-J)
how many marks lost if I integrated between 3 and 0 instead of 3 and 2?
does anyone know how many marks i would lose if for the question with 3i+4j i used 3 for vertical and 4 for horizontal but all my working out was right? out of all the 12 marks?
(edited 7 years ago)
Reply 452
Avatar for linaxq
linaxq
2
Original post by zandneger
Looking through the mark schemes, it looks like they all want the answer to 3sf OR BETTER, meaning you can't have rounded 10.44 any less than 3, rather than any more than 3


great ty for that
Original post by Cong
How did you got e in q7


Particle B is going towards the wall with speed 21u/10

Impulse received was 27mu/4

Using I = mv - mu

Which you get v = 3u/20

e = v/u

therefore e = (3u/20)/(21u/10) = 1/14

And for part c, both A and B are moving at the same speed, V_a = V_b therefore no further collisions will occur
(edited 7 years ago)
Reply 454
Avatar for TKA1998
TKA1998
Original post by TKA1998
For question 7 I made an arithmetic error and subsequently got the wrong value for the speed of B meaning I used the wrong value in part b and c, how many marks would i likely loose for this.


Anyone?
Reply 455
Avatar for SHJBHB
SHJBHB
1
Original post by zandneger
Looking through the mark schemes, it looks like they all want the answer to 3sf OR BETTER, meaning you can't have rounded 10.44 any less than 3, rather than any more than 3


Has to be 3sf in mechanics doesn't it, or can you get away with 4sf or greater for all questions?
Reply 456
Avatar for SHJBHB
SHJBHB
1
If A* is approx 68/69, would 100 UMS be any lower than 75/75?
Original post by linaxq
those are pretty low no??


They are, but looking at June 2012, the questions are quite similar. That year was 57 for 80 UMS. :eek:
Original post by angelitajin
I still don't understand how to do question 6 got completely wrong the lambda question
Since the exam ended I don't really care how to solve it (good bye m2 Let's never meet again ) but then again I want to know how many marks I can pick up hopefully few method marks
Can anybody please explain me how to do that question?


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
Original post by linaxq
those are pretty low no??


Paper was generally quite hard IMO.
Could be 67, 59, 52.

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