Edexcel Mathematics M2 15th June 2018 - Unofficial Mark Scheme & Discussion

7d) origional velocity = 21i/sqrt5 + 42j/sqrt5
This dot product with (21i/sqrt5 +xj)=0 when they are perpendicular therefore you can solve for x from that which is the verticle velocity at B it was around -0.9 then use suvat to find the time at B which was about 2.01s

Then just 21/sqrt5 multiplied by 2.01ish gave you the horizontal distance between o and B and it was about 18.9

Do you remember how many marks 7d was worth?

Was I on the right track if I said mew = (2cos^2theta -2)/2(costheta)(sintheta)

They usually like their 2 or 3 sig figs, regardless - you should be fine.

why is 6c -i - 149/3j isn't it in the direction of from A to B?

Haha yes of course. You know what I meant

Thank you

how many marks was question 6 for acceleration?

Please add, discuss, explain and correct in the comments

1)a) v =12 ms^-1
b) a = 0.47 ms^-2

2)a) I =0.875 Ns (Magnitude of 0.3468i + 0.803j)
b) Angle = 66.6 deg

3)a) Show that CoM = (20a) / 3(8+pi)
b) Angle (DE->Vertical) = 59.1 = 59 deg to nearest deg

4) Show that mu = (cosxsinx) / (2 - (cosx)^2)

5)a)i) Show that Vb = 7u/5
a)ii) Va = 2u/5
b) y = (10/22)x = (5/11)x

6) a) magnitude(F) = 5√2 N (=7.07)
b) a = 26.1 ms^-2 (Magnitude of -26i + 2j)
c) AB = -i - (149/3)j

7)a) show that y = 2x - x^2((5g) / (2u^2))
b) u = 21 ms^-1
KEmin = 13.2 J
OB = 22.5m (some people are saying 18.9 but I disagree lmao)

My predictions:

https://gyazo.com/cd341b1b620707bdd08a470246078038

anyone remember how many marks for question 6 acceleration

According to this I got 3b, 5b and the last part of 7 wrong. I got 86 degrees for 3b, y = 2x for 5b and 24 for last part of 7. Do you think I have a chance of getting a A still?

According to this I got 3b, 5b and the last part of 7 wrong. I got 86 degrees for 3b, y = 2x for 5b and 24 for last part of 7. Do you think I have a chance of getting a A still?

Definitely! You didn't mess up any questions that had follow through reliant on it, so you'll likely pick up plenty of method marks.

If we assume you got absolutely no marks for any of those parts, I wanna say that's 3 lost for 3b), 6 lost for 5b) and 5 lost for 7d). 75 - 14 = 61 marks, which was an A last year. I think worst case a very high B, but also your method marks might scooch you up.

Only thing I didn't get was y=(5/11)x, I ended up with y=(6/11)x instead :/ Shouldn't be a major issue if that's wrong but still annoying

7d) origional velocity = 21i/sqrt5 + 42j/sqrt5
This dot product with (21i/sqrt5 +xj)=0 when they are perpendicular therefore you can solve for x from that which is the verticle velocity at B it was around -0.9 then use suvat to find the time at B which was about 2.01s

I don't know what you did, but I got x as -4.69 using dot product which I also got from using the triangle method, which follows through to a distance of 22.5. Don't worry about it though sounds like several people made the same mistake as you.

****, there was a kinetic energy question???

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