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OCR MEI Maths: Paper 1, Unofficial mark scheme

Comment answers below and will add.

Section A:
It tips
a=1/2
theta= pi/6, 5pi/6

Section B:
Speed increasing
n=16 where sum is a maximum
Height around 91m
Range around 700-800m
Newton Raphson - 1.269
Stretch 1/sqrt53 in y-direction and translate across -0.2 something in column vector
for i and j: k=-4 and new velocity is 4i+9j

Differential equation
K= 1.4

Coeff. Of friction = 0.506
(edited 4 years ago)

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I got a different coefficient. About 0.2 ish
Was there an error on the paper? In part i of the parametric question
Reply 3
Original post by MrMoment
Comment answers below and will add.

Section A:
It tips
a=1/2

Section B:
Speed increasing
n=16 where sum is a maximum
Height around 91m
Range around 700-800m

Differential equation
K= 1.4

Coeff. Of friction = 0.506


Same as me for all of those.

Newton Raphson: 1.26 something
Reply 4
I did the magnitude of PR and PQ both equalling 4t^2 +4
Original post by honeydukes01
Was there an error on the paper? In part i of the parametric question

P =(4t^2, 4t) Q= (-4, 4t) Therefore distance = 4t^2 +4
P= (4t^2, 4t) R= (4,0) Therefore use Pythagoras with the change in x and change is y to work out magnitude to equal 4t^2 + 4
Therefore magnitudes equal, so equidistant.
(edited 4 years ago)
How did people show the last question? 5 marks felt like a lot for a pretty straightforward question.
Reply 6
Original post by Andrejcouture27
How did people show the last question? 5 marks felt like a lot for a pretty straightforward question.

I agree! Had me worried. I just drew a diagram and used N2L with F=ur
(edited 4 years ago)
Reply 7
It was a good paper tbh I liked it
Original post by Clark1111
I did the magnitude of PR and PQ both equalling 4t^2 +4

P =(4t^2, 8t) Q= (-4, 8t) Therefore distance = 4t^2 +4
P= (4t^2, 8t) R= (4,0) Therefore use Pythagoras with the change in x and change is y to work out magnitude to equal 4t^2 + 4
Therefore magnitudes equal, so equidistant.


Did something like this but couldn't get the equations to equal because of the 8t in PQ which wasn't in PR but I noticed when t=1 they do equal each other.
k=-4 on the vector one
Root 53= R and 0.2.... radians for a
I got coefficient of fricton to be 0.221
Original post by Mikeyhiggs
I got a different coefficient. About 0.2 ish
Reply 11
Original post by Callum130701
k=-4 on the vector one
Root 53= R and 0.2.... radians for a

yeah I got that
Reply 12
Yeah same the RcosX one, what did everyone get as the transformation, I got a stretch in y-direction scale factor 1/root53 and -0.2... translation in the x-direction
Original post by Callum130701
k=-4 on the vector one
Root 53= R and 0.2.... radians for a
Reply 13
I don't remember what I go for the coefficient of friction :confused:
I got 0.506 for the coefficient of friction.
Reply 15
Original post by Clark1111
Yeah same the RcosX one, what did everyone get as the transformation, I got a stretch in y-direction scale factor 1/root53 and -0.2... translation in the x-direction

same
Reply 16
Yeah pretty sure that's right tbh
Original post by Andrejcouture27
I got 0.506 for the coefficient of friction.
Reply 17
Original post by Andrejcouture27
I got 0.506 for the coefficient of friction.


I think I got that
Reply 18
seems like not a lot of people do OCR Mei
Reply 19
how did u guys find the paper overall ?

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