OCR M1 (not MEI) Official Thread 24/01/2011

Hopefully should have done okay, lost about 12 marks already, not counting the silly marks I made. Low boundaries and I'll be happy. Good luck everyone!

Reply 21

asabashoyuki

Original post by jd2010

That's really annoying, from that i have realised i made a silly mistake, i got
S= 4.9t +4.9t^2 having forgot to half it to get 2.45t^2

To prove T= 2/3 you would use the two equations S=9.8t^2 and S=4.9T +2.45T^2
and use simultaneous equations to get
9.8T^2 = 4.9T + 2.45T^2
Working that out you get that T = 2/3
DAMN. :angry:

It looks right but I doubt it...

Two Ss have different value (different length) so they aren't really a set of simultaneous equation....

It's interesting that it gives out 2/3 though...hmm..

Reply 22

Roy064

1

Original post by asabashoyuki

It looks right but I doubt it...

Two Ss have different value (different length) so they aren't really a set of simultaneous equation....

It's interesting that it gives out 2/3 though...hmm..

I used sine rule to get T=2/3
if you have the paper in front of you (I haven't so kinda guessing from memory), equation for P was the hypotenuse and Q was opposite the 30 degree angle.

P/Sin90 = Q/Sin30
P=Q/Sin30
solve what P and Q are because I can't remember, but I did something close to that with actual numbers obviously and got T=2/3 :smile:

EDIT: Remembered
P=2Q therefore 4.9T+2.45T^2=9.8T^2
4.9=7.35T
T=2/3 :smile:

hope that clears it up for you all

(edited 13 years ago)

Reply 23

asabashoyuki

Original post by Roy064

I used sine rule to get T=2/3
if you have the paper in front of you (I haven't so kinda guessing from memory), equation for P was the hypotenuse and Q was opposite the 30 degree angle.

P/Sin90 = Q/Sin30
P=Q/Sin30
solve what P and Q are because I can't remember, but I did something close to that with actual numbers obviously and got T=2/3 :smile:

Why would I have it with me? :confused:

Hmm, I think I did the same thing so let's hope it's right :cool:

Reply 24

chira07

5

I got:
Q1: -1.3kgms, 0.125
Q2: 36.9, 6N
Q3: 8.6, 1.39
Q4: 1.87, mu=0.681
Q5: proof, 30 degrees, proof, 6.53ms
Q6: 15.3m, -18m, -3.3m
Q7: mu=0.601, 139 degrees, 7.87N, 5.1N, 90 degrees

but I'm not sure about all of these answers :s-smilie:

Reply 25

Roy064

1

Original post by asabashoyuki

Why would I have it with me? :confused:

Hmm, I think I did the same thing so let's hope it's right :cool:

I did say "if", some people take the paper and ask teachers and such for peace of mind regarding their answers

Reply 26

kazaf

that was the worst exam paper that i have done in my entire life!!! cant believe i sat in the exam hall doing that paper ;(

Reply 27

abbas1994

1

Original post by kazaf

that was the worst exam paper that i have done in my entire life!!! cant believe i sat in the exam hall doing that paper ;(

i can't believe people actually are saying ti was alright?!?!?!?!

that was HELL
our teacher reckons its the hardest m1 paper done in a long time, so grade boundaries should be low, about 60% for an A :/ but still..

Reply 28

luck STA

6

Original post by chira07

I got:
Q1: -1.3kgms, 0.125
Q2: 36.9, 6N
Q3: 8.6, 1.39
Q4: 1.87, mu=0.681
Q5: proof, 30 degrees, proof, 6.53ms
Q6: 15.3m, -18m, -3.3m
Q7: mu=0.601, 139 degrees, 7.87N, 5.1N, 90 degrees

but I'm not sure about all of these answers :s-smilie:

we got some of the same answer

Reply 29

kazaf

Original post by abbas1994

i can't believe people actually are saying ti was alright?!?!?!?!

that was HELL
our teacher reckons its the hardest m1 paper done in a long time, so grade boundaries should be low, about 60% for an A :/ but still..

totally agree!!
is jus soooooo different compare to previous papers ;(((( especially the last question!!! wt was that all abt

Reply 30

Roy064

1

how was it different? was pretty straight forward for most of it except last question and the T=2/3 stuff (which just took some time to think about it). Everything else was the norm of past papers.

Reply 31

Freerider101

12

Original post by chira07

I got:
Q1: -1.3kgms, 0.125
Q2: 36.9, 6N
Q3: 8.6, 1.39
Q4: 1.87, mu=0.681
Q5: proof, 30 degrees, proof, 6.53ms
Q6: 15.3m, -18m, -3.3m
Q7: mu=0.601, 139 degrees, 7.87N, 5.1N, 90 degrees

but I'm not sure about all of these answers :s-smilie:

Q1: 1.3Ns 0.125
Q2: 36.9deg 6N
Q3: 8.6, 1.39, Parabola with y intercept of 2m, Linear with negative gradient crossing x axis
Q4: 2.16N (tension), 0.819(friction coefficient)
Q5: show that, 30 degrees, show T=2/3 12.25ms^-1, 14.7ms^-1
Q6: -15.3m, 18m, +17.3m from O
Q7: 0.601 (friction coefficient), 5.88N, 30 degrees, 5.09N, 30degrees
This is what I got in comparison, I could have some wrong though

(edited 13 years ago)

Reply 32

Freerider101

12

Original post by luck STA

we got some of the same answer

Compare answers to mine in the post above.

Reply 33

eanderson4

Original post by kazaf

totally agree!!
is jus soooooo different compare to previous papers ;(((( especially the last question!!! wt was that all abt

Original post by Roy064

how was it different? was pretty straight forward for most of it except last question and the T=2/3 stuff (which just took some time to think about it). Everything else was the norm of past papers.

I can see both points... though I DEFINITELY can't see the boundary being 60%. The lowest I've seen in the past is 50/72 for mechanics. Typically grade boundaries are between about 52 and 62. 60% would be 43 ish... it's not going to happen. Most of it was okay. The end bit was difficult... but it wasn't harder than the January 2010, which had a boundary of 52.

Reply 34

Roy064

1

Original post by eanderson4

I can see both points... though I DEFINITELY can't see the boundary being 60%. The lowest I've seen in the past is 50/72 for mechanics. Typically grade boundaries are between about 52 and 62. 60% would be 43 ish... it's not going to happen. Most of it was okay. The end bit was difficult... but it wasn't harder than the January 2010, which had a boundary of 52.

You studying Politics at university?

Reply 35

Jodie 123

Could anyone do the last part of question 7 or know what they were asking for, if so please post worked example. I am intriguied...

Reply 36

Student79

In reply to people worrying about the last question, you had to calculate the resultant of R and of Friction, can't remember the figures so cant tell you all the exact answers, from memory the angles for b were 121 and 60, magnitudes were around 5, if someone has a copy of the question, post it and :smile:

Reply 37