The Student Room Group

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

Scroll to see replies

Reply 40
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..


Samee, i was getting worried with everyone saying it was fine! I found it horrible, definitely the worst paper i've ever done. I hope the grade boundaries are low!
Reply 41
sorry i can remember exactly what the question was
how did people do the question with the block connected to a string and calculating the tension?!?!
Reply 43
i think it was the v=t^2 -9 question? from memory it wanted distance travelled from t=0 to when it changed direction. Well when it changed direction, v=0 so t^2=9 so t = 3. To calculate displacement we integrate t^2-9 to get t^3/3-9t+c. we worked out c=15 1/3 in part i. therefore just put in the numbers t=0 and t=3 into that formula, subtract them to calculate the difference, and you get the answer as 15 1/3 - -2 2/3 so the answer is 18
How the heck did you guys do the vertical motion one?

Particle is projected up with inital speed 5ms, from 2.5m above the ground

Find: speed it hits the ground with

and total time to reach the ground

can somebody work me through it?
Reply 45
For the question with the block, you basically just do f=ma, the pulling force = 2N, so the force acting left was 2-Tcos10, which was the component of T acting right. and thats equal to the mass times the acceleration, which i think was 0.8 and 0.2? i cant quite remember the exact figures, but from there just put it into your calculator to solve
ah yes i did that exactly:smile: so pleased i got it right! what did you for the second part of it? so annoying i cant remeember the exact decimal answers i got :s-smilie:
Reply 47
@milanmylove if we take the initial velocity to be positive then using suvats. s=-2.5 u=5 v=unknown a=-9.8 t= unknown. to calculate speed it hits the ground with, v^2=u^2+2as so v^2=25+(2x-9.8x-2.5)=74 so v=8.60ms^-1
Reply 48
@purplestarfish, I cant really remember the second part, i think it was involving friction of the particle? I think it was a similar approach, you know the acceleration, mass and all but one force, which should have allowed you to use exactly the same approach to solve as the first part. Hope that helps, if you post the question, i could solve with figures, but cant remember enough to help further otherwise :smile: hope you got it right though
Original post by Student79
@milanmylove if we take the initial velocity to be positive then using suvats. s=-2.5 u=5 v=unknown a=-9.8 t= unknown. to calculate speed it hits the ground with, v^2=u^2+2as so v^2=25+(2x-9.8x-2.5)=74 so v=8.60ms^-1


ah crap :frown:

i spent like 10 mins trying to work it out, and in the end used s= 2.5 +x

worked out x, but then got a final velocity of 20 or something lol :frown: complete fail
Reply 50
@milanmylove, to solve the total time taken, v=u+at so 8.6=-5+9.8t so 9.8t=13.6 so t=13.6/9.8=1.39s
Reply 51
ah sorry :frown: hope you did well on the rest of the paper, and at least we can all forget about it till may :smile:
Reply 52
march*
Reply 53
Original post by Roy064
You studying Politics at university?


Haha. It did sound quite diplomatic, but I'm not studying politics.
Original post by Student79
march*


i put 60 degrees for both angles on the last question, do you think that's right?
Original post by purplestarfish
i put 60 degrees for both angles on the last question, do you think that's right?


I put 30 degrees. Although I'm fairly sure they should both be that same.
Reply 56
from what i calculated, i got 60 for the last part of the last question, i got 131 for the first part
Reply 57
Praying I pick up some marks for working, I think I messed up quite a few final answers for things. Pretty disappointing considering it was a retake :frown:
Reply 58
they shouldnt be the same i dont think. You have to calculate the resultant force of the reaction force and of friction, when its in motion, friction acts down the slope, when it's stationary, friction acts up the slope, meaning the direction of the resultant should be completely different, one bigger than 90 degrees, one less than 90 degrees
Original post by YGB Jammy
Praying I pick up some marks for working, I think I messed up quite a few final answers for things. Pretty disappointing considering it was a retake :frown:


Mine was a retake too > < Still, I guess there's always retaking the retake in June. Actually I don't think I'll bother if I have to, I have 11 exams in June already without heaping on more! :biggrin:

Method marks make up the majority of the marks for a question anyway, I swear it's only the A1 mark at the end that's for the correct answer!

Quick Reply