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M1 OCR (Not MEI) Exam - 9/06/2015

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Reply 620
Original post by EconomicsSA
image.jpg Wouldn't this be above the horizontal but still a push


nope
draw the horizontal line from the start of the force, not the end
push.png

if your force was acting such that it had a vertical component acting in an opposite direction to the reaction force, it's wrong - I don't think ocr will allow two different answers on account of ambiguity because it very clearly states 'above the horizontal'.
(edited 8 years ago)
Original post by EconomicsSA
image.jpg Wouldn't this be above the horizontal but still a push


Yeah I think we all did the right drawings but I did a pull instead of a push
Wtf, this pull or push question is too polarised
I would just like to thank all of you who have helped me over the past few days. I have got from a U to a somewhat good grade. Wasn't the biggest fan of M1 as our teacher wasn't very good.

@MsFahima
@kawehi
@chloe-jessica
@Lilli1997


Hope the exam went well for you. Good luck to you if you are doing C3 and C4. :smile:
Could someone please take me through the whole of question 6? The whole integration and differentiation bit, I think I really screwed that bit up
(edited 8 years ago)
Reply 625
Original post by Shaagggyy
Could someone please take me through the whole of question 6? The whole integration and differentiation bit, I think I really screwed that bit up


I can go through it if someone has the question, the whole paper was posted before but I think it's been deleted, or at least I can't find it.
Reply 626
Original post by Super199
I would just like to thank all of you who have helped me over the past few days. I have got from a U to a somewhat good grade. Wasn't the biggest fan of M1 as our teacher wasn't very good.

@MsFahima
@kawehi
@chloe-jessica
@Lilli1997


Hope the exam went well for you. Good luck to you if you are doing C3 and C4. :smile:


That's so good to hear, I'm happy for you! :biggrin: I'm doing C3 but not C4, hope your exams all went/go okay!
Uh. For the last part of Q6, I added 68 to 54 and got 112 and then divided by 2 to get 61 :/ I didn't think the negative sign on the velocity mattered.
Also, can someone please do Q4 because I got 13.3 and 25.1 degrees I think. How did people get like 10 or something?
Original post by chloe-jessica
I can go through it if someone has the question, the whole paper was posted before but I think it's been deleted, or at least I can't find it.


I think it was to do with particle P passing through a point O and the velocity of the particle when it passes O is 2ms^-1 and the acceleration at the time it passes O is (4+12t)ms^-2


Then the questions were
Find the velocity of P when t = 3
Find the displacement of OP when t = 3
Reply 630
Original post by Shaagggyy
I think it was to do with particle P passing through a point O and the velocity of the particle when it passes O is 2ms^-1 and the acceleration at the time it passes O is (4+12t)ms^-2


Then the questions were
Find the velocity of P when t = 3
Find the displacement of OP when t = 3


That sounds familiar :smile:
(i) To get from acceleration to velocity, you integrate (as acceleration is change in velocity over time, a=dv/dt so to get back to v to integrate). So integrating 4+12t gives us 4t+6t2+c. When t=0, v=2 as it states in the question. Therefore 2=4(0)+6(0)2+c => c=2. You end up with v=4t+6t2+2. At t=3, sub in 3 to get v=4(3)+6(3)2+2 = 68m.
(ii) To get from velocity to displacement, you integrate again. Integrating 4t+6t2+2 gives you s=2t2+2t3+2t+c. When t=0, s=0 so c=0. Sub in t=3 to get s=2(3)2+2(3)3+2(3) = 78m.
What kind of grade do you reckon 55 marks would be?
Also, can someone please go through question 7iv)b. because I got 2.32 and I'm not sure where/why I went wrong
IMAG0196.jpg
Original post by chloe-jessica
That sounds familiar :smile:
(i) To get from acceleration to velocity, you integrate (as acceleration is change in velocity over time, a=dv/dt so to get back to v to integrate). So integrating 4+12t gives us 4t+6t2+c. When t=0, v=2 as it states in the question. Therefore 2=4(0)+6(0)2+c => c=2. You end up with v=4t+6t2+2. At t=3, sub in 3 to get v=4(3)+6(3)2+2 = 68m.
(ii) To get from velocity to displacement, you integrate again. Integrating 4t+6t2+2 gives you s=2t2+2t3+2t+c. When t=0, s=0 so c=0. Sub in t=3 to get s=2(3)2+2(3)3+2(3) = 78m.



Ah thank you, it all makes sense now, I forgot the plus C coefficient :///
Reply 633
Original post by artbreaker
What kind of grade do you reckon 55 marks would be?
Also, can someone please go through question 7iv)b. because I got 2.32 and I'm not sure where/why I went wrong
IMAG0196.jpg


Starting from (a) because it's easier. Using deceleration = 0.9 from part (iii).
(a) F=ma for P, gives T-0.3gsin30=0.3*-0.9 => T=0.3*-0.9+0.3gsin30 => T=1.2N.
(b) Tension must be constant throughout the string, so using F=ma for Q => 0.4gsin30-T-Fr=0.4*-0.9 => Fr=0.4gsin30-1.2-0.4*-0.9 (using T=1.2N from above) => Fr=1.12N.
Not sure about the grade sorry!
Reply 634
Original post by Shaagggyy
Ah thank you, it all makes sense now, I forgot the plus C coefficient :///


Sure you'll get method marks for attempting integration and substituting it, then you'll probably get ecf for the second bit :smile:
Original post by chloe-jessica
Starting from (a) because it's easier. Using deceleration = 0.9 from part (iii).
(a) F=ma for P, gives T-0.3gsin30=0.3*-0.9 => T=0.3*-0.9+0.3gsin30 => T=1.2N.
(b) Tension must be constant throughout the string, so using F=ma for Q => 0.4gsin30-T-Fr=0.4*-0.9 => Fr=0.4gsin30-1.2-0.4*-0.9 (using T=1.2N from above) => Fr=1.12N.
Not sure about the grade sorry!


Ooooh thank you! What a ridiculous mistake to make damn
Reply 636
Original post by Super199
I would just like to thank all of you who have helped me over the past few days. I have got from a U to a somewhat good grade. Wasn't the biggest fan of M1 as our teacher wasn't very good.

@MsFahima
@kawehi
@chloe-jessica
@Lilli1997


Hope the exam went well for you. Good luck to you if you are doing C3 and C4. :smile:


You're welcome! And good luck to you too! :biggrin: See you at the C3/C4 thread :tongue:
Original post by LukeBarnett
Thank you!


Could u post the link again please bcos I don't see any link
Original post by Amzii_b
Also, can someone please do Q4 because I got 13.3 and 25.1 degrees I think. How did people get like 10 or something?


4) i. I used the cosine rule:
R^2 = 10^2 + 6^2 -2 x 10 x 6 x cos(70)
which gives
R = 9.74 (3 s.f)
then I used the sine rule to find the angle:
sin(x)/6 = sin(70)/9.74
which gives x = 35.4 (3 s.f)
4) ii. I did:
20 - 9.74 = 10.3 (3 s.f)
because the particle is at rest so I figured the resultant force must be acting directly upward (90 degrees above the horizontal)
4) iii. I then did:
90 - 35.4 = 54.6

Hope that's right and I hope that helps!
For the vt graph question



For some reason I put 20km 4 hours (idk why) in part i)



For parts ii) and iii) I used the correct method to find T and the distance but I used 4 hours instead of 5.



Will I get ecf marks for parts ii) and iii) ?


(Sorry for reposting this but I did not get a reply the first time I posted it)


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