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# Mr M's OCR (not OCR MEI) M1 Answers May 2012

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1. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
Does anybody remember what the wording for Q4 was? Specifically part ii?
2. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by 4mar_ar5en4l)
Not a massive issue, but shouldn't it be 2.67m (3 s.f.)
No the paper says "give non-exact numerical answers to 3 s.f." - I gave the exact answer instead.

Accuracy greater than 3 s.f. is never penalised anyway.
3. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by Mr M)
Hard to say - lose two?
Hi Mr M I think you need to check your answer to the last part of the pulleys question. The answer should be 0.138m

First part before object hits ground

s=0.11025

Then final velocity = 0.735m/s
So initial velocity = 0.735m/s for object going up..

0^2 = 0.735^2 - 2x9.8x s

s= 0.0275625

overall s= 0.1378
therefore 0.138m
4. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by McMatty)
Hey,
For Q6iv: i showed it would stay in equilibrium by resolving down the plane using F=ma, came out with negative acceleration, therefore proved it was in equilibrium.
That a correct method or not?!
thanks
Yes that's fine.
5. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by Mr M)
No the paper says "give non-exact numerical answers to 3 s.f." - I gave the exact answer instead.

Accuracy greater than 3 s.f. is never penalised anyway.
Oh
6. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by the real viper)
Hi Mr M I think you need to check your answer to the last part of the pulleys question. The answer should be 0.138m

First part before object hits ground

s=0.11025

Then final velocity = 0.735m/s
So initial velocity = 0.735m/s for object going up..

0^2 = 0.735^2 - 2x9.8x s

s= 0.0275625

overall s= 0.1378
therefore 0.138m
No you have dropped 1 mark here.

P started 0.11025 m above the ground, rose another 0.11025 m and then finally went up 0.0275 m.
7. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by ILovePi)
Does anybody remember what the wording for Q4 was? Specifically part ii?
Anybody?
8. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by ILovePi)
Anybody?
4 (ii) Calculate the speed of the block at the instant the 14 N force was removed.
9. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by Mr M)
No you have dropped 1 mark here.

P started 0.11025 m above the ground, rose another 0.11025 m and then finally went up 0.0275 m.
Are you certain that he will drop 1 mark? I did the exact same thing. If you are correct I lost 3 marks max on the paper yayy!!

Not doubting your judgement just want to make sure because I'm ecstatic.
10. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by PreciseUrination)
Yes Kgms^-1 is fine.

Might be [-1], Answers are required to 3s.f unless otherwise specified.
They always say or better, you don't get penalised for giving more accurate answers!
11. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by Voglie)
Are you certain that he will drop 1 mark? I did the exact same thing. If you are correct I lost 3 marks max on the paper yayy!!

Not doubting your judgement just want to make sure because I'm ecstatic.
Yes. Well done.
12. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by ILovePi)
Anybody?
This isn't my paper, but found it on tsr.
Attached Thumbnails

13. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
14. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
How many marks would I get if I got 8m and 3.6m for the 2 runners on 3ii) but I subtracted them instead of adding them with the 1m? And also how many marks lost for the very last question as I got v=1.55ms^-1 from getting the left hand side as 3.1 x 0.5 = -0.5v + 1.5v
15. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
Hi Mr M,
For question 4, I stupidly thought that the block was on a slope rather than the force. For some reason I managed to get the coefficient exactly right but my frictional force was different and therefore the speed required was different but I think I used the right method (finding acceleration etc)
How many marks do you think I will lose and will there be any error carried forward?

Thanks
16. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
Would I get any marks from the 6 mark of the coefficient of friction if my value that I showed was 0.82 ?
17. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by Mr M)
No you have dropped 1 mark here.

P started 0.11025 m above the ground, rose another 0.11025 m and then finally went up 0.0275 m.

sneaky, cheers!!!!
18. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by Mr M)
This is a bit vague - lose 2 or 3?
sorry, my phone battery was dying :P.

I resolved perpendicular and parallel.

For paralell i did it correctly

However, for perp rather than 32cos30 - 6.4sin30. i did +6.4.sin30.

After this i did F=muR and found mu (i just gave my answer rather than the show that)

i was wondering how many i would lose for this?

thanks
19. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by ILovePi)
Does anybody remember what the wording for Q4 was? Specifically part ii?
Scanned paper from today if anyone wants to see actual questions:

Has the answers on there too (which match, thankfully, with Mr M's) which might be useful.
Last edited by tahseen1995; 31-05-2012 at 21:55.
20. Re: Mr M's OCR (not OCR MEI) M1 Answers May 2012
(Original post by As_Dust_Dances_)
How many marks would I get if I got 8m and 3.6m for the 2 runners on 3ii) but I subtracted them instead of adding them with the 1m? And also how many marks lost for the very last question as I got v=1.55ms^-1 from getting the left hand side as 3.1 x 0.5 = -0.5v + 1.5v
i got 1.55 m/s too do you know what we had to do ?

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