Mr M's OCR (not OCR MEI) M2 answers June 2011

Scroll to see replies

Oops my bad, I meant part ii. :redface:

Just use v^2 = u^2 + 2as to get a and then F = ma

Up slope T + 90

Down slope 686 x 3/20

so 686 x 3/20 - T - 90 = 70 x -0.06125

Reply 61

mrppaulo

Original post by Mr M

ah i didn't see you say this before

you will lose the accuracy marks - there will be at least 2 of them and maybe more

Okay thank you very much :smile:

Reply 62

AmroTT

Original post by Mr M

I didn't invent it!

I'm sure you've been stealing my Maths teacher's jokes. Here's another 'good' one:

\int \frac{1}{cabin} \; \mathrm{d} \, cabin \; = \mathrm{ln} cabin + c = \mathrm{Houseboat}

Try reading it out loud if you don't quite get it.

Reply 63

Mr M

Study Forum Helper

Original post by AmroTT

I'm sure you've been stealing my Maths teacher's jokes. Here's another 'good' one:

\int \frac{1}{cabin} \; \mathrm{d} \, cabin \; = \mathrm{ln} cabin + c = \mathrm{Houseboat}

Try reading it out loud if you don't quite get it.

Old as the hills.

Reply 64

Joe Wallace

I am really sorry to bother you, but it would mean a lot if you could give me a rough indication as to how many marks I would lose, even if you only look at question six.

1)ii) I did this question last and, whilst rushing, did not read the constant resistive force of 90N. If I had included this, my answer would have been correct.

6) This is horrible to admit, but I took the 30º to be the angle between the horizontal and the cone. It serves me right for making a sketch and not going back to the question paper properly. I have repeated my working using the new angle and I get the same answers as you. I am worried that, because it is such a simple mistake, hardly any method marks would be given. Would you agree with this?

7)ii) I have a value of 0.531 (3sig.fig.) but I cannot see where I have made a mistake.
Edit: I took the angle of T to be 30º and not 60º (again)! How many marks do you think I would lose?

Again, I am sorry to be such a pain, but realising what I have done for question six [and now seven] has really worried me. :frown:

(edited 12 years ago)

Reply 65

Spr-y

Anyone got a copy of the paper to upload? Might help when de-cyphering people's questions about questions / let me have another go at parts of questions 6/7 :biggrin:

Reply 66

Mr M

Study Forum Helper

Original post by Joe Wallace

you have just lost accuracy marks not method marks - this is usually one or two marks per part

Reply 67

wibletg

Original post by Mr M

you have just lost accuracy marks not method marks - this is usually one or two marks per part

What actually gains you method marks? I know it's for method, but is it correct method or do they allow you to make a mistake or two?

Reply 68

Mr M

Study Forum Helper

Original post by wibletg

What actually gains you method marks? I know it's for method, but is it correct method or do they allow you to make a mistake or two?

look at mark schemes, there are M marks for various steps leading to (what would be if no mistakes were made) a correct answer

Reply 69

wibletg

Original post by Mr M

look at mark schemes, there are M marks for various steps leading to (what would be if no mistakes were made) a correct answer

Right, and if you hadn't made the mistakes the accuracy marks would be awarded too?

Reply 70

Mr M

Study Forum Helper

Original post by wibletg

Right, and if you hadn't made the mistakes the accuracy marks would be awarded too?

yes

Reply 71

Joe Wallace

Original post by Mr M

you have just lost accuracy marks not method marks - this is usually one or two marks per part

Thank you for your response. I have never posted on this website before, but I often look at your answers after an exam. I just want to thank you for all of the times you have done this for everyone. :smile:

Reply 72

Mr M

Study Forum Helper

Original post by Joe Wallace

you are welcome

Reply 73

nikita_atikin

Original post by Mr M

Just use v^2 = u^2 + 2as to get a and then F = ma

Up slope T + 90

Down slope 686 x 3/20

so 686 x 3/20 - T - 90 = 70 x -0.06125

I got the right answer for 1)ii) but I did not do that :s-smilie:

In 1)i) I did (KE at start + PE at start) - (PE at end + KE at end + Work done by resistance to motion) I think and ended up with 344J so that's wrong.

So in 1)ii) I just divided 344J by 20m (the distance the object travelled by) and still got 17.2N

Will I get any marks at all?

Reply 74

snowboardinsteve

Thank you very much for the answers once again Mr M! :smile:

Looks like I got the same as you for everything :smile:

Original post by Mr M

Well getting into this debate is like trying to stand on an inclined surface with

\mu = 0

but it seemed reasonable to me.

LOL

Reply 75

mccain

Is anyone able to email me the paper? Thanks

Reply 76

Mr M

Study Forum Helper

Original post by snowboardinsteve

Thank you very much for the answers once again Mr M! :smile:

Looks like I got the same as you for everything :smile:

LOL

Good stuff. Are you Year 12 or 13 and, if 13, what are you doing next year?

Reply 77

Mr M

Study Forum Helper

Original post by nikita_atikin

I got the right answer for 1)ii) but I did not do that :s-smilie:

Well it is right so you will get full marks on part ii)

Reply 78

momories

Hi Mr M, could I ask a quick question?

For the question on finding the equation of trajectory, I left it in tan and cos, but made a mistake with my sec and basically i lost like a mark from finding the right equation. But the problem is, you had to use the equation you had produced to solve the rest of the question. Would I get error carried forward? Thank you very much

Reply 79

momories

Original post by Mr M

Well it is right so you will get full marks on part ii)

I think I got the same answer as you for part i) !
There must be some method marks involved. All the past papers I've done, theres been like marks to find the difference in KE, difference in GPE and work done by the frictional force. I think if we're lucky, we might only lose one! :smile: