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M2 OCR (not mei) JANUARY 2013 EXAM preperation thread

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Reply 40

JamesTaylorr

Which part of question 8 did you struggle with? All of it or just the last part?

Reply 41

Maphs

Q8) i am quite sure that in order to get 4F + 3R = 19.6, or something like that, friction would have to go upwards, but that does not make sense because the particle is trying to get higher up the slope hence friction should be going down, so was that a mistake?
Also that question on when a particle is projected up a slope and it wants you to find the speed at which it reaches Q, 0.3m away from the bottom?

Reply 42

jassi1

when the particle was projected up the slope I used the conservation of energy: I can't remember but I think the slope was frictionless so the loss in k.e. is the gain in p.e.

Reply 43

jassi1

For the first bit of the last question friction does go up the slope (which may seem weird) but think of it like this: If friction were not present the particle would go down the slope due to its weight therefore friction acts up the slope to oppose the motion.

Also for the last bit of this question were it asks for the highest possible angular speed I said this will happen when the friction acts down the slope and it is at its greatest value (limiting). This is because the highest possible resultant centripetal force corresponds to the highest angular speed as F=mw^2r where w is omega (angular speed).

Reply 44

Maphs

that is true but the particle is also moving with a speed hence causing it to move upward so friction acts downwards.
if at rest then friction will act upwards.

Reply 45

lewishnl

Original post by jassi1

Completely screwed up that last part, I like your idea, but did we have the co-efficient of friction?

Reply 46

jassi1

Original post by Maphs

that is true but the particle is also moving with a speed hence causing it to move upward so friction acts downwards.
if at rest then friction will act upwards.

I just realised we are both wrong.. Of course if the particle is going at a supersonic speed the friction will act down the slope, however if the particle is going at a snails pace the friction will act up the slope. And in fact there is a certain speed at which there is no friction.

and lewishnl:

I remember getting a coefficent of friction however I can't recall what question it was for..

Reply 47

Genesis2703

Hey guys!!! Hope the exam was ok for you guys. I personally don't want to be asked about what answers I put as I get very paranoid and worked up over it. It also means I won't be looking at the Mr M solutions thread :tongue:

I personally thought the exam was slightly harder than June 2012, based purely off the last question and a few weird questions here and there. Thought the collision question was strange, but do-able. The only question I completely guessed what to do at was the very last one :smile:

All in all a decent paper, I personally think the 80 UMS boundary willl be lower than last June (58/72)... maybe something like 56? The Jan 2012 paper was a lot harder than this one imo.

EDIT: Oh and on 7iii) I put the angle into the wrong equation, so I had to solve a quadratic for 2 marks, still should be right. Silly me overcomplicating things :redface:

(edited 11 years ago)

Reply 48

Jedicake

I thought this paper was impossible compared to previous papers...

I spent about 30 minutes on question 3 going "what? surely we need to know mass?" and then realised F=ma. And don't even get me started on spotting tan^2(x) = sec^2(x) + 1...

Reply 49

Little Wing

Apart from the last part of question 8 - this was a very straight forward paper, in my opinion ofcourse.

Reply 50

zhang.c

Did anyone get exactly 1.5 for the last 6 mark question for the last part of the last question

Reply 51

Mr.Guy

Original post by zhang.c

Did anyone get exactly 1.5 for the last 6 mark question for the last part of the last question

That was the angular speed of the particle in the previous question!

Reply 52

Genesis2703

Original post by Little Wing

Apart from the last part of question 8 - this was a very straight forward paper, in my opinion ofcourse.

Agreed, I was happy I predicted the last question would be a circular motion one, did not expect it to be like how it was though :redface:

Original post by Mr.Guy

That was the angular speed of the particle in the previous question!

Yeah, Mr.Guy is right, anyone who got 1.5 should be wrong, because that is the angular speed in the 1st case (slowest possible speed)

Reply 53

jassi1

Actually, now that I think of it, I'm not sure if 1.5 was my initial attempt or my second one, as I firstly did the last question completely wrong and then I realised what I had to do about 15 mins from the end so I did it properly on an additional sheet. So I cant guarantee that 1.5 is right.

From Genesis2703's post 1.5 must have been my first attempt because in that one I calculated for the friction up the slope instead of down so it was basically the same as the lowest speed situation..

(edited 11 years ago)

Reply 54

Maphs

also the question where you had to use energy, it said that Q is 0.3m from the bottom of the slope which means potential energy cannot be found as theta was not given, i too thought of energy but realised that h = 0.3sintheta? so that was strange.

Reply 55

Mr.Guy

Has anyone got the paper?

Reply 56

Genesis2703

Original post by Maphs

my interpretation was that the height (vertically) was 0.3m, since the question said the level of Q was 0.3m above the level of P.

So the PE = (0.5)g(0.3)

Reply 57

Little Wing

Original post by Genesis2703

my interpretation was that the height (vertically) was 0.3m, since the question said the level of Q was 0.3m above the level of P.

So the PE = (0.5)g(0.3)

That's what I did aswell, I'm pretty sure it was correct because it said 0.3m above the "level".