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OCR M1 - 31st May 2012 :)

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Reply 80
Can anyone help me on Jan 10 question 4? It was a toughie, still can;t understand it despite looking at it 4 times! :frown: Im screwed for this exam .
Reply 81
Original post by master y
Can anyone help me on Jan 10 question 4? It was a toughie, still can;t understand it despite looking at it 4 times! :frown: Im screwed for this exam .


i'll take a look at it now for you bud
Reply 82
Original post by master y
Can anyone help me on Jan 10 question 4? It was a toughie, still can;t understand it despite looking at it 4 times! :frown: Im screwed for this exam .


Ok so for the first part, its pretty simple. it just looks complicated than what it actually is (generally this is m1 all over). anyway, to resolve upwards...

Firstly, remember limiting friction is F = UR

ok so, it asks about P.

forget anything to do with the string. it is resting on the plane. so ignore the tension in the string.

resolving upwards,: R - 0.4gcos60 = 0
resolving across,: 0.4gsin60 - F = 0

it says in the question there is friction, and if P is more weight than Q, it's clearly gonna go down the plane when let go. so thats why friction on your diagram should be going UP the plane.

then to work out U its simple F / R by rearranging the 2 formulas i just gave you up there.

2. so if you add 0.2 to Q, its total mass is now 0.5 which is > than 0.4, so now the string will go the other way, and P will lift off the plane. So, we ignore the normal contact force and any other force acting on the plane, it just isn't there.

then resolving, remember just think logically where the particle is going to go, so for Q its 0.5g - T = 0.5a

for p its T - 0.4g = 0.4a

rearrange, to find A and T.

as soon as particles are in motion, F = MA needs to be used. as soon as you see acceleration anywhere, remember its F=MA :smile:
Reply 83
Original post by Tyles
Ok so for the first part, its pretty simple. it just looks complicated than what it actually is (generally this is m1 all over). anyway, to resolve upwards...

Firstly, remember limiting friction is F = UR

ok so, it asks about P.

forget anything to do with the string. it is resting on the plane. so ignore the tension in the string.

resolving upwards,: R - 0.4gcos60 = 0
resolving across,: 0.4gsin60 - F = 0

it says in the question there is friction, and if P is more weight than Q, it's clearly gonna go down the plane when let go. so thats why friction on your diagram should be going UP the plane.

then to work out U its simple F / R by rearranging the 2 formulas i just gave you up there.

2. so if you add 0.2 to Q, its total mass is now 0.5 which is > than 0.4, so now the string will go the other way, and P will lift off the plane. So, we ignore the normal contact force and any other force acting on the plane, it just isn't there.

then resolving, remember just think logically where the particle is going to go, so for Q its 0.5g - T = 0.5a

for p its T - 0.4g = 0.4a

rearrange, to find A and T.

as soon as particles are in motion, F = MA needs to be used. as soon as you see acceleration anywhere, remember its F=MA :smile:



Thank you so much! This was written very clearly, and i think i understand it now :smile:
Reply 84
Original post by master y
Thank you so much! This was written very clearly, and i think i understand it now :smile:


i remember the first time i did this question i was thrown off, np! :smile:
Reply 85
Original post by Tyles
Ok so for the first part, its pretty simple. it just looks complicated than what it actually is (generally this is m1 all over). anyway, to resolve upwards...

Firstly, remember limiting friction is F = UR

ok so, it asks about P.

forget anything to do with the string. it is resting on the plane. so ignore the tension in the string.

resolving upwards,: R - 0.4gcos60 = 0
resolving across,: 0.4gsin60 - F = 0

it says in the question there is friction, and if P is more weight than Q, it's clearly gonna go down the plane when let go. so thats why friction on your diagram should be going UP the plane.

then to work out U its simple F / R by rearranging the 2 formulas i just gave you up there.

2. so if you add 0.2 to Q, its total mass is now 0.5 which is > than 0.4, so now the string will go the other way, and P will lift off the plane. So, we ignore the normal contact force and any other force acting on the plane, it just isn't there.

then resolving, remember just think logically where the particle is going to go, so for Q its 0.5g - T = 0.5a

for p its T - 0.4g = 0.4a

rearrange, to find A and T.

as soon as particles are in motion, F = MA needs to be used. as soon as you see acceleration anywhere, remember its F=MA :smile:


Sorry i just wanted to ask , so if it was the other way around where p>Q , then would we need to use pgsin60-T=Pa ?
Reply 86
Hi guys me again, (soorry!), question 7iv June 2009, how do you know that its the smaller value which you use? ?
Reply 87
Original post by Rushika
Sorry i just wanted to ask , so if it was the other way around where p>Q , then would we need to use pgsin60-T=Pa ?


You wouldn't need to use T, you never see a question where it will slope down a plane and be taut to a string going verticly up. it would be pg-sin60 - F = p x a


(I think.. i'm not amazing at mechanics lol)
(edited 11 years ago)
Reply 88
Original post by Choppyy
Did Jan 2012 today as a mock, the last question was a real ball ache. Otherwise an ok paper - slightly more challenging than usual. Has anyone else already done this paper?

Edit: For anyone who hasn't sat this paper, I've attached both it and the mark scheme.


Thanks for this, I'll have a go at it today
Can anyone explain why on 7ii) b) (Particle is at rest) why friction is ignored when calculating the magnitude of the contact force? And why the angle is 60?
Thanks
http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61432_pp_11_jan_gce_472801.pdf?
Original post by hannah1994
Can anyone explain why on 7ii) b) (Particle is at rest) why friction is ignored when calculating the magnitude of the contact force? And why the angle is 60?
Thanks
http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61432_pp_11_jan_gce_472801.pdf?


I didn't do very well on that paper when I did it the other day, there are some quite tough questions on it.
Reply 91
Original post by hannah1994
Can anyone explain why on 7ii) b) (Particle is at rest) why friction is ignored when calculating the magnitude of the contact force? And why the angle is 60?
Thanks
http://pdf.ocr.org.uk/download/pp_11_jan/ocr_61432_pp_11_jan_gce_472801.pdf?


it's not ignored, at rest, friction is equal to the weight parallel to the plane and act upwards, so 0.6gsin30 = 2.95, and reaction force is 0.6gcos30 = 5.09, so it forms a right angled triangle to the plane , with the angle being tan^-1(5.09/2.95) = 59.9 degrees, remember it asks for angle between upward direction of the line of greatest slope.
Original post by PeterStoba
How would I solve this question guys?



I get a = 39.2 ms^-2 and T = 147 N but they don't seem right...


hey, ok uv tried but ur answers are wrng, look at it in this way, acceleration will b the same in the whole system, so get the equations for acceleration at both bodies lets call body A for the 5kg and body B for 3kg.

now a=f/m where the force would be the resultant force at each body

for bodyA it will be a= (5g-T)/5
for body B it will be a= (T-3g)/3

since a is equal for both bodies then;
(5g-T)/5 = (T-3g)/3

hence calculate tht ul get T=36.75 AND a=2.45
Reply 93
how i everyone revising? are you just sticking to past papers or doing exercises from the book?
Original post by h2shin
it's not ignored, at rest, friction is equal to the weight parallel to the plane and act upwards, so 0.6gsin30 = 2.95, and reaction force is 0.6gcos30 = 5.09, so it forms a right angled triangle to the plane , with the angle being tan^-1(5.09/2.95) = 59.9 degrees, remember it asks for angle between upward direction of the line of greatest slope.


But is R not just equal to weight perp to plane? so why is that the resultant contact force and not friction :/
Reply 95
has anybody got any vertical motion questions? or more questions i can revise from?! thanks!
Reply 96
Hey I need help on this question:

Its from the M1 Jan 2011 Paper question 5... :frown:

I can't seem to find T. It's 2/3... but how? :\

Can someone please go through it with me. :smile:
Reply 97
I think I should do a past paper now! This is my 3rd try at it amazingly and I do further maths! Last time I thought it went really well and got a C!! :smile:
Reply 98
Original post by ugk4life
how i everyone revising? are you just sticking to past papers or doing exercises from the book?


PP. I don't even have the book! :O
Reply 99
Original post by hannah1994
But is R not just equal to weight perp to plane? so why is that the resultant contact force and not friction :/


Contact force takes into account BOTH reaction and friction force, so you need to find the resultant of these two.

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