AQA M2- June 2011

Well what I just did was say that R=mg so F=0.3mg then 0.3mg=mv² /r and you pretty much do the same for the last part. I dont know if that answers what you asked but it gives you the right answer :biggrin:

Yeah I get that that's what you have to do but I don't know why friction is the centralpetal force :s-smilie:

Reply 81

Naathan_b

6

Original post by SmileyGurl13

Yeah I get that that's what you have to do but I don't know why friction is the centralpetal force :s-smilie:

errr try to put your mind in that of the coin! if you were flat on a turntable then the only thing keeping you on towards the centre would be the friction, don't know how to explain what I mean lol :s-smilie:

Reply 82

Katherine_G

im going to fail!! on the brightside it is my last exam :biggrin:

Reply 83

xX.Sabeel.Xx

Original post by SmileyGurl13

Yeah I get that that's what you have to do but I don't know why friction is the centralpetal force :s-smilie:

If this about the coin on a turntable?

The coin feels a force to the centre (centripetal force) and what opposes any motion/ force? the friction.

For it not to topple over or move away from it's path, the centripetal force must equal the frictional force, that way it's horizontally in equilibrium... hope it helps. I think this is right.

Reply 84

xX.Sabeel.Xx

I don't get why in the June 08 paper, question 7a why you assume there is no tension acting at the top, but when you look at Jan 11 paper question 6b, there is tension acting at the top.

Reply 85

Ollie901

1

Original post by xX.Sabeel.Xx

I don't get why in the June 08 paper, question 7a why you assume there is no tension acting at the top, but when you look at Jan 11 paper question 6b, there is tension acting at the top.

Link to Jan 11 paper please? :colondollar:

Nevermind, just seen it on previous page :smile:

(edited 12 years ago)

Reply 86

Misiak

what do you think the chances are that we get a spring in a vertical circle? :tongue:

Reply 87

xX.Sabeel.Xx

could do, then the tension=lambda X/L... ive seen one kind of like that in a textbook, but wasn't that hard, if they do put one in, it's going to be a nasty one though.

Reply 88

CoffeeStinks

11

Original post by SmileyGurl13

It's no problem at all! Could you help do you think? :smile:

http://www.billgrisdaledesign.co.uk/DCHS_mathspapers/images/AQA-MM2B-W-QP-Jan11.pdf

Q5

I don't understand why the centripetal force is equal to the friction :s-smilie:

Do you have the MS for Jan2011?

Reply 89

uxa595

21

I need a B in this, but it's not happening :frown:

I am only manageing 25-40 on these past papers, This has to be the worst maths module.

Reply 90

OllyHV

For 7b, I thought mv^2/r acts towards the centre of the circle, so surely R + mv^2/r = mgcos 60

Yet in the mark scheme it says R=mv^2/r + mgcos 60, well mgcos60 is definitely acting away from the centre, so how can mv^2/r be acting towards it?

Would really appreciate some help.
http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MM2B-W-MS-JAN09.PDF

Reply 91

Ollie901

1

Original post by OllyHV

For 7b, I thought mv^2/r acts towards the centre of the circle, so surely R + mv^2/r = mgcos 60

Yet in the mark scheme it says R=mv^2/r + mgcos 60, well mgcos60 is definitely acting away from the centre, so how can mv^2/r be acting towards it?

Would really appreciate some help.
http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MM2B-W-MS-JAN09.PDF

I had this problem the other week.

Bascially what you need to remember is that the centripetal force isnt a force in its own right, it's whats left after you radially resolve forces.

So Mv^2/r is actually R-mgcos60 as the forces acting towards the centre are R-mgcos60. This, incidently is the conclusion you would've reached if you'd have said mv^2/r is a new force completly acting away from the centre of the circle, which is always the case.

Reply 92

Misiak

Original post by OllyHV

For 7b, I thought mv^2/r acts towards the centre of the circle, so surely R + mv^2/r = mgcos 60

Yet in the mark scheme it says R=mv^2/r + mgcos 60, well mgcos60 is definitely acting away from the centre, so how can mv^2/r be acting towards it?

Would really appreciate some help.
http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MM2B-W-MS-JAN09.PDF

you're sort of treating the situation as if the particle is in equilibrium, which it's not
if you take radially towards the centre to be positive using simgaf=ma
R - mgcos60 = mv^2/r
hence R = mv^2/r +mgcos60

(edited 12 years ago)

Reply 93

Ollie901

1

Original post by CoffeeStinks

Do you have the MS for Jan2011?

http://www.billgrisdaledesign.co.uk/DCHS_mathspapers/images/AQA-MM2B-W-MS-Jan11.pdf

Reply 94

CoffeeStinks

11

Original post by Ollie901

http://www.billgrisdaledesign.co.uk/DCHS_mathspapers/images/AQA-MM2B-W-MS-Jan11.pdf

Thanks!

Reply 95

OllyHV

Thanks to both of you for the great explanations

I'd never heard of mv^2/r as being the resultant force, but it makes perfect sense cheers :smile:

However Ollie, you said mv^2/r acting away from the centre which is always the case, yet earlier somebody said mv^2/r is always acting towards the centre, which one is it ? Surely if its a resultant force its acting towards the centre.

Reply 96

Ollie901

1

Original post by OllyHV

Thanks to both of you for the great explanations

I'd never heard of mv^2/r as being the resultant force, but it makes perfect sense cheers :smile:

However Ollie, you said mv^2/r acting away from the centre which is always the case, yet earlier somebody said mv^2/r is always acting towards the centre, which one is it ? Surely if its a resultant force its acting towards the centre.

It is towards the centre, but what I'm saying is, if you treat it as a completley new force (i.e not the resultant of the rest) acting away from the circle, then you always get the same result as if you just resolve radially towards the centre. It's just another way of thinking about it.

Reply 97

Ollie901

1

on 7bii Jan 2011, why doesnt just using 196x/2 = 3g give the right extension?
Thanks.

EDIT: also on 7c, why is the maximum speed when T=mg?!

EDIT 2: never mind, I get them both now. Was being daft.

(edited 12 years ago)

Reply 98

OllyHV

Why does T=mg give max speed, we were just told this but it was never explained.

Edit: Also in jan 11, 7bii, I can get to the part where I get a quadratic where I solve for x = 0.7 or x=0.1, why is the answer 0.1 and not 0.7?

(edited 12 years ago)

Reply 99

Ollie901

1

Original post by OllyHV

Why does T=mg give max speed, we were just told this but it was never explained.

Its because on that particular question, you had reduced the mass of the block so that the block flies upwards, rests at the top of its flight and then begins to fall back down. Until T=mg, the only force acting is mg so it accelerates up to this point. After this point there is a tension force decellerating it.

Therefore max speed when T=mg

In answer to your edit, you want the position where it is next at rest. When x=0.7, it was when the gain in GPE is 0, so it was just exactly where it was before you took the 3kg block off.

(edited 12 years ago)

AQA M2- June 2011

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