AQA Physics A Unit 4- 11 June 2014

Can you help me with this one, question 11, I'm assuming it wants me to try to work out which answer would give g = 0, so I'm using the formula 2pi square root(l/g) but then I keep getting 24 hours which is obviously wrong, how do you do it?
http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JAN11.PDF

Can you help me with this one, question 11, I'm assuming it wants me to try to work out which answer would give g = 0, so I'm using the formula 2pi square root(l/g) but then I keep getting 24 hours which is obviously wrong, how do you do it?
http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JAN11.PDF

Hi, I took a quick look at this and got 1.4 hours, i'm not sure if its right because I didn't check the scheme and I know sometimes you get an answer thats an option but its not right and its there to trick people who make silly mistakes anyway...

Firstly I thought for it to feel no weight it has to be travelling a a high speed. If something is travelling at such a speed the the reaction force on the earths surface becomes 0 when it loses contact.

so mg - R = mv^2/r if R is 0 the mg =mv^2/r
rearranging gives v^2 = gr (m's cancel)
so v^2 = 9.81x6.4x10^6 = 6278400
root that for v = 7923
then T is distance/speed =2pie(6.4x10^6)/7923 =about 5075
thats in seconds so do 5075/60/60 for hours is 1.4

I think you forgot V^2= omega^2*r

Omega^2= 4pi^2/T^2

Anyone know how to do MC of june 2013 Q 13? The one with masses M and 4M, i try to add g=Gm/r^2 for both M and 4M and then equate to 0, but then I just end up with a long formula like this
5y^2 + d^2 -2dy=0
I'm sure it shouldn't be this hard for a MC question ..?

I didnt forget just did it a simpler way, try it and you will get exactly the same answer

Just do that GM/r^2 - G4M/(d-r)^2 = 0
Then you can cancel the GM's and get rid of denominators and you'll have
(d-r)^2 - 4r^2 =0
Expand brackets
D^2 - 2dr + r^2 -4r^2 =0
Simplify and times by -1
3r^2 + 2dr -d^2 = 0
Then make into quadratic
(3r-d)(r+d)=0
Reject r=-d as it's a distance and distances can't be negative,
Therefore r=d/3
Then as it asks for the ration of y/d (I used r, so r/d)
Then (d/3) / d = 1/3 so therefore the answer is B


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Hey guys, just wondering how many marks you would give me on this question (if any), have attached the mark scheme as well:




Sorry for my terrible handwriting

No-one?

Ah, I'll just give myself 0

it's hard to read. and sorry only saw it when you quoted yourself now.

and from what I could read you'd definitely get 1

Sorry, told you my handwriting was terrible

I would get 1? which one is that?

the spring force thing, if I read your work right.

and if I were you I'd brush up on my writing

ok thanks , that mark puts me on 120UMS

Yeah, I try to make my handwriting neater in exams. But in past papers, its (usually) just me who needs to read it, so its a lot messier as i don't bother so much

Hi, I took a quick look at this and got 1.4 hours, i'm not sure if its right because I didn't check the scheme and I know sometimes you get an answer thats an option but its not right and its there to trick people who make silly mistakes anyway...

Firstly I thought for it to feel no weight it has to be travelling a a high speed. If something is travelling at such a speed the the reaction force on the earths surface becomes 0 when it loses contact.

so mg - R = mv^2/r if R is 0 the mg =mv^2/r
rearranging gives v^2 = gr (m's cancel)
so v^2 = 9.81x6.4x10^6 = 6278400
root that for v = 7923
then T is distance/speed =2pie(6.4x10^6)/7923 =about 5075
thats in seconds so do 5075/60/60 for hours is 1.4