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AQA Physics Unit 4 Thread - JAN 28th 2010 (Post Exam) Watch

1. (Original post by Ram92)
7 is 18 - (-12) which is 30 then times that by the mass!
then you get the answer and you know it is going away from the racket
ok ive dun it all, wot questions did u need?
2. what paper are you guys doing??
do you think i'm wasting time revising HSM and forces? I should be doing induction shouldn't I?
3. (Original post by futuredoc77)
what paper are you guys doing??
do you think i'm wasting time revising HSM and forces? I should be doing induction shouldn't I?
Probably, most of the SHM stuff is all formula sheet, and forces aren't quite so revisable, again calculations. So if you are ok at maths, then I would/am revising more on fields and induction.
4. (Original post by Chriz M)
Probably, most of the SHM stuff is all formula sheet, and forces aren't quite so revisable, again calculations. So if you are ok at maths, then I would/am revising more on fields and induction.
ya u're right... i'll move onto fields...

btw, sorry its irrelevant to the thread, have you already done all 15 modules for Maths/fur.Maths?
5. The diagram shows two charges, +4 μC and –16 μC, 120 mm apart. What is the distance from the +4 μC
charge to the point between the two charges, where the resultant electric potential is zero?
A 24mm
B 40mm
C 80mm
D 96mm

Can someone help me with this please, which equations should I be using? I'm guessing electric potential will be 0 nearer +4uc but how do you actually work it out?
6. (Original post by Ram92)
http://store.aqa.org.uk/qual/gce/pdf...GPHYA4ASQP.PDF
hi i need help on the above link

Q 14 - where does the 2 come from? should it not be 2 on the top??

Q- 16 can someone help please

Q - 17 - I understand it accelerates as it changes direction but does it not also go in a circular motion?

THe answers are at the bottom

Also if you have not seen this paper before I hope it helps! and i hope you help me lol

for 14 you use v=2(pi)r/t and then square this and put in to 0.5mv^2. simplift the equation and you get the answer.

for 16 the volt is defined as so many joules of energy for every columb of charge, therefore it gets energy of ve when it enters this UNIFORM field. you set this equal to 0.5mv^2 and you get the answer.

for 17 it does not go in circular motion it follows a parabolic path, the force is not always perpendicular to the direction of motion rather it always act up. as there is a potential difference across it the electron will be accelerated.

and thanks i hadnt found these
7. can someone please explain no. 21 on this - http://store.aqa.org.uk/qual/gce/pdf...50-W-SP-10.PDF

thanks
8. (Original post by Mit321)
The diagram shows two charges, +4 μC and –16 μC, 120 mm apart. What is the distance from the +4 μC
charge to the point between the two charges, where the resultant electric potential is zero?
A 24mm
B 40mm
C 80mm
D 96mm

Can someone help me with this please, which equations should I be using? I'm guessing electric potential will be 0 nearer +4uc but how do you actually work it out?
find the potential for each of the charges (x is the point) then the other charge (120*10^-3-x) is the other .then add the two potentials so the resultant is 0
9. (Original post by MarkoVan)
for 14 you use v=2(pi)r/t and then square this and put in to 0.5mv^2. simplift the equation and you get the answer.

for 16 the volt is defined as so many joules of energy for every columb of charge, therefore it gets energy of ve when it enters this UNIFORM field. you set this equal to 0.5mv^2 and you get the answer.

for 17 it does not go in circular motion it follows a parabolic path, the force is not always perpendicular to the direction of motion rather it always act up. as there is a potential difference across it the electron will be accelerated.

and thanks i hadnt found these

Thanks for the help and your welcome for the paper!

Q 14 - If you square 2(pi)r/t you get 4pi^2r^2/t^2 and then you times it by m and 0.5 which would get you 2((pi)^2)r^2/t^2

16 thanks for that but why do you equal it to K.E??

17 - thanks ... so when does a particle go into circular motion in a field then??
10. When working out the gravitational force of a satellite orbiting the earth, is the radius, the radius of the orbit or the radius of the orbit + radius of the earth.
Thanks
11. (Original post by Ram92)
Thanks for the help and your welcome for the paper!

Q 14 - If you square 2(pi)r/t you get 4pi^2r^2/t^2 and then you times it by m and 0.5 which would get you 2((pi)^2)r^2/t^2

16 thanks for that but why do you equal it to K.E??

17 - thanks ... so when does a particle go into circular motion in a field then??
in question take r to be D/2 not and then it should work out.
in 16 you are asked for the velocity so you set the kinetic energy to the energy gained when it is accelerated into the field and then make v the subject of the equation.

you only get circular motion when the resultant force is always perpendicular, in this case it is always directed up. In electric fields the force is always directly upward but in magnetic fields the force is always perpendicular to the direction of motion thats why you get circular motion it that case. though i only learnt/realised yesterday when my teacher went over it.

btw sorry for the poor grammer im sooo tired lol
12. (Original post by Natsel)
When working out the gravitational force of a satellite orbiting the earth, is the radius, the radius of the orbit or the radius of the orbit + radius of the earth.
Thanks
the radius of the earth plus the radius of orbit from the surface of the earth. You would typically use the radius provided unless they say the radius is from the surface of the earth then you would have to adjust it to include the radius of the earth
13. (Original post by MarkoVan)
in question take r to be D/2 not and then it should work out.
in 16 you are asked for the velocity so you set the kinetic energy to the energy gained when it is accelerated into the field and then make v the subject of the equation.

you only get circular motion when the resultant force is always perpendicular, in this case it is always directed up. In electric fields the force is always directly upward but in magnetic fields the force is always perpendicular to the direction of motion thats why you get circular motion it that case. though i only learnt/realised yesterday when my teacher went over it.

btw sorry for the poor grammer im sooo tired lol

thank you so much! I finally understand that now!
Do not worry i am not one of those people who are like "your grammar is so bad you should be shot"!! lol

Good luck tommorow I can tell you will do good!
14. (Original post by Ram92)
thank you so much! I finally understand that now!
Do not worry i am not one of those people who are like "your grammar is so bad you should be shot"!! lol

Good luck tommorow I can tell you will do good!
no problem i know the feeling i get so wound up went i cant understand some thing lol so anything to help.

and thanks same to you, good luck man!
15. does anyone know a secret link to Jun 09 Unit 4 AQA past paper?
16. (Original post by Ram92)
http://store.aqa.org.uk/qual/gce/pdf...GPHYA4ASQP.PDF
hi i need help on the above link

Q 14 - where does the 2 come from? should it not be 2 on the top??

Q- 16 can someone help please

Q - 17 - I understand it accelerates as it changes direction but does it not also go in a circular motion?

THe answers are at the bottom

Also if you have not seen this paper before I hope it helps! and i hope you help me lol
hey. for Q14. the 2 comes from the 1/2 from kinetic energy = 1/2(mv^2)

hope that helps. as for the other questions. i ain't so sure with how to get it.
17. hey, does anybody know if we need to know about the cyclotron? i have no idea where the book got v = BQR/m from
18. (Original post by futuredoc77)
does anyone know a secret link to Jun 09 Unit 4 AQA past paper?
we are the first batch of people to sit it...so there is no past paper
19. I'm gulping at this exam later on...

20. am i the only idiot that does not feel ready at all?

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