Edexcel Physics Unit 4 - 21st June 2011
Physics exam discussion - share revision tips in preparation for GCSE, A Level and other physics exams and discuss how they went afterwards.
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Re: Edexcel Physics Unit 4 - 21st June 2011It will only ever be zero once the magnetic flux becomes zero, but if not, the area of flux swepped/cut by a wire of N coils (per second) is BvLN = emf. This area swepped will be constant, thus the emf induced will be constant.(Original post by englishman129)
Just had a question guys-
You know if you have a wire with N coils, moving through a magnetic field of constant flux density, at at a constant speed. Will the grpah of emf induced over time be 0? -
Re: Edexcel Physics Unit 4 - 21st June 2011Thanks(Original post by Summerdays)
It will only ever be zero once the magnetic flux becomes zero, but if not, the area of flux swepped/cut by a wire of N coils (per second) is BvLN = emf. This area swepped will be constant, thus the emf induced will be constant.
And for -
when matter is always accomponied by antimatter - what conservation laws are obeyed..
would it be sufficent to say - conservation of charge and momentum? -
Re: Edexcel Physics Unit 4 - 21st June 2011Yes that's sufficient. You could also say, if you're really feeling it, mass-energy is also conserved.(Original post by englishman129)
Thanks
And for -
when matter is always accomponied by antimatter - what conservation laws are obeyed..
would it be sufficent to say - conservation of charge and momentum?
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Re: Edexcel Physics Unit 4 - 21st June 2011awesome!(Original post by Summerdays)
Yes that's sufficient. You could also say, if you're really feeling it, mass-energy is also conserved.
sorry - another one loll -
ive attached an image of the electric field lines between two tubes of a linear linac. i dont understand why they are curving in? On most google searches they are curving outwards. And for the anwser they specifically say they cannot be straight - must be curved in that fashion.
OK - i actually dont why they are curved and then secondly why they curve in and not out lol.
Thanks
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Re: Edexcel Physics Unit 4 - 21st June 2011oh oh , omg! same here!(Original post by Samia99)
Argh! I'm hating this! I g0t c4 the day b4! And bio5 aftr this! -____- :/
Anyways hw's prep g0ing every1n?
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Re: Edexcel Physics Unit 4 - 21st June 2011Hello, I have the same doubt.(Original post by Summerdays)
They got the r measurement by literally measuring the distance between the cage and the beam (which is providing a tension; the tension is the centripetal force) using a ruler to measure the distance of the horizontal bar attached between the beam and the cage, because it says "drawn to scale".
Everything you've written above is correct.
I got like 4.3 cm after measuring it, I took from the middle of the cage (assuming the centre of mass) n TO the middle of the horiz bar
this is incorrect?
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Re: Edexcel Physics Unit 4 - 21st June 2011Much like the electric field lines between two oppositely charged plates, the electric field at the edges of the plates begin to curve. The reason for this is because the plates don't extend to infinity - they have a finite width. The electric field lines are only approximately constant towards the middle of the plates. It's a similar story with that picture. The electric field is constant towards the middle of the electrodes, but the electric field lines start to curve at the edges.(Original post by englishman129)
awesome!
sorry - another one loll -
ive attached an image of the electric field lines between two tubes of a linear linac. i dont understand why they are curving in? On most google searches they are curving outwards. And for the anwser they specifically say they cannot be straight - must be curved in that fashion.
OK - i actually dont why they are curved and then secondly why they curve in and not out lol.
Thanks
Last edited by Summerdays; 16-06-2011 at 17:38. -
Re: Edexcel Physics Unit 4 - 21st June 2011Not to the middle of the horizontal line but to the end of the horizontal line, I think(Original post by inspiringsoul)
Hello, I have the same doubt.
I got like 4.3 cm after measuring it, I took from the middle of the cage (assuming the centre of mass) n TO the middle of the horiz bar
this is incorrect?
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Re: Edexcel Physics Unit 4 - 21st June 2011(Original post by Summerdays)
Much like the electric field lines between two oppositely charged plates, the electric field at the edges of the plates begin to curve. The reason for this is because the plates don't extend to infinity - they have a finite length. The electric field lines are only approximately constant towards the middle of the plates. It's a similar story with that picture. The electric field is constant towards the middle of the electrodes, but the electric field lines start to curve at the edges.
seems like you're an expert on this unit
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Re: Edexcel Physics Unit 4 - 21st June 2011Is there any way I can help?(Original post by cpdavis)
Mainly me guessing and waffling
But really have improved on that though, do hoping for a much better grade
anyways here's a short check list of unit 4- (just skim through it)
Further Mechanics
Momentum = mass x velocity (vector quantity)
Units =Kgm/s or Ns
Momentum is always conserved in all types of collisions, explosions and interactions.
-The total momentum of a system before any interaction is equal to the total momentum after it PROVIDED no external forces act.
(as external forces would allow momentum to transfer to external bodies)
Change in 'p' =mv-mu
Rate of change of 'p' = (mv-mu)/t
(According to Newton's second law) , F is directly proportional to (mv-mu)/t
Hence F= ma AND Ft(impulse) = mv-mu
-Remember to use sign convention when you do momentum calculations
If u have solved the Jan 2011 paper, there was this 'newton cradle'-- and how to experimentally find the conservation of momentum, there are a lot of ways to do this- one being light gates, K.E.- P.E. formula, etc
Know the difference b/w elastic collision and elastic collision (In ELASTIC-K.E. is conserved, while INELASTIC- K.E. is not conserved)
- Know how to find the expression and learn how to derive it K.E =(p^2)/2m
Circular Motion-Know the basic formulae; v= r x omega ,T = (2pi)/omega , centripetal force F = (mv^2)/r , F = m x r x (omega)^2 etc
Remember - Centripetal force CANNOT do work on the particle, but it pulls the particle towards it centre.
Centripetal acceleration is at RIGHT angles to the velocity which is towards the centre of the circle, it DOES NOT cause an increase or decrease in speed of the body, only a change in direction.
btw, I prefer to understand how all the formulae are derived, as it's easier to remember, if there's any doubt, I could share with you.
Note- Centripetal force is NOT produced by circular motion, it is the resultant force needed for circular motion, without it the object would travel in a straight line.
In hodder book, page 27 , there is a list of situations and 'what gives rise to centripetal force', so make sure.. u learn them. eg, vehicles taking a curve (which is the frictional force b/w the tyres and road) etc -
Re: Edexcel Physics Unit 4 - 21st June 2011Thanks. JazakAllah khair.(Original post by inspiringsoul)
Hope to include fields, particle physics soon.
Hope this helps
and hope u get the grade yo want!
Very helpful.
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Re: Edexcel Physics Unit 4 - 21st June 2011(Original post by inspiringsoul)
Is there any way I can help?
anyways here's a short check list of unit 4- (just skim through it)
Further Mechanics
Momentum = mass x velocity (vector quantity)
Units =Kgm/s or Ns
Momentum is always conserved in all types of collisions, explosions and interactions.
-The total momentum of a system before any interaction is equal to the total momentum after it PROVIDED no external forces act.
(as external forces would allow momentum to transfer to external bodies)
Change in 'p' =mv-mu
Rate of change of 'p' = (mv-mu)/t
(According to Newton's second law) , F is directly proportional to (mv-mu)/t
Hence F= ma AND Ft(impulse) = mv-mu
-Remember to use sign convention when you do momentum calculations
If u have solved the Jan 2011 paper, there was this 'newton cradle'-- and how to experimentally find the conservation of momentum, there are a lot of ways to do this- one being light gates, K.E.- P.E. formula, etc
Know the difference b/w elastic collision and elastic collision (In ELASTIC-K.E. is conserved, while INELASTIC- K.E. is not conserved)
- Know how to find the expression and learn how to derive it K.E =(p^2)/2m
Circular Motion-Know the basic formulae; v= r x omega ,T = (2pi)/omega , centripetal force F = (mv^2)/r , F = m x r x (omega)^2 etc
Remember - Centripetal force CANNOT do work on the particle, but it pulls the particle towards it centre.
Centripetal acceleration is at RIGHT angles to the velocity which is towards the centre of the circle, it DOES NOT cause an increase or decrease in speed of the body, only a change in direction.
btw, I prefer to understand how all the formulae are derived, as it's easier to remember, if there's any doubt, I could share with you.
Note- Centripetal force is NOT produced by circular motion, it is the resultant force needed for circular motion, without it the object would travel in a straight line.
In hodder book, page 27 , there is a list of situations and 'what gives rise to centripetal force', so make sure.. u learn them. eg, vehicles taking a curve (which is the frictional force b/w the tyres and road) etc
Thanks! Hows everyone scoring in the three papers and the spec paper?
I havent done em yet - just been using other practice papers as i want to save them for later. -
Re: Edexcel Physics Unit 4 - 21st June 2011
I need some serious help guys
, I am doing Physics AS & A2, thank god AS went well, but I finished A2 really late
, I didn't have time to prepare for it with all the exams, I need some worksheets as there are only 2 past papers and one assessment paper to practice, I would appreciate it if anyone could provide some worksheets it would be more helpful they had answers to help me get familiar with the way of solving the questions. THANKS
please help me


seems like you're an expert on this unit