# AQA PHYA4 ~ 13th June 2013 ~ A2 Physics Watch

Announcements

Report

#121

(Original post by

Well work done is energy, so m*deltav, grav potential is work done per unit mass (in bringing a test mass from infinity to that point) so deltav=deltaW/m, and gravitational potential difference is just the difference in grav potential between two points, so if one point from earth is -36MJkg^-1 (A) and another is -12Jkg^-1 (B), the grav potential difference in moving from A to B is -12--36 = +24 Jkg^-1, which makes sense because the further you get from the planet, the less negative the potential gets and it's an attractive force so you are doing work against it to move further away from it.

**jonnyb123**)Well work done is energy, so m*deltav, grav potential is work done per unit mass (in bringing a test mass from infinity to that point) so deltav=deltaW/m, and gravitational potential difference is just the difference in grav potential between two points, so if one point from earth is -36MJkg^-1 (A) and another is -12Jkg^-1 (B), the grav potential difference in moving from A to B is -12--36 = +24 Jkg^-1, which makes sense because the further you get from the planet, the less negative the potential gets and it's an attractive force so you are doing work against it to move further away from it.

1

reply

Report

#122

(Original post by

so difference from work done and gravitational potential is that work done is for the whole mass whereas gravitation potential is based on 1kg of the mass?

**masryboy94**)so difference from work done and gravitational potential is that work done is for the whole mass whereas gravitation potential is based on 1kg of the mass?

0

reply

Report

#123

(Original post by

ok so basically we know that but we also know the distance (circumference) around a circle is so sub that in to make ( it is T and not t because after one whole cycle around the circle in this case the planet, that is one time period) .... we also know that the gravitational force and centripetal force would equal each other because it says

**masryboy94**)ok so basically we know that but we also know the distance (circumference) around a circle is so sub that in to make ( it is T and not t because after one whole cycle around the circle in this case the planet, that is one time period) .... we also know that the gravitational force and centripetal force would equal each other because it says

**it only just remains on the surface**, therefore make v the subect to get make the 2 v's equal each other , then make T the subject to get which gives youall under square root, just that i cant type it...

0

reply

Report

#124

(Original post by

hey, thanks for answering. ok, i get it upto the point where you make the two v's equal. but when i rearrange to make T the subject i keep getting T = 4π^2r / GM

all under square root, just that i cant type it...

**bobthebuilder**)hey, thanks for answering. ok, i get it upto the point where you make the two v's equal. but when i rearrange to make T the subject i keep getting T = 4π^2r / GM

all under square root, just that i cant type it...

so then

square both sides to get

this is the same as saying

square root both sides to give you

so you get

0

reply

Report

#125

Does anyone have good notes/videos on the theory of the simple pendulum and mass-spring system?

0

reply

Report

#126

Hope I don't f up the multiple choice section this time, I was consistently getting 20+ then I ended up with something like 14

0

reply

Report

#127

Just did two multiple choices one after the other and got 21/25 each time, getting my confidence

Posted from TSR Mobile

Posted from TSR Mobile

1

reply

Report

#128

(Original post by

Just did two multiple choices one after the other and got 21/25 each time, getting my confidence

Posted from TSR Mobile

**SDavis123**)Just did two multiple choices one after the other and got 21/25 each time, getting my confidence

Posted from TSR Mobile

0

reply

Report

#130

(Original post by

ok so

so then

square both sides to get

this is the same as saying

square root both sides to give you

so you get

**masryboy94**)ok so

so then

square both sides to get

this is the same as saying

square root both sides to give you

so you get

0

reply

Report

#131

(Original post by

whats the answer to this and why? help appreciated

**masryboy94**)whats the answer to this and why? help appreciated

Because electric potential is scalar and field strength is a vector.

So if you had 2 positive charges with equal charge, the midpoint would have 0 field strength because you take one from the other, but you always add potential, so the electric potential would actually be doubled at the midpoint.

Hope that helps

0

reply

Report

#132

(Original post by

It's C

Because electric potential is scalar and field strength is a vector.

So if you had 2 positive charges with equal charge, the midpoint would have 0 field strength because you take one from the other, but you always add potential, so the electric potential would actually be doubled at the midpoint.

Hope that helps

**The H**)It's C

Because electric potential is scalar and field strength is a vector.

So if you had 2 positive charges with equal charge, the midpoint would have 0 field strength because you take one from the other, but you always add potential, so the electric potential would actually be doubled at the midpoint.

Hope that helps

0

reply

Report

#133

Hey guys can someone help me on this question please.

So I answered this question as D, I was wrong the answer is C

So first off how would you obtain the acceleration of the ball at any given time from this graph ?

and why cant you obtain the speed of the ball as it leaves the racket ?

Thanks

So I answered this question as D, I was wrong the answer is C

So first off how would you obtain the acceleration of the ball at any given time from this graph ?

and why cant you obtain the speed of the ball as it leaves the racket ?

Thanks

0

reply

Report

#134

(Original post by

Hey guys can someone help me on this question please.

So I answered this question as D, I was wrong the answer is C

So first off how would you obtain the acceleration of the ball at any given time from this graph ?

and why cant you obtain the speed of the ball as it leaves the racket ?

Thanks

**Olive123**)Hey guys can someone help me on this question please.

So I answered this question as D, I was wrong the answer is C

So first off how would you obtain the acceleration of the ball at any given time from this graph ?

and why cant you obtain the speed of the ball as it leaves the racket ?

Thanks

I'll think about it, doing chemistry right now, anyone else know?

1

reply

Report

#135

(Original post by

To get the speed of it you need to know its mass, as v=F*deltaT/m

**jonnyb123**)To get the speed of it you need to know its mass, as v=F*deltaT/m

0

reply

Report

#136

(Original post by

To get the speed of it you need to know its mass, as v=F*deltaT/m...[edit]wait actually it says of known mass....hmm....

**jonnyb123**)To get the speed of it you need to know its mass, as v=F*deltaT/m...[edit]wait actually it says of known mass....hmm....

I guess you would need final vel then thanks for pointing the known mass bit out haha

0

reply

Report

#137

For anyone wondering, the paper in january 2013 topics examined were:

Q1 - 8 marks total

One Similarity & Difference of inelastic/elastic collisions

Momentum calc and kinetic energy "show that ..." question

Q2 - 12 marks total

Satellite in orbit - calculate the height above the earth of a satellite in geosynchronous orbit

Centripetal force calc.

Compare principle features of geosynchronous and polar orbits, explain consequences for possible uses of satellites in these orbits (6 marker)

Q3 - 5 marks total

Capacitors

Q4 - 12 marks total

Electric fields calculations and define Coulomb's law and drawing electric field lines surrounding charges

Q5 - 13 marks total

Define Lenz's law, electromagnetic induction q's - all written answers

So topics not on this paper were:

SHM

Magnetic fields

Transformers

So they might possibly come up next paper, transformers and SHM most likely, as they haven't been on for two papers now.

Q1 - 8 marks total

One Similarity & Difference of inelastic/elastic collisions

Momentum calc and kinetic energy "show that ..." question

Q2 - 12 marks total

Satellite in orbit - calculate the height above the earth of a satellite in geosynchronous orbit

Centripetal force calc.

Compare principle features of geosynchronous and polar orbits, explain consequences for possible uses of satellites in these orbits (6 marker)

Q3 - 5 marks total

Capacitors

Q4 - 12 marks total

Electric fields calculations and define Coulomb's law and drawing electric field lines surrounding charges

Q5 - 13 marks total

Define Lenz's law, electromagnetic induction q's - all written answers

So topics not on this paper were:

SHM

Magnetic fields

Transformers

So they might possibly come up next paper, transformers and SHM most likely, as they haven't been on for two papers now.

3

reply

Report

#138

Ok so I have a problem with this question

I managed to guess the right answer which is D

What I did was I compared the two time periods by using T=2pir/v

Then I put that Rr/Vr < Rs/Vs and tried to compare the two

I only managed to guess D because I know that the further away a planet is from the sun the longer it takes to make one full orbit. So a greater radius = greater PE

However how would you explain this like with equations ?

I managed to guess the right answer which is D

What I did was I compared the two time periods by using T=2pir/v

Then I put that Rr/Vr < Rs/Vs and tried to compare the two

I only managed to guess D because I know that the further away a planet is from the sun the longer it takes to make one full orbit. So a greater radius = greater PE

However how would you explain this like with equations ?

0

reply

X

### Quick Reply

Back

to top

to top