AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread]

hey physics people if I have two planets and want to work out the potential at a point between the two planets what do i do? Calculate the potentials due to each planet separately and add right? Also in the formula book gravitational potential is negative, do you sum the magnitude, or sum the negatives to get an even more negative value. And if the point is in the centre between the two planets?

Original post by philo-jitsu

Been trying to get an answer for this....Im still at a loss lol

Gravitational potential is a scalar quantity. When considering a point between two planets, work has to be done in both directions to move it to that point, so you essentially add the potentials at a particular point.

So basically, at the midpoint between two planets of equal mass M, separated by distance r

the potential in moving from one planet is -GM/(0.5r) = -2GM/r
the potential in moving from the other is identical, -2GM/r

so the net potential is the sum, V = -4GM/r

hope this helps, if you need more ask :smile:

Reply 721

particlestudent

13

Original post by Ayaz789

Well is the picture you attached on page 22?

Yes it is.

Reply 722

bag2gies

Hi I was wondering if anybody could help me with question 13 on the june 2013 multiple choice??

cheers

Reply 723

philo-jitsu

2

Original post by alevelstresss

Gravitational potential is a scalar quantity. When considering a point between two planets, work has to be done in both directions to move it to that point, so you essentially add the potentials at a particular point.

So basically, at the midpoint between two planets of equal mass M, separated by distance r

the potential in moving from one planet is -GM/(0.5r) = -2GM/r
the potential in moving from the other is identical, -2GM/r

so the net potential is the sum, V = -4GM/r

hope this helps, if you need more ask :smile:

Thanks for this, I have a question though :-)

If the potential at distance r from earth is Xjoules, and then you add another earth at a distance of double r, so r is now the midpoint.

why would the potential add? because wouldnt you now use less energy to get to point r, becaus now you have an attractive force from the other earth, what I mean is no work would have to be done to move from a distance towards a single earth because the earth is attracting the mass, so when there are 2 earths wouldnt the amount of energy required to get to the mid point be less because you now have a force working 'with' the mass that is moving away from earth 1.

If you can make sense of what I just said you are quite brilliant lol

Reply 724

alevelstresss

2

Original post by philo-jitsu

Thanks for this, I have a question though :-)

If the potential at distance r from earth is Xjoules, and then you add another earth at a distance of double r, so r is now the midpoint.

why would the potential add? because wouldnt you now use less energy to get to point r, becaus now you have an attractive force from the other earth, what I mean is no work would have to be done to move from a distance towards a single earth because the earth is attracting the mass, so when there are 2 earths wouldnt the amount of energy required to get to the mid point be less because you now have a force working 'with' the mass that is moving away from earth 1.

If you can make sense of what I just said you are quite brilliant lol

The attractive gravitational force is not the same thing as potential. Potential is the work done per unit mass in moving an object from infinity to a particular point in the field.

The effect of adding another Earth essentially just adds another gravitational field - work still needs to be done for this field, even if the gravitational force acts in the other direction, so the potential is doubled.

Just consider that potential is a scalar quantity, they will never cancel out like vectors do, its a hard concept to understand - but basically potential is kind of like the work done in the field, but there are now two fields so the work done has do be doubled.

Reply 725

D_Boyce

List of probable six marker topics?

Reply 726

Dann_a

2

Original post by WillRose

Right hand rule he meant, that's left :' :wink:

don't think iv ever seen right come up in an exam tbh

My bad 😂
In answer to the actual question I have had to use it once in section B to gain a mark and for two questions in section A before

Reply 727

alevelstresss

2

Original post by bag2gies

Hi I was wondering if anybody could help me with question 13 on the june 2013 multiple choice??

cheers

Gravitational field strength is a vector quantity, given by g = GM/r²

Make the two field strengths equal and find the distance:

let x be the distance of equal field strength from M

GM/x² = 4GM/(d-x)²
4x² = (d-x)²
4x² = d² -2xd +x²
3x² + 2xd - d² = 0
d = 1/3, d = -1 (using quadratic formula)

but d > 0, so d = 1/3 only

the distance is d/3 from the planet of mass M

so the answer is y/d = 1/3

Reply 728

philo-jitsu

2

Original post by bag2gies

Hi I was wondering if anybody could help me with question 13 on the june 2013 multiple choice??

cheers

nvm answer above me makes waaayyy more sense

(edited 7 years ago)

Reply 729

spanisholive

1

Was wondering if anyone could help me out with the question, for (a)(i) the answer is from P to S, but wouldn't that make QR remain still, as it is parallel to the magnetic field?

Reply 730

Yo12345

2

ImageUploadedByStudent Room1466348745.206486.jpg

Just to clarify. Is no work done when a charge moves perpendicular to a uniform electric field? Is work only done when the charge moves parallel to the field lines?

Posted from TSR Mobile

Reply 731

alevelstresss

2

Original post by spanisholive

Was wondering if anyone could help me out with the question, for (a)(i) the answer is from P to S, but wouldn't that make QR remain still, as it is parallel to the magnetic field?

Yes, but only a section of the wire experiences the force, not all of it, the force is experienced by the two longer sides

Reply 732

spanisholive

1

Original post by alevelstresss

Yes, but only a section of the wire experiences the force, not all of it, the force is experienced by the two longer sides

But it says in the question that PQ and RS are parallel to B? The mark scheme also says that these two sides don't experience a force :confused:

Reply 733

SirRaza97

7

Original post by bag2gies

Hi I was wondering if anybody could help me with question 13 on the june 2013 multiple choice??

cheers

Reply 734

asdf0323

Does anyone have any predictions as to what the six marker might me?

Reply 735

SirRaza97

7

Original post by spanisholive

But it says in the question that PQ and RS are parallel to B? The mark scheme also says that these two sides don't experience a force :confused:

Hey can i see the paper?

Reply 736

spanisholive

1

Original post by SirRaza97

Hey can i see the paper?

Sure, here it is :smile:

Question 6 btw

Unit_4_-_Section_B__Waves__Fields_and_Nuclear_Energy_Question_Paper-3.pdf217.3KB

Reply 737

philo-jitsu

2

Original post by alevelstresss

The attractive gravitational force is not the same thing as potential. Potential is the work done per unit mass in moving an object from infinity to a particular point in the field.

The effect of adding another Earth essentially just adds another gravitational field - work still needs to be done for this field, even if the gravitational force acts in the other direction, so the potential is doubled.

Just consider that potential is a scalar quantity, they will never cancel out like vectors do, its a hard concept to understand - but basically potential is kind of like the work done in the field, but there are now two fields so the work done has do be doubled.