---> Physics Paper 4 Past Paper Discussion <---

[vivo990]

The most thing I hate about physics P4 is that some questions are freaking looong, and sometimes 70% of the paper are structured questions and the rest are calculations. Initially, I took physics to get away from writing long responses, and enjoy solving quantitative problems. Now, I feel the exam is more English than physics!!

 

[Msbh22]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w10_qp_41.pdf

Question no. 1 (c) (iii) what is the shape of the graph??

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_41.pdf

Question no. 1 (c) (iii) what is the shape of the graph??

The graph starts off quite high on the y axis (but not at x = 0, at "surface of Earth") and decreases as a curve till a point closer to the "surface of Moon" than the "surface of Earth", at which point it goes to zero. Then it changes polarity (positive to negative of negative to positive, both are correct) and increases until it reaches a final value when the graph ends at "surface of Moon"; however, this maximum is not as high as that at the beginning of the curve at the "surface of Earth".

The explanation is this:

The gravitational field strength at the surface of the Earth is pretty high, and as you move further from Earth, this field strength decreases. At approximately the point you calculated in the previous part of the question, the net field strength (the sum of the gravitational pull per unit mass from both the Moon and the Earth) goes to zero and you can represent this on your graph by having the line decrease to zero.

If your initial potential is positive, the potential after that point then becomes negative; if the initial potential at Earth was negative, then the potential at the Moon is positive. The reason for this is that the force per unit mass is in opposite directions; the Moon pulls in one direction, the Earth pulls in the other direction; that is why the potential can be zero at one point; it is the point where the pull from the Moon is the same as the pull from the Earth.

However, since the Moon has a much lower mass than the Earth, this maximum is not as high as it is at the surface of the Earth.
Since the field strength decreases according to the inverse square law (Strength is proportional to 1/r^2), the gradient of the graph slowly decreases until the
zero-strength point and then the gradient increases(in the negative sense; literally, the gradient decreases). Hopefully the attachment will clear this up at least a little bit!

(The line is really scratchy, I hope it's not too bad!)

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w08_qp_4.pdf​

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Hey guys, could someone please help me out in the above paper? ​

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firstly i dont get in question 3 why are we multiplying v by 11? shouldnt the amplitude be 8 mm above at max?​

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secondly in question 7, could someone please explain b part 2 to me?​

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Thanks alot ​

We did the same question in class, it's a pretty confusing one!

In the question, it is mentioned that the needle undergoes S.H.M, and i'm quite sure we're just supposed to take it up blindly and go along - For all you know, it might be the cloth that is provided the force required for S.H.M! It might be that the cloth is needed for S.H.M and since it is a form of motion, the cloth shouldn't interfere with it, because that's what the paper is telling us!

It is given in writing that the oscillates between 22 mm, so there has to be a center of oscillation, and two extremes where velocity reduces to zero.
Considering these two points, we could say that while the distance between the needle and the cloth is 8 mm, the total distance between the two extremes of the needles motion is 22 mm, and thus the amplitude has to be 11 mm - even though the cloth is there, the needle simply moves through the cloth, and since the needle is carrying out this kind of oscillation, it's only what the needle does that we have to consider at this moment, and what it does is going 22 mm from one extreme to the other, resulting in an amplitude of 11 mm.

Are you still out there with the second question or have you got it down?

Hope this helped!
Good Luck for all your exams!

 

[Jinkglex]

We did the same question in class, it's a pretty confusing one!

In the question, it is mentioned that the needle undergoes S.H.M, and i'm quite sure we're just supposed to take it up blindly and go along - For all you know, it might be the cloth that is provided the force required for S.H.M! It might be that the cloth is needed for S.H.M and since it is a form of motion, the cloth shouldn't interfere with it, because that's what the paper is telling us!

It is given in writing that the oscillates between 22 mm, so there has to be a center of oscillation, and two extremes where velocity reduces to zero.
Considering these two points, we could say that while the distance between the needle and the cloth is 8 mm, the total distance between the two extremes of the needles motion is 22 mm, and thus the amplitude has to be 11 mm - even though the cloth is there, the needle simply moves through the cloth, and since the needle is carrying out this kind of oscillation, it's only what the needle does that we have to consider at this moment, and what it does is going 22 mm from one extreme to the other, resulting in an amplitude of 11 mm.

Are you still out there with the second question or have you got it down?

Hope this helped!
Good Luck for all your exams!

Ahh alright, so we have to take the total movement of the needle as one complete deviation to both sides of the mean position! Thanks so much!

 

[1357913579]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_41.pdf
number-4 please can u help iam totally confused about this question
thanks alot

 

[MaxStudentALevel]

Hey sagar65265
Thats really great of you! And you're welcome!
Could you please explain MJ2010 41 / Question 9 about the op amp? And its calculations?
Its the only one im stuck on! And a SIMPLE to the point exp is wha i need

 

[1357913579]

another doubt is
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_42.pdf
number-4
in this queston iam confused like in the middle wouldnt the potential be 0 like when the electron would be at 6micro second so would the potential be 0?
and what if they ask the same question but now with a proton then how to deal with the question??
thanks alot for your cooperation

 

[MaxStudentALevel]

Also sagar65265 in MJ 2010 42 Question one part b (ii) how do we find change in potential? All i need to know is where that value come from??

 

[backtodev]

Wow sagar65265 you really know your physics well mashallah!
Quick question to anyone - what does your diagram look like in 10 b)
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_42.pdf

 

[backtodev]

Also sagar65265 in MJ 2010 42 Question one part b (ii) how do we find change in potential? All i need to know is where that value come from??

change in k.e. = change in p.e
so k.e. which is 1/2 x m x v squared is equal to m x gravitational potential energy right?

use the graph to find the gravitational potential energy at 3R
because if it's at a distance of 2R above Earth's surface, it's R + that 2R distance right? so you use the 3R value from the graph
[hope that made sense]

 

[backtodev]

and that question I was asking earlier but wrote the question wrong
sagar65265
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_4.pdf
Q1 b) ii)

this one
(ii) The cord and mass rotate so that the cord is vertically below C, as shown in Fig. 1.3. Calculate the length L of the cord, assuming it obeys Hooke’s law.

 

[MaxStudentALevel]

change in k.e. = change in p.e
so k.e. which is 1/2 x m x v squared is equal to m x gravitational potential energy right?

use the graph to find the gravitational potential energy at 3R
because if it's at a distance of 2R above Earth's surface, it's R + that 2R distance right? so you use the 3R value from the graph
[hope that made sense]

OHHHHHH THATS WHERE 2.1 COMES FROM!!!!
All i was doing wrong was missing out the 3R! Kept using 2R -_______-

I hope I dont get a silly confusion like this in the paper :'(

Anyways THANKYOUSOMUCH!!! I feel like im under so much oressure and every little bit helps!!! <3

 

[Zari]

can anyone help me i have a couple of dbts
-M/J 2010 varianr 42 Q4 c) ?
-O/N 2010 variant 41 Q2 c) iii) for R to P how do we find work done on gas ?
-O/N 2010 variant 43 Q4) c) i really dont understand how do we get 0.15 ?
-M/J 2011 variant 41 Q4)i )

i will be glad if someone would be able to clear my dbts i know they r simple but i really dont get them

 

[sagar65265]

Hey sagar65265

Thats really great of you! And you're welcome!​

Could you please explain MJ2010 41 / Question 9 about the op amp? And its calculations?​

Its the only one im stuck on! And a SIMPLE to the point exp is wha i need ​

​

9) a) i) The point is at the connection of the wire coming from the output to the inverting input - this is the point at which the net potential is virtually zero.​

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ii) An op amp has a huge gain, and if the op amp is not saturated, the difference between the inverting potential and the non - inverting potential is minimal - practically zero. So, if the difference in practically zero, the inverting input should have the same potential across it as the non-inverting input. But since the non inverting input is connected to the ground, it has a potential of zero, so the inverting input should also have a practically zero potential to prevent the op amp from being saturated. Therefore the inverting input is the virtual Earth, where the input potential is practically zero.​

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b) i) The total input resistance is 1200 ohms, so the gain is given by the expression -Rout/Rin = -4200/1200 = - 3.5. Since this is the gain, the output voltage is the gain times the input voltage. Now take a look at the battery. The battery is reversed - it's not how we would expect it to be, so this pushes the inverting input towards a negative potential = - 1.5 V. Therefore the output = -3.5 * -1.5 = 5.25 V.​

ii) When the light intensity on an LDR decreases, it's resistance increases. This means that the value of the closed - loop gain decreases and thus the value of the output voltage decreases too, because the gain * input voltage also decreases.​

​

Hope this helped!​

Good Luck for all your exams!​

 

[sagar65265]

another doubt is
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdf
number-4
in this queston iam confused like in the middle wouldnt the potential be 0 like when the electron would be at 6micro second so would the potential be 0?
and what if they ask the same question but now with a proton then how to deal with the question??
thanks alot for your cooperation

Yes, that is right, in the middle an electron would experience equal attraction from the two positive charges and a proton would experience equal repulsion from the same two charges and thus the net force (and the net field strength) would be equal to zero.

If they ask the question with a proton, I assume you're talking about part (c), then the work done should be the positive value of the work done there, because the potential at a point is the work done by an external force in moving unit test charge from infinity to that point; an electron would accelerate towards the protons, and to ensure that it goes at a constant velocity a negative amount of work has to be done. However, a proton will not move there due to the repulsion so an external applied force would have to do positive work in moving the proton to any point in the diagram.

The question says the work done by the electron, not the external force, so the work done is positive. If a proton were involved, I would guess that the answer would be the negative version of the same value since both the proton and the electron have the same charge magnitudes.

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

Nope. The potential wouldn't be zero because the charges on point A and B are both positive. The electron is attracted to both the charges, hence the two potentials are added.

But since they're both in opposite directions, wouldn't that mean that they cancel each other out? Because that's what the top answer on this Yahoo Answers page says:

http://answers.yahoo.com/question/index?qid=20120322031104AAaEyQJ

 

[sagar65265]

and that question I was asking earlier but wrote the question wrong
sagar65265
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf
Q1 b) ii)

this one
(ii) The cord and mass rotate so that the cord is vertically below C, as shown in Fig. 1.3. Calculate the length L of the cord, assuming it obeys Hooke’s law.

I'm pretty sure you wrote it right earlier but I answered the wrong question... silly me!

Anyways, here's the solution:

The mass is undergoing circular motion at the bottom of the path. At this time, there are two forces acting on it, its weight and the tension in the rope. For the mass to undergo circular motion, these two forces have to add up to provide the centripetal force. However the problem is that they are in opposite directions, so because the centripetal force needs to be directed towards the center, the tension in the rope has to be greater than 5.0 N.

Considering the forces on the ball,

Tension - Weight = m * Angular Frequency^2 * Radius

Radius = Original length of rope + extension (which we need to find)

So Tension = Weight + (5/9.81) * (8.7^2) * (0.13 + extension)

The tension in the rope causes it to stretch, so if the rope indeed does obey Hooke's Law, the tension in the rope = kx

We can figure out the force constant by using the information they gave us earlier, that the length of rope when the mass is suspended with no motion occurring is equal to 1.8 cm = 0.018 m:

5.0 N = k * 0.018 m
k = 5.0N/0.018 m

Substituting these values,

k * extension = 5.0 + (0.509 * 75.69) * (0.13 + extension)
(5.0/0.018) * extension = 5.0 + 5.015 + 38.577 * extension

Transposing the equation and solving for extension we get

extension =10.01 /239.199 = 0.0418 m = 4.18 cm

So the final length = 13cm + 4.18cm = 17.2 cm

Hope this helped!
Good Luck for all your exams!

 

[1357913579]

Yes, that is right, in the middle an electron would experience equal attraction from the two positive charges and a proton would experience equal repulsion from the same two charges and thus the net force (and the net field strength) would be equal to zero.

If they ask the question with a proton, I assume you're talking about part (c), then the work done should be the positive value of the work done there, because the potential at a point is the work done by an external force in moving unit test charge from infinity to that point; an electron would accelerate towards the protons, and to ensure that it goes at a constant velocity a negative amount of work has to be done. However, a proton will not move there due to the repulsion so an external applied force would have to do positive work in moving the proton to any point in the diagram.

The question says the work done by the electron, not the external force, so the work done is positive. If a proton were involved, I would guess that the answer would be the negative version of the same value since both the proton and the electron have the same charge magnitudes.

Hope this helped!
Good Luck for all your exams!

thanks alot bro for your cooperation
this is whats written in the examiner report:
Very few candidates were able to cope with the two charges and determine the potential change
between the two points. A large number did not realise that the potential at each point was the
sum of the potentials for each charge. A number of candidates assumed the midpoint was at zero
potential or that there was only one charge. A significant number tried to determine the work done
using W = Fd or Vq. Very few were able to determine the change in potential between the two
points for both charges and then determine the work done on an electron.
Bro can you please solve this question to get the value of work done as i have tried for hours and hours still getting the wrong answer.

 

[1357913579]

But since they're both in opposite directions, wouldn't that mean that they cancel each other out? Because that's what the top answer on this Yahoo Answers page says:

http://answers.yahoo.com/question/index?qid=20120322031104AAaEyQJ

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_41.pdf
number-4 please can u help iam totally confused about this question
thanks alot
iam only able to a(i) part nothing other than this iam totally 0 about how to solve the rest of the parts of this question