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

What do you find most difficult in physics P4?

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sagar65265

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seneria said:
can someone solve question 4 C(!!)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_42.pdf
Click to expand...

The capacitance of the sphere, as calculated in the previous part of the same question, is 7 * 10^-11 Farads.
When the capacitor discharges, the energy stored in the sphere decrease to 25 % of the original, so the final energy is 1/4th of the initial energy:

0.5 * C * V(final)^2 = 0.5 * C * V(initial)^2 * 0.25

V(initial) is equal to 1.2 * 10^6 Volts, as is shown in the question.
C = capacitance = 7.0 * 10^-11 Farads

Therefore,

V(final)^2 = V(initial)^2 * 0.25
V(final) = 6.0 * 10^5 Volts

Hope this helped!

Good Luck for all your exams!
 
B

backtodev

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sagar65265 said:
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!
Click to expand...
seneria
sagar already answered one of your questions here
 
blueberryyums

blueberryyums

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seneria said:
for the first option work is being done on the system , so there must be an increase in internal energy reight? this is confusing me.. as water is BEING freezed

however, for the third option the water is evaporating , meaning work is done BY the system... so internal energy must decrease... because increae in the internal energy is work done on the system and heat is supplied to the system....

can youclear my confusion?
Click to expand...

Hello,
I am finding it difficult to find the question again. But if I remember correctly, Internal energy of the body is increasing
Internal energy = Total potential energy + Total kinetic energy

Water freezing at const. temp x
There is no temp change so therefore no change in KE. However, the motion of the particles decrease leading to a decrease in potential energy.
Water evaporating at const. temp tick
Evaporation means the molecules gain higher velocity which leads to higher KE and leads to an increase in Internal energy

I hope this helped a little :)
 
F

Freaked out

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okay, you mean that if we want to increar the voltage then we must increse the capacitors in series?
yes
 
Anika Raisa

Anika Raisa

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sagar65265 said:
For 3 b) 1, I guess you could just average out the maximum and minimum potential across the resistor, because the minimum is 4.8 V on the graph and the maximum is 6 V on the graph. Taking the average, it gives us 5.4 V, which should be the right way of getting the value (at least I hope so!).

Hope this helped!
Good Luck for all your exams!
Click to expand...
Thank you very much!! :)
 
F

Freaked out

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Q1.c(i):​
First of all imagine a mass in the middle of the earth and moon . Now imagine a force on the mass from the moon pulling it towards itself, and then imagine a force on the mass by the earth pulling it towards itself. as the forces are in opposite direction from both the fields of the earth and moon. They cancel each other out at that specific point.​
c(ii):​
Simply equate both the fields from the moon and the earth. label the distance of the imagined mass as x and as the overall distance is 60Re, the distance from the moon as 60Re-x. Equal them and simply find x.:)​
c(iii): confused myself in it.​
Q8.(c):​
As the neutrons are moving with speed in the fission reaction, they will also have kinetic energy. As the boron sod absorbs the neutron its movement stops hence, all of its kinetic energy is converted into heat energy making the rods become hot.​
 
5

50sum

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is it just me or the winter 2012 qp 42 is too easy? I have no idea how the grade boundary is still around 50 for A in this paper.
 
M

MaxStudentALevel

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50sum said:
is it just me or the winter 2012 qp 42 is too easy? I have no idea how the grade boundary is still around 50 for A in this paper.
Click to expand...

IKNOWWWW
I literally was like :O what is this???
Every single intuitive easy question! Whats gonna cme in our paper?? Lol!
 
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