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

What do you find most difficult in physics P4?

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MaxStudentALevel

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Hey guys!
This thread is specifically for tricky past paper questions!
After thoroughly revising th syllabus, its time to tackle thepastpapers!
Whether or not you like it, they form an important part of preparation,
So lets get past paper traps sorted out before they can catch us!

-Please update your findings of questions that caught you unsuspecting and weak!
So that together we can slay this beast called physics!!
(or we can just invest our time in inventing a time machine to go back and show albert einstein and other physicists the TV)
 
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sagar65265

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backtodev said:
Salam/hi
I've been stuck on this question for a long time despite doing it a couple of times in a class (I guess it just doesn't make sense to me)
Nov 07 Q1 b) ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf
Help would be great :)
Click to expand...


For this question, the requirement is to calculate the angular speed ω, and to do this, there are some things we need to realize first before diving into the calculations:

i) The elastic cord stretches whenever there is tension in it. Now, when the cord is rotating, it's pretty difficult to find out how much tension is in the cord (at the very least it will take some time and ink, but since time is important in these papers, we may have to find a way out!), but there is one point where there is no tension in the cord; when the mass is at the top of it's motion in the vertical plane.

ii) We can say this because they have told us that the cord is unstretched at the top of the circular motion of the mass. Going one step further, we can now see that the only force acting on the mass at the top of the rotation is it's weight.

iii) Lastly, the radius of it's motion at this stage, at the top of the rotation, is the unstretched length of the cord, which is equal to 13 cm = 0.13 m. Also, the mass of the rotating object = 5/9.81 = 0.509 kg

If we take the equation:

F = mω^2 * r

We have, for the motion at the top of the rotation,

5.0 = (5/9.81) * ω^2 * 0.13

Rearranging the equation, we get:

ω^2 = 75.46

ω = 8.686 = 8.7 rad s^-1

If you use the SI units for all the variables in this equation, your final answer should be in rad s^-1

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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backtodev said:
Oh and small question
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_4.pdf
How would you do Q3 b) i)
Click to expand...

The total energy of an oscillating system during the said oscillations remains constant;

i) At the maximum displacement the kinetic energy goes down to zero, but potential energy is stored in whatever object(s) provide the force that acts towards the center of the motion (e.g. energy is stored in the spring in this case that gets compressed and put under tension at each alternative displacement.)
ii) At the center, displacement = 0, there is no potential energy stored since the spring is not stretched at that point, but the reason the oscillation carries on is because the oscillating mass has kinetic energy and this kinetic energy then gets converted into the potential energy of the spring, etc.

So, since the total energy of the system is conserved and the kinetic energy is at its maximum when the potential energy is zero, the energy at any one point is the same as the energy at any other point; namely, the energy at zero displacement where k.e. is maximum and p.e. is zero, is equal to the maximum k.e. of the oscillation, so the answer is:

The line representing the total energy is the oscillations is a straight, horizontal line parallel to the x - axis passing through the peak of the
kinetic energy - displacement curve (it is a flat line tangent to the top of the curve, so to say.)

Hope this helped!
Good Luck for all your exams!
 
blueberryyums

blueberryyums

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Can someone please help me with Paper 4? I am self taught and finding some concepts hard to grasp. Even those that seem very very easy. For example, when we have an altitude, how do we incorporate it in the -GM/R equation to find the potential energy.
 
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sagar65265

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backtodev said:
sagar65265
Thanks for the Nov 08 explanation!

And ooops I meant Nov 07 3 b) ii) as in the next part of the question lol
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf
[skipped out the i when I was typing, my bad, but that was great explanation anyway!]
Click to expand...

Sorry, about that, didn't see it....

Okay, so for part 1. of the question, the frequency should remain constant;

damping in a system does not change the frequency of the system; the process of damping basically reduces the speed/ velocity of the oscillation of the system; when this happens, the amplitude also decreases ( I guess following a similar pattern) and so due to that, the time period remains constant. Since frequency is directly dependent on time period, the frequency doesn't change when damping is applied. So the frequency should be unchanged, so it should be 4.0 Hz.

For part (ii), using the graph to try and obtain an answer gives something way off the marking scheme - the examiners report implies that the question says "it decreases the energy BY 1.0 mJ" rather than "TO 1.0 mJ" because the examiners report says that, in it's exact words,

Candidates did not appear to realise that the graph would have the same basic shape, regardless of total energy. Thus, reading the value of x at (2.56 – 1.00) mJ would give the amplitude for a total energy of 1.00 mJ.

This implies that the energy was decreased BY 1.0 mJ rather than TO 1.0 mJ. Well, it's kinda messed up, ain't it?
Either ways, the messed up thing is that calculating the value of the amplitude, rather than reading it off the graph, gives the correct value according to the marking scheme even though the assumption (TO 1.0 mJ instead of BY 1.0 mJ) that we then have to make may be wrong! Anyways, here it is:

Since the frequency remains constant, we can use the following formula to calculate amplitude:

Vmax = 2 * pi * frequency * amplitude

We have the frequency (approximately 4.0 Hz) so

Vmax = 8 * pi * amplitude

If the maximum total energy is 1.0 mJ, the maximum kinetic energy is also 1.0 mJ and this occurs at the maximum velocity, so

10^-3 = 0.5 * m * Vmax^2
0.001 = 0.5 * 0.13 * (8 * pi * amplitude)^2
8 * pi * amplitude = sqrt(0.015) = 0.124
Amplitude = 0.0049 meters
Amplitude = 0.5 cm

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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backtodev said:
sagar65265
Thanks for the Nov 08 explanation!

And ooops I meant Nov 07 3 b) ii) as in the next part of the question lol
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf
[skipped out the i when I was typing, my bad, but that was great explanation anyway!]
Click to expand...

Damn, damn, damn, so, so sorry - which question did I reply on? I'm just really confused now, I just checked my earlier post, and it concerns a question you asked about 08, and I replied about 07 - sorry!!
Are you good on that one or still out there for that question?
Really, really, sorry!
 
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sagar65265

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Anika Raisa said:
PLEASE HELP!!!

Question: 3 b i 1

QP: http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_nos_sp_4.pdf
MS: http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_nos_sm_4.pdf

Thank u 4 helping in advance!!
Click to expand...

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!
 
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sagar65265

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MaxStudentALevel said:
Hey sagar65265 !!

Could you please help me unsderstand how
To use the formula for KE in order to calculate Energy of an oscillation??!?
Click to expand...

Sure, here's the method:

There's a formula that relates the amplitude, displacement and angular frequency of the mass undergoing simple harmonic motion to its velocity:

V = Angular Frequency * sqrt(Amplitude^2 - Displacement^2)

Note that the root can be positive or negative, so there are two points/displacements in every cycle of oscillation when the velocity has a particular value.
According to this equation, the velocity will be at its maximum value when the value under the root is maximum.
Since a square can never be negative, the minimum value it can take is zero.
If we set the displacement of the particle undergoing S.H.M as zero (it's at the equilibrium position), the equation becomes:

V = Angular Frequency * Amplitude

Since this is the maximum value the velocity can have, the equation then becomes

Vmax = Angular Frequency * Amplitude

Because Angular Frequency = 2 * pi * frequency,

Vmax = 2 * pi * F * Amplitude

So if you have the values of the frequency/the time period and the value of the Amplitude of the S.H.M, you can calculate the maximum velocity that the oscillating mass can have using the above formula. When the kinetic energy of an oscillation is at its greatest, the potential energy of the oscillation will be zero, so the maximum kinetic energy is equal to the total energy of the oscillation at any point in time.
Once you've found the value of the maximum velocity of the oscillating mass, you can use the standard KE formula:

KE = 0.5 * mass * Vmax^2

to calculate the maximum energy of the oscillation.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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1357913579 said:
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_43.pdf
1(c)(ii) what do we write for this?
because what i think is that answer should be is he hydrogen molecules dont have enough kinetic energy to move in mars so they escape they dont have enough k.e becuase they dont have enough temp as both ase proportional whats wrong in this answr??
please help
Click to expand...

The question has stated that the mean kinetic energy of an ideal gas molecules at a particular temperature is given by that equation. In any sample of a gas, at any temperature, there will be a massive number of molecules/atoms interacting with each other by means of random collisions - there is no way to predict how many of these collisions occur, where they occur, etc (at least at A level!). Due to all these collisions, energy transfers also take place.

There will be some molecules going at a very low velocity, almost not moving, which will then gain kinetic energy when they collide with faster moving molecules. There will be a large percentage of molecules that have an average amount of energy and will not transfer so much energy when they collide, so their velocities remain approximately the same. Then again, there will be a small proportion of molecules that have way, way, higher kinetic energy than the others, but this number of molecules is comparatively small.

This distribution of molecular velocities is known as the Boltzmann Distribution, so the equation they've given speaks only about the average kinetic energy; it averages out all the kinetic energy values that a sample of gaseous molecules can have, so in reality, in a sample with that average velocity, there will be a small proportion of molecules that have more kinetic energy than the average and a small proportion of molecules with less kinetic energy than the calculated average.

So, the bottom line should be that, at the temperature below the value you've calculated earlier, a large proportion of molecules with the average amount of energy will not be able to move fast enough to escape from the atmosphere of Mars, but since the calculated value is only an average, there will still be a group of molecules in the same sample that have a high enough, above average, velocity to enable them to escape from the atmosphere of Mars.

Sorry for the long explanation, sorry if I bored you!

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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blueberryyums said:
Can someone please help me with Paper 4? I am self taught and finding some concepts hard to grasp. Even those that seem very very easy. For example, when we have an altitude, how do we incorporate it in the -GM/R equation to find the potential energy.
Click to expand...

If you are considering the altitude above the surface above a planet/moon/other celestial body, the matter can be simplified pretty quickly

In any individual body, there are a huge number of particles; suppose you have to find out the force of attraction due to gravity between the Earth and a Football of some mass. There are a huge, huge number of particles in the football, and almost an uncountable number of particles in the Earth, and each one of those particles attracts another - how is it possible to calculate these values, when we have to add up so may forces?

Newton made a huge step in simplifying the calculations required to obtain these values by proposing a theorem:

Any spherical object of a particular mass will behave like a point mass placed at the center of this spherical shape with the same mass as before.

(Of course, this isn't the way Newton would have said it, i'm just paraphrasing!)
So if you consider the Earth to be a particle held at its center with the mass the same as that of the Earth, you can make the calculation much simpler.

Say the Earth has a radius of 6400 kilometers. This means that an object placed on the surface of the Earth is about 6400 kilometers from the center, so the value of R that you use is the radius.

If, instead, the object is kept at a height of a 1000 kilometers above the Earth's surface, the distance of this particle from the center of the Earth would be the radius of the Earth (6400 km) plus the height above the surface (1000 km), so the value of R you should use would be equal to

6400 + 1000 = 7400 km = 7, 400, 000 meters

Hope this helped!
Good Luck for all your exams!
 
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