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

[Ariel Robert]

Which sensor is used to measure the length?

 

[sagar65265]

Which sensor is used to measure the length?

Measure the length of what?

 

[Ariel Robert]

Measure the length of what?

well for pressure its piezo electric, whats for length in metres...?

 

[blueberryyums]

If you are considering the altitude above the surface above a planet/moon/other celestial body, the .....!

Thank you so much. I roughly understand but I will fully understand if you help using an example if it is not too much work

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s03_qp_4.pdf

Number 1 c(i)

I just don't get what the R should be. It is quite frustrating..

 

[sagar65265]

Thank you so much. I roughly understand but I will fully understand if you help using an example if it is not too much work

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_4.pdf

Number 1 c(i)

I just don't get what the R should be. It is quite frustrating..

Not at all too much work, here it is:

First things first: all meters! Practically every formula in the syllabus would be defined for SI units alone, so they've given you values in kilometers; these need to be converted!
Anyways, it is said that the object is moved from the surface of the Earth to a point further away from it. They have asked you to calculate the change in gravitational potential of the object - this is just the difference between the gravitational potential at that point and at the surface. Since the value of G and the value of M (mass of the Earth) are constants, this only depends on R.

If we take the Earth to be a sphere with all its mass concentrated at its center, the object is 6.4 * 10^3 kilometers from the center when its at the surface. Thus, the value of R at the surface is 6.4 * 10^3 kilometers = 6.4 * 10^6 meters.

Now, once the object has been moved to an altitude (height above the Earths surface) of 1.3 * 10^4 kilometers, it is at a distance of:

13000000 + 6400000 = 19,400,000 meters from the center of the planet. Therefore, the value for R at the point where it has been taken to is 1.94 * 10^7 m.

Basically, when an object has been taken a distance away from the surface of a planet, the value of R is basically the distance from the center of the planet to the object, whether it is on the surface or above it.

Hope this helped!
Good Luck for all your exams!

 

[blueberryyums]

Not at all too much work, here it is:

First things first: all meters! Practically every formula in the syllabus would be defined for SI units alone, so they've given you values in kilometers; these need to be converted!
Anyways, it is said that the object is moved from the surface of the Earth to a point further away from it. They have asked you to calculate the change in gravitational potential of the object - this is just the difference between the gravitational potential at that point and at the surface. Since the value of G and the value of M (mass of the Earth) are constants, this only depends on R.

If we take the Earth to be a sphere with all its mass concentrated at its center, the object is 6.4 * 10^3 kilometers from the center when its at the surface. Thus, the value of R at the surface is 6.4 * 10^3 kilometers = 6.4 * 10^6 meters.

Now, once the object has been moved to an altitude (height above the Earths surface) of 1.3 * 10^4 kilometers, it is at a distance of:

13000000 + 6400000 = 19,400,000 meters from the center of the planet. Therefore, the value for R at the point where it has been taken to is 1.94 * 10^7 m.

Basically, when an object has been taken a distance away from the surface of a planet, the value of R is basically the distance from the center of the planet to the object, whether it is on the surface or above it.

Hope this helped!
Good Luck for all your exams!

DAAAAAAAMMMN!!! THANK YOU SO SO SO SO SO SO MUCH!

 

[sagar65265]

well for pressure its piezo electric, whats for length in metres...?

A piezoelectric crystal coupled with a receiver is also used to measure distances; it's the centerpiece of ultrasound production and detection and it's also used in SONAR (or, at the very least, the same principle is used in SONAR as for Ultrasound).

The pulse is released, and if the machine is connected to a CRO or any similar device, the pulse shows up on the screen. When the sound wave produced is reflected and subsequently arrives at the crystal/detector, another pulse is registered. The time taken for the pulse to travel there and back again is twice the distance between the wave producer and the object that the wave has been reflected off.

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

DAAAAAAAMMMN!!! THANK YOU SO SO SO SO SO SO MUCH!

I should be thanking you - it's the first time I had seen the question, and I got it wrong when I first tried it - I had forgotten to include the 10^4 thing, and had thus gotten it wrong, so thank you for reminding me how important powers of 10 are!

Good Luck for all your exams!

 

[1357913579]

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!

Thanks alot brother for your explanation
i have another doubt can u please clear out is that can u please draw the graphs in electroc field chapters of electric potential, potential energy, electric field strength for all of these in inifrom electric field as well as normal electric field please iam really confused at them with explanation please.

 

[MaxStudentALevel]

sagar65265

Oh goodness thank you SO much for all your help!
May yoube successful in your endeavors in life!!!
You really seem like a gift from God at thw moment!
THANKS AGAIN <3

 

[sagar65265]

sagar65265

Oh goodness thank you SO much for all your help!
May yoube successful in your endeavors in life!!!
You really seem like a gift from God at thw moment!
THANKS AGAIN <3

Thanks a lot for the kind words, but I really owe all of you guys a load, most of the questions posted here came from papers I probably wouldn't have checked - the 2007 paper, the 2008 paper, the specimen paper, 2011 paper 43 (our school does variant 42) - checking out these questions has really, really helped me loads, so I should be thanking you guys a lot, so Thank you so, so much!

Good Luck for all your exams!

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf
Help, in question 5 part b (ii). How does area equal to 21.2 cm^2 ?

I'm not really sure, but the examiners report said that the students who obtained the answer counted squares (individual small squares or 1 cm^2 squares) or drew a series of strips that covered the entire area under the curve. The answer is 3300 microCoulombs, and it would be pretty difficult to get an exact value, so the mark scheme does have an allowance of half a square centimeter.

I guess by approximating the graph from x = 0 to x = 3 as a parallelogram and approximating the rest of the graph using similar parallelograms or by approximating the rest of the graph to a triangle, a value of 21.2 or thereabouts can surely be obtained.

I tried counting the individual squares (which, as it turns out, is really annoying to do on a computer screen!) and got a value of 3600 microCoulombs. This value is WAY off the mark scheme, but it came somewhere close, so I guess you just have to do some measurements and calculate the area, it's worth the 4 marks to get the answer!

Good Luck for all your exams!

 

[Freaked out]

divide the area in under the graph into small boxes of 5 horizontal and 5 vertical and then count the number of boxes which might be 21 and multiply it by the area of one box giving you an answer about 3280 microcolumbs

 

[Freaked out]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s03_qp_4.pdf
can some one please explain and solve Q2???!!!

 

[blueberryyums]

AL

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_4.pdf
can some one please explain and solve Q2???!!!

ALRIGHT!!! I GOT THIS! (but not everything though)

2ai x x tick tick

x - because molecules are not moving therefore P.E is zero.

x - because K.E energy is increasing and P.E is decreasing

tick- because molecules are moving which means PE has increased / gained energy

tick - stretching involves moving the molecules therefore PE is increased.

 

[blueberryyums]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_4.pdf
can some one please explain and solve Q2???!!!

b.)

SHC eqn is

C= Q/mT

the mass of gold is 0.95 of the mass because the other 5 percent of it contains silver

Heat lost by the liquid gold is

0.95m x 129 x T (this is basically rearranging the above eqn to get Q which is energy)

Heat gained by the silver is

0.05m x 235 x (1340 - 300) + 0.05m x 10500

Equate both,

122.5mT=17470m
T = 143 K

Total temperature = 143 + 1340 = 1483 K

c.) Thermocouple (because it has a large sensitivity range)

 

[blueberryyums]

0.05m x 235 x (1340 - 300) + 0.05m x 10500

I'll explain this better.

Why is it ((1340 - 300) you may ask?

Because of the change in temperature. You convert 27 C into Kelvins therefore, (27 + 273 = 300)


What does this equation mean, it looks soooo foreign to me.

0.05m x 10500

This is the Specific Capacity Heat of Fusion for Silver. Seeing as it is in kJ we change it to J which is 105,00

You have to play with the equation of specific latent heat and specific heat capacity.

Hope I helped PM me for any further questions or you can just send it here. I'll be here 24/7 till Paper 4 on Wednesday!!

 

[blueberryyums]

TIPS FOR OTHERS,

Make use of the examiner reports. They are our friends and when we have trouble, it opens the door a little for us.

 

[Freaked out]

TIPS FOR OTHERS,

Make use of the examiner reports. They are our friends and when we have trouble, it opens the door a little for us.

Thankyou!!!!!

 

[blueberryyums]

Thankyou!!!!!

You are most welcome.