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# Official OCR A2 G484 June 2016 [Module 1] [90UMS] watch

1. (Original post by PotAuFeu)
To add onto what Revision King said...

2) Trap smoke in smoke cell
Cover top of cell with cover slip
Focus light onto cell using glass rod and bulb
Focus microscope on smoke particles in smoke cell

3) E=kT, E is kinetic energy, k is Boltzman constant, T is absolute temperature

4) I'm not sure if this is technically in the specification but you can look at the AQA formula sheet to see what affects period. i.e. for SHM on a spring, period is proportional to root of mass (but NOT length). For SHM on a string, period is proportional to root of length of string (but NOT mass).

5) Used in circular motion.
Not sure if this exactly what you want but for weightlessness for motion in a vertical motion, a=g at the top of the circle. Hence weight provides the centripetal force so no contact force between you and the vehicle so you feel "weightless".
Thank you so much. This will sound stupid but to clarify for 1), is the light shone through the glass rod to focus it?

For 5) could you use a = v^2/r when they want you to work out g/orbital speed/radius, having given you one/two of the others?

Also, I have absolutely no clue about the attached June 2015 question. I used
SUVAT simply using v^2 = u^2 + 2as where u = (ucostheta)^2.

Any similar manipulation/derivation questions that you think could come up?
2. How likely do you think it is that there will be a question on an experiment about Boyle's law? I don't think there has been a question on it from recent past papers.
3. (Original post by 89911998a)
Thank you so much. This will sound stupid but to clarify for 1), is the light shone through the glass rod to focus it?

For 5) could you use a = v^2/r when they want you to work out g/orbital speed/radius, having given you one/two of the others?

Also, I have absolutely no clue about the attached June 2015 question. I used
SUVAT simply using v^2 = u^2 + 2as where u = (ucostheta)^2.

Any similar manipulation/derivation questions that you think could come up?
Yes, look at p62 in the textbook for the full method.

Yes, though I don't think I've ever used a=v^2/r directly on a paper yet. You're more likely to use g=GM/r^2, F=GMm/r^2 or F=mv^2/r

You've forgotten to attach it I believe, but assuming it is the same question check the attached picture. If my method is a bit unclear just tell me.
Attached Images

How likely do you think it is that there will be a question on an experiment about Boyle's law? I don't think there has been a question on it from recent past papers.
You might be right but i have no clue about any of ideal gas experiments 😒 (not mentioned in neither cgp or physics 2 official book) what type of of experiments u think they might ask about?
5. (Original post by Tajwar786)
You might be right but i have no clue about any of ideal gas experiments 😒 (not mentioned in neither cgp or physics 2 official book) what type of of experiments u think they might ask about?
If you go on physicsandmathstutor- a website which is very useful for physics revision, they describe an experiment on Boyle's law in detail. Look over it just in case it appears in tomorrows paper!

the experiment is used to provide evidence for the idea that pressure is inversely proportional the volume provided the temperature remains constant.
How likely do you think it is that there will be a question on an experiment about Boyle's law? I don't think there has been a question on it from recent past papers.
Hopefully unlikely, I didn't even realise the textbook explained one. Although, reading through it, there's only really one piece of equipment so if they did ask a question, it wouldn't be a 6-mark one. At the top of the page it also doesn't say "you should be able to... describe an experiment demonstrating Boyle's Law" so I assume that means we don't have to know it off by heart.
7. (Original post by PotAuFeu)
Yes, look at p62 in the textbook for the full method.

Yes, though I don't think I've ever used a=v^2/r directly on a paper yet. You're more likely to use g=GM/r^2, F=GMm/r^2 or F=mv^2/r

You've forgotten to attach it I believe, but assuming it is the same question check the attached picture. If my method is a bit unclear just tell me.
Thank you again - yes, I did forget to attach it but your working makes it perfectly clear.
8. (Original post by Danimtz)
For part ii on the first one, you calculate the force of the engine in 1 second (F = rate of change of momentum), so the mass of the all the xenon ions ejected in 1 second x their initial speed all over the time(1 second) . Then using Newton's third law you know the force on the spaceship and you can use F=ma to find acceleration.

For the laser question, you are given the energy of a single photon and the rate of emission of the photons. (Joules and "per second" --> Joules per second)
Then using he E=mc x change in temperature, you can find out the energy necessary to heat the titanium.

Joules/Joules per second = time
https://gyazo.com/33a1467023e54647e0703d5c9f04e82f

the third line I don't understand

To me it looks like

P = W/t

W x P/t x t = W

Am i right that they are using this or something else
9. Can anyone help with 1b on the June 2015 paper please? About the golf ball going for miles and miles?
10. (Original post by smartsy)
Can anyone help with 1b on the June 2015 paper please? About the golf ball going for miles and miles?
Its to do with suvat's, vertical/horizontal component of u.... quite long winded manipulation i suggest you watch the g484 2015 paper walkthrough on youtube by "cowen physics" he goes through the steps
11. (Original post by kingLAWZA)
https://gyazo.com/33a1467023e54647e0703d5c9f04e82f

the third line I don't understand

To me it looks like

P = W/t

W x P/t x t = W

Am i right that they are using this or something else
The third line in a word equation:

Energy of each photon(Joules) x photons emitted per second x t(time taken a.k.a. the value you are trying to find) = Energy required to raise the temperature of titanium.

This is basically doing: (answer from second line on MS)/(energy per photon * rate of photons) = time taken

If you think about it you are doing Joules/Joules per second = seconds.

Idk about your equation, I always get confused with the P and W on the formula sheet XD
12. (Original post by Artmanlikesart)
Can you explain why my answer is incorrect?
Surface area of Earth = 4pi/3(6.4 x 10^5)^3 = 1.56 x 10^21 m^2

Area captured by camera = 14(orbits) x 2pi x (6.4 x 10^5) x (3 x 10^3) = 1.9 x 10^15 m^2
Surely the satellite can't capture the entire surface then?

I
Surface area is 4pi×r^2
13. (Original post by smartsy)
Can anyone help with 1b on the June 2015 paper please? About the golf ball going for miles and miles?
Resolve vertically:
s=0
u= usin(theta)
a= g (this is isn't 9.81 since it's on the moon so just write g little m or something)
t = ?

Using: s= ut + 1/2 at^2

0 = usin(theta)t - 1/2 gt^2
1/2 gt^2 = usin(theta)t
t^2= 2usin(theta)t/g
t=2usin(theta)/g

Resolve horizontally( using speed = distance/time since there is no horizontal force):
Distance = x

x = speed * time(t)
x= ucos(theta) * time(t)
x= ucos(theta) * 2usin(theta)/g
x = (2u^2*sin(theta)cos(theta))/g
x=u^2 * sin(theta)cos(theta)/g

so x is proportional to u^2.

And you only get 3 marks for that :P (It looks complicated but it really isn't that bad, my working are just long)
14. how to answer the question: how a gas exerts a pressure on the wall of its container. i literally throw everything in each time, and still manage to miss out something out.
15. (Original post by lai812matthew)
how to answer the question: how a gas exerts a pressure on the wall of its container. i literally throw everything in each time, and still manage to miss out something out.
Large number or molecules each traveling at a speed v, when collide with a wall the particle has a change in momentum of 2mv , time taken to collide with the wall again is 2d/v so force = 2mv/2d/v. So pressure = sum of forces/ area. Area of one wall = dsquared
16. (Original post by Revision King98)
Large number or molecules each traveling at a speed v, when collide with a wall the particle has a change in momentum of 2mv , time taken to collide with the wall again is 2d/v so force = 2mv/2d/v. So pressure = sum of forces/ area. Area of one wall = dsquared
q 6b) june 2014
http://www.ocr.org.uk/Images/236084-...world-june.pdf
why do you have to mention newton's third law?
17. (Original post by lai812matthew)
q 6b) june 2014
http://www.ocr.org.uk/Images/236084-...world-june.pdf
why do you have to mention newton's third law?
N2L as since there is a change in momentum and time of collision there is a rate of change of momentum. N2L says since rate of change of momentum, there most be a force applied. Use this force to find pressure.
18. (Original post by lai812matthew)
q 6b) june 2014
http://www.ocr.org.uk/Images/236084-...world-june.pdf
why do you have to mention newton's third law?
Sorry, miss read N3L: Since force on molecule, must be force on wall too
19. guys, you know when dealing with specific heat capacity problems, you leave temperature in celcius
but when it is ideal gases question, you convert into Kelvins.

just want to confirm this
20. http://www.ocr.org.uk/Images/79737-q...nian-world.pdf
http://www.ocr.org.uk/Images/64639-m...ld-january.pdf
q3aii) jan 2012 how to answer it. i cannot understand what the answer is saying.

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