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# 2016 | OCR A2 Advancing Physics B | G494 &amp; G495 | 20th &amp; 28th June watch

1. (Original post by terpanter)
i thought its out of 60??!!
It is out of 60 the person I replied to had made a mistake.
2. (Original post by NamelessPersona)
I got 170 to 2 s.f.
How could the bullet velocity be 170? that would lead to a ke value of around 14.5J for a 0.001kg mass, which isn't even close to 0.2J
3. (Original post by kennz)
ah ok, i didnt see that you had to compare it to C, i knew it would increase but convinced myself it was 1/4 as it had to have 4 in it
do we have to calculate boatsman factor? they didn't give us activation energy to temperature did they?
4. (Original post by jajajaaaaa)
How could the bullet velocity be 170? that would lead to a ke value of around 14.5J for a 0.001kg mass, which isn't even close to 0.2J
A lot of other people I spoke to also got around 170. Energy was lost, as you had to explain why kinetic energy was not conserved.
5. (Original post by terpanter)
do we have to calculate boatsman factor? they didn't give us activation energy to temperature did they?
If you are talking about the question with the helium atoms then you worked out the activation energy earlier.
6. (Original post by kennz)
was there an original line?
yes original line as in the line C; the line for the original conditions before things were halved and doubled
7. (Original post by jajajaaaaa)
How could the bullet velocity be 170? that would lead to a ke value of around 14.5J for a 0.001kg mass, which isn't even close to 0.2J
Exactly - a lot of kinetic energy is lost.

0.001kg * 170 m/s = 0.17 Ns
0.081kg * 2.1m/s = 0.17 Ns

Momentum is conserved
8. (Original post by jajajaaaaa)
How could the bullet velocity be 170? that would lead to a ke value of around 14.5J for a 0.001kg mass, which isn't even close to 0.2J
that is correct - that was the case. A lot of KE was lost in the collision.
9. (Original post by kennz)
ah ok, i didnt see that you had to compare it to C, i knew it would increase but convinced myself it was 1/4 as it had to have 4 in it

we have to calculate bolzman factor for atmosphere losing helium discuss question?
i thought they didn't give us activation energy and temperature of atmosphere?!?!
**** 3 marks gone
10. (Original post by Simonium1010)
average mark for an A is 41-43
So it is. I'm used to using higher boundaries in my class as my teacher would say that between 75% and 80% raw marks was an A. I still stand by my point of around 46 being an A for this paper though as it wasn't outrageously hard.
11. (Original post by NamelessPersona)
So it is. I'm used to using higher boundaries in my class as my teacher would say that between 75% and 80% raw marks was an A. I still stand by my point of around 46 being an A for this paper though as it wasn't outrageously hard.
no, true. Although I'd like to think the gulf between A and A* isn't as big as usual, i.e. only like 51 for 81ums = 90%ums...
12. (Original post by terpanter)
we have to calculate bolzman factor for atmosphere losing helium discuss question?
i thought they didn't give us activation energy and temperature of atmosphere?!?!
**** 3 marks gone
they gave you the minimum energy needed for one helium atom to escape (part a)
then they gave you the temperature for the boltzmann question you're talking about
13. (Original post by M12345689)
did anyone get 172 for the bullet velocity?
Yes !

Posted from TSR Mobile
14. Not too bad a paper tbh, found it lot more wordy than previous papers and thought the vacuum questions was a real curveball
15. (Original post by uk_shahj)
they gave you the minimum energy needed for one helium atom to escape (part a)
then they gave you the temperature for the boltzmann question you're talking about

Im so sorry, can anyone please tell me what are earlier questions for bolzman/helium about? Im scared i miss a page
16. How did you work out the velocity of the pellet?
17. (Original post by Miracle1)
How did you work out the velocity of the pellet?
Although I don't remember exact numbers you use conservation of momentum:
mass*velocity of pellet = total mass*velocity of tack and pellet. Using the energy from the previous part and equating to KE you can work out the velocity of the tack and pellet, allowing you to solve for u, giving around 170 ms^(-1) to 2 s.f.
18. i thought the gradient of the line on the first question remained unchanged, as the mass was doubled, number of molecules in the box was half, but half the pressure volume doubles.

question wasn't clear about the mass.
19. (Original post by danthebox)
i thought the gradient of the line on the first question remained unchanged, as the mass was doubled, number of molecules in the box was half, but half the pressure volume doubles.

question wasn't clear about the mass.
Double the mass implied double the number of particles.
20. (Original post by NamelessPersona)
Double the mass implied double the number of particles.
So the answer was line A? Because in that case I got it right without even reading the question properly… Just now I realised that I wasn't even looking at the numbers, hopefully I don't do that in the G495 exam

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