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

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49-51 A* , 56 full UMS.
Original post by Xabier
I struggled for so long on the very first question trying to understand the graph I rushed everything else. Can someone explain the graph in question 1? And how you got the energy for the lead pellets?


If you think about what the gradient represented, it was the change in velocity/ time, and (v-u)/t = acceleration, and therefore the gradient was equivalent to g

And therefore as the object is free falling, the acceleration is the same after both bounces and is equal to g.

And for the second part of the question, the area represented total displacement travelled, and as the ball lost energy after it bounced, it did not travel as far after the bounce, hence why the second triangle had a smaller area.
(edited 7 years ago)
Original post by Danny_L
That's what I used, but no, because there was residual epe, it was epe max - epe min = ke max


Fair enough, ecf please save me :P
Original post by AlistairMcDonald
I got 50 because max PE was 80 and min PE was 30.

Why did you get 80?


Because people slipped up thinking ke max is equal to epe max like it nearly always is. Myself included, was just rushing through fml
(edited 7 years ago)
Reply 324
Avatar for 1z3
1z3
1
What's the answer for mass of S2? I got 6.4x10^31 kg
Original post by screwedmofo
me too, no idea where they got 50 from.


Apparently residual potential energy, which now I think about it makes sense because the potential curve never reached 0.
Original post by Jackhawkins21
Max KE is 50 Because the graph started at 30 mj so 80-30=50


so how did you draw ur KE graph?
A year is the time taken for a complete orbit, so was unsure to use Earth year for mass calculation (4 x 365.25 x 24 x 60 x 60) or whether to use the time period :L in the end i assumed it meant earth year and got a value of something like 1.7x10^33 but overall i think that its bad how they are not specific on the questions. I also got the phase difference as 5pi/3 but people are saying mixed answers for it?? i counted the phase of the wave as 6 big squares and the difference as 5 squares hence phase diff = 5/6 x pi = 5pi/3
Original post by m00ngrk
E=mc theta
C inversely proportional to mass
E=mgh
E proportional to mass
C proportional to E, so proportional to m and inversely proportional to m
m/m is 1 so stays the same


how many marks was this?
Screwed up the KE graph..., thought the maximum was 80mJ. Such a silly error... It was two marks though, what would the other mark be for? Also will there be ecf for maximum speed with 80mJ, or will they only give ecf for the time period?
Original post by screwedmofo
so how did you draw ur KE graph?


Starts at Max displacement 30 up and is and opposite curve to the PE
Original post by President J
I got power of 41.


Fortunately for me, I know the Sun is 10^30. I reckon there was a power error somewhere there.
Original post by WilliamMoore
49-51 A* , 56 full UMS.


What do you guys think? :wink:
Original post by WilliamMoore
49-51 A* , 56 full UMS.


Yeah, nah last couple years they've kept 100%und in low 50's or there about, I reckon today was no exception, I presume that's where they're trying to set the bar. I guess 53 for 100% ums
Original post by wispaedits
A year is the time taken for a complete orbit, so was unsure to use Earth year for mass calculation (4 x 365.25 x 24 x 60 x 60) or whether to use the time period :L in the end i assumed it meant earth year and got a value of something like 1.7x10^33 but overall i think that its bad how they are not specific on the questions. I also got the phase difference as 5pi/3 but people are saying mixed answers for it?? i counted the phase of the wave as 6 big squares and the difference as 5 squares hence phase diff = 5/6 x pi = 5pi/3


I think most are saying 5/4pi, I think you might have miscounted :frown:
Original post by WilliamMoore
What do you guys think? :wink:


For such a mathematical paper I actually think that would be fair!
Original post by Danny_L
Yeah, nah last couple years they've kept 100%und in low 50's or there about, I reckon today was no exception, I presume that's where they're trying to set the bar. I guess 53 for 100% ums


I think it might be lower boundaries because of the damn binary star system question, threw a load of people over here.
Reply 337
Avatar for Jm098
Jm098
2
also when talking about the energy conversions, I said GPE into KE and frictional forces in the tube and between pellets?
Reply 338
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GWLZ1
Original post by screwedmofo
me too, no idea where they got 50 from.


When the slider is at the amplitude, the kinetic energy is 0J. But the total energy in the system is always the same which is 80mJ(kE is 0 PE is 80). Therefore when the PE is minimum 30mJ, the kE should be 80mJ - 30mJ =50mJ
Original post by Jm098
Does the temperature change half if mass doubles?


I don't think so because the energy is gained by mgh and m is doubled. Then Temperature change=E/mc before hand then 2E/2mc=E/mc so it is the same?

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