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

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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


Fair enough, lower the amount of marks required the better! :wink: :P
Original post by President J
I got power of 41.


lol same, seems impossible tho
Original post by Jm098
also when talking about the energy conversions, I said GPE into KE and frictional forces in the tube and between pellets?


Between pellets wouldn't be right I don't think, but heat loss to the tube, air resistance, anything like that should be fine.
Reply 343
Avatar for BriO
BriO
1
Anyone remember was the the initial height the thing was dropped from in question 1.
http://i.imgur.com/Ogi2A0R.jpg Here is the workings for the S2 mass question if people want to see it =p
Original post by Jackhawkins21
Starts at Max displacement 30 up and is and opposite curve to the PE


i think ur probs right, but wat doesnt make sense is that from ur graph the glider wud never have zero kinetic energy, and i dont think thats right, thats why i drew it down to zero for max displacement.
Reply 346
Avatar for Reda2
Reda2
14
Someone make a poll please.

WHAT I DONT understand, why 2 questions about gravitational fields, and they missed almost HALF of the whole specification....They want us to work with NASA or what..
Original post by tobybes
For such a mathematical paper I actually think that would be fair!


For the people who take Maths and Further Maths it makes it easier for them so that might cancel out?
For sketching graph I got this
1466420245136.jpg53.8KB
Original post by BriO
Anyone remember was the the initial height the thing was dropped from in question 1.

1.7m bro
Original post by BriO
Anyone remember was the the initial height the thing was dropped from in question 1.


1.7m
Teeeeeeaaaaaachheeer cooooool? Wheeeerreeeee aaaaaaarreeeee yoooouuuuu????
Anyone do simultaneous equations to find out R1 and R2? I thought I was in a maths mechanics exam rather than a physics exam..
eZZZZ
Original post by Zonothene
here are the workings for the S2 question


I got 1.73*10^33
M = (4pi^2*radius^3)/(G * T^2)
= (4pi^2*(1.2*10^12)^3)/((6.67*10^-11)*(86400*365*4)^2)
Original post by Howshotmyexamis
Guys can u tell me why is it 1070?
(92+8.25sin55)×9.81 then i got 969


Because the force from the hose is actually from Newtons 2nd law. So F=mv/t with t=1 m=8.25 v=25. Then you resolve vertically and times by sin55. Then add the weight of the man.
http://www.thestudentroom.co.uk/showthread.php?t=4176590

Made a poll guys, votevotevote
Original post by BriO
Anyone remember was the the initial height the thing was dropped from in question 1.


I think 1.7m?
Original post by Danny_L
I got 1.73*10^33
M = (4pi^2*radius^3)/(G * T^2)
= (4pi^2*(1.2*10^12)^3)/((6.67*10^-11)*(86400*365*4)^2)


As its a binary system you have to use this if you wish to do it using kepler's 3rd law (i.e adding the two masses together in the equation).
Keepo.gif0.7KB
Original post by AlexeiLipov
For sketching graph I got this


**** sake. I had KEmax as 80mj. So that means my second part of the answer is wrong too. How many marks would I lose for that?

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