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OCR Physics A G484: 20th June 2016

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It's basically when the acceleration of the plate when it moves downwards is faster than free falls (9.81ms-2) so when the plate moves down faster than the particles ontop can fall then they will lose contact for a brief time
Make sure you remember the Joules->W->KWh conversions. I didnt find it in the textbook in this unit but it shows up in the papers
do we need to know : change in energy= mass* specific latent heat
ive never seen it come up
Reply 43
Anyone know what experiments could possibly come up in the paper?
Anyone got the 2015 past paper?
Reply 45
Original post by Stahri
Anyone know what experiments could possibly come up in the paper?


I think the specific heat capacity is coming up. It's on the 2014 June paper if you want to know it.
Mass of S2 - go!
Reply 47
That was not bad for me apart from that ****ING alphabet question, pisses me off! I almost lost all those marks. But that was not a normal paper, although they were all answerable but literally NO easy questions, even defining internal energy was hard.

For the energy of the pellets, did you guys do mgh * 50? or what did you do to get the E for the specific heat capacity.

And for the forces similarity I said,
- Same magnitude, same type of force
- Opposite direction, does not act on the same object.

Who agrees??

For the Kepler's law question, I did T2/r3 = T2/r3, right?

For the first question for the first part where you had to get speed. I did I = Ft, and then I = change in momentum to get speed. For the next part I used SUVAT, but last part of that question, finding the height I could not do, I tried but got 7m which is the initial height, and it can't be that.

Let me know what you guys did, and what you did for those questions.
(edited 7 years ago)
Original post by Reda2
That was not bad for me apart from that ****ING alphabet question, pisses me off! I almost lost all those marks. But that was not a normal paper, although they were all answerable but literally NO easy questions, even defining internal energy was hard.

For the energy of the pellets, did you guys do mgh * 50? or what did you do to get the E for the specific heat capacity.

And for the forces similarity I said,
- Same magnitude, same type of force
- Opposite direction, does not act on the same object.

Who agrees??

For the Kepler's law question, I did T2/r3 = T2/r3, right?


All correct
Original post by Reda2
...


yes for kepler's law you do that..
specific heat capacity you times 1.2 by 50 and find gpe
then use formula to find c
messed up last question so bad though
Reply 50
For the first question for the first part where you had to get speed. I did I = Ft, and then I = change in momentum to get speed. For the next part I used SUVAT, but last part of that question, finding the height I could not do, I tried but got 7m which is the initial height, and it can't be that.

Anyone?
Original post by anonymous_veg
Mass of S2 - go!


Something x10^33??
Reply 52
Original post by cookie_raider01
yes for kepler's law you do that..
specific heat capacity you times 1.2 by 50 and find gpe
then use formula to find c
messed up last question so bad though


1.2 was the height right?
Reply 53
Very heavy on the maths this year. All in all a very fair and nice paper. Nothing too crazy IMO. The impulse question (at least I hope it was an impulse question!) threw me a little.
Reply 54
Original post by cookie_raider01
yes for kepler's law you do that..
specific heat capacity you times 1.2 by 50 and find gpe
then use formula to find c
messed up last question so bad though


I was doing the last question as the time ended, I got to the part where I'm putting the values in the calculator. Due to not having time, I wrote R instead of the actual value, do you think that will cost me the mark?
Reply 55
Original post by AlexeiLipov
Something x10^33??


How did you do that question?
How did people do the KE graph? Was your max KE at 50? And start from 0?
Original post by Reda2
How did you do that question?

Not sure if it's right but I used Kepler's third law equation, we know time period, and radius (found out in previous part) so we can find out the mass.
About same difficulty as last year maybe a little easier. So
52 - 90 UMS
48 - A*
44 - A


wasn't the graph like this???

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