OCR Physics A newtonian World 20/6/16 Unofficial Mark Scheme

I got 1.02*10^33, anyone else get this?

I got 1.02*10^33, anyone else get this?

Yep and it's right this teacher guy who uploaded is wrong


Posted from TSR Mobile

I think it will be similar to last years...

FULL 51
A* 47
A 43
B 39
C 35

Then again, that's probably me being a bit too hopeful. A lot of people seemed to find it much easier than I did

Teachercol is a real life Physics teacher who gets to see the actual paper, so I would wager he is right. Good luck though?

I got 1.02*10^33, anyone else get this?

Sorry can someone tell me why for Q2 it was mg + Fsin55 rather than mg - Fsin55 when both forces are acting in opposite directions of each other?

Other than that, it looks like I've done better than expectedo

Posted from TSR Mobile

If you use mg with g as (-9.81), then youd be right. The hose is placed at an angle of 55 to the horizontal, meaning that Fsin 55 would give you the horizontal force (acting downwards) from the hose. You also have mg acting downwards, meaning they add up to form R. Newtons 2nd and 3rd law, as the change in momentum/t is equal to the force, and there is an equal and opposite force downwards.


TL;DR The water is pushed upwards from the hose, meaning it pushes the hose downwards, with an equal and opposite force.

Ignore the number in the brackets, thats me tallying how many marks i dropped worst case.
If anyone could help put my mind at ease for some of these that'd be great.
2)
c)ii)
R = mg + Fsin55 = 92 x 9.81 + (8.25 x 25)sin 55
= 902.5+169 = 1071N (3)
I omitted the Multiple by 25 by 8.25 and instead did Weight plus 8.25sin55.
Is this likely to get 1 or 2 marks out of the 3? (-2/3)

Q3 a)iii)
Left in Pi :/
b) i) I just flipped the EPE curve, is that both marks out the window?
ii)&iii) I then calculated these using 80mJ opposed to 50, would I drop all of these marks, or would I get e.c.f? (-3/5)


Q5e) Used keplers third law and got 1.73*10^33 :/
Is that 3 marks out the window? Or could I (hopefully) get 2 for this method? (-2/3)

Q6)a) Possibly didnt answer in enough detail. (-1/2)
d) So stupid, was rushing at this point, stated 2x the energy input, but not 2x mass so said temperature would increase by 2. Ridiculously stupid. Would I get 1 mark for stating that energy input is doubled, or is it 1 mark for saying constant and 1 mark for correct explanation? (-2/2)

Q7)a) Not enough detail (-1/2)

Worst case 49/60. Very disappointed considering I was getting 58+/60 on my past papers. Absolutely kicking myself.

I don't know if you've seen my earlier post but you will find that he messed up on equating the formulae. Not much point repeating what i said (you can read it yourself if you want)

but essentially, yes he is wrong (surprise!), and 1.0x10^33 is the correct answer.

@teachercol For the binary star question, could you treat the centre of the two circular motions to be a body of mass (M1 + M2 all divided by two) as the question said the two stars orbit about their centre of masses. By doing this could one then say : (v1)^2/R1 = (G)(M1+M2/all divided by two)/ (R1)^2 or (V2)^2/R2 = G(M1+M2/all divided by two)/ (R2)^2? If unclear I'll attach a diagram

Teachercol is a real life Physics teacher who gets to see the actual paper, so I would wager he is right. Good luck though?

I don't know if you've seen my earlier post but you will find that he messed up on equating the formulae. Not much point repeating what i said (you can read it yourself if you want)

but essentially, yes he is wrong (surprise!), and 1.0x10^33 is the correct answer.

Oh God man you're looking at a A/B grade in that paper from what I can infer from your mistakes, sorry I hope I'm wrong. I probably dropped similar amount of marks as you. The only *mathematical* mistakes I made was the mass of S2 and the hosepipe, and goddamn phase difference had to be in number radians not in terms of 'pi'. other than that I think got all the other maths bits right :O So glad I got the shm question right tho xd not bragging