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AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread]

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Original post by Datta10
For the six marker my graph was T^2 against l so 4pi^2/gradient equals g, I got g is 16 times bigger, will I lose all 3 marks?
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same here :frown: !!!
Original post by XxswagmasterxX
cbaddddbbbccccababacaadbd


holy **** dude, the only answers i can remember from multiple choice is 1,2,3 and question 23 and u got them all wrong, dont use ur marks for a guide o.0
Thinking 60 for an A*. Anyone else agree?
I got momentum to be like 2 times 10 to the power of something? Anyone got the same ?
Section B was nice. Section A felt a little rushed so I've probably made loads of mistakes there.
Original post by shuu00
Omg was N the mass of the original nucleus or the daughter nucleus?


:Original.:

EDIT- Apparently it was daughter.
(edited 7 years ago)
How many marks was the momentum question out of?


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Original post by CourtlyCanter
Original. 100% positive.


No, daughter one


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Original post by Questioner1234
yeah around about that


Original post by marioman
Less than that for an A* probably. 58-60?


Thanks guys :smile:
Feel sorted now, sorta counted up what I was likely to lose marks on and I'm looking at about 65. Hopefully a comfortable A*
Original post by spmenz123
Thinking 60 for an A*. Anyone else agree?

I'm feeling 58 tbh. 4 higher than last year makes sense. Plus multiple choice last year was probs easier than this year too imo.
Original post by CourtlyCanter
Does it matter if I'd drawn mine a little pointy rather than as curvy as yours?My 'pivot' point was just a little over the middle bit closer to Venus. Do you think I will get all three marks on that?


it should be curvy but i doubt they'll care that much. point is meant to be closer to venus but not clear on that diagram. yes you'll get at least 2/3
Original post by Mango Milkshake
What did you graph look like for g potential?

I can't draw but 1/x graph shape down from to 0 near Venus and then up to 8 whatever
Original post by Cadherin
If I showed all my working?

Idk lol maybe?
Original post by Questioner1234
Oh and what was the answer to when the current is the least in the tick box???? I put when the rate of dissipation of energy is greatest??

I put V across the resistor is small
Original post by CourtlyCanter
Original. 100% positive.


It couldn't have been b/c I couldn't do the proof. People that got N as the daughter nucleus ended up proving it . . .
For the energy of the capicitor question i used the equation instead of the area under the graph and got 3.5 times by ten to the something, will i still get the marks?!?!!
Original post by Rid123
Last question on adv and disad of the different braking??
I put advan is the EM brakes wont wear down as much as the other ones with the pads
And dis EM brakes nore expensive?? Idk i dont think got either of them ahaha

Ye
I wrote something along the lines of-
Advantage-
The electric-magnet brake thing doesn't wear out as quickly as the pads so it wouldn't need to be replaced as frequently and so less cost on the long run.
Disadvantage-
The electric-magnet brake thing uses EMF which means electrical energy from the car, costs more as fuel is used up quicker.

I am a bit sceptical on the disadvantage one though.
Original post by C0balt
I can't draw but 1/x graph shape down from to 0 near Venus and then up to 8 whatever

Idk lol maybe?

I put V across the resistor is small


if pd across the resistor is small then doesn't it mean that the pd across the capacitor is big? so how would the current be greatest at that point? (since V is proportional to I)

i got that one wrong probably
Original post by rxns_00
Thanks guys :smile:
Feel sorted now, sorta counted up what I was likely to lose marks on and I'm looking at about 65. Hopefully a comfortable A*


Well done! 65 is solid and will be near 100% imo.

I got around 60-64 so hopefully (fingers crossed) anything below 60 is an A*. Ofc the grade doesn't matter and ums does which is something everyone needs to remember.
I'm sure in previous papers the correct answer for the energy stored in a capacitor was given by the area under the graph BUT WITH V ON THE Y AXIS RATHER THAN ON THE X AXIS. By this logic shouldn't the energy stored in Q1 be given by the area of the trapezium to the left of the line, rather than the area 'under' the graph?
What was Q2 in section B?


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Original post by C0balt
I can't draw but 1/x graph shape down from to 0 near Venus and then up to 8 whatever

Idk lol maybe?

I put V across the resistor is small


1/x^2

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