# AQA AS Physics Unit 2 (PHYA2) June 9th 2016 Resit paperWatch

#261
(Original post by Sufyxyz)
It was a better reflector, it had a higher refferactive index and sinØcrit=n2/n1
As n1 for the otter material was higher sinø was smaller, to the critical angle was smaller. So more light hits at angle greater than critical angle so it's a better reflector.
I got it was worse because the critical handle was smaller meaning less chance of total internal reflection? Brilliant there go more marks...
0
2 years ago
#262
For the support force i got 196N
Show 12ms-1 = 11.831....ms-1
Work out horizontal speed = 10.9ms-1
A few other answers I got:
7.3
15.6
0.21ms-2
3.3ms-1
0.18J
0
2 years ago
#263
(Original post by Sufyxyz)
It was a better reflector, it had a higher refferactive index and sinØcrit=n2/n1
As n1 for the otter material was higher sinø was smaller, to the critical angle was smaller. So more light hits at angle greater than critical angle so it's a better reflector.
ah i think ur right but caught up in the moment i put it the other way round which is very annoying
0
2 years ago
#264
No bro you get like to marks for saying the critical angle was smaller, I think you only lost the last and first mark.
0
#265
(Original post by mangoli)
What a ****ing piss take, 8 questions each of which was harder than any other past paper question (and i've done them all). Managed to just squeeze in all the questions in the time. I swear they just have it in for us because they knew we were all retakers.
YES! You'd think they'd be nice knowing we're all resisting because we need better grades but noooooo AQA are absolute ********s once again, nice one, another D for me looks like. There goes my place at uni
0
2 years ago
#266
(Original post by rubberduckies321)
"did anyone get liek 2.34 or something like that for J for the spring like adding 2.2j + the energy required to overcome the friction"

Yesssss, glad someone got the same as me
Did you get 1.8n for the resistive force?

Then i did e=fe/2....plus the 2.2....probably wrong though
0
#267
(Original post by the_chosen_one97)
For the support force i got 196N
Show 12ms-1 = 11.831....ms-1
Work out horizontal speed = 10.9ms-1
A few other answers I got:
7.3
15.6
0.21ms-2
3.3ms-1
0.18J
How did you get the projectiles ones because i couldn't do any of them and legit nearly started crying
0
2 years ago
#268
(Original post by ScalyJake)
Since acceleration is inversely proportional to mass, the decreasing mass meant that the lorry would accelerate at a greater rate. This means the initial gradient on the graph would be steeper.

The lorry reaches its terminal velocity when it is in equilibrium. The driving force and the restitive forces are equal. Due to the higher acceleration, this lorry reaches its terminal velocity sooner than that in figure 7.

Because the air resistance is independent of the mass of the lorry, the change in mass would not effect the maximum speed.

This is roughly what I put, not sure what's right or wrong though.

i thought it couldn't maintain a constant speed though cause its mass is constantly decreasing? or is that wrong
0
2 years ago
#269
(Original post by McDerdactyl)
YES! You'd think they'd be nice knowing we're all resisting because we need better grades but noooooo AQA are absolute ********s once again, nice one, another D for me looks like. There goes my place at uni
Exactly what I thought! F*** SAKE!!
0
2 years ago
#270
I'm happy everyone struggled with this, gives me confidence

47 for an A maybe?

The 6 marker and spring questions are the ones I probably lost the most marks on.
0
2 years ago
#271
A-47
B-42
C-38
D-33
E-28
? just my predictions.... any thoughts?
0
2 years ago
#272
Unfortunately, I think the grade boundaries will be pretty high since there was a lot of simplistic mechanics maths involved.The main gimmick of this paper was that it was really difficult to do in the allocated time period.I personally didn't finish all parts of question 1 or the 8 marker, so I'm pretty worried about that, but all of the calculations were pretty straightforward so I could've picked up marks there.This was a horrible paper for slow writers and less good mathematicians!
0
2 years ago
#273
(Original post by McDerdactyl)
How did you get the projectiles ones because i couldn't do any of them and legit nearly started crying
They were quite tricky I thought but managed to get them luckily:

The first one when showing 12ms-1 you knew the vertical component is 5ms-1. To work out this component you do a value (say x) multiplied by the sin of the angle as it was vertical - this will equal the vertical component, 5. So you get xsin25 = 5, solve for x. The others were suvat and still not sure about the others until I hear what other people got
0
2 years ago
#274
(Original post by McDerdactyl)
I got it was worse because the critical handle was smaller meaning less chance of total internal reflection? Brilliant there go more marks...
A smaller critical angle means that there are more angles above the critical angle so there is more chance of the incidence able being above the critical angle and then TIR would happen
0
2 years ago
#275
(Original post by Physicsretake)
did anyone get liek 2.34 or something like that for J for the spring like adding 2.2j + the energy required to overcome the friction
Yes I think I did it like this-
Friction force= 1.8
W= Fs

1.8 x 0.2 = 0.36J

2.2 (Initial energy of block when it leaves the spring) + 0.36 = 2.56 J

The follow-up question about the force thing, I just used W = Fs again. Since the energy needed to push the block away was the same as that needed to compress the spring, the energy needed to compress the spring must've been 2.56j.

So...
2.56 = F x 0.2
F = 12.8 N

Anyone else got the same answer?
0
#276
(Original post by ryandaniels2015)
Exactly what I thought! F*** SAKE!!
I thought they'd forgotten we were the old spec and just gave us new stuff instead tbh because i couldn't do half of it ****ing hell
0
2 years ago
#277
Didnt the landing shuttle thing end up at 1.48ms-1 so it could land?? A lot of the clever people in my class remembered that it was below 3ms-1 so it could land (cant remember actual numbers though)

I did all the past papers and this one was the hardest by miles, the grade boundaries will be very low indeed
0
2 years ago
#278
(Original post by ScalyJake)
Since acceleration is inversely proportional to mass, the decreasing mass meant that the lorry would accelerate at a greater rate. This means the initial gradient on the graph would be steeper.

The lorry reaches its terminal velocity when it is in equilibrium. The driving force and the restitive forces are equal. Due to the higher acceleration, this lorry reaches its terminal velocity sooner than that in figure 7.

Because the air resistance is independent of the mass of the lorry, the change in mass would not effect the maximum speed.

This is roughly what I put, not sure what's right or wrong though.
I agree with the acceleration at a greater rate thing but I also talked about how the rate of change of acceleration is also decreasing as the pressure in the water tank decreases as the water leaves the tank so the rate of water loss would be decrease with time.

I messed up the terminal velocity bit. Said they'd level out at the same time.
0
#279
(Original post by mbh16)
A smaller critical angle means that there are more angles above the critical angle so there is more chance of the incidence able being above the critical angle and then TIR would happen
for some reason my brain told me that if its smaller, it's harder to match ???? idek man i swear my brain abandoned ship right after question 1
0
2 years ago
#280
(Original post by davgen7)
Didnt the landing shuttle thing end up at 1.48ms-1 so it could land?? A lot of the clever people in my class remembered that it was below 3ms-1 so it could land (cant remember actual numbers though)

I did all the past papers and this one was the hardest by miles, the grade boundaries will be very low indeed
i got around 2.8ms-1 myself
0
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