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

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

Reply 262

Physicsretake

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

Reply 263

Sufyxyz

3

No bro you get like to marks for saying the critical angle was smaller, I think you only lost the last and first mark.

Reply 264

OP

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

Reply 265

philo-jitsu

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 :smile:

Did you get 1.8n for the resistive force?

Then i did e=fe/2....plus the 2.2....probably wrong though

Reply 266

OP

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

Reply 267

txnilxnur

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

Reply 268

ryandaniels2015

11

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!!

Reply 269

Timotheus

I'm happy everyone struggled with this, gives me confidence :smile:

47 for an A maybe?

The 6 marker and spring questions are the ones I probably lost the most marks on.

Reply 270

splashywill

17

A-47
B-42
C-38
D-33
E-28
? just my predictions.... any thoughts?

(edited 7 years ago)

Reply 271

FluffyLoader

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!

Reply 272

the_chosen_one97

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

(edited 7 years ago)

Reply 273

mbh16

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

Reply 274

CourtlyCanter

8

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?

Reply 275

OP

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 :mad:

Reply 276

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

Reply 277

CourtlyCanter

8

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.

Reply 278

OP

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

Reply 279

Timotheus