2016 Official AQA New Spec AS Level Physics Paper 1 - 24th of May 2016

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aeroline1999

Totally agree because like someone wrote early the melted chocolates were at nodes however they don't vibrate so how on earth does heat up. Ridiculous and that harmonic frequency was hard because was it 6cm or 12cm argh

i thought 6cm at first but then i changed it to 12cm, no starting to think it was 6 again! ugh

Reply 461

username2488725

Did anyone get a small force for the stress/young's modulus question?

Reply 462

bowliez

Original post by CrazyFool229

Did anyone get a small force for the stress/young's modulus question?

23N? or something similar?

Reply 463

Georgia.C123

What did people get for the airplane question?

Reply 464

Cinna21

Original post by KD1024

Nah, 3 antinodes means 3rd harmonic, so 1 and a half wavelength. If I can remember correctly it was 120mm for 3 antinodes so 2/3 x 120 = 80mm was wavelength.

Argh I knew I should of tried to re do that question I'm sure i got some marks though

Reply 465

119881

Original post by MimiH04

I thought this was paper was pretty difficult, messed up the 6 marker! What property of material A did everyone put? Many people said elastic

Brittle, cos it didnt show any plastic deformation upto its breaking stress

Reply 466

username2488725

Original post by bowliez

23N? or something similar?

Bugger.... Somehow I like 0.023 or whatever,... Must have buggered up the conversions. Actually yes I did just realised I accidentally did 10^6 rather than 10^9 so I'm out by a bit... Stupid mistake

Reply 467

11234

Original post by aeroline1999

i thought 6cm at first but then i changed it to 12cm, no starting to think it was 6 again! ugh

I put 12 because distance between two antinodes is 6 and thats lambda/2.

Reply 468

Cinna21

Original post by 11234

Really I put antinodes because theyre points of max energy/amplitude

Yes so did I but if you remember the diagram in the exams the chocolates were at nodes which is extremely odd unless i'm not reading the diagram properly

Reply 469

SaltandSugar

Original post by bowliez

23N? or something similar?

i thought i got a really big force ;/

Reply 470

Exavior

Original post by 11234

Nah, 3 antinodes means 3rd harmonic, so 1 and a half wavelength. If I can remember correctly it was 120mm for 3 antinodes so 2/3 x 120 = 80mm was wavelength.

I forgot that....I got wavelength as 120mm. How many marks would I lose

But the question had 2 antinodes didn't it? and 3 nodes? So the second harmonic so it equals 1 wavelength? I did it both ways then I crossed out the third harmonic one as realised it was the second.

Reply 471

119881

Original post by Cinna21

Nah, 3 antinodes means 3rd harmonic, so 1 and a half wavelength. If I can remember correctly it was 120mm for 3 antinodes so 2/3 x 120 = 80mm was wavelength.

Argh I knew I should of tried to re do that question I'm sure i got some marks though

I thought Node to Node / Antinode to Antinode is always half a wavelength? so the wavelength was around 120mm (6x2)

Reply 472

11234

Original post by Cinna21

Nah, 3 antinodes means 3rd harmonic, so 1 and a half wavelength. If I can remember correctly it was 120mm for 3 antinodes so 2/3 x 120 = 80mm was wavelength.

Argh I knew I should of tried to re do that question I'm sure i got some marks though

No wait, if it STARTED at an antinode and finished at another with one inbetween is it not one wavelength because the third harmonic starts with nodes. I did by the distance between antinodes being lambda/2

Reply 473

haes

Do they tell us if we get marks for significant figures?

Reply 474

aeroline1999

Original post by 11234

I put 12 because distance between two antinodes is 6 and thats lambda/2.

well i assumed the dots were the nodes so it would be half a wavelength therefore 12cm. i just hope I'm right since i now know they were antinodes not nodes

Reply 475

11234

What did people get for the percentage change in diameter. Did anyone get 0.79%

Reply 476

Mom0

Im sure its 12- it is like the 3rd one on this picture. 1 wavelength would be the distance between the 1st and 3rd antinode

Reply 477

bowliez

Original post by SaltandSugar

i thought i got a really big force ;/

It was the cross sectional area, which was (d/2)^2 pi, then multiplied by the stress which was something Giga, x10^9? Thats what I did not sure if its right though.