Post by: silverpixeli on November 12, 2014, 04:07:20 am
There aren't many hours left till the start of the physics exam, good luck everyone!
Feel free to use this thread for discussion after the exam, too. Once I see a paper I'll be happy to discuss with you all.
Edit: itute made some answers! thanks for the original post
Edit 2: Physics teacher got me a copy of the paper, there's now a scan available HERE (5.17MB)
Post by: Targor on November 12, 2014, 07:39:08 am
Just finished my brother's brand new personalised 5-hours-of-work cheat sheet, while he's getting a good night's sleep.
There aren't many hours left till the start of the physics exam, good luck everyone!
Feel free to use this thread for discussion after the exam, too. Once I see a paper I'll be happy to discuss with you all.
Good on you man, not many people would do that for someone else.
Post by: rui97 on November 12, 2014, 11:59:58 am
Post by: Marrogi12 on November 12, 2014, 12:18:04 pm
Post by: liamh96 on November 12, 2014, 12:18:47 pm
Post by: speedy on November 12, 2014, 12:24:11 pm
Post by: Plitzer on November 12, 2014, 12:26:33 pm
That modulation question though... what did your graphs look like?
Post by: shroomer101 on November 12, 2014, 12:28:04 pm
Post by: KayKay on November 12, 2014, 12:29:34 pm
Spring question?I believe it was:
1. 0.8 m
2. Didn't take kinetic energy into account. Ek+Ep+Es=constant.
3. 2 ms-1
Post by: rui97 on November 12, 2014, 12:31:19 pm
I believe it was:
1. 0.8 m
2. Didn't take kinetic energy into account. Ek+Ep+Es=constant.
3. 2 ms-1
looking around, yeah that seems to be the general consensus
Post by: shroomer101 on November 12, 2014, 12:40:35 pm
I believe it was:
1. 0.8 m
2. Didn't take kinetic energy into account. Ek+Ep+Es=constant.
3. 2 ms-1
yaay i think i got all of that too
for the modulation i drew them translated up and did something weird for it i dont even know myself...it was two marks from my memory
also for the light globe question with the maximum of 6V what did everyone get??
Post by: KayKay on November 12, 2014, 12:41:44 pm
yaay i think i got all of that tooI got 26V, may not be right tho.
for the modulation i drew them translated up and did something weird for it i dont even know myself...it was two marks from my memory
also for the light globe question with the maximum of 6V what did everyone get??
Post by: rui97 on November 12, 2014, 12:44:39 pm
I got 26V, may not be right tho.yep same
Post by: Plitzer on November 12, 2014, 12:44:49 pm
I got 26V, may not be right tho.Yeah 26 V, the Voltage drop along the transmission lines was 20V and 6V had to be supplied to the globe, so a maximum supply of 26V was needed
For the modulation question I basically copied the carrier wave graph from 0 to 4.5ms but translated a little bit, then repeated the pattern at a lower point for after 5ms
Post by: speedy on November 12, 2014, 12:47:13 pm
I believe it was:
1. 0.8 m
2. Didn't take kinetic energy into account. Ek+Ep+Es=constant.
3. 2 ms-1
Hmm, I got 0.8 -> but then I said 0.4 for "extension", which I took as length that it has been extended from its initial length.
Post by: shroomer101 on November 12, 2014, 12:47:46 pm
I got 26V, may not be right tho.
yes i got 26V as well :)
okay this is making me feel better than how i did getting out of the exams
Post by: KayKay on November 12, 2014, 12:53:33 pm
Post by: Hiesenberg on November 12, 2014, 12:53:40 pm
I put 0 because an AC is needed to generate an output voltage. Right?
Post by: KayKay on November 12, 2014, 12:54:22 pm
for the first electric power question, something like 10 V DC is supplied to a transformer whats the output?Yep, that's correct.
I put 0 because an AC is needed to generate an output voltage. Right?
Post by: speedy on November 12, 2014, 12:55:48 pm
3. 2 ms-1
To do this, do you take h and x as 0.2? Ie. halfway between the max's and min's of the oscillation?
Post by: samroberts on November 12, 2014, 12:56:27 pm
Sounds like that's the correct answer
Post by: KayKay on November 12, 2014, 12:57:55 pm
To do this, do you take h and x as 0.2? Ie. halfway between the max's and min's of the oscillation?If I remember correctly, halfway between max and min of oscillation was an extension of 0.4 m. So 1.6=mgh + 1/2kx^2 + 1/2mv^2. h=x=0.4, sub and solve. But yeah, you do find the halfway point.
Post by: Plitzer on November 12, 2014, 12:58:54 pm
To do this, do you take h and x as 0.2? Ie. halfway between the max's and min's of the oscillation?The halfway point is when the speed will be a maximum, yes, but the extension found from an earlier part is 0.8m. The value you found IS the extension as you are calculating from the unstretched length, so the 0.4m length of the unstretched spring doesn't apply.
Then the halfway point will be at x = 0.4m and you use conservation of energy principles
Post by: Marrogi12 on November 12, 2014, 01:01:18 pm
Post by: wanigara on November 12, 2014, 01:03:25 pm
Post by: SeanC on November 12, 2014, 01:03:58 pm
Post by: chuck981996 on November 12, 2014, 01:04:56 pm
I don't think time was a problem as most of the students in my class finished early , I finished 45 mins early ahah exactly at 11.00 ahah
Yeah time was fine. I mean, I took the whole 2.5hours, but that's because I check as I go and go slowly hahaha
Did other people get 20Ω for the variable resistor when the voltage across the photodiode is 1V? Or maybe I'm thinking of the resistance in the coil… Everything is slipping away…
Post by: Plitzer on November 12, 2014, 01:08:22 pm
Yeah time was fine. I mean, I took the whole 2.5hours, but that's because I check as I go and go slowly hahahaYeah I think so. One was 20ohm and another question had 60ohm
Did other people get 20Ω for the variable resistor when the voltage across the photodiode is 1V? Or maybe I'm thinking of the resistance in the coil…
What did everyone get for the voltage divider circuit? I think i had 400ohm for Rv
Post by: shroomer101 on November 12, 2014, 01:11:50 pm
I think you guys are right with 26v but did something else. I assumed (probably incorrectly) that the power output of the power supply must be the same if you increased the voltage so you would decrease the current. this led to something very complicated and i ended up using the quadratic formula to solve and find v to equal like 17.something i think. any chance of getting any marks for this? it was a 3 mark question
Hmm I'm not too sure because when I calculated it I found that the power output at the generator was 104W which is different from the original one :/
Post by: KayKay on November 12, 2014, 01:17:48 pm
Yeah I think so. One was 20ohm and another question had 60ohmFor the 1V photodiode one I think I got a resistance of 187.5ohm... :/
What did everyone get for the voltage divider circuit? I think i had 400ohm for Rv
Post by: the_analyser on November 12, 2014, 01:19:59 pm
Post by: Plitzer on November 12, 2014, 01:22:47 pm
I got 187.5 as well, to find voltage across it was 1.3-1 or something similarThe photodiode is acting as the battery in the circuit, isn't it? So if it was producing how ever many amps that corresponded to 1V, then the resistor would also have 1V across it?
Post by: SeanC on November 12, 2014, 01:27:56 pm
Vtotal=vloss+6
vtotal=iline*rline+6
vtotal=5i+6 if power output must be the same then p=26 so i=26/v
v=130/v +6
v^2-6v-130=0 and solve with quadratic formula
Post by: Chalkhous on November 12, 2014, 01:30:40 pm
But looking at the replies it looks like 0.8m was right which is great cause i guessed that way.
Post by: patel.aayush on November 12, 2014, 01:31:15 pm
Post by: RKTR on November 12, 2014, 01:33:49 pm
I got 187.5 as well, to find voltage across it was 1.3-1 or something similargot 187.5 as well
Post by: Ya Habibi on November 12, 2014, 01:36:11 pm
2b was really odd cause if you used f=kx you got 0.4m as the extension. if you use mgx=1/2kx^2 you got 0.8m
But looking at the replies it looks like 0.8m was right which is great cause i guessed that way.
Was it a three mark question? I got an answer of 0.8 but wasn't sure if the process was worth the three marks.
Post by: KayKay on November 12, 2014, 01:36:43 pm
got 187.5 as wellI put 187.5ohm as well but I think it might not be right. If you take the photodiode to be the power source, if it produces 1V, then the resister would also have 1V. The current is 1.6x10^-3A, which means the resister is 625ohm :/ I have no idea if this is right or not, I don't even understand how the current was changing when the light intensity on the photodiode was kept constant. It was a really odd question.
Post by: SeanC on November 12, 2014, 01:39:55 pm
I put 187.5ohm as well but I think it might not be right. If you take the photodiode to be the power source, if it produces 1V, then the resister would also have 1V. The current is 1.6x10^-3A, which means the resister is 625ohm :/ I have no idea if this is right or not, I don't even understand how the current was changing when the light intensity on the photodiode was kept constant. It was a really odd question.I think i put 625ohm
Post by: Rishi97 on November 12, 2014, 01:40:32 pm
got 187.5 as well
same wooo!!!
Post by: evogamerz on November 12, 2014, 01:43:54 pm
a,b,c,d,a,c,b,b,c,d,a
Post by: d4m0 on November 12, 2014, 01:56:03 pm
I think i put 625ohmI got 625 ohms as well.
Post by: theshunpo on November 12, 2014, 01:56:56 pm
Post by: S61778 on November 12, 2014, 02:01:53 pm
Post by: bpearson on November 12, 2014, 02:02:38 pm
Post by: shroomer101 on November 12, 2014, 02:02:50 pm
What did you guys get for the '' eV question?
I think I got 8.9 ev
Post by: speedy on November 12, 2014, 02:04:10 pm
The halfway point is when the speed will be a maximum, yes, but the extension found from an earlier part is 0.8m. The value you found IS the extension as you are calculating from the unstretched length, so the 0.4m length of the unstretched spring doesn't apply.
Then the halfway point will be at x = 0.4m and you use conservation of energy principles
Uhh, you sure? Because I'm pretty sure it oscillates between 0.4 and 0.8?
Post by: sani8274 on November 12, 2014, 02:05:33 pm
A
B
B
D
A
B
B
A
B
A
C
Post by: Sono69 on November 12, 2014, 02:09:16 pm
I think I got 8.9 ev
Same here, also got 625 ohms for that resistor question
Post by: KayKay on November 12, 2014, 02:10:15 pm
Could you determine the mass of the planet?You couldn't because according to Kepler's Law: T^2/R^3=4pi^2/GM all objects at that radii will orbit the sun in the same time period, given the mass of the Sun, no matter what the mass is. So from that info, the mass could be anything.
Post by: shroomer101 on November 12, 2014, 02:10:22 pm
Uhh, you sure? Because I'm pretty sure it oscillates between 0.4 and 0.8?
The x found which is 0.8 is the length of extension so it oscillates from 0.4 to 1.2m
Post by: Bronzebottom64 on November 12, 2014, 02:10:58 pm
Meterial n strctures
A
B
B
D
A
B
B
A
B
A
C
I don't think the last question was c. I originally thought this too but if the inside three were all cables then the outer two would fall down. As such I put A.
Post by: SeanC on November 12, 2014, 02:11:43 pm
Post by: S61778 on November 12, 2014, 02:12:13 pm
You couldn't because according to Kepler's Law: T^2/R^3=4pi^2/GM all objects at that radii will orbit the sun in the same time period, given the mass of the Sun, no matter what the mass is. So from that info, the mass could be anything.
It was 2 marks, so would saying 'no' be 1 mark? I did that questions at the very end and just said no, didn't have time to explain
Post by: NathanGauci on November 12, 2014, 02:12:24 pm
yes i got 26V as well :)
I treated the circuit as a voltage divider and got that. So relieved
Post by: KayKay on November 12, 2014, 02:15:19 pm
It was 2 marks, so would saying 'no' be 1 mark? I did that questions at the very end and just said no, didn't have time to explainYeah saying no will give you 1/2 I think
Post by: 3fold on November 12, 2014, 02:16:23 pm
Post by: theshunpo on November 12, 2014, 02:18:22 pm
Same here, also got 625 ohms for that resistor question
Sounds good to me!
Post by: Sono69 on November 12, 2014, 02:18:45 pm
Post by: S61778 on November 12, 2014, 02:18:52 pm
Yeah saying no will give you 1/2 I think
Also for the max speed of mass spring question, I got 2m/s but I think my working out was wrong. I did elastic potential energy is converted to kinetic energy.
So from the previous question, I got x=0.4m
Then, calculated the Spring potential energy with 1/2kx2
Then made that equal to kinetic energy and solved for v
Got v=2m/s
No marks or?
Post by: NathanGauci on November 12, 2014, 02:22:17 pm
Edit:the 2008 one.
Post by: KayKay on November 12, 2014, 02:24:14 pm
Anyone do the sound detailed study and remember what they got for the resonance questions?100Hz and 150Hz
Post by: the_analyser on November 12, 2014, 02:29:08 pm
Post by: KayKay on November 12, 2014, 02:29:38 pm
Also for the max speed of mass spring question, I got 2m/s but I think my working out was wrong. I did elastic potential energy is converted to kinetic energy.Mm not too sure. Well the previous question the answer was 0.8 m. For the next one though... if you're lucky and the examiner isn't paying attention to working out you might get full marks, but otherwise 1/3 because 1 for answer and 2 for explanation.
So from the previous question, I got x=0.4m
Then, calculated the Spring potential energy with 1/2kx2
Then made that equal to kinetic energy and solved for v
Got v=2m/s
No marks or?
Post by: Sono69 on November 12, 2014, 02:30:27 pm
100Hz and 150Hz
Damn I think I got 50Hz and 100Hz
Post by: S61778 on November 12, 2014, 02:33:27 pm
Mm not too sure. Well the previous question the answer was 0.8 m. For the next one though... if you're lucky and the examiner isn't paying attention to working out you might get full marks, but otherwise 1/3 because 1 for answer and 2 for explanation.
Hope I'm lucky, in fact I've been told by teachers, one I think is close in rank to the chief assessor, that it's full marks if the answer in the box is correct. She even said that if it says 'show your working' it's full marks if the answer in the box is right...but not too sure on this
Post by: patel.aayush on November 12, 2014, 02:38:11 pm
Hope I'm lucky, in fact I've been told by teachers, one I think is close in rank to the chief assessor, that it's full marks if the answer in the box is correct. She even said that if it says 'show your working' it's full marks if the answer in the box is right...but not too sure on this
its full marks if u get the ans right if it doesnt say show your working, if it does u have show working
btw did anyone do synchrotron? I got A A B D A C B D B C C
Post by: Jnf17 on November 12, 2014, 02:44:12 pm
Hope I'm lucky, in fact I've been told by teachers, one I think is close in rank to the chief assessor, that it's full marks if the answer in the box is correct. She even said that if it says 'show your working' it's full marks if the answer in the box is right...but not too sure on this
Can't exactly remember what I have heard about physics but I thought the sciences and maths (methods anyway) had designated method and answer marks? Like I didn't think you could get full marks on a physics question with only an answer in the box with no working?
Post by: Sono69 on November 12, 2014, 02:45:29 pm
Can't exactly remember what I have heard about physics but I thought the sciences and maths (methods anyway) had designated method and answer marks? Like I didn't think you could get full marks on a physics question with only an answer in the box with no working?
That's my understanding of it as well
Post by: chuck981996 on November 12, 2014, 02:51:32 pm
For the 1V photodiode one I think I got a resistance of 187.5ohm... :/
Oh yeah that sounds right! I think I got that. I think 20Ω was for the coil.
I got 187.5 as well, to find voltage across it was 1.3-1 or something similar
Yeah we were given the current, and voltage was 0.3 so you just used ohm's law I think. That sounds right.
I think i put 625ohm
The photodiode produces a current, not a voltage.
Post by: d4m0 on November 12, 2014, 02:53:43 pm
yes i got 26V as well :)
okay this is making me feel better than how i did getting out of the exams
I filled a page making it more complicated then it really was and then right in the last two minutes I did it really quickly and got 26V yay
Post by: shroomer101 on November 12, 2014, 02:56:47 pm
Post by: chuck981996 on November 12, 2014, 02:59:24 pm
mc answers for materials and structures
a,b,c,d,a,c,b,b,c,d,a
I got a,c,c,d,a,b,b,a,b,d,d…
Post by: thanman123 on November 12, 2014, 03:00:20 pm
its full marks if u get the ans right if it doesnt say show your working, if it does u have show workingI got D for 10, which is probably wrong, but I got the same for everything else.
btw did anyone do synchrotron? I got A A B D A C B D B C C
Post by: the_analyser on November 12, 2014, 03:00:53 pm
Oh yeah that sounds right! I think I got that. I think 20Ω was for the coil.Yep, 20 ohm sounds good, thanks for the feedback
Yeah we were given the current, and voltage was 0.3 so you just used ohm's law I think. That sounds right.
The photodiode produces a current, not a voltage.
Post by: skeletalclown on November 12, 2014, 03:15:02 pm
for the reverse biased photodiode. I THINK it was for that at least, I know I got 60 ohms for something at least.
Post by: patel.aayush on November 12, 2014, 03:18:45 pm
I got D for 10, which is probably wrong, but I got the same for everything else.
Hmm, do you remember what it was?
Post by: Brayden_Partridge on November 12, 2014, 03:20:53 pm
Post by: MoeyLeh on November 12, 2014, 03:28:13 pm
Post by: skeletalclown on November 12, 2014, 03:37:19 pm
Fuck vcaa..................
Same.
Post by: bovawatkins on November 12, 2014, 04:17:35 pm
Post by: ala0001 on November 12, 2014, 04:23:09 pm
Post by: Brunette15 on November 12, 2014, 04:24:23 pm
What do you think the cut off will be this year?? lower than last years or around the same?
I found the exam a lot easier than last year I think the cut off will be lower :/
Post by: joecantwell on November 12, 2014, 04:24:56 pm
Did anyone notice the projectile motion question which was almost identical to a previous exam one?
Edit:the 2008 one.
Yeah i had the question on my cheat sheet...winning!
Post by: bovawatkins on November 12, 2014, 04:39:44 pm
I found the exam a lot easier than last year I think the cut off will be lower :/
last year a 48/150 got ya a c, hopefully its something like that again cos if not.......
Post by: Ya Habibi on November 12, 2014, 04:40:20 pm
I found the exam a lot easier than last year I think the cut off will be lower :/
You are contradicting yourself. It was easy, yet the cuttoff for A+ will be lower?
Post by: shroomer101 on November 12, 2014, 04:45:28 pm
You are contradicting yourself. It was easy, yet the cuttoff for A+ will be lower?
haha maybe she meant higher?
Post by: floyd.everest on November 12, 2014, 04:48:08 pm
Post by: samroberts on November 12, 2014, 04:50:06 pm
[/quote]
I got the same for all!!
Post by: Ya Habibi on November 12, 2014, 05:03:29 pm
Hey what did everyone say for the Diffraction patterning of the Electrons and the X rays?
Yeah, they have the energy but their momentum is different, hence they have a different de Broglie wavelength, hence they diffract to different extents.
Post by: chuck981996 on November 12, 2014, 05:08:18 pm
Hey what did everyone say for the Diffraction patterning of the Electrons and the X rays?
I said:
- For the diffraction patterns to be the same, the electron must have a de Broglie wavelength equal to the wavelength of the photon.
- Since
, the de Broglie wavelength is dependent on both the mass and the energy of the electron.
- Therefore, even though their energies are the same, the photon and electron have a different mass and thus have a different wavelength and different diffraction patterns.
Post by: floyd.everest on November 12, 2014, 05:14:20 pm
I said:
- For the diffraction patterns to be the same, the electron must have a de Broglie wavelength equal to the wavelength of the photon.
- Since
- Therefore, even though their energies are the same, the photon and electron have a different mass and thus have a different wavelength and different diffraction patterns.
Hey wouldn't you be incorrect just in implying the photon has mass?
Post by: chuck981996 on November 12, 2014, 05:16:42 pm
Yeah I had the same jist with the De Broglie wavelength! I did rearrange the equation with the Square root of 2mE to just mv, just to simplify things
Oh yeah, I just used that form of the equation (I had like 5 on my cheat sheet lel) because it shows clearly that wavelength is dependent on more than just energy. That sort of becomes lost when you use the standard lambda=h/p
Post by: chuck981996 on November 12, 2014, 05:26:41 pm
Hey wouldn't you be incorrect just in implying the photon has mass?
Hm I don't know. Do you think I was implying that? It depends how mean the examiners are I guess. The important point was the an electron has mass, so it's momentum and energy aren't directly proportional, unlike for a photon. Perhaps that would have been a better way to word it.
Post by: Stew_822 on November 12, 2014, 05:29:09 pm
Post by: chuck981996 on November 12, 2014, 05:33:50 pm
Does anyone remember what they got for the height of the projectile in the second question? I think I got like 3.75 but then I tried it again two different ways and ended up with like three different answers and then I tried working backwards from my original answer to see if it was right and like I'm pretty sure it's not because I ended up with a totally different time than what I used to calculate the 3.75.....
What was the distance to the screen?
Post by: Stew_822 on November 12, 2014, 05:36:47 pm
Post by: TheFlyingWeed on November 12, 2014, 05:46:39 pm
Does anyone remember what they got for the height of the projectile in the second question? I think I got like 3.75 but then I tried it again two different ways and ended up with like three different answers and then I tried working backwards from my original answer to see if it was right and like I'm pretty sure it's not because I ended up with a totally different time than what I used to calculate the 3.75.....
I'm pretty sure I had 3.75 as well
Post by: Stew_822 on November 12, 2014, 05:49:15 pm
I'm pretty sure I had 3.75 as wellCool! Maybe I got it right then!
Post by: FSH7 on November 12, 2014, 05:50:26 pm
Post by: chuck981996 on November 12, 2014, 05:51:13 pm
Pretty sure it was 26 meters, but we had to find the height that the projectile hit a board. I think it had an initial velocity of 20m/s at an angle of 30 degrees from the ground. Never done a question like this before so not sure if I had the right method
If it was 26m, the formula I used yields 3.75m too:
http://www.wolframalpha.com/input/?i=26*tan%2830%C2%BA%29-%2810*26%5E2%29%2F%282*%2820*cos%2830%C2%BA%29%29%5E2
Post by: isuru.e on November 12, 2014, 05:59:40 pm
Who did the sound detailed study? I got A C C D D C B C B A B.Got them except for 8 and 11. Im probably wrong though.
Post by: FSH7 on November 12, 2014, 06:05:09 pm
Got them except for 8 and 11. Im probably wrong though.
I'm not quite sure about question 8 or 11 either to be honest. I thought that the person would hear the low frequency instruments for longer because of greater diffraction.
Post by: swaggydawg on November 12, 2014, 06:07:42 pm
Post by: jumcakes on November 12, 2014, 06:17:02 pm
Which question was Question 8, if you can remember?
Post by: FSH7 on November 12, 2014, 06:23:20 pm
Post by: isuru.e on November 12, 2014, 06:24:32 pm
I'm not quite sure about question 8 or 11 either to be honest. I thought that the person would hear the low frequency instruments for longer because of greater diffraction.
Ye i thought i circled that? I'm worried now :P
Post by: jumcakes on November 12, 2014, 06:25:09 pm
Do you mean on the sound detailed study? It was something like: Joe plays a standing wave of 80Hz into a tube, he attaches a meter which measures pressure variation from the normal air pressure. It asks you which of the choices are correct. I believe 'C' was the meter needle will swing from negative to positive 80 times a second. Another choice was the needle swings from negative to zero 80 times a second.
I think I wrote D, as it was a pressure node, and the pressure would be the same as that at the open end of the tube, where the pressure is equal to atmospheric pressure. I'm not 100% though.
Post by: Brunette15 on November 12, 2014, 06:26:52 pm
haha maybe she meant higher?
Yeah I meant higher haha ::)
Post by: KayKay on November 12, 2014, 06:27:24 pm
I think I wrote D, as it was a pressure node, and the pressure would be the same as that at the open end of the tube, where the pressure is equal to atmospheric pressure. I'm not 100% though.It's definitely D
Post by: isuru.e on November 12, 2014, 06:29:26 pm
I think I wrote D, as it was a pressure node, and the pressure would be the same as that at the open end of the tube, where the pressure is equal to atmospheric pressure. I'm not 100% though.
Yee I got D too :D
Post by: Stew_822 on November 12, 2014, 06:35:45 pm
Although I don't think it will be too much of a hassle, I believe if you used exact/non truncated values the projectile question turned out to be 3.74 metres (3.744something).
Which question was Question 8, if you can remember?
Yeah thinking about it I think I got 3.74 as well.
I got D for that question to, it asked for a pressure node and a node is where the pressure remains constant, so D. An antinode is where the pressure varies, so since they used the word "node" I hoped they just didn't mean an antinode
Post by: wanigara on November 12, 2014, 07:30:14 pm
Post by: xdtash on November 12, 2014, 08:35:19 pm
Yeah thinking about it I think I got 3.74 as well.Thank god someone else got that, I thought I rounded the wrong way
Post by: Targor on November 12, 2014, 08:35:58 pm
Post by: rhys.oo on November 12, 2014, 08:44:08 pm
was there friction
and for P=mv what was the mass before and after the collision
Post by: theshunpo on November 12, 2014, 08:46:05 pm
in question 1b in motion where the truck had 40 tonne and there were 2 trailers of 10 tonne each. calculate the tension in the rope between the truck and the trailer.I don't think there was any friction.
was there friction
Post by: xdtash on November 12, 2014, 08:48:44 pm
was there frictionI didn't think about friction because we knew their acceleration. We didn't have to subtract forces to sub into F=ma to get 'a'. So T=mass*acceleration=0.2*20000?
and for P=mv what was the mass before and after the collision
For the momentum question, the locomotive was 40000kg, and each carriage was 10000kg, so after the collision, total mass was 80000kg. I got a speed of 2m/s
Post by: FSH7 on November 12, 2014, 08:51:21 pm
Post by: xdtash on November 12, 2014, 08:53:36 pm
how did u get 80000kgWell the locomotive was 40000kg, and four carriages of 10000kg each. 40000+4(10000)=80000
I really hope I didn't misread the question haha
Post by: theshunpo on November 12, 2014, 08:54:27 pm
Post by: xdtash on November 12, 2014, 08:56:21 pm
Speaking of the locomotive question, what did you guys get for the tension in the coupling?I got 4000N. I treated the coupling as kind of the 'thrust' of the two carriages, so a=0.2, and the two masses equal 20000. Multiply them together to get 4000.
Post by: theshunpo on November 12, 2014, 08:59:04 pm
I got 4000N. I treated the coupling as kind of the 'thrust' of the two carriages, so a=0.2, and the two masses equal 20000. Multiply them together to get 4000.Yeah 4000N is what i got as well. I worked from the 2nd carriage first to find the tension between the 1st and the 2nd. Then proceeded to resolve forces on the second carriage and set it to ma.
Post by: FSH7 on November 12, 2014, 09:00:22 pm
Post by: jumcakes on November 12, 2014, 09:03:00 pm
What did everyone else get for the graph of the modulated signal? It kinda threw me off because I've never seen one involving light pulses before.
I think it looks like a tall column between t = 1 and 2, 3 and 4, and then a short column (just like the signal wave) between 5 and 6, all above the x-axis (since intensity is always non-negative), but again I'm not completely sure.
Post by: rhys.oo on November 12, 2014, 09:03:39 pm
how many marks was 1a and the first projectile motion question
Post by: FSH7 on November 12, 2014, 09:04:44 pm
was momentum elastic or inelasticInelastic since KE was lost according to KE=1/2 mv^2
Post by: d4m0 on November 12, 2014, 09:06:00 pm
Does no one have a copy of the physics exam? Strange because after the methods one, the solutions were put up almost straight away. How do people get their hands on the exams afterwards anyway?
They get given out to teachers, presumably for methods the people posting the solutions asked their old teachers for access to their copies, or something. Seems no one cares enough about physics to do it. Side note: you go to my old school :p
also I got what jumcakes did for the modulation question
Post by: xdtash on November 12, 2014, 09:06:35 pm
What did everyone else get for the graph of the modulated signal?I left that one until the very end. I ended up just making it like an "addition of ordinates" that you learn in maths.
Post by: theshunpo on November 12, 2014, 09:09:18 pm
Question 1a) was the distance traveled question..i'm pretty sure that was 2 marks.
how many marks was 1a and the first projectile motion question
Post by: FSH7 on November 12, 2014, 09:12:58 pm
was momentum elastic or inelasticIf by the first projectile motion question you mean the cricket questions then I distinctly remember that both parts were worth 2 marks, each that is.
how many marks was 1a and the first projectile motion question
Post by: rhys.oo on November 12, 2014, 09:21:35 pm
how many marks was that question
Post by: xdtash on November 12, 2014, 09:25:38 pm
for the circular motion question did people get a mass using a formula such as r3 / T2 = G.M / 4.ᴨ2Yeah, I ended up getting like 1.something*10^31? Something around that
Post by: chuck981996 on November 12, 2014, 09:25:50 pm
Does no one have a copy of the physics exam? Strange because after the methods one, the solutions were put up almost straight away. How do people get their hands on the exams afterwards anyway?
lol that's because I don't do methods. Teachers don't like to give you an exam that you just sat.
Post by: FSH7 on November 12, 2014, 09:33:26 pm
Yeah, I ended up getting like 1.something*10^31? Something around thatYes! I got 1.08 x 10^31.
Post by: xdtash on November 12, 2014, 09:38:10 pm
Post by: jumcakes on November 12, 2014, 09:41:41 pm
For the light component, was the answer to "why it can absorb the photon but not emit it" because it's not part of emission spectrum or something like that? Or did we have to make reference to electrons jumping
A photon can be absorbed only if there exists an energy level above its current energy state where the difference in energies corresponds to the photon energy, so I thought that it could absorb 1.8 eV as there was a stable energy level 1.8eV above the first excited state - the first excited state was of energy 4.9eV, and it could successfully jump to 6.7eV, the second excited state.
Emitting 1.8eV however wasn't possible since there were no two energy levels below the 4.9 eV state that differed by 1.8eV - only a 4.9eV photon could be emitted (in the transition from 4.9-->0)
Post by: xdtash on November 12, 2014, 09:46:12 pm
A photon can be absorbed only if there exists an energy level above its current energy state where the difference in energies corresponds to the photon energy, so I thought that it could absorb 1.8 eV as there was a stable energy level 1.8eV above the first excited state - the first excited state was of energy 4.9eV, and it could successfully jump to 6.7eV, the second excited state.Nice
Emitting 1.8eV however wasn't possible since there were no two energy levels below the 4.9 eV state that differed by 1.8eV - only a 4.9eV photon could be emitted (in the transition from 4.9-->0)
Post by: jasoN- on November 12, 2014, 09:59:44 pm
Post by: Mujteba on November 13, 2014, 02:57:39 am
mc answers for materials and structures
a,b,c,d,a,c,b,b,c,d,a
Last one is d
Post by: Mujteba on November 13, 2014, 03:02:01 am
Hm I don't know. Do you think I was implying that? It depends how mean the examiners are I guess. The important point was the an electron has mass, so it's momentum and energy aren't directly proportional, unlike for a photon. Perhaps that would have been a better way to word it.
Have to use E=hc/wavelength to calculate energy of the photon as it has no mass. Then you can conclude that since the formulae are different then the wavelengths would be different. Hence, different patterns show different spacing
Post by: Mujteba on November 13, 2014, 03:05:52 am
What did everyone else get for the graph of the modulated signal? It kinda threw me off because I've never seen one involving light pulses before.
Have to fit the carrier wave in the input signal whilst maintaining its frequency. Also must NOT touch the axis because it is light intensity cannot be zero.
Post by: patel.aayush on November 13, 2014, 07:09:11 am
Have to fit the carrier wave in the input signal whilst maintaining its frequency. Also must NOT touch the axis because it is light intensity cannot be zero.
Wait what? why cant light intensity be zero?
Post by: Stew_822 on November 13, 2014, 07:22:32 am
Wait what? why cant light intensity be zero?Imagine if you're on the recieving end of this signal, say a kilometer away and it's transmitted through like those light wire things, whatever they're called, imagine if the light turned off. It could mean two things; that the signal is zero or the power was lost. So, adding the carrier wave means that there should always be light, and a zero signal might be transmitted through a 50% intensity. So you know if the light goes out, it isn't because the signal is 0, it's because something's gone haywire on the other end.
I just added the two graphs together for my answer, but I thought it was a weird question, because I thought signals were usually positive an negative. And wasn't the carrier wave just pulses of light? Or was that the signal? I can't remember
Post by: SweetBrown on November 13, 2014, 10:41:46 am
Post by: thebigleague on November 13, 2014, 11:26:15 am
I got a,c,c,d,a,b,b,a,b,d,d…
how confident are you on these because i got the exact same
Post by: examscoreslol on November 13, 2014, 12:05:34 pm
Post by: chuck981996 on November 14, 2014, 09:02:35 am
how confident are you on these because i got the exact same
Very confident. Some will say Q8 is b, however I think that if the wall wasn't there, the cantilever would wing left and down, so the wall must provide a force right and up, hence a. However, it's probably the only answer I'm unsure about. The rest are definitely right.
Post by: Seanos on November 14, 2014, 12:37:40 pm
Post by: Plitzer on November 14, 2014, 02:36:15 pm
For the arch question in materials and structures, did people get that no steel reinforcing was needed because the blocks were all in compression?Yeah, that's right.
Very confident. Some will say Q8 is b, however I think that if the wall wasn't there, the cantilever would wing left and down, so the wall must provide a force right and up, hence a. However, it's probably the only answer I'm unsure about. The rest are definitely right.There are a set of itute solutions up that say the answer to Q6 is D.... I got B as well but I dont remember what question that was. http://www.itute.com/board/viewtopic.php?f=6&t=2660&sid=803f70b44af0fc5bf87c392ff31981a3
Post by: Ya Habibi on November 14, 2014, 06:48:52 pm
For the arch question in materials and structures, did people get that no steel reinforcing was needed because the blocks were all in compression?The arch needed no steel reinforcements. Arches can support themselves because everything is in compression, and usually the material is strong against compressive stress.
Post by: silverpixeli on November 15, 2014, 12:44:00 am
Very confident. Some will say Q8 is b, however I think that if the wall wasn't there, the cantilever would wing left and down, so the wall must provide a force right and up, hence a. However, it's probably the only answer I'm unsure about. The rest are definitely right.
Someone described this question to me and I think you're right.
If you consider torques about the point of connection to the wall there's the torque from the weight of the cantilever itself and also from the tension in the cable, or effectively its vertical component. I understand that these were at different distances from the pivot. That being the case, their magnitude must have been different.
If the vertical component of tension up through the cable and the vertical weight force of the cantilever don't have the same magnitude, then there must be some other vertical force acting on the cantilever else it would not be in horizontal equilibrium. That force will be the vertical component of the force from the wall, meaning that overall it will point diagonally (with a horizontal component that cancels out the horizontal component of tension).
Only caveat is that I'm assuming the cable was not connected at the centre of mass of the bar, else the forces would cancel and the force at the wall would be horizontal only.
Post by: Plitzer on November 15, 2014, 02:00:33 pm
Someone described this question to me and I think you're right.You would be correct however the cable was supporting the bar at the centre of it's mass, so yes the force from the wall would be horizontal on the beam
If you consider torques about the point of connection to the wall there's the torque from the weight of the cantilever itself and also from the tension in the cable, or effectively its vertical component. I understand that these were at different distances from the pivot. That being the case, their magnitude must have been different.
If the vertical component of tension up through the cable and the vertical weight force of the cantilever don't have the same magnitude, then there must be some other vertical force acting on the cantilever else it would not be in horizontal equilibrium. That force will be the vertical component of the force from the wall, meaning that overall it will point diagonally (with a horizontal component that cancels out the horizontal component of tension).
Only caveat is that I'm assuming the cable was not connected at the centre of mass of the bar, else the forces would cancel and the force at the wall would be horizontal only.
Post by: silverpixeli on November 15, 2014, 04:17:59 pm
You would be correct however the cable was supporting the bar at the centre of it's mass, so yes the force from the wall would be horizontal on the beam
Cheers for clearing that up, I'm still keen to see a paper and I'll make up some worked solutions if someone can get me a scan!
Post by: silverpixeli on November 21, 2014, 10:36:01 am
Edit: done (5.17MB)