Post by: kenhung123 on June 08, 2010, 01:59:04 pm
Post by: bomb on June 08, 2010, 02:01:39 pm
Pretty easy but missed out on a few questions so don't think I got a B even.
Post by: tkjnz on June 08, 2010, 02:02:13 pm
Post by: fdsfsgdfgdf on June 08, 2010, 02:03:33 pm
Post by: 99.95 on June 08, 2010, 02:03:56 pm
Post by: VCE123456789 on June 08, 2010, 02:05:03 pm
Post by: fdsfsgdfgdf on June 08, 2010, 02:05:40 pm
Post by: rubiks on June 08, 2010, 02:06:24 pm
Post by: dirty_commie on June 08, 2010, 02:06:36 pm
Post by: bomb on June 08, 2010, 02:06:46 pm
what did you guys get for the spring question, with the total energy lost
I got 4J, it should have been 0 I think, but I couldnt show the working out to prove it.
Post by: curious111 on June 08, 2010, 02:07:45 pm
didnt get the spring one either. what was the difference in energy lost?
what did you guys get for the spring question, with the total energy lost
8J initial, and 4J final for a loss of 4J is what i got. The Initial energy was the GPe at a height of 0.4m, and the final energy was the Spring Potential. Not sure about this though.
Post by: fdsfsgdfgdf on June 08, 2010, 02:07:56 pm
Post by: kenhung123 on June 08, 2010, 02:08:47 pm
Well are we supposed to equate mg=kx? I got x=50n/m using that method
Yea I got 4J as well is that wrong?
Post by: Greggler on June 08, 2010, 02:09:13 pm
and a spring constant of 50 but it think i did it wrong.
For spring constant q i did F=kx whereas a friend of mine was certain you were meant to find the change in energy or something?
What about banked angle anyone get 5.7degrees?
Post by: fdsfsgdfgdf on June 08, 2010, 02:09:16 pm
Post by: dirty_commie on June 08, 2010, 02:09:20 pm
what did you guys get for the spring question, with the total energy lost
I got 4J, it should have been 0 I think, but I couldnt show the working out to prove it.
I'm guessing that was the one where the spring was held and then released? I got 0, because the GPE was converted so SPE, I'm guessing that was right, I hope.
Post by: kenhung123 on June 08, 2010, 02:09:43 pm
i got 0J but i didnt explain it properly had like 5 minutes. i just said energy is conservedHow'd you calculate 0? Or you just wrote 0 assuming its conserved?
Post by: fdsfsgdfgdf on June 08, 2010, 02:10:20 pm
1 got 1J of energy lost.i used Us=1/2 kx2, and i got 5.7 degrees dunno why
and a spring constant of 50 but it think i did it wrong.
For spring constant q i did F=kx whereas a friend of mine was certain you were meant to find the change in energy or something?
What about banked angle anyone get 5.7degrees?
Post by: sasha on June 08, 2010, 02:10:44 pm
was fairly decent, finished in time, made a few mistakes i know i shouldnt have. i literally walked out of the exam facepalming myself :( hoping for B-A
Post by: kenhung123 on June 08, 2010, 02:10:51 pm
1 got 1J of energy lost.Yea I got 5.7 using theta=tani^-1(v^/rg) (derived formula)
and a spring constant of 50 but it think i did it wrong.
For spring constant q i did F=kx whereas a friend of mine was certain you were meant to find the change in energy or something?
What about banked angle anyone get 5.7degrees?
Post by: fdsfsgdfgdf on June 08, 2010, 02:11:18 pm
i got 0J but i didnt explain it properly had like 5 minutes. i just said energy is conservedHow'd you calculate 0? Or you just wrote 0 assuming its conserved?
i calculated GPE and spring , since GPE is lost it is negative so Total Energy is zero
Post by: sasha on June 08, 2010, 02:11:35 pm
Post by: curious111 on June 08, 2010, 02:11:47 pm
Post by: fdsfsgdfgdf on June 08, 2010, 02:12:54 pm
Did anyone else get A, B, B for the three graphs. Not sure about this one
i got A, the straight line, and straight line, since momentum is conserved in elastic and inelastic
Post by: sasha on June 08, 2010, 02:13:08 pm
Did anyone else get A, B, B for the three graphs. Not sure about this one
pretty sure it was A B C
Post by: Greggler on June 08, 2010, 02:13:20 pm
apart from the spring q on motion though. i think i went alright. Only other concern are explain questions in electronics. they always confuse the hell out of me.
structures was the easiest i have ever seen lol. no angles or anything lol
Post by: kenhung123 on June 08, 2010, 02:13:26 pm
First is one that goes down then up to same level
Second stays same
Third goes down and back to lower level
Post by: dirty_commie on June 08, 2010, 02:13:45 pm
Did anyone else get A, B, B for the three graphs. Not sure about this oneYep, sure did.
Post by: sasha on June 08, 2010, 02:13:55 pm
Did anyone else get A, B, B for the three graphs. Not sure about this one
i got A, the straight line, and straight line, since momentum is conserved in elastic and inelastic
thrid one was asking if it was inelastic, i got C
Post by: Greggler on June 08, 2010, 02:14:19 pm
third w was about momentum.. trickkkyy asssss
Post by: fdsfsgdfgdf on June 08, 2010, 02:14:30 pm
yeah just derived that bad boy.yeah none of that difficult torque questions (i dont get torque still :()
apart from the spring q on motion though. i think i went alright. Only other concern are explain questions in electronics. they always confuse the hell out of me.
structures was the easiest i have ever seen lol. no angles or anything lol
Post by: kenhung123 on June 08, 2010, 02:16:10 pm
When you want the light to go on earlier LDR would increase causing Vout to increase so R must increase to reduce the influence of LDR increase?
Post by: bomb on June 08, 2010, 02:16:43 pm
Post by: jeff213 on June 08, 2010, 02:17:06 pm
Post by: kenhung123 on June 08, 2010, 02:17:17 pm
ABBthird was if the collision was inelastic so energy is lost..
third w was about momentum.. trickkkyy asssss
Post by: fdsfsgdfgdf on June 08, 2010, 02:17:37 pm
The explain questions for photonics when darker is increase and then increase yea? Coz less light=more resistance=more Vouti agree with the first ,
When you want the light to go on earlier LDR would increase causing Vout to increase so R must increase to reduce the influence of LDR increase?
but the second one doesnt the light increase since it is at sun set so early would be after noon eg.
Post by: kenhung123 on June 08, 2010, 02:17:46 pm
What do you think the A+ cut off will be for this exam? Lower than last year's 96?Hopefully...........
Post by: dirty_commie on June 08, 2010, 02:18:33 pm
What do you think the A+ cut off will be for this exam? Lower than last year's 96?
Without a doubt, lower.
Post by: Greggler on June 08, 2010, 02:19:53 pm
ABBthird was if the collision was inelastic so energy is lost..
third w was about momentum.. trickkkyy asssss
however it was asking about momentum. i thought too that it was asking about the total energy or kinetic energy or something. but it was about momentum. which is conserved despite being inelastic or elastic
Post by: sasha on June 08, 2010, 02:19:59 pm
this year's exam was alot harder, but still possible :D hoping cut of mark would be around 75ish
Post by: kyzoo on June 08, 2010, 02:21:51 pm
and LOL my calculator was in radians moed so i got 6.3 degrees for the banked curve one even though I had the correct working out
And I am really really confused about the 0J vs 4J question. I got 4J, but physics teacher says 0J =/ Such a confusing question =/
And haha I was going really slow at first...I saw i was at question 8 for motion when 30 minutes already passed >.< So I quickened up and finished with 20 mins to go lol...then i spent the last 20 minutes checking motions questions i had asterixed (there weren't really any contentious question in electricity or strctures)
Post by: kenhung123 on June 08, 2010, 02:23:37 pm
I got ABB for the graphs questionIs it 0J!!! Damn!!!!!!!!!!! I don't get why the graph is B for last one?
and LOL my calculator was in radians moed so i got 6.3 degrees for the banked curve one even though I had the correct working out
And I am really really confused about the 0J vs 4J question. I got 4J, but physics teacher says 0J =/ Such a confusing question =/
And haha I was going really slow at first...I saw i was at question 8 for motion when 30 minutes already passed >.< So I quickened up and finished with 20 mins to go lol...then i spent the last 20 minutes checking motions questions i had asterixed (there weren't really any contentious question in electricity or strctures)
Post by: tkjnz on June 08, 2010, 02:23:53 pm
what did you guys get for the spring question, with the total energy lost
I got 4J, it should have been 0 I think, but I couldnt show the working out to prove it.
i got 4J too, i was going to go back to recheck it but ran out of time
Post by: Greggler on June 08, 2010, 02:24:10 pm
spring q was a bit of a lolwut moment.
what about the projectile one./ did everyone get around 63m the 63m/s or something like that. i wasnt too sure about that one
Post by: sasha on June 08, 2010, 02:24:15 pm
Post by: fdsfsgdfgdf on June 08, 2010, 02:24:52 pm
I got ABB for the graphs questionIs it 0J!!! Damn!!!!!!!!!!! I don't get why the graph is B for last one?
and LOL my calculator was in radians moed so i got 6.3 degrees for the banked curve one even though I had the correct working out
And I am really really confused about the 0J vs 4J question. I got 4J, but physics teacher says 0J =/ Such a confusing question =/
And haha I was going really slow at first...I saw i was at question 8 for motion when 30 minutes already passed >.< So I quickened up and finished with 20 mins to go lol...then i spent the last 20 minutes checking motions questions i had asterixed (there weren't really any contentious question in electricity or strctures)
in my text book it says in a inelastic collision the KE is converted but momentum is conserved. lucky i read the book this morning
Post by: sasha on June 08, 2010, 02:25:00 pm
Post by: Cthulhu on June 08, 2010, 02:25:40 pm
Post by: kyzoo on June 08, 2010, 02:26:52 pm
Post by: curious111 on June 08, 2010, 02:27:06 pm
So yeah, 63.25 m, and 64.something for the final velocity.
Post by: sasha on June 08, 2010, 02:27:45 pm
The projectiles one confused me for a sec, when i calculated the final vertical velocity to be the EXACT same as the distance lol (63.25). Then i realised it makes sense because the initial velocity was 10m/s horizontal and the accelration was obviously 10.
So yeah, 63.25 m, and 64.something for the final velocity.
yea haha confused me for a sec aswell, but the horiztonal velocity was 10m/s :D
Post by: kenhung123 on June 08, 2010, 02:27:58 pm
I thought momentum is conserved despite elastic or not and kinetic energy is not conserved in inelastic collisionsI got ABB for the graphs questionIs it 0J!!! Damn!!!!!!!!!!! I don't get why the graph is B for last one?
and LOL my calculator was in radians moed so i got 6.3 degrees for the banked curve one even though I had the correct working out
And I am really really confused about the 0J vs 4J question. I got 4J, but physics teacher says 0J =/ Such a confusing question =/
And haha I was going really slow at first...I saw i was at question 8 for motion when 30 minutes already passed >.< So I quickened up and finished with 20 mins to go lol...then i spent the last 20 minutes checking motions questions i had asterixed (there weren't really any contentious question in electricity or strctures)
in my text book it says in a inelastic collision the KE is converted but momentum is conserved. lucky i read the book this morning
Post by: fdsfsgdfgdf on June 08, 2010, 02:28:07 pm
Momentum is conserved regardless of whether the collision is inelastic or not. And I spent my last 15 minutes on that 0J vs4J spring question >.< Omg like the principles don't match the calculations.what did you get for K? because my calculations were -8J[GPE]+8J[Spring]
Post by: Greggler on June 08, 2010, 02:28:35 pm
i kept ending up with 63.25 or sometrhing on my calc for both questinos and i was like....... this cant be right lol but i coudltn think of any other method to use
Post by: Blakhitman on June 08, 2010, 02:32:51 pm
I thought momentum is conserved despite elastic or not and kinetic energy is not conserved in inelastic collisionsI got ABB for the graphs questionIs it 0J!!! Damn!!!!!!!!!!! I don't get why the graph is B for last one?
and LOL my calculator was in radians moed so i got 6.3 degrees for the banked curve one even though I had the correct working out
And I am really really confused about the 0J vs 4J question. I got 4J, but physics teacher says 0J =/ Such a confusing question =/
And haha I was going really slow at first...I saw i was at question 8 for motion when 30 minutes already passed >.< So I quickened up and finished with 20 mins to go lol...then i spent the last 20 minutes checking motions questions i had asterixed (there weren't really any contentious question in electricity or strctures)
in my text book it says in a inelastic collision the KE is converted but momentum is conserved. lucky i read the book this morning
Exactly why it was B.
Post by: m@tty on June 08, 2010, 02:33:37 pm
Post by: Blakhitman on June 08, 2010, 02:34:30 pm
Post by: Blakhitman on June 08, 2010, 02:35:05 pm
Post by: m@tty on June 08, 2010, 02:35:35 pm
EDIT: Yeah. For structures:
1.D
2.B
3.C
4.D
5.C
6.B
7.D
8.A
9.C
10.D
11.D
12.B
Post by: kyzoo on June 08, 2010, 02:35:57 pm
'Momentum is conserved regardless of whether the collision is inelastic or not. And I spent my last 15 minutes on that 0J vs4J spring question >.< Omg like the principles don't match the calculations.what did you get for K? because my calculations were -8J[GPE]+8J[Spring]
k = 50N/m
If you use F = kx you get k = 50
If you use Us = 0.5kx^2 you get k = 100
I would have got this wrong if the same question hadn't been on CSE 2010 lol
Post by: rubiks on June 08, 2010, 02:36:54 pm
Post by: Twenty10 on June 08, 2010, 02:37:24 pm
And I put 0J aswell...I hope to God I'm right.
this... completely frkin confusing. made me go back and change my spring constant to 100 :(...will i get a mark for having the right answer then crossing it out?
Post by: coletrain on June 08, 2010, 02:37:40 pm
'Momentum is conserved regardless of whether the collision is inelastic or not. And I spent my last 15 minutes on that 0J vs4J spring question >.< Omg like the principles don't match the calculations.what did you get for K? because my calculations were -8J[GPE]+8J[Spring]
k = 50N/m
If you use F = kx you get k = 50
If you use Us = 0.5kx^2 you get k = 100
I would have got this wrong if the same question hadn't been on CSE 2010 lol
yeah i got this :)
what did u get for the question after this, the change in energy?
Post by: mkl3 on June 08, 2010, 02:38:24 pm
Post by: coletrain on June 08, 2010, 02:38:31 pm
Post by: rubiks on June 08, 2010, 02:38:51 pm
Post by: googoo on June 08, 2010, 02:39:10 pm
Post by: coletrain on June 08, 2010, 02:39:29 pm
what did the circuit look like
the final electronics question?
Post by: Greggler on June 08, 2010, 02:39:39 pm
although i remmebr to f=kx to find spring constant. lol, like kyzoo after doing cse2010 i only ever find spring constants with f=kx haha
Post by: kyzoo on June 08, 2010, 02:39:49 pm
Why would there be discrepancy? It said nothing about loss to anything...
EDIT: Yeah. For structures:
1.D
2.B
3.C
4.D
5.C
6.B
7.D
8.A
9.C
10.D
11.D
12.B
=/ I don't remember what I put down for structures, only that there were no confusing hard questions in there.
'Momentum is conserved regardless of whether the collision is inelastic or not. And I spent my last 15 minutes on that 0J vs4J spring question >.< Omg like the principles don't match the calculations.what did you get for K? because my calculations were -8J[GPE]+8J[Spring]
k = 50N/m
If you use F = kx you get k = 50
If you use Us = 0.5kx^2 you get k = 100
I would have got this wrong if the same question hadn't been on CSE 2010 lol
yeah i got this :)
what did u get for the question after this, the change in energy?
I put down 4J because that's what the maths said, but I have no idea what the correct answer is maent to be -.- I spent like 15 minutes trying to figure out if there was any trick in the wording and I couldn't decide if them saying "total energy" instead of "total mechanical enregy" was meant to be a trick or not
Post by: rubiks on June 08, 2010, 02:40:05 pm
not too hard. itute has the answers up.can you link us to it?
Post by: m@tty on June 08, 2010, 02:40:18 pm
'Momentum is conserved regardless of whether the collision is inelastic or not. And I spent my last 15 minutes on that 0J vs4J spring question >.< Omg like the principles don't match the calculations.what did you get for K? because my calculations were -8J[GPE]+8J[Spring]
k = 50N/m
If you use F = kx you get k = 50
If you use Us = 0.5kx^2 you get k = 100
I would have got this wrong if the same question hadn't been on CSE 2010 lol
Serious? Where does the energy go then ?? ?
Post by: Twenty10 on June 08, 2010, 02:40:38 pm
not too hard. itute has the answers up.how do u access them?
Post by: taiga on June 08, 2010, 02:41:20 pm
Multichoice motion was ABB i'm pretty sure of that. 64m/s for the final speed. I can't remember what the distance I put was though =\ I don't think I put 63 though...
I got different answers to matty for multichoice though. That's based on memory.
Post by: Blakhitman on June 08, 2010, 02:41:50 pm
Post by: m@tty on June 08, 2010, 02:42:29 pm
Post by: curious111 on June 08, 2010, 02:42:42 pm
Didn't have time to check them though, very high chance of a stupid calculation error.
Post by: rubiks on June 08, 2010, 02:43:39 pm
Post by: Greggler on June 08, 2010, 02:44:04 pm
Post by: Blakhitman on June 08, 2010, 02:44:15 pm
D
B
C
C
C
B
D
A
C
D
D
A
Post by: kyzoo on June 08, 2010, 02:44:23 pm
The formula Us = 0.5kx^2 doesn't work for obtaining the spring constant k, therefore it doesn't work for getting the potential spring energy stored in the spring after extension.
But...i dunno still >.< I still think it's ambiguous question
Post by: rubiks on June 08, 2010, 02:45:12 pm
Post by: m@tty on June 08, 2010, 02:45:19 pm
Post by: Blakhitman on June 08, 2010, 02:45:23 pm
Actually now that I think about it, it should be 0J.
The formula Us = 0.5kx^2 doesn't work for obtaining the spring constant k, therefore it doesn't work for getting the potential spring energy stored in the spring after extension.
But...i dunno still >.< I still think it's ambiguous question
hopefully marks for both!!! :D
Post by: coletrain on June 08, 2010, 02:45:43 pm
Post by: kyzoo on June 08, 2010, 02:46:34 pm
why doesnt .5kx^2 work for getting the spring constant?
Because that formula assumes that the initial spring energy stored in the spring is 0J, but if it doesn't start at 0J, then the formula doesn't find the area under the graph anymore.
But then again, the question did say the spring was initially under no extension...so it must be a question in which the maths contradicts the wording =/
Post by: googoo on June 08, 2010, 02:47:42 pm
http://www.itute.com/board/viewforum.php?f=6not too hard. itute has the answers up.can you link us to it?
Post by: rubiks on June 08, 2010, 02:48:29 pm
Post by: coletrain on June 08, 2010, 02:48:48 pm
http://www.itute.com/board/viewforum.php?f=6not too hard. itute has the answers up.can you link us to it?
TYYYYYYYYYYYYY
Post by: Blakhitman on June 08, 2010, 02:49:40 pm
63.something m for the projectile, and 64 m/s for velocity at ground. 20 m/s for the roller coaster one and 5.71 degrees for the banked angle.
Post by: kyzoo on June 08, 2010, 02:51:50 pm
Quote
Q8 Correct, force of gravity
That's wrong, you need to distinguish between true weightlessness and apparaent weightlessness in that question
Post by: Blakhitman on June 08, 2010, 02:52:32 pm
Post by: Greggler on June 08, 2010, 02:52:42 pm
hopefully make the a/a+ cutoff
Post by: Twenty10 on June 08, 2010, 02:53:37 pm
Post by: fdsfsgdfgdf on June 08, 2010, 02:53:53 pm
i think he meant thatQuoteQ8 Correct, force of gravity
That's wrong, you need to distinguish between true weightlessness and apparaent weightlessness in that question
Post by: m@tty on June 08, 2010, 02:54:03 pm
QuoteQ8 Correct, force of gravity
That's wrong, you need to distinguish between true weightlessness and apparaent weightlessness in that question
It was only two marks. I don't think you did. All you had to do was explain why the statement was correct/incorrect.
Post by: kyzoo on June 08, 2010, 02:54:17 pm
Post by: sasha on June 08, 2010, 02:54:38 pm
NOOOOOOOOOOOOO last of structures was B according to the Itutes guy.can you link to the itute physics solutions please?
Post by: confusedperson123 on June 08, 2010, 02:54:56 pm
Yea the 4J question is really ambiguous...
Post by: coletrain on June 08, 2010, 02:55:09 pm
Post by: Blakhitman on June 08, 2010, 02:55:19 pm
Post by: m@tty on June 08, 2010, 02:55:47 pm
I got D 0.o ... it's a footbridge, therefore it sags over a gap, therefore the steelcables should be at the bottom
That was question 11, the itute answers have D...
2.0 m/s for the rollercoaster??? lol gurantee you that's wrong, 1000kg from 20 m high will NOT have 2m/s speed.
The itute answers say 20..
Post by: Blakhitman on June 08, 2010, 02:57:22 pm
NOOOOOOOOOOOOO last of structures was B according to the Itutes guy.can you link to the itute physics solutions please?
http://www.itute.com/board/viewtopic.php?f=6&t=1653&sid=8119a1c2fafd324760c30716dea82382
Post by: fdsfsgdfgdf on June 08, 2010, 02:57:29 pm
I got D 0.o ... it's a footbridge, therefore it sags over a gap, therefore the steelcables should be at the bottomtrue crap i got that wrong :(
Post by: Blakhitman on June 08, 2010, 02:57:56 pm
I got D 0.o ... it's a footbridge, therefore it sags over a gap, therefore the steelcables should be at the bottom
That was question 11, the itute answers have D...2.0 m/s for the rollercoaster??? lol gurantee you that's wrong, 1000kg from 20 m high will NOT have 2m/s speed.
The itute answers say 20..
nevermind me then...lol
Post by: kyzoo on June 08, 2010, 02:58:09 pm
And the correct speed is 28.3m/s im pretty sure
Post by: Galvanic Cell on June 08, 2010, 02:58:23 pm
Post by: m@tty on June 08, 2010, 02:59:09 pm
Question 12 was the one about permanent deformation => past its elastic limit.
Post by: Blakhitman on June 08, 2010, 02:59:17 pm
Post by: confusedperson123 on June 08, 2010, 02:59:46 pm
Post by: Greggler on June 08, 2010, 02:59:54 pm
Post by: m@tty on June 08, 2010, 03:00:18 pm
Post by: taiga on June 08, 2010, 03:00:35 pm
[edit]ohwait that was 11 nvm :D
I thought it was 2.5m/s^2 for the block
Post by: confusedperson123 on June 08, 2010, 03:00:45 pm
Post by: fdsfsgdfgdf on June 08, 2010, 03:02:56 pm
2.5 is wrong you have to use mass of both objects.no you use m1 as the only force acting is the Force gravity on M1, and m2 has no force acting on it (overal force) force net
Post by: taiga on June 08, 2010, 03:03:28 pm
Post by: rubiks on June 08, 2010, 03:04:32 pm
Post by: kyzoo on June 08, 2010, 03:05:58 pm
Think about it. The two blocks must accelerate at the same rate if they are attached like that. If the small block is attached to a large truck, then obviously the small blocks will travel slower than if the small block was attached to nothing.
Hence you must use the mass of the entire system, rather than of one block.
We can work backwards too...
Assume the answer is 2m/s^2
Net force on 0.4kg block = ma = 0.8N = Tension in string
If the 0.1kg block is undergoing acceleration of 2m/2^s, then net force on 0.1kg block is ma = 0.2N
Net force on 0.1kg block = Weight force - tension force = 1.0 - 0.8 = 0.2N
Post by: confusedperson123 on June 08, 2010, 03:06:49 pm
Post by: Blakhitman on June 08, 2010, 03:07:01 pm
Post by: rubiks on June 08, 2010, 03:07:40 pm
Post by: fdsfsgdfgdf on June 08, 2010, 03:07:43 pm
xD I am sure its 2 m/s^2, I used two methods to work it out.but the mass is on a frictionless surface, so where does the force come from to oppose the acceleration of m1?
Think about it. The two blocks must accelerate at the same rate if they are attached like that. If the small block is attached to a large truck, then obviously the small blocks will travel slower than if the small block was attached to nothing.
Hence you must use the mass of the entire system, rather than of one block.
Post by: taiga on June 08, 2010, 03:08:13 pm
F....M.....L
Post by: Yoshi on June 08, 2010, 03:11:07 pm
Anyone else do relativity? I was so pissed off with the question about the 'direct' result of the Michelson Morley experiment, selected the one about no evidence of ether but I reckon they probably wanted the one to do with speed of light.
Post by: ghadz7 on June 08, 2010, 03:11:22 pm
Post by: Blakhitman on June 08, 2010, 03:12:22 pm
Post by: confusedperson123 on June 08, 2010, 03:12:45 pm
but it was on a frictionless surface, so it doesn't matter what the mass of the other block was because there is nothing that will oppose the motion
uhm so youre assuming the mass of the other block is 0? lol
Post by: Stroodle on June 08, 2010, 03:13:01 pm
Post by: rubiks on June 08, 2010, 03:13:19 pm
Post by: rubiks on June 08, 2010, 03:14:08 pm
^ its 4J 1million percent surebut it was on a frictionless surface, so it doesn't matter what the mass of the other block was because there is nothing that will oppose the motionyou're disregarding the mass of the other block because its on a frictionless surface. well thats what i did anyway
uhm so youre assuming the mass of the other block is 0? lol
Post by: ghadz7 on June 08, 2010, 03:14:35 pm
Post by: confusedperson123 on June 08, 2010, 03:15:04 pm
xD I am sure its 2 m/s^2, I used two methods to work it out.
Think about it. The two blocks must accelerate at the same rate if they are attached like that. If the small block is attached to a large truck, then obviously the small blocks will travel slower than if the small block was attached to nothing.
Hence you must use the mass of the entire system, rather than of one block.
We can work backwards too...
Assume the answer is 2m/s^2
Net force on 0.4kg block = ma = 0.8N = Tension in string
If the 0.1kg block is undergoing acceleration of 2m/2^s, then net force on 0.1kg block is ma = 0.2N
Net force on 0.1kg block = Weight force - tension force = 1.0 - 0.8 = 0.2N
Post by: fdsfsgdfgdf on June 08, 2010, 03:16:22 pm
^ its 4J 1million percent surebut it was on a frictionless surface, so it doesn't matter what the mass of the other block was because there is nothing that will oppose the motion
uhm so youre assuming the mass of the other block is 0? lol
no it doesnt matter because a=F\M (m2) there is zero opposing force, so it wouldnt matter what the mass is
Post by: strathyboy23 on June 08, 2010, 03:17:58 pm
This was definately the hardest exam since the '06 exam, and A+ cutoff for that year was
79.45% (71.5/90) so looking at it like that it makes me feel slightly better.
Post by: kyzoo on June 08, 2010, 03:18:25 pm
Then net force on 0.4kg block = ma = 1N = Tension force
Net force on 0.1kg block = ma = 0.25N
Net force on 0.1kg block = weight force - tension force = 1 - 1 = 0N
But the net force on the 0.1kg block can't be 0N. Also, 0N does not equal 0.25N
Post by: taiga on June 08, 2010, 03:19:34 pm
Post by: Fyrefly on June 08, 2010, 03:25:58 pm
Dividing by 0? Dangerous territory there.
(http://i74.photobucket.com/albums/i249/seviman/divide_by_zero.jpg)
(http://lgo.mit.edu/blog/drewhill/files/divide-by-zero-blog-safe.jpg)
Post by: confusedperson123 on June 08, 2010, 03:26:27 pm
Post by: Stroodle on June 08, 2010, 03:27:30 pm
dropped so many marks...looking at the answers now i probably shouldnt have dropped them.
This was definately the hardest exam since the '06 exam, and A+ cutoff for that year was
79.45% (71.5/90) so looking at it like that it makes me feel slightly better.
Seriously? I thought it was about the same as last year. Just with a couple of tricks.
My teacher reckons A+ cutoff will be lower than last year..
The mass of the tennis ball stumped a few ppl in my class.
Post by: fdsfsgdfgdf on June 08, 2010, 03:28:33 pm
And if you assume 2.5 m/s^2 is the right answerhmm there is no tension force since the .4kg is on a frictionless surface so where does the force opposingn it come from?
Then net force on 0.4kg block = ma = 1N = Tension force
Net force on 0.1kg block = ma = 0.25N
Net force on 0.1kg block = weight force - tension force = 1 - 1 = 0N
But the net force on the 0.1kg block can't be 0N. Also, 0N does not equal 0.25N
Post by: kyzoo on June 08, 2010, 03:29:38 pm
Post by: confusedperson123 on June 08, 2010, 03:30:52 pm
Post by: fdsfsgdfgdf on June 08, 2010, 03:33:03 pm
If there's no tension force...then how does the 0.4kg block move?good point . lol ,but why did you subract shouldnt the direction be the same, its accelerating down, and there is no opposing force.
ok becasue there is an action reaction?
Post by: m@tty on June 08, 2010, 03:34:20 pm
Post by: kyzoo on June 08, 2010, 03:34:54 pm
Weight force = downwards
Tension force pulls objects towards the middle of the string, so its upwards.
Hence you subtract tension force
Post by: Stroodle on June 08, 2010, 03:36:47 pm
Chief examiner must have got hammered for making it too easy last time around, what an increase in difficulty.
Anyone else do relativity? I was so pissed off with the question about the 'direct' result of the Michelson Morley experiment, selected the one about no evidence of ether but I reckon they probably wanted the one to do with speed of light.
Yeah, it was the speed of light, according to my teacher; and he's a relativity nut.
Post by: rubiks on June 08, 2010, 03:50:00 pm
Post by: kenhung123 on June 08, 2010, 03:52:18 pm
If there's no tension force...then how does the 0.4kg block move?I think there was tension but they cancelled and the net force was just weight of hanging mass?
Post by: Blakhitman on June 08, 2010, 03:52:52 pm
Post by: fdsfsgdfgdf on June 08, 2010, 03:55:38 pm
umm there is no tension becayse both masses were moving at the same speed, and the mass M2 doesnt provide any forces on M1 [down], therefore it can be considered as one objectIf there's no tension force...then how does the 0.4kg block move?I think there was tension but they cancelled and the net force was just weight of hanging mass?
edit: nvm too hard to prove, if it was really; probaly not
Post by: dyaner on June 08, 2010, 03:57:58 pm
Post by: Twenty10 on June 08, 2010, 03:58:46 pm
Post by: superflya on June 08, 2010, 04:00:33 pm
where are the solutions i see nothing on itute :P
Post by: Blakhitman on June 08, 2010, 04:01:55 pm
Post by: fdsfsgdfgdf on June 08, 2010, 04:02:37 pm
umm there is no tension becayse both masses were moving at the same speed, and the mass M2 doesnt provide any forces on M1 [down], therefore it can be considered as one objectIf there's no tension force...then how does the 0.4kg block move?I think there was tension but they cancelled and the net force was just weight of hanging mass?
edit: nvm too hard to prove, if it was really; probaly not
for a tension to occur there has to be an object pulling at one end and the opposite at the other. however object M2 doesnt pull on it so there is no tension, just an overall motion
Post by: appianway on June 08, 2010, 04:04:11 pm
Post by: rubiks on June 08, 2010, 04:06:20 pm
umm there is no tension becayse both masses were moving at the same speed, and the mass M2 doesnt provide any forces on M1 [down], therefore it can be considered as one objectIf there's no tension force...then how does the 0.4kg block move?I think there was tension but they cancelled and the net force was just weight of hanging mass?
edit: nvm too hard to prove, if it was really; probaly not
for a tension to occur there has to be an object pulling at one end and the opposite at the other. however object M2 doesnt pull on it so there is no tension, just an overall motion
i hope you're right
Post by: schnappy on June 08, 2010, 04:12:10 pm
I got 4J for the '4Jvs0J' question... however I'm 99% (arbitary numbers hooray) sure it is 0J. It did say it was an ideal spring... Also, they gave writing space for your answer. Not white space for calculations. I believe the 'lowered slowly' part if just to stop people from introducing stuff to do with kinetic energy.
Post by: Blakhitman on June 08, 2010, 04:15:09 pm
Post by: superflya on June 08, 2010, 04:16:44 pm
Post by: coletrain on June 08, 2010, 04:17:18 pm
Post by: m@tty on June 08, 2010, 04:17:39 pm
Post by: rubiks on June 08, 2010, 04:18:09 pm
omg if it really is 0J then my life may not be over after all.
Post by: confusedperson123 on June 08, 2010, 04:19:39 pm
Post by: schnappy on June 08, 2010, 04:20:56 pm
Also about the blocks, an afterthought just occured. The rope that binds the blocks... the acceleration of the top block must be the same as the falling black. a=g...
Post by: coletrain on June 08, 2010, 04:22:05 pm
The energy is conserved BECAUSE IT WAS A FUCKING IDEAL SPRING. Rightio then.
Also about the blocks, an afterthought just occured. The rope that binds the blocks... the acceleration of the top block must be the same as the falling black. a=g...
i totally screwed this question up :(
Post by: tha1ne on June 08, 2010, 04:22:27 pm
what you think.. i still have no idea y
Post by: Greggler on June 08, 2010, 04:24:09 pm
Post by: m@tty on June 08, 2010, 04:24:55 pm
Post by: superflya on June 08, 2010, 04:25:13 pm
Post by: kenhung123 on June 08, 2010, 04:27:04 pm
The graphs were definitely NOT all A. The first one is A. The second one is about momentum. Momentum is conserved throughout a collision, hence B. Also, the third one was apparently asking about momentum also(I'm not sure though; I read it as kinetic energy...), so that should have also been B.I thought it was kinetic energy too...
Post by: coletrain on June 08, 2010, 04:27:14 pm
Post by: Juddinator on June 08, 2010, 04:27:24 pm
Post by: rubiks on June 08, 2010, 04:27:55 pm
if thats the case then spring constant should be 100?I will be so happy if this was true, along with 0J
Post by: Stroodle on June 08, 2010, 04:28:02 pm
Spring constant was 50, and difference was 4J.
Post by: Blakhitman on June 08, 2010, 04:29:49 pm
if thats the case then spring constant should be 100?I will be so happy if this was true, along with 0J
My life will be looking good.
Post by: m@tty on June 08, 2010, 04:30:03 pm
Post by: coletrain on June 08, 2010, 04:30:13 pm
I'm pretty sure I aced the S&M imo, that was a joke this year, so easy haha.
i know right... i finished the core at around 10 past and i was like "F**K i dont have enough time for M&S" but it was really straightforward
Post by: confusedperson123 on June 08, 2010, 04:31:06 pm
Post by: schnappy on June 08, 2010, 04:31:39 pm
Post by: bomb on June 08, 2010, 04:31:54 pm
And, if all energy was conserved, how did you show it through calculations?
Post by: schnappy on June 08, 2010, 04:32:32 pm
For the third graph, it did saysome stuff about kinetic energy. It said the collision was now elastic or inelastic... to make you think EK. But, the question did ask for the graph of P-t. Ie. they wanted to see if you understood how momentum changes in an elastic collision.
No. the hand slowly let it down, the force acting to bring the mass down was weight/gravity.
Post by: schnappy on June 08, 2010, 04:33:27 pm
Ideal spring doesn't have to mean it can't lose energy to heat/sound. It could just mean it obeys Hooke's law.
And, if all energy was conserved, how did you show it through calculations?
Ideal spring does mean energy is conserved. That's what's so nice about it.
You don't show it through calculations. Was i the only person that was given about 5 lines to write my answer on? It seams obvious now... although I did say 4J.
Post by: TrueTears on June 08, 2010, 04:33:53 pm
Post by: rubiks on June 08, 2010, 04:34:03 pm
Post by: bufumufu_ on June 08, 2010, 04:34:18 pm
the energy goes into the hand becoz it applies a force when lowering the spring.
godlike explanation right there
Post by: Stroodle on June 08, 2010, 04:34:29 pm
Post by: taiga on June 08, 2010, 04:35:18 pm
I'ld assume you'ld scrape some consequential for a 10, 0 pair though
Post by: Juddinator on June 08, 2010, 04:37:27 pm
Yeah I know I did most of it in my head during reading time haha.I'm pretty sure I aced the S&M imo, that was a joke this year, so easy haha.
i know right... i finished the core at around 10 past and i was like "F**K i dont have enough time for M&S" but it was really straightforward
Post by: coletrain on June 08, 2010, 04:38:12 pm
anyone have the exam?
Post by: coletrain on June 08, 2010, 04:38:36 pm
the only way to know for sureanyone have the exam?
Post by: TrueTears on June 08, 2010, 04:39:28 pm
Post by: schnappy on June 08, 2010, 04:39:59 pm
Post by: bomb on June 08, 2010, 04:40:38 pm
Ideal spring doesn't have to mean it can't lose energy to heat/sound. It could just mean it obeys Hooke's law.
And, if all energy was conserved, how did you show it through calculations?
Ideal spring does mean energy is conserved. That's what's so nice about it.
You don't show it through calculations. Was i the only person that was given about 5 lines to write my answer on? It seams obvious now... although I did say 4J.
There was also a fair bit of white space above that....Which is where I did my calculations.
Besides you cant just say it's all conserved with no calculations, you need to show some proof.
Post by: Stroodle on June 08, 2010, 04:41:14 pm
Anyone else do further elec? Hardest further elec questions I've ever seen... regret doing F.E. for sure. Pricks doing structures!
I think they'll scale them according to the results. At least, that's what my teacher reckons they do.
Post by: schnappy on June 08, 2010, 04:43:16 pm
Ideal spring doesn't have to mean it can't lose energy to heat/sound. It could just mean it obeys Hooke's law.
And, if all energy was conserved, how did you show it through calculations?
Ideal spring does mean energy is conserved. That's what's so nice about it.
You don't show it through calculations. Was i the only person that was given about 5 lines to write my answer on? It seams obvious now... although I did say 4J.
There was also a fair bit of white space above that....Which is where I did my calculations.
Besides you cant just say it's all conserved with no calculations, you need to show some proof.
The law of conservation of energy. We weren't given the resources or time to proof that. And it was 2 marks wasn't it? 1 mark for 0J, 1 mark for stating that law.
Post by: Blakhitman on June 08, 2010, 04:44:48 pm
the energy goes into the hand becoz it applies a force when lowering the spring.
If you think of it this way...how did you know what F=? you're saying the hand applied the force, yet you use mg as the force?
Post by: bomb on June 08, 2010, 04:58:32 pm
Ideal spring doesn't have to mean it can't lose energy to heat/sound. It could just mean it obeys Hooke's law.
And, if all energy was conserved, how did you show it through calculations?
Ideal spring does mean energy is conserved. That's what's so nice about it.
You don't show it through calculations. Was i the only person that was given about 5 lines to write my answer on? It seams obvious now... although I did say 4J.
There was also a fair bit of white space above that....Which is where I did my calculations.
Besides you cant just say it's all conserved with no calculations, you need to show some proof.
The law of conservation of energy. We weren't given the resources or time to proof that. And it was 2 marks wasn't it? 1 mark for 0J, 1 mark for stating that law.
Or one mark for 4J and one mark for calculation. It did say 'show your working' rather than 'explain', which is what stopped me from writing 0J. I think that the answer was 0J, but I wrote 4J because I didn't have time to change it.
Post by: kyzoo on June 08, 2010, 04:59:10 pm
Post by: TrueTears on June 08, 2010, 05:01:19 pm
Post by: kyzoo on June 08, 2010, 05:02:05 pm
Post by: TrueTears on June 08, 2010, 05:02:37 pm
Post by: rubiks on June 08, 2010, 05:02:46 pm
Post by: lachymm on June 08, 2010, 05:03:50 pm
Post by: Twenty10 on June 08, 2010, 05:04:24 pm
Further Elec was pre hard i thought
Post by: Cthulhu on June 08, 2010, 05:05:53 pm
Post by: Greggler on June 08, 2010, 05:06:14 pm
lol i dunno. confusing and ambiguous
Post by: lachymm on June 08, 2010, 05:06:30 pm
Post by: qshyrn on June 08, 2010, 05:07:54 pm
Er what exactly is the question?well there was a spring (vertical) when u put a 2 kg mass on it, it extends by 0.4m
Post by: Cthulhu on June 08, 2010, 05:08:34 pm
and?Er what exactly is the question?well there was a spring (vertical) when u put a 2 kg mass on it, it extends by 0.4m
Post by: qshyrn on June 08, 2010, 05:10:12 pm
it said wats the spring kand?Er what exactly is the question?well there was a spring (vertical) when u put a 2 kg mass on it, it extends by 0.4m
Post by: fdsfsgdfgdf on June 08, 2010, 05:13:38 pm
what is the magnitude of change in energy of figure a [unstreched spring] and figure b [spring moves downwards and is stretched by .4m]it said wats the spring kand?Er what exactly is the question?well there was a spring (vertical) when u put a 2 kg mass on it, it extends by 0.4m
Post by: lachymm on June 08, 2010, 05:14:42 pm
Post by: schnappy on June 08, 2010, 05:14:52 pm
You needed to first find k, the spring constant, then explain the change in energy as the 2kg mass is lowered from the spring causing it to extend 0.4m.
The second question did say the change in the system or the change in the spring?
Post by: Cthulhu on June 08, 2010, 05:17:48 pm
I made a negative for good measure to cancel out the - in the RHS
then just plug and play and you'd get
Whats so hard about that?
Then for the change in energy:
Edit: Sorry about that folks.
Post by: Greggler on June 08, 2010, 05:18:20 pm
Post by: schnappy on June 08, 2010, 05:21:30 pm
Post by: fdsfsgdfgdf on June 08, 2010, 05:21:59 pm
Well if it were me I'd set it outthe problem is some people including my self used Us=GPE to find the constant, hence the variation in the change in energy
I made a negative for good measure to cancel out the - in the RHS
then just plug and play and you'd get
Whats so hard about that?
Then for the change in energy:
Or am I missing something?
Post by: schnappy on June 08, 2010, 05:24:11 pm
Post by: Cthulhu on June 08, 2010, 05:26:10 pm
Err.. I made a woopsie.... REVISED BELOW:Well if it were me I'd set it outthe problem is some people including my self used Us=GPE to find the constant, hence the variation in the change in energy
I made a negative for good measure to cancel out the - in the RHS
then just plug and play and you'd get
Whats so hard about that?
Then for the change in energy:
Or am I missing something?
I made a negative for good measure to cancel out the - in the RHS
then just plug and play and you'd get
Whats so hard about that?
Then for the change in energy:
Sorry about that folks.
Post by: rubiks on June 08, 2010, 05:32:27 pm
I believe the alternate response was that the change is 0J since that EGP must become strain energy, and since it's an ideal spring there are no losses.
thats what I did. Maybe VCAA will be nice and say both answers are right!
Post by: appianway on June 08, 2010, 05:34:37 pm
Post by: Blakhitman on June 08, 2010, 05:36:28 pm
Post by: fdsfsgdfgdf on June 08, 2010, 05:37:21 pm
Post by: appianway on June 08, 2010, 05:38:04 pm
Post by: Blakhitman on June 08, 2010, 05:38:45 pm
Post by: schnappy on June 08, 2010, 05:39:42 pm
I don't even see how you can use 0.5kx^2 when you don't know the energy when x = 0.4...
Post by: appianway on June 08, 2010, 05:40:28 pm
Post by: kyzoo on June 08, 2010, 05:41:00 pm
will I lose marks for putting
"Energy is lost to heat??" as the end sentence for explanation?
Post by: m@tty on June 08, 2010, 05:41:47 pm
So you can't use mg=kx?
I think you can, actually. In the end the spring IS supporting the whole weight of the 2.0 kg mass (mg). So you can equate it to kx...
Why didn't I realise this before the exam finished ... :[
Post by: Juddinator on June 08, 2010, 05:42:26 pm
Post by: appianway on June 08, 2010, 05:42:40 pm
Post by: fdsfsgdfgdf on June 08, 2010, 05:43:26 pm
...depends on the wording of the question, but i thought it was an ideal situation, anyways we need the exam
will I lose marks for putting
"Energy is lost to heat??" as the end sentence for explanation?
Post by: schnappy on June 08, 2010, 05:43:41 pm
Post by: sasha on June 08, 2010, 05:43:53 pm
Post by: Cthulhu on June 08, 2010, 05:45:40 pm
The hand is going down, but as the weight isn't accelerating at near the acceleration of gravity+ acceleration that the spring would cause, the hand is also providing a force.But after the spring has been lowered the mass is hanging freely and is being acted upon by gravity?
Is there anyway people can get their hands on the paper? Will your teachers let you have it for a night or something? :PWhen I sat mine the nice ladies in the exam room let me take a spare one.
Idk. Maybe you need to be more specific, but it'd be a dissipative conversion of energy, so it'd probably go into thermal energy (rising entropy! ;D).
Post by: jeff213 on June 08, 2010, 05:47:51 pm
Post by: rubiks on June 08, 2010, 05:49:34 pm
Post by: SixWinged on June 08, 2010, 05:50:45 pm
If the "ideal spring" was allowed to fall at full speed, the gravitational energy would be transformed into both elastic potential and kinetic energy (oscillations). At the very bottom of the first oscillation (past the point where the mass is stationary in the second picture) all of the energy would have been transformed into elastic potential. In a real life scenario, air friction would do work to reduce the oscillation distance until it was 0, it's clear that energy is lost in this situation.
However when the hand lowers the mass, it is effectively doing the same work as air friction would do, and hence energy is lost.
All in all, pretty tricky exam, dropped one due to neglect on the electricity section, pretty confident with everything else though, having said that, I did spend a good 20 minutes staring at this question, wasn't until I left the exam that I came to this conclusion though.
Post by: appianway on June 08, 2010, 05:53:38 pm
The hand is going down, but as the weight isn't accelerating at near the acceleration of gravity+ acceleration that the spring would cause, the hand is also providing a force.But after the spring has been lowered the mass is hanging freely and is being acted upon by gravity?
Let me explain.
The spring must be at its equilibrium point when the hand lets go and it is at 0.4 m. This is as it DOES NOT oscillate, and if it were at any other point, it would experience a net force and hence oscillate. However, let's assume that the hand provides no force whatsoever while lowering the object. This would mean that if conservation of energy were to hold, Eg = kx^2/2. kx = 10N, and x = 4, so k=2.5 (have I subbed in the wrong numbers?) Eg = 4J, and kx^2 is significantly lower. This indicates that another form of energy would be needed - there'd be kinetic energy. However, as there is no kinetic energy as it does not oscillate, a force must do work on the object to diminish this form of energy.
Post by: bblovee on June 08, 2010, 06:17:54 pm
Well i put k=50 and 0J >_<
Post by: Kennybhoy on June 08, 2010, 06:22:58 pm
Post by: schnappy on June 08, 2010, 06:24:41 pm
Post by: Killerkob on June 08, 2010, 06:24:51 pm
Anyone else think these were the hardest ESR questions on an exam?
Star Wars reference ftw.
Post by: kyzoo on June 08, 2010, 06:27:30 pm
Post by: Stroodle on June 08, 2010, 06:29:06 pm
woohoo my physics teacher just said answer was 0.4J ~~
Really? Mine said it was 4J :(
*Edit*
woohoo my physics teacher just said answer was 4J ~~
Woot ;D
Post by: ks04 on June 08, 2010, 06:33:50 pm
woohoo my physics teacher just said answer was 4J ~~My teacher looked at it and was like.... it should be 0J... but he hadn't looked at it closely, this was when we bombarded him immediatly following the exam.
Post by: samuch on June 08, 2010, 06:37:03 pm
yayy :) i put 4J ;)woohoo my physics teacher just said answer was 4J ~~My teacher looked at it and was like.... it should be 0J... but he hadn't looked at it closely, this was when we bombarded him immediatly following the exam.
Post by: appianway on June 08, 2010, 06:38:20 pm
Post by: kyzoo on June 08, 2010, 06:39:21 pm
Post by: a.zirek on June 08, 2010, 07:05:13 pm
1. You can't get 0.5 marks for working out. The VCAA ONLY gives 1,2,3... marks. They're always whole numbers.
2. The acceleration of the n1 and m2 system is 2 m/s. You MUST take the total mass into account.
3. For the graph questions I got A,B,B but my teacher says it was A,A,A as when the collision begins, momentum is lost before it is gained again. I still think it is A,B,B.
4. I AM SCREWED FOR CHEMISTRY TOMORROW.
And.. If I didn't get that graph question wrong like my teacher said.. hello full marks! :)
Post by: jasopan on June 08, 2010, 07:09:16 pm
Marks for almost completely wrong working out and answers but you wrote something? (1 out of 3 or something)
Why was the gain in the amplifier 50? i got 33.33 =$
and please tell me they give consequential marks?! Especially that amplifier one...
Post by: m@tty on June 08, 2010, 07:10:06 pm
Just a few points I felt the need to make:
1. You can't get 0.5 marks for working out. The VCAA ONLY gives 1,2,3... marks. They're always whole numbers.
2. The acceleration of the n1 and m2 system is 2 m/s. You MUST take the total mass into account.
3. For the graph questions I got A,B,B but my teacher says it was A,A,A as when the collision begins, momentum is lost before it is gained again. I still think it is A,B,B.
4. I AM SCREWED FOR CHEMISTRY TOMORROW.
And.. If I didn't get that graph question wrong like my teacher said.. hello full marks! :)
Momentum is conserved throughout the collision. B is correct. There is no fluctuation.
Post by: coletrain on June 08, 2010, 07:16:56 pm
Do they give marks when u sub in wrong numbers? (out of 2 marks)
Marks for almost completely wrong working out and answers but you wrote something? (1 out of 3 or something)
Why was the gain in the amplifier 50? i got 33.33 =$
and please tell me they give consequential marks?! Especially that amplifier one...
i got 33.33 too, could someone explain how itute got (-)50
Post by: Blakhitman on June 08, 2010, 07:18:21 pm
use any vout and use the corresponding vin. I remember using 10/200x10^3.
Post by: samuch on June 08, 2010, 07:18:35 pm
voutput = -10V
therefore .2 x gain = 10
gain = -50
Post by: ilovecake on June 08, 2010, 07:20:24 pm
Post by: Davoo! on June 08, 2010, 07:20:54 pm
Post by: samuch on June 08, 2010, 07:24:18 pm
Post by: schnappy on June 08, 2010, 07:29:30 pm
Vin*gain=Vout. Same as above but this makes more intuitive sense to me :)
Did you up the mV value to V?
with that bike question, will i get the mark off for drawing the gravity arrow starting at the bottom of the wheel instead of at the centre of mass?
I'd say yes. I draw my mg correct, the part I wasn't sure about though was the normal. I draw that coming from the wheel... I figured that's right because that is after all, where the force is acting?
Post by: Davoo! on June 08, 2010, 07:32:48 pm
Post by: samuch on June 08, 2010, 07:34:34 pm
I just got a point on the graph, wher ethe clipping occured was at the axis nicely so i used that. I said;
Vin*gain=Vout. Same as above but this makes more intuitive sense to me :)
Did you up the mV value to V?with that bike question, will i get the mark off for drawing the gravity arrow starting at the bottom of the wheel instead of at the centre of mass?
I'd say yes. I draw my mg correct, the part I wasn't sure about though was the normal. I draw that coming from the wheel... I figured that's right because that is after all, where the force is acting?
yeah for my graph, the input volts i put were in volts not mV
but ahhhhhhhhhhh nooooooooooo :( what a stupid mistake
Post by: SixWinged on June 08, 2010, 08:27:44 pm
I just got a point on the graph, wher ethe clipping occured was at the axis nicely so i used that. I said;
Vin*gain=Vout. Same as above but this makes more intuitive sense to me :)
Did you up the mV value to V?with that bike question, will i get the mark off for drawing the gravity arrow starting at the bottom of the wheel instead of at the centre of mass?
I'd say yes. I draw my mg correct, the part I wasn't sure about though was the normal. I draw that coming from the wheel... I figured that's right because that is after all, where the force is acting?
yeah for my graph, the input volts i put were in volts not mV
but ahhhhhhhhhhh nooooooooooo :( what a stupid mistake
now that you mention it I'm not sure I put any labels on my axis.... :(
Post by: kenhung123 on June 08, 2010, 08:32:14 pm
Post by: Blakhitman on June 08, 2010, 08:33:21 pm
Post by: m@tty on June 08, 2010, 08:35:38 pm
Post by: Stormer on June 08, 2010, 08:36:53 pm
Post by: m@tty on June 08, 2010, 08:38:58 pm
Post by: Blakhitman on June 08, 2010, 08:39:48 pm
Yeah, that's all that was needed. And it specifically asked for it, so you had to...
I thought it only specifically asked for the scales...as in 5, 10, 15 etc.
Post by: Stormer on June 08, 2010, 08:40:36 pm
I don't know, maybe? Did you also label the V out marks with 2V etc. ? If you did then depends on examiner. They may miss them though...Yeah, I didnt label the end of the axis but labelled each individual point.
Post by: m@tty on June 08, 2010, 08:44:26 pm
Yeah, that's all that was needed. And it specifically asked for it, so you had to...
I thought it only specifically asked for the scales...as in 5, 10, 15 etc.
It said
Quote
On the grid below sketch the voltage characteristic of the amplifier.
Include a scale on both axis.
I interpreted that as V and mV. The others(100, 200, 300 and 2, 4, 6) are like inseparable from the characteristic curve... I thought they were assumed ... IDK ...
I don't know, maybe? Did you also label the V out marks with 2V etc. ? If you did then depends on examiner. They may miss them though...Yeah, I didnt label the end of the axis but labelled each individual point.
Probably then, I guess. The only thing is the examiner *might* miss it, late at night.. Hopefully, though, you'll get it...
Post by: Blakhitman on June 08, 2010, 08:45:45 pm
someone should seriously suspend my account till after the chem exam....
Post by: schnappy on June 08, 2010, 08:54:12 pm
Post by: m@tty on June 08, 2010, 08:55:28 pm
Post by: superflya on June 08, 2010, 08:57:39 pm
Post by: Ball Lightning on June 08, 2010, 10:18:13 pm
I equated the energy at the top unstretched with the energy .4m below, the spring energy.
This gave:
mgh=(1/2)*k*x^2
which came out to k=100
As there is no loss in energy (as the hand does not take away from either sides energy)
8=8 Therefore there is a 0 change in energy, and it is conserved.
From a brief discussion we had with our teacher he agreed with us about this answer.
Also I got ABB.
Post by: Akirus on June 08, 2010, 10:20:44 pm
About the 0J/4J question..
I equated the energy at the top unstretched with the energy .4m below, the spring energy.
This gave:
mgh=(1/2)*k*x^2
which came out to k=100
As there is no loss in energy (as the hand does not take away from either sides energy)
8=8 Therefore there is a 0 change in energy, and it is conserved.
From a brief discussion we had with our teacher he agreed with us about this answer.
Also I got ABB.
You made the assumption that all gravitational potential energy is completely converted into elastic potential energy, which is not necessarily true.
Post by: Ball Lightning on June 08, 2010, 10:25:44 pm
You made the assumption that all gravitational potential energy is completely converted into elastic potential energy, which is not necessarily true.
Well if some energy is transferred into the hand then i guess you could say that it isn't true, but how do you calculate that out?
Post by: m@tty on June 08, 2010, 10:27:24 pm
Equate mg and kx => k=50Nm^{-1}
Figuring out the energy indicates a loss of 4 joules
Then to explain this it goes to work done by the hand, heat etc.
Post by: Akirus on June 08, 2010, 10:29:35 pm
You made the assumption that all gravitational potential energy is completely converted into elastic potential energy, which is not necessarily true.
Well if some energy is transferred into the hand then i guess you could say that it isn't true, but how do you calculate that out?
It doesn't have to transfer to the hand; it's not momentum. We're talking about energy conversions.
You know that the weight force causes an extension of 0.40m; using F=kx, it is possible to calculate the elastic potential energy in the spring.
The fact that the force applied to stretch it may not have been constant is irrelevant; once released, if the weight force is insufficient to maintain an 0.40m extension, it will simply contract back to the appropriate length (as per the force-extension characteristics).
Post by: googoo on June 08, 2010, 10:43:30 pm
Post by: Killerkob on June 08, 2010, 11:09:16 pm
Post by: Davoo! on June 08, 2010, 11:14:32 pm
Fail at missing the labeling of cyclist/diagram question. :-[
Same! That question was so BS! As if wedge the question in between two fat paragraphs... 2 marks gone instantly. Now I'm sitting on the borderline of an A and A+ because of that. Pissed big time.
Post by: Stroodle on June 08, 2010, 11:26:34 pm
Fail at missing the labeling of cyclist/diagram question. :-[
Same! That question was so BS! As if wedge the question in between two fat paragraphs... 2 marks gone instantly. Now I'm sitting on the borderline of an A and A+ because of that. Pissed big time.
Haha. I lost two (or hopefully one) of my 4-5 marks by mislabelling that bike. I dunno what I was thinking. I drew the centripetal force and the nr force...
Post by: schnappy on June 08, 2010, 11:50:59 pm
Did anyone else get really confused when they asked for the graph of the voltage across the zener diode, and none of them were stable DC voltages? I think they must have really fucked it.
Post by: olly_s15 on June 08, 2010, 11:52:19 pm
Post by: Stroodle on June 08, 2010, 11:53:34 pm
Further electronics:
Did anyone else get really confused when they asked for the graph of the voltage across the zener diode, and none of them were stable DC voltages? I think they must have really fucked it.
We had someone come into our exam and tell us about an error in the further electronics section, None of us were doing further electronics though.
They gave us an extra minute at the end because of it :)
Post by: KRISTOFFA on June 08, 2010, 11:54:25 pm
did everyone get notified of the mistake in the further electronics section?We got told, but we don't do it. What was the mistake that they made?
Post by: olly_s15 on June 08, 2010, 11:56:25 pm
did everyone get notified of the mistake in the further electronics section?We got told, but we don't do it. What was the mistake that they made?
two of the exact same graphs for a multi-choice
lol fail vcaa
Post by: dyaner on June 09, 2010, 12:15:36 am
did everyone get notified of the mistake in the further electronics section?We got told, but we don't do it. What was the mistake that they made?
two of the exact same graphs for a multi-choice
lol fail vcaa
Hahaha. Same!
Post by: schnappy on June 09, 2010, 01:31:02 am
I know about that error... but... really, the further electronics this year was a ridiculous amount harder than any practice or previous VCAA exam. And yet I heard the S&M was easy... this is why I think the whole choose your own detailed study is a bad idea. They should just make electronics harder, it's not like it's hard at the moment. Not that I really care any more...
Post by: Killerkob on June 09, 2010, 11:34:14 am
Fail at missing the labeling of cyclist/diagram question. :-[
Same! That question was so BS! As if wedge the question in between two fat paragraphs... 2 marks gone instantly. Now I'm sitting on the borderline of an A and A+ because of that. Pissed big time.
Haha. I lost two (or hopefully one) of my 4-5 marks by mislabelling that bike. I dunno what I was thinking. I drew the centripetal force and the nr force...
Looking at it now though, that question is more about how pedantic and particular you are during reading time. Not how good you are at physics at all.