ATAR Notes: Forum
Archived Discussion => 2010 => Mid-year exams => Exam Discussion => Victoria => Physics => Topic started by: appianway on June 08, 2010, 03:59:24 pm
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Thought I'd start this thread in case anyone has any specific questions they'd like explained or answered... it's a bit hard to trawl through the plethora of posts in the other thread :)
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Question 3\4 about the two masses, the other thread is all round the place.
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Question 3\4 about the two masses, the other thread is all round the place.
That question messed me up, just 5 minutes before the exam finished.
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do u think the examiners will give say... 1.5 marks out of 4 if u show working out?
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I'm not quite sure what the actual question is, so I'll try and explain it without really knowing... might make a few careless mistakes but this is the general gist
So if you have two masses, and one's over a ledge, this is what will happen.
Both masses must move with the same acceleration as the length of the string is conserved. Denote this acceleration as a.
Denote the mass hanging over the ledge as M. Denote the mass resting on a tabletop as m.
Resolve the forces on M:
Mg-T=Ma
However, the tension force is the only force acting on m, so T=ma
Mg-ma=Ma
Mg = a(M+m)
Thus a = Mg/(M+m)
Then to find the force, multiply m by a. Tension should be the same or something.
I may have misunderstood the question though... I haven't seen it or a diagram...
And yes, marks are allocated for working.
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And yes, marks are allocated for working.
HOPE REMAINS
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What about consequential marks? Is Physics the same as Chem in that you can get full marks by using an already punished wrong answer from a previous question? So you get a wrong, then b uses a . You get the wrong answer for b because of the mistake in a. Do you get full marks for b?
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I think you should, as you won't be penalised twice for the same mistake.
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I think you should, as you won't be penalised twice for the same mistake.
MORE HOPE
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Cool. I think maths is different though, that's why I was unsure...
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Maths is maths though :P
Oh, and if you guys have any more questions, please post them! It's better to do them properly than to see what the most common response is... just make sure that you explain the question properly, and I'm sure someone (myself or someone else) will answer :)
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If your workings are correct... or you state something that's correct. They don't just hand out mark's cos you know how to write down arbitary numbers.
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The banked corner question!
you had to find the angle of banking if v = 10m/s and r = 100m i think.
anyway apparently the answer is 5.7 degrees
someone explain?
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If your workings are correct... or you state something that's correct. They don't just hand out mark's cos you know how to write down arbitary numbers.
I was speaking of cases where the ONLY thing wrong is the value previously found(erroneously). So your working and everything is exact.
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The acceleration contributed by the banking is gtan(theta). You can work this out by finding the component of the normal force acting towards the centre. From there, you just rearrange to find inverse tan.
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Kai
You could use the banking formula of Tan^-1 (v^2 / gr)
Tan^-1 (10^2 / (10)(100))
Tan^-1 (100/1000)
Tan^-1 (1/10)
5.7 degrees
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"The banked corner question!
you had to find the angle of banking if v = 10m/s and r = 100m i think.
anyway apparently the answer is 5.7 degrees
someone explain?"
theres a formula. TAN(THETHA) = V^2/rg
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Spring of mass 2 kg is hanging resting on a hand, it's then lowered 0.4 metres. What is the spring constant?
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The banked corner question!
you had to find the angle of banking if v = 10m/s and r = 100m i think.
anyway apparently the answer is 5.7 degrees
someone explain?
For friction to be irrelevant, Force (Centripetal) = Weight force x Tan(Theta)
mv^2/r = mg x tan (theta)
Simplifies to: tan (theta) = v^2/rg
v was 10m/s, radius was 100.
Therefore tan (theta) = 0.1
Hence Theta = tan^-1 (0.1)
=5.71 degrees
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The banked corner question!
you had to find the angle of banking if v = 10m/s and r = 100m i think.
anyway apparently the answer is 5.7 degrees
someone explain?
I got this answer as well.
Edit: Sorry, I read this wrong. People above me have explained it well.
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I did the working correctly, then was like wtf it's only 0.1 degrees. Left it cause I thought it may be possible, since the radius was 100m, then at the end realised my calc was in radian mode :(
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I did the working correctly, then was like wtf it's only 0.1 degrees. Left it cause I thought it may be possible, since the radius was 100m, then at the end realised my calc was in radian mode :(
Make sure it's on degree mode for the end of year exam!
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I wrote bloody 57o .. :(
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thanks everyone. i think i cant use my calc properly.
but now i understand stuff! cheers guys
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Spring of mass 2 kg is hanging resting on a hand, it's then lowered 0.4 metres. What is the spring constant?
Does the hand push on it to lower it? I'm going to presume that it just falls down and oscillates around this point before coming to rest.... although I'm not sure how the question was worded.
Say that when the spring is in the hand, x = 0. Say that when it is lowered 0.4 m, no other forces act on it aside from gravity - in other words, it's not pushed down. F = kx, and thus k = F/x, where F = mg as you want Fnet to be 0. In this case, you can't use conservation of energy - although energy is conserved, what they've neglected to tell you is that the mass won't be completely at rest when F(net) = 0- it'll come to rest eventually because of dampened oscillations (ie frictional/air resistance forces reducing the oscillation). Thus if it's just lowered because of gravity, you can't use conservation of gravitational energy as you're also going to have kinetic energy at that point.
Might have misunderstood the question, but hopefully that helps.
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wtfff my calculator gave me 0.57
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I wrote bloody 57o .. :(
Now I am unsure whether I put 5.7 or 57.
Great!
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I'm pretty sure the question was like this:
Spring with 2.0kg weight attached is lowered gently to a position where the extension is 0.4 m. There is no oscillation.
Don't remember exactly, but that was basically it...
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Spring of mass 2 kg is hanging resting on a hand, it's then lowered 0.4 metres. What is the spring constant?
Does the hand push on it to lower it? I'm going to presume that it just falls down and oscillates around this point before coming to rest.... although I'm not sure how the question was worded.
Say that when the spring is in the hand, x = 0. Say that when it is lowered 0.4 m, no other forces act on it aside from gravity - in other words, it's not pushed down. F = kx, and thus k = F/x, where F = mg as you want Fnet to be 0. In this case, you can't use conservation of energy - although energy is conserved, what they've neglected to tell you is that the mass won't be completely at rest when F(net) = 0- it'll come to rest eventually because of dampened oscillations (ie frictional/air resistance forces reducing the oscillation). Thus if it's just lowered because of gravity, you can't use conservation of gravitational energy as you're also going to have kinetic energy at that point.
Might have misunderstood the question, but hopefully that helps.
Thanks, ok then the next part is "what is the magnitude of difference in the total energy in the mass-spring system from figure a and b" firgure a was the spring at rest and b was extended 0.4 metres.
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Spring of mass 2 kg is hanging resting on a hand, it's then lowered 0.4 metres. What is the spring constant?
Does the hand push on it to lower it? I'm going to presume that it just falls down and oscillates around this point before coming to rest.... although I'm not sure how the question was worded.
Say that when the spring is in the hand, x = 0. Say that when it is lowered 0.4 m, no other forces act on it aside from gravity - in other words, it's not pushed down. F = kx, and thus k = F/x, where F = mg as you want Fnet to be 0. In this case, you can't use conservation of energy - although energy is conserved, what they've neglected to tell you is that the mass won't be completely at rest when F(net) = 0- it'll come to rest eventually because of dampened oscillations (ie frictional/air resistance forces reducing the oscillation). Thus if it's just lowered because of gravity, you can't use conservation of gravitational energy as you're also going to have kinetic energy at that point.
Might have misunderstood the question, but hopefully that helps.
its gently lowered to that position were it remains at rest. Gently was in bold on the exam.
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If it's lowered to rest (so that no further oscillation can take place), it'd mean that Fnet = 0 when it stops as otherwise the mass would experience a force and would oscillate. The lowering gently would just provide enough force so that oscillations don't occur, however, energy would be transferred to whatever is doing the lowering.
I'm not sure how complicated they want the next part to be :/ Do they want to consider the change of the centre of mass of the spring or do we presume it to be massless? I'd say it'd be the initial spring energy (=0) + gravitational energy (say that h=0.4) - the energy in the spring.
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Yeah it didn't oscillate, the hand moves with the spring.
Ok that clears it thanks...not happy though...
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I used 9.8 for g throughout, so I got 49 for spring constant and 3.9J for the spring questions. Any marks lost here?
As far as I know, I only bunged up the mass question with 0.4kg and 0.1kg blocks. I had no idea... But I'm sure I had more docked somewhere along the way. How much was the block question worth in total?
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Don't worry about the two mass question. I can only think of about 2 people in Victoria who I'm sure would've gotten full marks for it because they've seen similar ones before - there might be a few more who do uni level stuff for fun, but you never know...
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na u shudnt lose marks for using 9.8 examiners are human, they will allow 9.8.
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appianway, im not sure if you know but the question included a frictionless pulley, will that change anything?
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No. The pulley won't rotate because it's completely frictionless, so even its moment of inertia won't change anything. The pulley just changes the direction of the force.
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Ok, the mass on the table was 0.4kg and the other was 0.1, so how would you calculate
a) the acceleration of the table mass
and
b) its kinetic energy after 1m?
Thanks.
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The kinetic energy you need to use constant acc formula to find v then use 1/2mv^2, Consequentials there for sure...I hope.
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.8J ???
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if you used the right acceleration (from question before)...which I didn't....
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how do you find the velocity? I'm sure I did this question, I just can't remember how I did it lol
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yeah there would deffs be consequentials.
considering you use accel to find v, then use v to find energy. i got 1J
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Same. But I just used work lol. Force 1N left. Displacement 1 m left. All work becomes kinetic(no friction) Therefore Ek=1x1=1J lol.
But I got the wrong force :|
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Why is that wrong force?
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For kinetic energy you can also say that energy is conserved, so the work done on the block is its kinetic energy (ie ma * 1m = Ek)
I think a lot of people didn't look at how the tension influences the acceleration of the 0.1kg box, so a lot of people got the wrong acceleration from that and hence the wrong force.
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for the box i got the acceleration as 2.5ms2
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ahh yea true.
hooray for consequentials.
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Why is that wrong force?
Net force on the 0.4kg block is meant to be 0.8 N..
That's basically what the whole argument was about before.
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i got 1J in that question too..
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Why is that wrong force?
Net force on the 0.4kg block is meant to be 0.8 N..
That's basically what the whole argument was about before.
lol nevermind me, I'm trying to do chem but also really pissed off so I'm kinda losing it.