ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: appianway on January 06, 2010, 11:06:21 am
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Hi guys,
I know that a lot of you are only just beginning the course, but I thought I'd post a link to the National Qualifying Examination papers for the Physics Olympiad. Despite not being VCE papers (and being regarded as very difficult!), they contain a plethora of mechanics questions, ranging from multiple choice akin to slightly more difficult (or in some rare cases, standard) VCE questions to much harder extended response. In saying that, they're definitely doable... if you think about what they're asking! Have a look through the papers and attempt anything to do with forces if you feel ready, and if you want, have a go at the other questions too. They're quite enjoyable and they'll get you thinking in a physics-y way! This is also a good resource for students in year 11 (or under, but I think this year all of the students who made it into the summer school were in year 11 when they sat the paper for physics) who are considering sitting the National Qualifying Exam and want to practise the questions.
Anyway, have fun! :)
http://www.aso.edu.au/www/index.cfm?itemid=31&CFID=5544640&CFTOKEN=5035285bc2b0ffbd-00F13507-B30D-FED8-30006C2417BB550C
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I'm bored, it's late and physics is fun. Here we go...
http://www.aso.edu.au/www/docs/2007PhysicsNQEPaper-FINALWeb.pdf
Question 13:
In any collision the sum of the momenta of all the bodies involved should be the same immediately
before and after the collision. In an elastic collision the same is true for the sum of the kinetic
energies of the bodies.
a. If a small blob with a small mass m has a velocity v directly towards an extremely heavy
wall, and the ensuing collision is elastic, what is the approximate final velocity of the blob?
(1 mark)
As indicated in the opening paragraph, no kinetic energy will be lost. The ball has a small mass and the wall is extremely heavy so it is unlikely that the kinetic energy of either will change, thus: ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?E_{k - initial} = E_{k - final})
However, the velocity of the ball is a vector quantity, and the elastic collision will result in a reversal of direction. We can therefore model this with the equation:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?v_{final} = -v)
There is a box with lots and lots and lots of these blobs with mass m, all moving with different
speeds in different directions so that the total momentum is initially (approximately) zero. The box
is held firmly in place. Assume that all collisions are elastic.
b. What happens to the total kinetic energy of all the blobs? Why? (2 marks)
c. What happens to the total momentum of the all the blobs? Why? (2 marks)
b. The total kinetic energy will remain constant, as per the law of conservation of energy.
c. The total momentum of the blobs will remain constant, as per the law of conservation of momentum.
If there were a hole in the side of the box that the blobs could escape through and the initial
distribution of velocities of the blobs were the same as before:
d. What happens to the total kinetic energy of the blobs left in the box? Why? (1 mark)
e. What happens to the total momentum of the blobs left in the box? Why? (2 marks)
d. The total kinetic energy of the blobs will decrease as blobs escape from the hole. This is because the box ceases to be a closed system; kinetic energy is transferred to other external bodies by the escaped balls.
e. The total momentum of the blobs will decrease as blobs escape; this can also be considered an extension of the loss of total kinetic energy, since they're both linked directly to the mass and velocity of the blobs. For the same reason, as the box is no longer a closed system, the total momentum will decrease as blobs escape.
When a blown up balloon is released before it is tied up it shoots off away from the hand of the
person holding it.
f. Why does the balloon shoot off in the direction that it does? (2 marks)
f. When the balloon is held in the hand, the force exerted by the hand balances the forces of the particles pushing against the balloon from within. However, when it is released, the equilibrium of the balloon is lost and it will "shoot off away from the hand" in a seemingly random direction (due to the various forces of the gas molecules within).
I know some of these are wrong and some are just overly brief.
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I think 13 a's correct. b and c look fine, as does d. However, with a, be careful with your wording. The kinetic energy's DEFINITELY going to change, but it's going to be very small. With b and c, try and give a bit more of an elaboration rather than just quote the laws - that's already shown in the question. Instead, perhaps relate back to a and show why the energy's conserved once again.
e should be correct, but you need to give a proper justification and make it a bit detailed. Be more specific about how the momentum changes and in what direction. Maybe give a mathematical explanation as well.
f is a little bit correct, but relate it not just to equilibrium, but Newton's laws. Be more specific about how the balloon influences the air particles, and perhaps relate it back to how this affects the motion of the the balloon.
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Yeah, I'm well aware the written responses are really brief. It was late and I was hoping for more calculations, so I got lazy.
The one regarding momentum with the escaped blobs is wrong in that I didn't consider its vector properties, and for the balloon I didn't mention the effect of the change in momentum due to the escaped particles.
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Yup. In my opinion though, the most interesting part is actually elaborating on the explanation to cement why it happens and understand the interactions between different things. Fun!
I'm actually leaving for the Summer School for this in 45 minutes, but I'll try and formulate some of my own questions like these (perhaps I'll be so exposed to them that I'll be able to? Here's hoping :)) in 2.5 weeks when I come back :)
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I actually prefer doing tricky mathematical proofs rather than written theory. Of course, that's not to say I don't enjoy that as well, just to a lesser extent.
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That's the opposite of me! I don't like the mathematical manipulation (although 3 dimension integration and 4-vectors are fun), but I like looking at a problem and considering the physical principles at play.
Although the maths IS useful for proving things.
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Suppose that the radius of the Sun were increased to
(the average radius of orbit for pluto), that the density of the expanded sun was uniform and that the planets revolved within it. Calculate earths orbital speed within the 'new' sun and what would be the earths new period of revolution.
Errata: When I say the sun increases it's radius I also meant to say it's volume and mass. Volume is implied however. Uniform density as well.
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Ahh, would you do it something like this? I'm not going to compute it, but yes...
- Use Gauss' law to calculate the gravitational field strength at the point of the earth by considering the mass enclosed within the orbit (it should be the original mass x Volume of earth's orbit as a sphere/new volume of the sun)
-v^2/r = g because circular motion occurs, and solve for v
However, the current radius of the orbit is unknown, as is the mass of the sun. At present (without the new expansion), g = v^2/r, and v can be found by considering the period of the orbit and the distance (2 pi r). g also = G(mass of sun)/r^2, so if the two expressions for g are equated, one of the values (either the mass or the radius) can be found in terms of the other.
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Note that I said the density was (should be is) uniform?
so if the density is uniform you could find the original density of the sun
to help you find
which can then help you find the new mass of the sun.
Hint:![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \rho_{orig} = \rho_{new})
You're on the right track with
but think in terms of forces...
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Ah, just wondering, why are both densities equal? Shouldn't one be much smaller (seeing as the same mass is distributed over a larger volume)?
Oh, and edit: Just making sure I've understood the forces acting. So, is there a gravitational force and a buoyancy force?
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But it isn't the same mass it is the same density. The mass and volumes can be different as long as the density is the same.
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Ah, it's not as thought the sun itself was expanded to fill the same volume (rather, additional matter is added to expand the volume).
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yes! something like that.
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If the only two forces acting are the buoyancy force and the weight force, I thinkkkk I've got an expression... but I can't be bothered typing it up (it could be wrong anyway!)
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Suppose that the radius of the Sun were increased to
(the average radius of orbit for pluto), that the density of the expanded sun was uniform and that the planets revolved within it. Calculate earths orbital speed within the 'new' sun and what would be the earths new period of revolution.
To solve this I am going to first derive an expression for g
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \oint_{S} \bf{g} . d \bf{A} = - 4 \pi G M )
since rEarth < rSun, then let:
MSun =
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore \oint_{S} \bf{g} . d \bf{A} = - 4 \pi G \dfrac{4}{3} \pi \rho_{Sun} r^{3} )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore - \bf{g} . 4 \pi r^{2} = - 4 \pi G \dfrac{4}{3} \pi \rho_{Sun} r^{3} )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore \bf{g} = \dfrac{4}{3} G \pi \rho_{Sun} r )
I love using Guass's law on spheres! :)
Edit: Now, r = 1.5x109m,
= 1.41x103 kg/m3
implying g = 591 m/s2
Now,
vEarth = 941 km/s (31x faster than what speed it orbits at)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? g = \dfrac{ 4 \pi ^{2} r}{T^{2} } )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore T = \sqrt{ \dfrac{4 \pi^{2} r}{g} } )
implying Edit: T = 1.00x104 s = 2 hours 47 minutes!
So if this were to happen, in theory a year would only be about 2 hours 47 minutes!
Am I right?
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Hmm, that's essentially what I think I described earlier, but I think you have to consider the effect of the buoyancy force (although I could be mistaken, seeing as I've only had it mentioned for around 20 seconds in one lecture), as earth essentially displaces a portion of the sun.
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Sorry quantumJG, way off :( and the question is to find the orbital velocity of earth... >.>
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Hmm, that's essentially what I think I described earlier, but I think you have to consider the effect of the buoyancy force (although I could be mistaken, seeing as I've only had it mentioned for around 20 seconds in one lecture), as earth essentially displaces a portion of the sun.
hmm... I never considered that!
I calculated the buoyancy force to be 9.02x1026 N and it's weight to be 3.53x1027 N, implying a force of 2.63x1027 N is pulling the Earth to the Sun's centre.
So
vEarth = 812km/s
T = 11,601 s = 3 hours 13 minutes
Sorry quantumJG, way off :( and the question is to find the orbital velocity of earth... >.>
What is the answer?
Can I have a hint?
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The earths orbital radius still comes into play here.
I suppose I could tell you that there is of course a centripetal force acting on the planet within in the new expanded sun and that gravitational force may still play a part in all of it.
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The earths orbital radius still comes into play here.
I suppose I could tell you that there is of course a centripetal force acting on the planet within in the new expanded sun and that gravitational force may still play a part in all of it.
I'm really confused?
:(
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I don't understand where I have gone wrong? :(
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IF YOU DON'T WANT TO SEE THE ANSWER DO NOT READ THIS POST
fuck this forums lack of spoiler tags
Sorry. It's a bit rushed. It might not make sense but its the answer none the less.
You're on your own to find the period of orbit though. This is just the major part of the problem.
First lets define some variables:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?R = 5.90 \times 10^{12} m)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?r = radius\ of\ earth\ orbit = 1.50 \times 10^{11} m)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?M = mass\ of\ sun\ within\ earths\ orbit )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?M_s = original\ mass\ of\ sun = 1.988435 \times 10^{30} kg)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?V = \frac{4}{3}\pi r^{3})
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?V_{orig} = 1.412 \times 10^{27} m^3)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?V_{new} = 8.603 \times 10^{38} m^3)
Since the density of the expanded sun is uniform we can use
to find the mass of the sun within earths orbit.
is the density of the sun when the radius is the orbital radius of earth
or
while
is the density of the sun at radius ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?R)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?<br />\rho_{in} = \rho_{out}<br />)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{3M}{4 \pi r^3} = \frac{3M_s}{4 \pi R^3})
solving for
we get ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? M = \left( \frac{r}{R} \right)^3 M_s)
and so:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?<br />M = \left(\frac{1.50 \times 10^{11} m}{5.90 \times 10^{12} m}\right) \times 1.988435 \times 10^{30} kg = 3.27 \times 10^{25}kg<br />)
Now using the equations
and using
because the earth is experiencing a centripetal force. setting these equal to each other we can solve for
:
and therefore
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?<br />v = \sqrt \frac{\left(6.67 \times 10^{-11}\right)\left(3.27 \times 10^{25}\right)}{1.50 \times 10^{11}} = 1.21 \times 10^{2} m/s<br />)
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IF YOU DON'T WANT TO SEE THE ANSWER DO NOT READ THIS POST
fuck this forums lack of spoiler tags
Sorry. It's a bit rushed. It might not make sense but its the answer none the less.
You're on your own to find the period of orbit though. This is just the major part of the problem.
First lets define some variables:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?R = 5.90 \times 10^{12} m)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?r = radius\ of\ earth\ orbit = 1.50 \times 10^{11} m)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?M = mass\ of\ sun\ within\ earths\ orbit )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?M_s = original\ mass\ of\ sun = 1.988435 \times 10^{30} kg)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?V = \frac{4}{3}\pi r^{3})
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?V_{orig} = 1.412 \times 10^{27} m^3)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?V_{new} = 8.603 \times 10^{38} m^3)
Since the density of the expanded sun is uniform we can use
to find the mass of the sun within earths orbit.
is the density of the sun when the radius is the orbital radius of earth
or
while
is the density of the sun at radius ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?R)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?<br />\rho_{in} = \rho_{out}<br />)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{3M}{4 \pi r^3} = \frac{3M_s}{4 \pi R^3})
solving for
we get ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? M = \left( \frac{r}{R} \right)^3 M_s)
and so:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?<br />\bf{M = \left(\frac{1.50 \times 10^{11} m}{5.90 \times 10^{12} m}\right) \times 1.988435 \times 10^{30} kg = 3.27 \times 10^{25}kg }<br />)
Now using the equations
and using
because the earth is experiencing a centripetal force. setting these equal to each other we can solve for
:
and therefore
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?<br />v = \sqrt \frac{\left(6.67 \times 10^{-11}\right)\left(3.27 \times 10^{25}\right)}{1.50 \times 10^{11}} = 1.21 \times 10^{2} m/s<br />)
I) There is a contradiction here! You have stated that the original Sun (I.e. Our real Sun) was not being expanded (I.e. the Sun's expanding it's volume and mass).
M = 1.99x1031 kg
rEarth's orbit = 1.5x109 m
vEarth's orbit = 941km/s
I rest my case!
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You have no case though. Your first two answers were both different to that and the first answer didn't even answer the question I asked. I worded the question poorly and I apologise; I will re-word it for future generations.
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You have no case though. Your first two answers were both different to that and the first answer didn't even answer the question I asked. I worded the question poorly and I apologise; I will re-word it for future generations.
Okay please word the question so we know our data and explain specifically what you want from us.
Okay first I made some mistakes, but fixed them up. Then I investigated appianway's buoyancy idea. It's almost like you want us to fail at this question.
I'm obsessed with getting on my soapbox
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I don't want you guys to fail. I actually thought it was a pretty good question had I worded it right I'm sure it would have all made sense. I apologise for that and next time I write a question I'll word it carefully to make sure you guys know what I want.
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If the buoyancy force isn't considered (hence the only force causing circular motion being the gravitational force), can't it just be found using g like I originally described (equating the two g expressions, although I admittedly didn't know that the density was constant so used the GM/r^2 to find the mass in terms of the radius, but it's still the same general idea *coughcough*).
Good question though. Wording's always a bit tricky. It'd also be interesting to consider any drag forces acting on the body...
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I might as well do a survey while I'm here.
Are there any particular areas people are interested in getting questions in?
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Thermodynamics, physical optics, special relativity, electromagnetism. :)
Oh, and how can I forget... quantum. :)
I probably need practise in geometrical optics too, seeing as it's my least favourite area...
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I'm allowed to ask questions right? I need help with this:
Magnetic resonance imaging (MRI) is a noninvasive technique for visualising the living human body. It depends on a phenomenon called nuclear magnetic resonance (NMR) which occurs when certain atoms are place in very strong magnetic fields. All nuclei that have odd mass numbers exhibit NMR as do all nuclei that have even mass numbers but odd atomic numbers. Nuclei having both even mass numbers and even atomic numbers do not exhibit NMR.
The number of neutrons in the nucleus of an atom in which NMR occurs is always
A odd
B even
C odd if the number of protons is even
D even if the number of protons is odd
Also explain why
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n(protons) +n(neutrons) = mass number
n(protons) = atomic number
Therefore if n(protons) + n(neutrons) is odd, NMR is exhibited.
This staement means that if n(protons) is odd, n(neutrons) is even and vice versa.
Taking the second statement
n(protons) + n(neutrons) = even for NMR to occur
n(protons) = odd
Thus n(neutrons) = odd.
Therefore A is always true... I think.
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Unfortunately that's wrong :( (Took me a while to figure out though)
A The number of neutrons in the nucleus of an atom in which NMR occurs is always odd
Not always true. Look up lithium on the periodic table. Mass number 19 and atomic number (proton) 9, so even number of neutrons (10). NMR occurs in this nucleus, and has even neutrons ;)
According to the information provided, odd mass number and even atomic number exhibit NMR. An example is Beryllium (Mass 9, atomic 4). W can deduce that the number of neutrons is 5 and this is odd. We can then conclude that the number of neutrons in the nucleus of an atom in which NMR occurs is always odd if the number of protons is even.
D can't be the answer because fluorine can exhibit NMR (mass of 19 and atomic of 9 - both odd)
Not much of Physics, but perhaps logic? :P Any better way of coming to this conclusion?
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Thermodynamics, physical optics, special relativity, electromagnetism. :)
Oh, and how can I forget... quantum. :)
I probably need practise in geometrical optics too, seeing as it's my least favourite area...
lol same here.
I wouldn't mind some special relativity or some tricky electromagnetism questions.
Also sorry for attacking you (cthulhu), I just got annoyed at the question. Also what are you studying at uni?
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Find necessary and sufficient conditions for a region in
to be the unit ball for some norm.
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mark_alec here:
Do you accept university level questions in those areas? That is, questions that require the use of more advanced techniques than are available in the VCE curriculum.
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I'm allowed to ask questions right? I need help with this:
Magnetic resonance imaging (MRI) is a noninvasive technique for visualising the living human body. It depends on a phenomenon called nuclear magnetic resonance (NMR) which occurs when certain atoms are place in very strong magnetic fields. All nuclei that have odd mass numbers exhibit NMR as do all nuclei that have even mass numbers but odd atomic numbers. Nuclei having both even mass numbers and even atomic numbers do not exhibit NMR.
The number of neutrons in the nucleus of an atom in which NMR occurs is always
A odd
B even
C odd if the number of protons is even
D even if the number of protons is odd
Also explain why
Awesome question (we looked at this in nuclear physics).
So a nucleon such as a proton or a neutron has a spin of +/- 1/2. Let's first look at a proton which can take two quantum states (+/- 1/2), now let's say we have two hydrogen atoms together where one proton is -1/2 and the other is +1/2. Now if I put in a magnetic field the -1/2 proton gains energy by being anti-aligned with the magnetic field, whilst the other drops to a lower energy level and this energy difference can be used to find the resonant frequency.
Sorry for the digression to physics, but it is a physics forum.
Now basically as long as you have either an odd number of neutrons or protons or both, you get NMR.
e.g. C-11(p-spin,n-spin) could be C-11(-,-1/2) & C-11(-,+1/2).
So the answer is C, why?
C says that the number of neutrons must be odd if Z is even.
NMR occurs for,
odd n & even p, odd n & odd p and even n & odd p, but it doesn't happen if even n & even p. C basically states this.
A is wrong since NMR can occur if n is even (iff p is odd)
B is wrong since NMR can't occur if n is even and p is even
D is wrong since NMR can occur if n is odd and p is odd.
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Find necessary and sufficient conditions for a region in
to be the unit ball for some norm.
what???
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QuantumJG: I study Physics
mark_alec here:
Do you accept university level questions in those areas? That is, questions that require the use of more advanced techniques than are available in the VCE curriculum.
I'm wondering the same thing as mark. The thread is "harder" physics questions... but how hard do you want them....
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QuantumJG: I study Physics
Cool!
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Whoops, my logic was obviously very flawed for the NMR question. I just treated it as a logic question, but just realised that I misread it as NMR always occurs when... rather than the number of neutrons is always.
Oh, and to marc_alec: Perhaps things that use 1st year knowledge but don't use complicated maths (so that eager VCE students can attempt them).
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Heres a quantum-type question:
Through what angle must a 200keV photon be scattered by a free electron so that the photon loses 10% of its energy?
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Presuming an elastic collision?
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Since the photon loses energy, isn't it an inelastic collision (Compton Scattering)?
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It's compton scattering, and I presume that energy is conserved, as energy is transferred to the electron... however, it's more of a relativistic transfer of energy (that doesn't make much sense; I just meant that you use the relativistic equations for energy), I think.
So under non-relativistic guidelines, the energy isn't conserved (ie the energy of the photon + energy of the electron is less after the collision), but if you consider relativity, the energies should be the same.
I guess that's not really elastic, but that's what I was getting at.
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Heres a quantum-type question:
Through what angle must a 200keV photon be scattered by a free electron so that the photon loses 10% of its energy?
Ah, a Compton scattering problem!
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \lambda ' - \lambda = \dfrac{h}{ m_{e} c } (1 - cos( \theta ) ) )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \lambda = \dfrac{hc}{E} )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore \dfrac{1}{E'} - \dfrac{1}{E} = \dfrac{1}{ m_{e} c^{2} } (1 - cos( \theta ) ) )
let E' = 0.9E
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\theta = arcos( \dfrac{ 9E - m_{e} }{9E} ) )
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Heres a quantum-type question:
Through what angle must a 200keV photon be scattered by a free electron so that the photon loses 10% of its energy?
Ah, a Compton scattering problem!
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \lambda ' - \lambda = \dfrac{h}{ m_{e} c } (1 - cos( \theta ) ) )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \lambda = \dfrac{hc}{E} )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore \dfrac{1}{E'} - \dfrac{1}{E} = \dfrac{1}{ m_{e} c^{2} } (1 - cos( \theta ) ) )
let E' = 0.9E
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\theta = arcos( \dfrac{ 9E - m_{e} }{9E} ) )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore \theta = 44.33^{o} )
Yeah I did exactly the same thing, did you have to convert the energy into Joules or something?
/me bangs keyboard :(
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Heres a quantum-type question:
Through what angle must a 200keV photon be scattered by a free electron so that the photon loses 10% of its energy?
Ah, a Compton scattering problem!
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \lambda ' - \lambda = \dfrac{h}{ m_{e} c } (1 - cos( \theta ) ) )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \lambda = \dfrac{hc}{E} )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore \dfrac{1}{E'} - \dfrac{1}{E} = \dfrac{1}{ m_{e} c^{2} } (1 - cos( \theta ) ) )
let E' = 0.9E
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\theta = arcos( \dfrac{ 9E - m_{e} }{9E} ) )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore \theta = 44.33^{o} )
Yeah I did exactly the same thing, did you have to convert the energy into Joules or something?
/me bangs keyboard :(
Yeah you 'can' convert eV to J aslong as mec2 is in the same units.
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Heres a quantum-type question:
Through what angle must a 200keV photon be scattered by a free electron so that the photon loses 10% of its energy?
Ah, a Compton scattering problem!
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \lambda ' - \lambda = \dfrac{h}{ m_{e} c } (1 - cos( \theta ) ) )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \lambda = \dfrac{hc}{E} )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore \dfrac{1}{E'} - \dfrac{1}{E} = \dfrac{1}{ m_{e} c^{2} } (1 - cos( \theta ) ) )
let E' = 0.9E
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\theta = arcos( \dfrac{ 9E - m_{e} }{9E} ) )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore \theta = 44.33^{o} )
Yeah I did exactly the same thing, did you have to convert the energy into Joules or something?
/me bangs keyboard :(
Yeah you 'can' convert eV to J aslong as mec2 is in the same units.
I missed out the 'k' in 200keV :( :(
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QuantumJG is right.
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QuantumJG is right.
YAY!!!
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This is probably the best place to put my 900th post.
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Good work! I was going to do the question and derive compton scattering from first principles, but I didn't end up having time.
If anyone feels like doing some relativity, use Lorenz transformations to derive the formulae for time and length dilation (without consulting any type of textbook/other internet page... they're probably very easy to find!).
If anyone feels like mechanics, here's a relatively straightforward question from the Resnick-Halliday book. A block is placed on the inside surface of the cone, and the cone is spun. Find the maximum and minimum frequencies of rotation of the cone so that the block does not move relative to the cone.
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I feel like Quantum Mechanics.
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How about something from the Electromagnetism field (no pun intended)?
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OK.
A solenoid with a current i and n loops is placed on a ramp inclined at an angle theta to the horizontal. Consider the magnetic field of the earth and calculate the acceleration of the solenoid down the slope in respect to position.
(You'll need to introduce a few basic constants with the algebra)
Edit: To make life easier and to simplify the maths, presume that the solenoid can immediately roll without slipping.
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I've already done the Lorentz stuff last year why can't I look in my notebook? ;.; why do you hate me appianway? Is it my hair? I can change it!
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Whoa who said anything about hating?
OK, I'll try to think of a quantum question. This one's quite straightforward, and it's qualitative, so I apologise for the fact that it's so simple.
How can a spectrum indicate the lifetime of a photon in an energy level?
I really need to think of something harder...
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mark_alec here:
I hope you mean the lifetime of a proton, not of a photon.
Something to research about Quantum Mechanics. An interpretation of observing the wavefunction of a particle is that the wave-function collapses, so if other observations are made immediately afterwards, the system will be in an identical state. If a radioactive nucleus is observed, can it ever decay?
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OK.
A solenoid with a current i and n loops is placed on a ramp inclined at an angle theta to the horizontal. Consider the magnetic field of the earth and calculate the acceleration of the solenoid down the slope in respect to position.
(You'll need to introduce a few basic constants with the algebra)
Edit: To make life easier and to simplify the maths, presume that the solenoid can immediately roll without slipping.
This sounds like a fun question, but I'll do it in the morning (I.e. 9am).
Btw: If my name stays on a thread, its because I'll just keep threads up that look interesting and look over them every oncewhile.
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Hi everyone, sorry but I've just joined this conversation. could we go back a bit to the gravity problem?
I'm going to ignore the need for the buoyancy force. Using gauss' law (as QuantumJG did)....
I get a value of g = 5.91*10^4 m/s^2
Orbital velocity of earth: 9.4*10^7 m/s
Orbital period: 1.00*10^4 seconds.
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OK.
A solenoid with a current i and n loops is placed on a ramp inclined at an angle theta to the horizontal. Consider the magnetic field of the earth and calculate the acceleration of the solenoid down the slope in respect to position.
(You'll need to introduce a few basic constants with the algebra)
Edit: To make life easier and to simplify the maths, presume that the solenoid can immediately roll without slipping.
Been a while since I've done (few) umep problems, hope it's close
If we assume the magnetic field is perpendicular to the direction of the slope. The current will not affect the acceleration due to the right hand slap rule cancelling out over the circular loops. The problem is then reduced to a rolling tube.
The same thing happens if the magnetic field is parallel to the slope.
Take the contact point between the solenoid and the ramp to be the pivot.
By the parallel-axis theorem ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?I = I_{cm}+mR^2=2mR^2)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tau=I\alpha)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?R \times mg\sin{\theta}=2mR^2(\frac{a}{R}))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?mg\sin\theta=2ma)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?a=\frac{g\sin\theta}{2})
(I know this solution probably isn't what you had in mind and I might have misunderstood but oh well...
In the meantime, have a go at this problem. I hope it isn't pushing the math limits, but it is a variation on a classic electromagnetism problem.
We have a line of positive charge of length L and total charge Q. At one end, a certain distance 'z' (or call it whatever you like)
above the line of charge is a point P. Find the Electric field at that point. Express it in terms of vectors in the x and y directions.)
[There are several ways to get the answer]
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Oops, I think I wrote solenoid when I meant wires wrapped around a cardboard tube (so not like a coil like a solenoid, wrapped lengthwise around...). So yes, that changes it significantly...
And I think I've done that question before. I'll try and find my solution or just do it again :)
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I've done a question like that. I used logic to answer it though and the rod was semi-infinite and I ended up getting it wrong. so......... *facepalm*
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Oops, I think I wrote solenoid when I meant wires wrapped around a cardboard tube (so not like a coil like a solenoid, wrapped lengthwise around...). So yes, that changes it significantly...
And I think I've done that question before. I'll try and find my solution or just do it again :)
Sorry I still don't think I really understand the problem, but it definitely sounds like an interesting one, and it would be great if you could post a diagram and/or solution to it.
If anyone is interested in the solution to my previous problem, it's
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\mathbf{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{zL}\left[\left(-1+\frac{z}{\sqrt{z^2+L^2}}\right)\mathbf{\hat{i}}+\left(\frac{L}{\sqrt{z^2+L^2}}\right)\mathbf{\hat{j}}\right])
Anyway,
POST MORE PROBLEMS!
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Yeah any new problems to share? :-\
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The pressure underwater at a depth
is
,
where
is atmospheric pressure,
and
is density.
A circular pool has radius of 10m, and the water depth is 1.5m. Find the total force exerted on the walls.
Extension: Prove the formula
(i.e. that pressure depends on height)
[Let's try to get this into a Physics Marathon, so if you solve a question you have to post another one!]
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Now I'm no physicist and I haven't done fluids in a long time but intuition and total guessing tells me that the area/volume of the pool doesn't come into play here its just
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? P = P_0 + \rho g \left(\Delta h \right) )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? P = P_0 + \rho g \left(h_1 - h_2 \right))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? P = 1.1 \times 10^5 + 1000 * 9.8 * \left(0 - 1.5\right))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? P = 1.1 \times 10^5 - 14700)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \therefore P = 95300Pa)
or maybe it's just
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?P = -14700Pa)
*shrug*
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Ugh, my bad, it's meant to read total 'Force' exerted on the walls. (Of course, since Pressure is per unit area)
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ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh I sort of know what to do then
*goes back to work*
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Again I'm only speculating the answer/working/method
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? P = \frac{F}{A} )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? P = P_0 + \rho g \Delta h )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? \frac{F}{A} = P_0 + \rho g \Delta h )
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? F = \left(P_0 + \rho g \Delta h\right) \times A )
From above
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?P_0 + \rho g \Delta h = 95300Pa)
and ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? A = \pi r^2 = \pi \times 10^2 \approx 314.16m^2)
So
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? F = P \times A = 95300Pa \times 314.16m^2 \approx 2.9939\times10^7 N )
Or as before using -14700Pa
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi? F = P \times A = -14700Pa \times 314.16m^2 \approx 4.618\times10^6 N )
Again. Only speculation I have no idea what I'm doing/where I'm going with this. I suck at fluids :-[
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The key thing here is that P is not constant, it increases with depth. Also, the area we're seeking is the surrounding walls, not the bottom. But apart from that you've got mostly the right idea.
Try graphing P against h and see if you can draw any conclusions :P
Hint: average!
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you have to use a pressure trapezoid
and multiply that by the surface area of the cylinder which will be 2pi*r*h
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The pressure underwater at a depth
is
,
where
is atmospheric pressure,
and
is density.
A circular pool has radius of 10m, and the water depth is 1.5m. Find the total force exerted on the walls.
Extension: Prove the formula
(i.e. that pressure depends on height)
[Let's try to get this into a Physics Marathon, so if you solve a question you have to post another one!]
Cool, let's do some calculus;
F = p x A
dF = p x dA
dF = (Po + ρgh) x 2πr x dh
F =
(Po + ρgh) x 2πr x dh
= 2πrPo
dh + 2πrρg
hdh
= 2πrPo h|01.5 + πrρgh2|01.5
F = 1.106x107N
I'll try the challenge question now!
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Correct QuantumJG! (at least I think it is)
There is another solution that doesn't involve calculus.
The pressure at the bottom is
, and the pressure at the top is just ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?P = 1.1 \times 10^5\ \mbox{Pa})
Since
is a linear function of
, the average pressure is
.
Thus we have
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Pressure at a certain point depends on the total downward force acting on it
Force exerted by atmosphere on water surface: ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?F = P_o \times A)
Force exerted by some mass of water occupying some volume: ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?F = ma = \rho \times V \times g = \rho \times (A \times h) \times g)
Total force is hence ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?F = A (P_o + \rho \times g \times h) \implies P = P_o + \rho \times g\times h)
EDIT, it took some self-convincing that the first statement is true. I'm still a little unconvinced.
Gravity force at that point = 0 [mass is rho * A * dh ~ 0]
Normal reaction force is the same magnitude as the force downwards [it's in translational equilibrium], however, if you were a plate, you would only measure pressure on one side, hence calculating either of these forces would give the pressure
No idea how sideways pressure works,
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Correct QuantumJG! (at least I think it is)
There is another solution that doesn't involve calculus.
The pressure at the bottom is
, and the pressure at the top is just ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?P = 1.1 \times 10^5\ \mbox{Pa})
Since
is a linear function of
, the average pressure is
.
Thus we have ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?F=P_{ave}A = (117350)(2\pi (10) \times 1.5) = 1.106 \times 10^7\ \mbox{N})
That's neat!
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Nice derivation Mao, I'm not really sure how sideways pressure works either, I was hoping someone could tell me :P
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*facepalm*
When I used calculus I used the depth of the water as the radius of the pool instead of 10m.... thats what you get for doing calculus at 3am though -.-
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From Dresden Codak.
[img]http://dresdencodak.com/comics/2005-06-14-lil_werner.jpg[/img]
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LOL Cthulhu xD
Heisenberg is a legend!
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Oh, and just in terms of sideways pressure, I think this is how you'd go about figuring it out - similar derivations are often used when looking at pressure differences and the speed of sound. This ignores gravity and the like, and it's probably only useful for considering ideal gases (not liquids!), seeing as I'm not accounting for how the particles with velocities on the z axis (defined as the "upwards" direction) undergo a change in momentum (this would require the surface of the water to exert a force downwards, and for the surface to be static).
Consider a particle moving at the average velocity of a substance v. When it hits a wall, a change in momentum occurs as it changes direction, and hence a force is exerted. The magnitude of this force = 2mv/t. However, as pressure is F(av)/t, you can take the time interval between hits, which should be 2x/v, where x is the length from one end of whatever it's in to the other. If you multiply this by the number of particles in the mixture, you should get the total force exerted, and if you then divide by the area, you'll find the pressure.
However, how can the sideways velocity be calculated? If the temperature is known, we can then find out the kinetic translational energy, as only the translational energy contributes to the temperature. Relationships between internal energy and degrees of freedom could then be used to find out the velocity of the water, and presuming the equipartition of velocity, the x direction of the velocity^2 = v(total)^2/3. Or something like that.
I don't think this even holds for liquids, because the intermolecular forces are too large and there's the whole problem of the downwards force at the top (it could be caused by surface tension, but I don't actually know anything at all about that, so I'm not going to make any assumptions). You could probably make a model taking these into account, but it seems like it'd end up being really fiddly...
Edit: I don't even know if gravity needs to be taken into account, because the whole pressure equation for downwards pressure takes F(net) = 0 (dp/dx = -rho * g or something...)
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I remember appianway mentioned thermodynamics earlier on and I found this question in a book(it's shortened a bit):
Calculate the temperature at which many hydrogen atoms will be in the first excited state (n=2).
(
average thermal energy.)
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Bumpin it up.
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Well, I presume you plug in the formula for the energies of the hydrogen atom with n=2 (I don't know it off the top of my head), and equate that to the internal energy of 3/2kT before solving for T. Although I could be wrong.
If anyone wants to have a look, IPhO 2009 has a nice, neat gravitational question. If I could do it, I'm sure you can too. I prefer the quantum question though :)
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This isn't that hard but I can't remember how to do it:
A piece of timber 2 metres long and of retangular cross section 30cm x 20 cm floats vertically in water of density 1000kg/m3 with 20cm protruding above the surface. Which of the following is closest to the average density of the timber?
800kg/m3, 850kg/m3, 900kg/m3, 950kg/m3
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OK.
Fnet = 0, because it's not accelerating upwards. You can find out the volume of water displaced, and the buoyancy force is equal to the weight of this (density * volume * g). This force has to be of equal magnitude to the weight force of the log, and then the weight force is the density times the volume times g. You just solve for the density.
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So I did
Buoyancy force = 1000kg/m3 x (0.10*0.20*2) x 9.8
and
Weight force = density of log x (0.30*0.20*2.00) x 9.8
and make them equal to solve for density, is this correct?
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Don't quite think so. The amount of water displaced is given by cross sectional area * (length - 0.2 m).
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Yes that is correct:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?F_b = F_g)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\rho_f LWhg = \rho LWHg)
where h = the height of the block that is submerged (1.8m) and H is the total height of the block(2m) L and W are the length and width respectivley
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\rho = \frac{\rho_f h}{H})
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\rho = \frac{1000kgm^{-3} \times 1.8m}{2m} = 900kgm^{-3})
I SURE HOPE THIS IS RIGHT.
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Aah thanks, I calculated the density with the timber floating horizontally =/
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GAMSAT over! Came across some interesting questions I might like to share :P
For 1 mark (1.5 minutes),
The amount of energy needed to raise 1 mL of water by 1°C is 4.2 J. One meal served in a restaurant is equivalent to 8000 kJ. If all the energy is converted to heat, how much ice water with a temperature of 0°C must a diner (body temperature =37°C) drink to maintain his body temperature (or counteract this increase in energy input)?
Feel free to clarify since I might not remember the question too well.
The options were 50L, ~70L, cant remember the other two options though... ;)
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Energy that the water gains = energy converted to heat.
Energy that water gains = volume of water * change in temperature * 4.2J
Change in temperature = 37 degrees
37 * 4.2 * V = 8 * 10^6
V = 8* 10^6 / (37*4.2)
And yeah, just plug it in.
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Well that looks right, I probably forgot the formula and guessed it wrong
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From memory, I'm pretty sure i got 50L. Can't remember the exact question though. :S
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A nice problem back from my first quantum problem set:
Show that, if a photon collides with a stationary electron, they cannot both emerge
from the collision along the same trajectory.
Useful Formulae:
Relativistic momentum: ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\gamma mv)
Rest energy: ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?mc^2)
Relativistic total energy: ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\gamma mc^2)
Photon Energy: ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?E = pc)
Where ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}})
Hint: Reductio ad Absurdum
(I'll post up a really good question after I hand in my current thermo assignment xD)
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GAMSAT over! Came across some interesting questions I might like to share :P
For 1 mark (1.5 minutes),
The amount of energy needed to raise 1 mL of water by 1°C is 4.2 J. One meal served in a restaurant is equivalent to 8000 kJ. If all the energy is converted to heat, how much ice water with a temperature of 0°C must a diner (body temperature =37°C) drink to maintain his body temperature (or counteract this increase in energy input)?
Feel free to clarify since I might not remember the question too well.
The options were 50L, ~70L, cant remember the other two options though... ;)
Trick question the water is in a solid state so it can't be readily consumed :P :P :P :P
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How long do black holes last?
1) The Schwarzchild radius of a Black Hole is the radius of the 'event horizon', past which even light cannot escape.
Find an expression for the Schwarzchild radius of a black hole using the notion of escape velocity (non-rigorous derivation :-o)
2) Over time, black holes lose energy and hence mass via "Hawking Radiation". The power emitted by a Black hole is:
, where
, and
is Planck's constant,
is Boltzmann's constant, and
is the speed of light,
is the surface area,
is emissivity, and
is temperature.
Since we may assume a black hole to be a perfect blackbody (perfectly absorbs and emits radiation),
.
Given that the temperature of a black hole is:
, where
is the Gravitational constant, and the energy of a black hole is
, form a differential equation to find the lifetime of a black hole in terms of its initial mass.
3) How long would a black hole the mass of the sun last (
)? How about the mass of a proton (
)?