# 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

Title: Harder Physics Questions
Post by: appianway on January 06, 2010, 11:06:21 am
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

Title: Re: Harder Physics Questions
Post by: Akirus on January 06, 2010, 11:57:06 pm
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:

Quote
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: $E_{k - initial} = E_{k - final}$

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:

$v_{final} = -v$

Quote
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.
Quote
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.

Quote
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.
Title: Re: Harder Physics Questions
Post by: appianway on January 07, 2010, 02:53:31 pm
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.
Title: Re: Harder Physics Questions
Post by: Akirus on January 07, 2010, 03:15:56 pm
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.
Title: Re: Harder Physics Questions
Post by: appianway on January 07, 2010, 03:19:16 pm
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 :)
Title: Re: Harder Physics Questions
Post by: Akirus on January 07, 2010, 10:34:08 pm
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.
Title: Re: Harder Physics Questions
Post by: appianway on January 23, 2010, 11:19:16 am
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.
Title: Re: Harder Physics Questions
Post by: Cthulhu on January 31, 2010, 07:45:19 pm
Suppose that the radius of the Sun were increased to $5.90 * 10^{12}m$ (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.
Title: Re: Harder Physics Questions
Post by: appianway on January 31, 2010, 07:51:53 pm
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.
Title: Re: Harder Physics Questions
Post by: Cthulhu on January 31, 2010, 08:08:01 pm
Note that I said the density was (should be is) uniform? $\rho = \frac{m}{V}$ so if the density is uniform you could find the original density of the sun $\rho_{orig}$ to help you find $\rho_{new}$ which can then help you find the new mass of the sun.
Hint:$\rho_{orig} = \rho_{new}$
You're on the right track with $g = \frac{v^{2}}{r}$ but think in terms of forces...
Title: Re: Harder Physics Questions
Post by: appianway on January 31, 2010, 08:11:11 pm
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?
Title: Re: Harder Physics Questions
Post by: Cthulhu on January 31, 2010, 08:25:37 pm
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.
Title: Re: Harder Physics Questions
Post by: appianway on January 31, 2010, 08:27:12 pm
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).
Title: Re: Harder Physics Questions
Post by: Cthulhu on January 31, 2010, 08:31:49 pm
yes! something like that.
Title: Re: Harder Physics Questions
Post by: appianway on January 31, 2010, 08:37:48 pm
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!)
Title: Re: Harder Physics Questions
Post by: QuantumJG on January 31, 2010, 10:08:01 pm
Suppose that the radius of the Sun were increased to $5.90 * 10^{12}$ (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

$\oint_{S} \bf{g} . d \bf{A} = - 4 \pi G M$

since rEarth < rSun, then let:

MSun = $\dfrac{4}{3} \pi \rho_{Sun} r^{3}$

$\therefore \oint_{S} \bf{g} . d \bf{A} = - 4 \pi G \dfrac{4}{3} \pi \rho_{Sun} r^{3}$

$\therefore - \bf{g} . 4 \pi r^{2} = - 4 \pi G \dfrac{4}{3} \pi \rho_{Sun} r^{3}$

$\therefore \bf{g} = \dfrac{4}{3} G \pi \rho_{Sun} r$

I love using Guass's law on spheres! :)

Edit: Now, r = 1.5x109m, $\rho_{Sun}$ = 1.41x103 kg/m3

implying g = 591 m/s2

Now,

vEarth = 941 km/s (31x faster than what speed it orbits at)

$g = \dfrac{ 4 \pi ^{2} r}{T^{2} }$

$\therefore T = \sqrt{ \dfrac{4 \pi^{2} r}{g} }$

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?
Title: Re: Harder Physics Questions
Post by: appianway on January 31, 2010, 10:13:40 pm
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.
Title: Re: Harder Physics Questions
Post by: Cthulhu on January 31, 2010, 10:38:36 pm
Sorry quantumJG, way off :( and the question is to find the orbital velocity of earth... >.>
Title: Re: Harder Physics Questions
Post by: QuantumJG on January 31, 2010, 11:15:47 pm
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... >.>

Can I have a hint?
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 01, 2010, 11:15:26 am
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.
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 01, 2010, 01:15:47 pm
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?

:(
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 01, 2010, 02:35:05 pm
I don't understand where I have gone wrong? :(
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 01, 2010, 06:58:38 pm
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:
$R = 5.90 \times 10^{12} m$
$r = radius\ of\ earth\ orbit = 1.50 \times 10^{11} m$
$M = mass\ of\ sun\ within\ earths\ orbit$
$M_s = original\ mass\ of\ sun = 1.988435 \times 10^{30} kg$
$V = \frac{4}{3}\pi r^{3}$
$V_{orig} = 1.412 \times 10^{27} m^3$
$V_{new} = 8.603 \times 10^{38} m^3$
Since the density of the expanded sun is uniform we can use $\rho = \frac{m}{V}$ to find the mass of the sun within earths orbit. $\rho_{in}$ is the density of the sun when the radius is the orbital radius of earth
or $r$ while $\rho_{out}$ is the density of the sun at radius $R$
$
\rho_{in} = \rho_{out}
$

$\frac{3M}{4 \pi r^3} = \frac{3M_s}{4 \pi R^3}$
solving for $M$ we get $M = \left( \frac{r}{R} \right)^3 M_s$
and so:
$
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
$

Now using the equations $F = G \frac{Mm}{r^2}$ and using $F = \frac{mv^2}{r}$ because the earth is experiencing a centripetal force. setting these equal to each other we can solve for $v$:
$v = \sqrt \frac{GM}{r}$ and therefore
$
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
$

Title: Re: Harder Physics Questions
Post by: QuantumJG on February 01, 2010, 07:59:30 pm
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:
$R = 5.90 \times 10^{12} m$
$r = radius\ of\ earth\ orbit = 1.50 \times 10^{11} m$
$M = mass\ of\ sun\ within\ earths\ orbit$
$M_s = original\ mass\ of\ sun = 1.988435 \times 10^{30} kg$
$V = \frac{4}{3}\pi r^{3}$
$V_{orig} = 1.412 \times 10^{27} m^3$
$V_{new} = 8.603 \times 10^{38} m^3$
Since the density of the expanded sun is uniform we can use $\rho = \frac{m}{V}$ to find the mass of the sun within earths orbit. $\rho_{in}$ is the density of the sun when the radius is the orbital radius of earth
or $r$ while $\rho_{out}$ is the density of the sun at radius $R$
$
\rho_{in} = \rho_{out}
$

$\frac{3M}{4 \pi r^3} = \frac{3M_s}{4 \pi R^3}$
solving for $M$ we get $M = \left( \frac{r}{R} \right)^3 M_s$
and so:
$
\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 }
$

Now using the equations $F = G \frac{Mm}{r^2}$ and using $F = \frac{mv^2}{r}$ because the earth is experiencing a centripetal force. setting these equal to each other we can solve for $v$:
$v = \sqrt \frac{GM}{r}$ and therefore
$
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
$

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

$\therefore$ vEarth's orbit = 941km/s

I rest my case!
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 01, 2010, 08:08:13 pm
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.
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 01, 2010, 08:49:31 pm
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

Title: Re: Harder Physics Questions
Post by: Cthulhu on February 01, 2010, 09:10:27 pm
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.
Title: Re: Harder Physics Questions
Post by: appianway on February 01, 2010, 09:27:01 pm
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...
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 01, 2010, 09:36:44 pm
I might as well do a survey while I'm here.
Are there any particular areas people are interested in getting questions in?
Title: Re: Harder Physics Questions
Post by: appianway on February 01, 2010, 09:37:58 pm
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...
Title: Re: Harder Physics Questions
Post by: Edmund on February 01, 2010, 09:59:21 pm
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
Title: Re: Harder Physics Questions
Post by: appianway on February 01, 2010, 10:04:34 pm
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.
Title: Re: Harder Physics Questions
Post by: Edmund on February 01, 2010, 10:28:46 pm
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?
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 01, 2010, 10:49:09 pm
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?
Title: Re: Harder Physics Questions
Post by: kamil9876 on February 01, 2010, 11:04:24 pm
Find necessary and sufficient conditions for a region in $\mathbb{R}^2$ to be the unit ball for some norm.
Title: Re: Harder Physics Questions
Post by: polky on February 01, 2010, 11:35:40 pm
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.
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 02, 2010, 12:06:18 am
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.
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 02, 2010, 12:08:46 am
Find necessary and sufficient conditions for a region in $\mathbb{R}^2$ to be the unit ball for some norm.

what???
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 02, 2010, 12:12:48 am
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....
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 02, 2010, 01:31:01 am
QuantumJG: I study Physics

Cool!
Title: Re: Harder Physics Questions
Post by: appianway on February 02, 2010, 05:02:10 pm
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).
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 02, 2010, 06:55:59 pm
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?
Title: Re: Harder Physics Questions
Post by: appianway on February 02, 2010, 06:57:31 pm
Presuming an elastic collision?
Title: Re: Harder Physics Questions
Post by: Edmund on February 02, 2010, 10:04:00 pm
Since the photon loses energy, isn't it an inelastic collision (Compton Scattering)?
Title: Re: Harder Physics Questions
Post by: appianway on February 02, 2010, 10:08:16 pm
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.
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 02, 2010, 10:17:15 pm
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!

$\lambda ' - \lambda = \dfrac{h}{ m_{e} c } (1 - cos( \theta ) )$

$\lambda = \dfrac{hc}{E}$

$\therefore \dfrac{1}{E'} - \dfrac{1}{E} = \dfrac{1}{ m_{e} c^{2} } (1 - cos( \theta ) )$

let E' = 0.9E

$\theta = arcos( \dfrac{ 9E - m_{e} }{9E} )$

$\therefore \theta = 44.33^{o}$
Title: Re: Harder Physics Questions
Post by: Edmund on February 02, 2010, 10:24:15 pm
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!

$\lambda ' - \lambda = \dfrac{h}{ m_{e} c } (1 - cos( \theta ) )$

$\lambda = \dfrac{hc}{E}$

$\therefore \dfrac{1}{E'} - \dfrac{1}{E} = \dfrac{1}{ m_{e} c^{2} } (1 - cos( \theta ) )$

let E' = 0.9E

$\theta = arcos( \dfrac{ 9E - m_{e} }{9E} )$

$\therefore \theta = 44.33^{o}$
Yeah I did exactly the same thing, did you have to convert the energy into Joules or something?
/me bangs keyboard :(
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 02, 2010, 10:27:39 pm
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!

$\lambda ' - \lambda = \dfrac{h}{ m_{e} c } (1 - cos( \theta ) )$

$\lambda = \dfrac{hc}{E}$

$\therefore \dfrac{1}{E'} - \dfrac{1}{E} = \dfrac{1}{ m_{e} c^{2} } (1 - cos( \theta ) )$

let E' = 0.9E

$\theta = arcos( \dfrac{ 9E - m_{e} }{9E} )$

$\therefore \theta = 44.33^{o}$
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.
Title: Re: Harder Physics Questions
Post by: Edmund on February 02, 2010, 10:37:14 pm
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!

$\lambda ' - \lambda = \dfrac{h}{ m_{e} c } (1 - cos( \theta ) )$

$\lambda = \dfrac{hc}{E}$

$\therefore \dfrac{1}{E'} - \dfrac{1}{E} = \dfrac{1}{ m_{e} c^{2} } (1 - cos( \theta ) )$

let E' = 0.9E

$\theta = arcos( \dfrac{ 9E - m_{e} }{9E} )$

$\therefore \theta = 44.33^{o}$
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 :( :(
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 02, 2010, 11:21:56 pm
QuantumJG is right.
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 02, 2010, 11:50:43 pm
QuantumJG is right.

YAY!!!
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 02, 2010, 11:52:20 pm
This is probably the best place to put my 900th post.
Title: Re: Harder Physics Questions
Post by: appianway on February 03, 2010, 05:41:21 pm
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.
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 03, 2010, 06:44:46 pm
I feel like Quantum Mechanics.
Title: Re: Harder Physics Questions
Post by: Edmund on February 03, 2010, 07:21:20 pm
How about something from the Electromagnetism field (no pun intended)?
Title: Re: Harder Physics Questions
Post by: appianway on February 03, 2010, 08:15:34 pm
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.
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 03, 2010, 09:00:55 pm
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!
Title: Re: Harder Physics Questions
Post by: appianway on February 03, 2010, 09:18:49 pm
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...
Title: Re: Harder Physics Questions
Post by: polky on February 03, 2010, 11:26:18 pm
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?
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 04, 2010, 12:12:57 am
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.
Title: Re: Harder Physics Questions
Post by: full of electrons on February 04, 2010, 12:39:01 am
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.
Title: Re: Harder Physics Questions
Post by: /0 on February 04, 2010, 01:44:53 am
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 $I = I_{cm}+mR^2=2mR^2$

$\tau=I\alpha$

$R \times mg\sin{\theta}=2mR^2(\frac{a}{R})$

$mg\sin\theta=2ma$

$a=\frac{g\sin\theta}{2}$

(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]
Title: Re: Harder Physics Questions
Post by: appianway on February 04, 2010, 08:26:10 pm
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 :)
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 04, 2010, 10:48:40 pm
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*
Title: Re: Harder Physics Questions
Post by: /0 on February 12, 2010, 10:49:39 pm
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

$\mathbf{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{zL}\left[\left(-1+\frac{z}{\sqrt{z^2+L^2}}\right)\mathbf{\hat{i}}+\left(\frac{L}{\sqrt{z^2+L^2}}\right)\mathbf{\hat{j}}\right]$

Anyway,

POST MORE PROBLEMS!
Title: Re: Harder Physics Questions
Post by: Edmund on February 17, 2010, 09:27:58 pm
Yeah any new problems to share?  :-\
Title: Re: Harder Physics Questions
Post by: /0 on February 18, 2010, 01:03:01 pm
The pressure underwater at a depth $h$ is $P = P_0+\rho g h$,
where $P_0=1.1 \times 10^5 \ \mbox{Pa}$ is atmospheric pressure,
and $\rho=1000\ \mbox{kgm}^{-3}$ 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 $P=P_0 +\rho gh$ (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!]
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 18, 2010, 05:00:11 pm
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
$P = P_0 + \rho g \left(\Delta h \right)$
$P = P_0 + \rho g \left(h_1 - h_2 \right)$
$P = 1.1 \times 10^5 + 1000 * 9.8 * \left(0 - 1.5\right)$
$P = 1.1 \times 10^5 - 14700$
$\therefore P = 95300Pa$
or maybe it's just
$P = -14700Pa$
*shrug*

Title: Re: Harder Physics Questions
Post by: /0 on February 18, 2010, 05:29:59 pm
Ugh, my bad, it's meant to read total 'Force' exerted on the walls. (Of course, since Pressure is per unit area)
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 18, 2010, 05:38:22 pm
ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh I sort of know what to do then
*goes back to work*
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 18, 2010, 07:38:10 pm
Again I'm only speculating the answer/working/method
$P = \frac{F}{A}$
$P = P_0 + \rho g \Delta h$
$\frac{F}{A} = P_0 + \rho g \Delta h$
$F = \left(P_0 + \rho g \Delta h\right) \times A$
From above
$P_0 + \rho g \Delta h = 95300Pa$
and $A = \pi r^2 = \pi \times 10^2 \approx 314.16m^2$
So
$F = P \times A = 95300Pa \times 314.16m^2 \approx 2.9939\times10^7 N$
Or as before using -14700Pa
$F = P \times A = -14700Pa \times 314.16m^2 \approx 4.618\times10^6 N$
Again. Only speculation I have no idea what I'm doing/where I'm going with this. I suck at fluids :-[
Title: Re: Harder Physics Questions
Post by: /0 on February 19, 2010, 12:30:49 am
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!
Title: Re: Harder Physics Questions
Post by: enwiabe on February 19, 2010, 02:16:23 am
you have to use a pressure trapezoid

and multiply that by the surface area of the cylinder which will be 2pi*r*h
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 19, 2010, 10:39:12 am
The pressure underwater at a depth $h$ is $P = P_0+\rho g h$,
where $P_0=1.1 \times 10^5 \ \mbox{Pa}$ is atmospheric pressure,
and $\rho=1000\ \mbox{kgm}^{-3}$ 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 $P=P_0 +\rho gh$ (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 = $\int _{0} ^{1.5}$ (Po + ρgh) x 2πr x dh

= 2πrPo $\int _{0} ^{1.5}$ dh + 2πrρg $\int _{0} ^{1.5}$ hdh

= 2πrPo h|01.5 + πrρgh2|01.5

$\therefore$ F = 1.106x107N

I'll try the challenge question now!
Title: Re: Harder Physics Questions
Post by: /0 on February 19, 2010, 11:23:13 am
Correct QuantumJG! (at least I think it is)

There is another solution that doesn't involve calculus.

The pressure at the bottom is $P=1.1 \times 10^5 + (1000)(9.8)(1.5) = 124700\ \mbox{Pa}$, and the pressure at the top is just $P = 1.1 \times 10^5\ \mbox{Pa}$

Since $P$ is a linear function of $h$, the average pressure is $\frac{124700+1.1 \times 10^5}{2} = 117350 \ \mbox{Pa}$.

Thus we have $F=P_{ave}A = (117350)(2\pi (10) \times 1.5) = 1.106 \times 10^7\ \mbox{N}$
Title: Re: Harder Physics Questions
Post by: Mao on February 19, 2010, 11:30:51 am
Pressure at a certain point depends on the total downward force acting on it

Force exerted by atmosphere on water surface: $F = P_o \times A$

Force exerted by some mass of water occupying some volume: $F = ma = \rho \times V \times g = \rho \times (A \times h) \times g$

Total force is hence $F = A (P_o + \rho \times g \times h) \implies P = P_o + \rho \times g\times h$

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,
Title: Re: Harder Physics Questions
Post by: QuantumJG on February 19, 2010, 12:04:16 pm
Correct QuantumJG! (at least I think it is)

There is another solution that doesn't involve calculus.

The pressure at the bottom is $P=1.1 \times 10^5 + (1000)(9.8)(1.5) = 124700\ \mbox{Pa}$, and the pressure at the top is just $P = 1.1 \times 10^5\ \mbox{Pa}$

Since $P$ is a linear function of $h$, the average pressure is $\frac{124700+1.1 \times 10^5}{2} = 117350 \ \mbox{Pa}$.

Thus we have $F=P_{ave}A = (117350)(2\pi (10) \times 1.5) = 1.106 \times 10^7\ \mbox{N}$

That's neat!
Title: Re: Harder Physics Questions
Post by: /0 on February 19, 2010, 12:47:52 pm
Nice derivation Mao, I'm not really sure how sideways pressure works either, I was hoping someone could tell me :P
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 19, 2010, 03:09:58 pm
*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 -.-
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 22, 2010, 08:53:51 pm
From Dresden Codak.
[img]http://dresdencodak.com/comics/2005-06-14-lil_werner.jpg[/img]
Title: Re: Harder Physics Questions
Post by: /0 on February 23, 2010, 11:36:59 pm
LOL Cthulhu xD
Heisenberg is a legend!

Title: Re: Harder Physics Questions
Post by: appianway on February 24, 2010, 01:36:50 pm
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...)
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 25, 2010, 05:29:50 pm
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).
($\frac{3}{2}k_B T =$ average thermal energy.)
Title: Re: Harder Physics Questions
Post by: Cthulhu on February 27, 2010, 01:41:13 am
Bumpin it up.
Title: Re: Harder Physics Questions
Post by: appianway on February 28, 2010, 09:45:29 pm
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 :)
Title: Re: Harder Physics Questions
Post by: Edmund on March 07, 2010, 10:30:04 pm
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
Title: Re: Harder Physics Questions
Post by: appianway on March 07, 2010, 10:40:50 pm
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.
Title: Re: Harder Physics Questions
Post by: Edmund on March 07, 2010, 10:52:25 pm
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?
Title: Re: Harder Physics Questions
Post by: appianway on March 07, 2010, 10:58:42 pm
Don't quite think so. The amount of water displaced is given by cross sectional area * (length - 0.2 m).
Title: Re: Harder Physics Questions
Post by: Cthulhu on March 07, 2010, 10:59:26 pm
Yes that is correct:
$F_b = F_g$
$\rho_f LWhg = \rho LWHg$
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
$\rho = \frac{\rho_f h}{H}$
$\rho = \frac{1000kgm^{-3} \times 1.8m}{2m} = 900kgm^{-3}$
I SURE HOPE THIS IS RIGHT.
Title: Re: Harder Physics Questions
Post by: Edmund on March 07, 2010, 11:04:29 pm
Aah thanks, I calculated the density with the timber floating horizontally =/
Title: Re: Harder Physics Questions
Post by: Edmund on March 20, 2010, 09:07:51 pm
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... ;)
Title: Re: Harder Physics Questions
Post by: appianway on March 20, 2010, 09:16:04 pm
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.
Title: Re: Harder Physics Questions
Post by: Edmund on March 20, 2010, 09:20:20 pm
Well that looks right, I probably forgot the formula and guessed it wrong
Title: Re: Harder Physics Questions
Post by: iamdan08 on March 20, 2010, 10:12:10 pm
From memory, I'm pretty sure i got 50L. Can't remember the exact question though. :S
Title: Re: Harder Physics Questions
Post by: /0 on May 07, 2010, 01:33:17 pm
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: $\gamma mv$
Rest energy: $mc^2$
Relativistic total energy: $\gamma mc^2$
Photon Energy: $E = pc$

Where $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

(I'll post up a really good question after I hand in my current thermo assignment xD)
Title: Re: Harder Physics Questions
Post by: schnappy on May 10, 2010, 08:22:36 pm
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
Title: Re: Harder Physics Questions
Post by: /0 on May 11, 2010, 07:13:00 pm
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:
$P = \sigma e AT^4$, where $\sigma = \frac{2\pi^5k^4}{15hc^3}$, and $h$ is Planck's constant, $k$ is Boltzmann's constant, and $c$ is the speed of light, $A$ is the surface area, $e$ is emissivity, and $T$ is temperature.
Since we may assume a black hole to be a perfect blackbody (perfectly absorbs and emits radiation), $e = 1$.
Given that the temperature of a black hole is: $T = \frac{hc^3}{16\pi^2kGM}$, where $G$ is the Gravitational constant, and the energy of a black hole is $Mc^2$, 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 ($2 \times 10^{30}kg$)? How about the mass of a proton ($1.67 \times 10^{-27}kg$)?