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Oxbridge Physics Prelims Revision
Hey, guys, I know that most of you probably aren't even thinking about your prelims which are unfortunately looming, but being the worrier that I am, I'm starting to panic and feel I'll be asking lots of questions again! As the other thread was getting quite full (we almost reached 400posts!) I thought I'd start a new one dedicated entirely to prelims. Hopefully, people can post any problems they have and likewise help solve others (probably primarily mine!) all in an effort to improve our understanding before 7th week!
So anyways here it goes...(I'm going to post the first question in a separate post!)
So anyways here it goes...(I'm going to post the first question in a separate post!)
Scroll to see replies
Q. A meson of rest mass M is produced at rest in the lab in an excited state with excitation energy
======
My workings so far (NB. M and m are different)
Let energy in ground state be
Momentum is same; using invariant:
->
1 INTO 2
From an energy level diagram:
thus
Thus
However, this isn't the correct answer. Any ideas? Thanks
Unparseable latex formula:
. The excited state decays to the ground state by emitting a photon with an energy \fontsize{3} E_{o}
Unparseable latex formula:
. Show that the different in energy \fontsize{3} E_{\gamma}
Unparseable latex formula:
is given by:\fontsize{3} \Delta = E_{o} - E_{\gamma}
Unparseable latex formula:
\fontsize{3} \Delta = E_{o^{2}} \div 2(Mc^{2} + E_{o} )
======
My workings so far (NB. M and m are different)
Unparseable latex formula:
1\fontsize{3} Mc^{2} = E_{A} + E_{\gamma}
Let energy in ground state be
Unparseable latex formula:
\fontsize{3} E_{A}
Momentum is same; using invariant:
Unparseable latex formula:
\fontsize{3} E_{A}^{2} - (pc)^{2} = (mc^{2})^{2}
Unparseable latex formula:
\fontsize{3} E_{\gamma}^{2} - (pc)^{2} = 0
->
Unparseable latex formula:
\fontsize{3} E_{\gamma}^{2} = (pc)^{2}
Unparseable latex formula:
2\fontsize{3} E_{A}^{2} E_{\gamma} = (mc^{2})^{2}
1 INTO 2
Unparseable latex formula:
\fontsize{3} E_{\gamma} = [(Mc^{2})^{2} - (mc^{2})^{2}] \div 2Mc^{2}
From an energy level diagram:
Unparseable latex formula:
\fontsize{3} mc^{2} = Mc^{2} - E_{o}
thus
Unparseable latex formula:
\fontsize{3} E_{\gamma} = [(E_{o}^{2} + 2Mc^{2}E_{o}] \div 2Mc^{2}
Thus
Unparseable latex formula:
\fontsize{3} E_{o} - E_{\gamma} = \Delta = -E_{o} \div 2Mc^{2}
However, this isn't the correct answer. Any ideas? Thanks
i've got a feeling it's something to do with your energy level diagram. You see, surely the meson recoils and so isnt at rest, and so you must use the 1/root(1-v^2/c^2) factor in that part of the question to factor that. I'll have a little think, I normally like these types of questions
Hey guys, you know how mugh I hate EM and I honestly don't know how to do the following question (yes you can all say I am officially s**t at physics...I already know!)....any suggestions?
A long solenoid is uniformly wound with 10000 turns per metre. A long metal tube of mean diameter 1cm and thickness 0.1mm is placed coaxially inside the solenoid. The resistivity of the metal of the tube is
A long solenoid is uniformly wound with 10000 turns per metre. A long metal tube of mean diameter 1cm and thickness 0.1mm is placed coaxially inside the solenoid. The resistivity of the metal of the tube is
Unparseable latex formula:
An alternating current of r.m.s value at 1A at a frequency of 100Hz flows in the solenoid. Calculate the power dissipated in the tube per metre length.\fontsize{3} 5 \times 10^{-7} \sigma m
Well I dont think people should try and just do these for you. I'm sure you can do more than you know.
Start from the B-field inside the solenoid. Then calculate the B field flux through a cross section of the tube. Find the rate of change of that and you have the EMF round the tube. Have a go
what is the final answer btw if you know, so I can check I'm on the right lines
Start from the B-field inside the solenoid. Then calculate the B field flux through a cross section of the tube. Find the rate of change of that and you have the EMF round the tube. Have a go
what is the final answer btw if you know, so I can check I'm on the right lines
Ok:
B-Field inside a solenoid
Then do I define
where D = 1A (even though that's rms rather than maximum value?) and
Then I can diffentiate the flux wrt time to get
B-Field inside a solenoid
Unparseable latex formula:
\fontsize{3} B = \mu_{o}nI
Unparseable latex formula:
\fontsize{3} B = \mu_{o} \times 10000I
Unparseable latex formula:
\fontsize{3} A = \pi \times [(1.01)^{2} - (0.99)^{2}]
Unparseable latex formula:
\fontsize{3} A = 0.04\pi cm^{2}
Unparseable latex formula:
\fontsize{3} A = 4\pi \times 10^{-6} m^{2}
Unparseable latex formula:
\fontsize{3} \phi = \mu \times 10000 \times I \times 4\pi\times10^{-6}
Unparseable latex formula:
\fontsize{3} \phi = 0.04\pi \times \mu \times I
Then do I define
Unparseable latex formula:
\fontsize{3} I = Dcos\omega t
where D = 1A (even though that's rms rather than maximum value?) and
Unparseable latex formula:
\fontsize{3} \omega=2\pi \times 100Hz
Then I can diffentiate the flux wrt time to get
Unparseable latex formula:
? Am I right so far? I'm afraid I don't have the answers to this.\fontsize{3} \frac{d\phi}{dt} = -\epsilon
well since my post previously I'm currently contemplating the last step myself (i like a challenge and these questions certainly are!). If we work together on this one I'm sure it can be solved.
Basically i was thinking about the nature of the induced currents, they would surely be circular around the tube, cos of the B field direction. So then I started thinking about taking a thin disc of radius r and thickness dr, which you know would contain pi*r^2*B flux. So if you take the derivative of that you have the EMF round your thin tiny disc.
So then I was looking at trying to do some form of integral. over the different radii but I got a bit stuck in how to introduce the dr.
Perhaps then you could do some bit of a fudge and say that the average EMF is just the EMF through the central part of the metal part of the tube. So take that radius and find the EMF from pi*r^2*B
Either way, you have made a mistake in your above working....at least I think you have. It's to do with your calculating flux. Cos you've taken A to be the cross section of the tube, but you have to consider the area contained by the loop as well surely?
any ideas....I like this question cos it seems accessible, there's a leap of imagination that it requires, but we're already quite clsoe i reckon!
edit: i also worry about how we are gonna approach a value of power dissiptated per metre, cos these currents seem to be circular. Hmm, tis a toughy! I need food, ergh ergs!
Basically i was thinking about the nature of the induced currents, they would surely be circular around the tube, cos of the B field direction. So then I started thinking about taking a thin disc of radius r and thickness dr, which you know would contain pi*r^2*B flux. So if you take the derivative of that you have the EMF round your thin tiny disc.
So then I was looking at trying to do some form of integral. over the different radii but I got a bit stuck in how to introduce the dr.
Perhaps then you could do some bit of a fudge and say that the average EMF is just the EMF through the central part of the metal part of the tube. So take that radius and find the EMF from pi*r^2*B
Either way, you have made a mistake in your above working....at least I think you have. It's to do with your calculating flux. Cos you've taken A to be the cross section of the tube, but you have to consider the area contained by the loop as well surely?
any ideas....I like this question cos it seems accessible, there's a leap of imagination that it requires, but we're already quite clsoe i reckon!
edit: i also worry about how we are gonna approach a value of power dissiptated per metre, cos these currents seem to be circular. Hmm, tis a toughy! I need food, ergh ergs!
Willa
Either way, you have made a mistake in your above working....at least I think you have. It's to do with your calculating flux. Cos you've taken A to be the cross section of the tube, but you have to consider the area contained by the loop as well surely?
I think I have indeed made a small error in my working with the area. I still however think that we only need to consider the flux through the solid part of the tube and therefore the calculation should read:
Unparseable latex formula:
\fontsize{3} A = \pi \times [(1.005)^{2} - (0.995)^{2}]
I've also included a picture showing my intepretation of what it looks like.
I would have honestly thought that once we've got the emf in terms of
Unparseable latex formula:
and t we just then calculate the resistance from the resistivity question and then can use Power = V^2/R....but then why did we need to calculate the emf to even begin with as we could have just used P=I^2R?! This question is confusing me, and I don't think the others have done it yet...it was in our collection paper and I got as far as I have above, but we'll be getting it back tomorrow so I'll find out from my tutor. If anyone else has any ideas....\fontsize{3} \omega
I need food, ergh ergs!
hehe...my coach seems to have forgotten we were going to do ergs today and I'm certainly not going to remind him when I have all this work to do! He probably hasn't forgotten and it's just because we're still having loads of problems sorting out the crews (ppl quiting/moving around etc) and I still don't know who will be rowing with me (but at least I'm about 80% sure I'm going to be stroke this term!
)
Hoofbeat
I think I have indeed made a small error in my working with the area. I still however think that we only need to consider the flux through the solid part of the tube and therefore the calculation should read:
I've also included a picture showing my intepretation of what it looks like.
I would have honestly thought that once we've got the emf in terms of
Unparseable latex formula:
\fontsize{3} A = \pi \times [(1.005)^{2} - (0.995)^{2}]
I've also included a picture showing my intepretation of what it looks like.
I would have honestly thought that once we've got the emf in terms of
Unparseable latex formula:
and t we just then calculate the resistance from the resistivity question and then can use Power = V^2/R....but then why did we need to calculate the emf to even begin with as we could have just used P=I^2R?! This question is confusing me, and I don't think the others have done it yet...it was in our collection paper and I got as far as I have above, but we'll be getting it back tomorrow so I'll find out from my tutor. If anyone else has any ideas....\fontsize{3} \omega
Well you cant use I^2R straight off cos you dont know that the current in the tube is the same as the current in the solenoid. In fact it's probably quite different.
And I dont think you can just take the cross sectional area so simply because remember that it is EMF round the loop = rate of change of flux through the loop. If you really can just use the cross sectional area it's a bit of a fluke, because think about it: If you take a very thing loop right on the inside of the tube, then that contains, say, pir^2 area. Then take the next thin loop, that contains pi(r+dr)^2 area. So this would imply that the EMF round the tube is gradually increasing with radius.....and I consider it a fluke if you can average this out to being the rate of change of flux through the metal part of the cross section alone...but in fact I consider that to probably be wrong.
As I said, what concerns me is that the currents induced will be circular about the axis of the tube, wheras the questions asks for power dissipitated per unit length, implying currents flow along the tube (in the direction of the axis).
I am very keen to know what the true answer is, cos I reckon this question seems unfairly complex.
hehe...my coach seems to have forgotten we were going to do ergs today and I'm certainly not going to remind him when I have all this work to do! He probably hasn't forgotten and it's just because we're still having loads of problems sorting out the crews (ppl quiting/moving around etc) and I still don't know who will be rowing with me (but at least I'm about 80% sure I'm going to be stroke this term!
)If only my captain was as forgetful. Thankfully we've now got most of our land training done for the week: 10am circuits yesterday, 1 hour erg at 5pm. Then circuits AGAIN at 7:30am this morning. Only one more erg to do in the middle of the week.
The state of our lower boats is stupid atm though. We barely have a second boat it seems, and cos of the lack of commitment they are only doing two outtings a week atm. It's ridiculous, they're heading for spoons again this term! I really want to go up at least one place this summer, we're far enough down to do so!
And have you ever stroked a boat before, it's quite an alarming experience when I did it at training camp. You become very sensitive to the stability of the boat. I didn't like it cos I always like having someone to follow in front of me, maybe being bow pair all last term has made me that way, cos I was fine at 7-seat when i noviced.
Good luck!
Willa
And have you ever stroked a boat before, it's quite an alarming experience when I did it at training camp. You become very sensitive to the stability of the boat. I didn't like it cos I always like having someone to follow in front of me, maybe being bow pair all last term has made me that way, cos I was fine at 7-seat when i noviced.
Good luck!
Good luck!
Yeah, I've done it once before in the very first proper outing I had with my Novice crew in the first term...they decided that I rushed too much to be given the iportant job of stroke, but my new coach thinks I've clearly improved and there's no1 else suitable to do it! I've been rowing at 7 the last few times (in the first's bowrigged boat as we broke ours! oops) and I've started to realise how uncomfortable it is when someone in the bows catches out of time but despite this stroke and myself managed to keep in time and try and keep a steady rhythm ... i think I'm more scared about getting a good start and setting everyone up correctly. Anyways our boat has now been fixed which should mean I'll be stroking tomorrow (how wrong does that sound?!).
I'll take a good look at the EM question a bit later. thanks for your help. Nick, considering you've already done it perhaps you can shed some light on the topic?
shiny
Dude! Don't help her unless she buys you a cookie! 
heehee, i've heard all about your and chloés cookie swapping - it's like rep but tastier and less frowned upon!
The reason I haven't helped you with the question, chloe, is that I didn't get any further than you have already in this thread and I haven't gone through it with my tutor yet! (Wouldn't say no to a cookie though
) I actually have to go into town now anyway so might get 1 on my way back Triple choc chip, obviously!edit: according to hoofbeat its closed today
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