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May 2014 IB Physics Paper 3 HL TZ2

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So How did Everyone do? I did Astro and EM waves. Astro was a bit harder than usual but overall not too bad and EM Waves was OK as well. So what did you guys think about it?

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I did astrophysics and relativity. It was pretty ok, especially compared to the horrible paper 2 on Wednesday. There were some tricky questions though. I remembered the one in relativity about simultaneity, which I’m sure I got it wrong, cuz I said in the end that the left lamp must emit light first.

Also do you guys remember the one about Cepheid Variable graph in Astrophysics? The one that we had to use m-M=5log(d/10) to find the star’s distance from Earth. To calculate M (the average absolute magnitude) was quite easy. But how do we find m (the average apparent magnitude) from the graph?. I just took the (max+min)/2 (i.e. I got m=(15.9+15.2)/2=15.55), but I’m not really sure it was the correct method. Anybody who knows how to do this question?

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I did astrophysics and relativity. It was pretty ok, especially compared to the horrible paper 2 on Wednesday. There were some tricky questions though. I remembered the one in relativity about simultaneity, which I’m sure I got it wrong, cuz I said in the end that the left lamp must emit light first.

Also do you guys remember the one about Cepheid Variable graph in Astrophysics? The one that we had to use m-M=5log(d/10) to find the star’s distance from Earth. To calculate M (the average absolute magnitude) was quite easy. But how do we find m (the average apparent magnitude) from the graph?. I just took the (max+min)/2 (i.e. I got m=(15.9+15.2)/2=15.55), but I’m not really sure it was the correct method. Anybody who knows how to do this question?

I did exactly the same, hopefully it's right because that question was 5 Marks!!!

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I also did the exactly same thing! Do you remember how much did you get in the end? I think was 7.6x10^6 pc ( or something like that)

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I did astrophysics and relativity. It was pretty ok, especially compared to the horrible paper 2 on Wednesday. There were some tricky questions though. I remembered the one in relativity about simultaneity, which I’m sure I got it wrong, cuz I said in the end that the left lamp must emit light first.

Also do you guys remember the one about Cepheid Variable graph in Astrophysics? The one that we had to use m-M=5log(d/10) to find the star’s distance from Earth. To calculate M (the average absolute magnitude) was quite easy. But how do we find m (the average apparent magnitude) from the graph?. I just took the (max+min)/2 (i.e. I got m=(15.9+15.2)/2=15.55), but I’m not really sure it was the correct method. Anybody who knows how to do this question?

I did exactly the same, hopefully it's right because that question was 5 Marks!!!

I also did the exactly same thing! Do you remember how much did you get in the end? I think was 7.6x10^6 pc ( or something like that)

Guys they were talking about the VARIATION. 15.55 it a very large number for apparent magnitude... So you do maximum - minimum.... And the answer should approximately equate to about 73 pc... which is then converted into meters... 7.6x10^6 pc is a very large number.... but you should get marks for the working out don't worry :D

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Didn't do as well as I would have hoped, but I did ok. I did astro and medical, messed up a bit on the cepheid variable and calculating intensity level (I took 10^-2 instead of 10^-12 for threshold....) but both I should get method marks on. I also forgot one of the properties of the CMB radiation :( but other than that it was good lol

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I did astrophysics and relativity. It was pretty ok, especially compared to the horrible paper 2 on Wednesday. There were some tricky questions though. I remembered the one in relativity about simultaneity, which I’m sure I got it wrong, cuz I said in the end that the left lamp must emit light first.

Also do you guys remember the one about Cepheid Variable graph in Astrophysics? The one that we had to use m-M=5log(d/10) to find the star’s distance from Earth. To calculate M (the average absolute magnitude) was quite easy. But how do we find m (the average apparent magnitude) from the graph?. I just took the (max+min)/2 (i.e. I got m=(15.9+15.2)/2=15.55), but I’m not really sure it was the correct method. Anybody who knows how to do this question?

I did exactly the same, hopefully it's right because that question was 5 Marks!!!

I also did the exactly same thing! Do you remember how much did you get in the end? I think was 7.6x10^6 pc ( or something like that)

Guys they were talking about the VARIATION. 15.55 it a very large number for apparent magnitude... So you do maximum - minimum.... And the answer should approximately equate to about 73 pc... which is then converted into meters... 7.6x10^6 pc is a very large number.... but you should get marks for the working out don't worry :D

No I'm pretty sure you were supposed to get the mean, I even looked it up on the internet and it said the same thing. http://www.atnf.csiro.au/outreach/education/senior/astrophysics/variable_cepheids.html. I don't remember the answer that I got but I remember it being a very large number, and if you look at the example in this website which had values close to the ones in the exam, you will find that the answer was also big.

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For the cepheid question, here was exactly what I did:

I found M to be M = -3.61 (from the graph and from the given equation), and m to be m=15.55 (as I have explained earlier). Plug these into m-M=5log(d/10) yields d=6.81*10^4 pc

Guys they were talking about the VARIATION. 15.55 it a very large number for apparent magnitude... So you do maximum - minimum.... And the answer should approximately equate to about 73 pc... which is then converted into meters... 7.6x10^6 pc is a very large number.... but you should get marks for the working out don't worry :D

Not to say that you are wrong Abdelbari because im not even sure myself. But don’t you think that 73pc is a little bit too small for a Cepheid star? I mean if you google it, you’ll see that many Cepheid variables are at around 50000 pc from Earth. Also, 15.55 isn’t large at all because the bigger the magnitude, the dimmer the star. So I would say that 15.55 is a fairly reasonable value.

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I did astrophysics and relativity. It was pretty ok, especially compared to the horrible paper 2 on Wednesday. There were some tricky questions though. I remembered the one in relativity about simultaneity, which I’m sure I got it wrong, cuz I said in the end that the left lamp must emit light first.

Also do you guys remember the one about Cepheid Variable graph in Astrophysics? The one that we had to use m-M=5log(d/10) to find the star’s distance from Earth. To calculate M (the average absolute magnitude) was quite easy. But how do we find m (the average apparent magnitude) from the graph?. I just took the (max+min)/2 (i.e. I got m=(15.9+15.2)/2=15.55), but I’m not really sure it was the correct method. Anybody who knows how to do this question?

I did exactly the same, hopefully it's right because that question was 5 Marks!!!

I also did the exactly same thing! Do you remember how much did you get in the end? I think was 7.6x10^6 pc ( or something like that)

Guys they were talking about the VARIATION. 15.55 it a very large number for apparent magnitude... So you do maximum - minimum.... And the answer should approximately equate to about 73 pc... which is then converted into meters... 7.6x10^6 pc is a very large number.... but you should get marks for the working out don't worry :D

No I'm pretty sure you were supposed to get the mean, I even looked it up on the internet and it said the same thing. http://www.atnf.csiro.au/outreach/education/senior/astrophysics/variable_cepheids.html. I don't remember the answer that I got but I remember it being a very large number, and if you look at the example in this website which had values close to the ones in the exam, you will find that the answer was also big.

For the cepheid question, here was exactly what I did:

I found M to be M = -3.61 (from the graph and from the given equation), and m to be m=15.55 (as I have explained earlier). Plug these into m-M=5log(d/10) yields d=6.81*10^4 pc

Guys they were talking about the VARIATION. 15.55 it a very large number for apparent magnitude... So you do maximum - minimum.... And the answer should approximately equate to about 73 pc... which is then converted into meters... 7.6x10^6 pc is a very large number.... but you should get marks for the working out don't worry :D

Not to say that you are wrong Abdelbari because im not even sure myself. But don’t you think that 73pc is a little bit too small for a Cepheid star? I mean if you google it, you’ll see that many Cepheid variables are at around 50000 pc from Earth. Also, 15.55 isn’t large at all because the bigger the magnitude, the dimmer the star. So I would say that 15.55 is a fairly reasonable value.

Oh that made me see it in another angle, but how can we also say that calculating the mean would also give an accurate number? If I got the apparent magnitude wrong would that be penalized more than once throughout a 5 mark question?

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Oh that made me see it in another angle, but how can we also say that calculating the mean would also give an accurate number? If I got the apparent magnitude wrong would that be penalized more than once throughout a 5 mark question?

I'm not really sure how the examiners mark the tests. I know that you also got M=-3.61 (as you said your final answer was 73pc), and if M is indeed -3.61, then i'm sure you can get at least 2/5. But again, let's hope that the examiners will be more sympathetic than that. I personally think that this question is quite tricky though

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Oh that made me see it in another angle, but how can we also say that calculating the mean would also give an accurate number? If I got the apparent magnitude wrong would that be penalized more than once throughout a 5 mark question?

I'm not really sure how the examiners mark the tests. I know that you also got M=-3.61 (as you said your final answer was 73pc), and if M is indeed -3.61, then i'm sure you can get at least 2/5. But again, let's hope that the examiners will be more sympathetic than that. I personally think that this question is quite tricky though

I think this is the 5 mark layout:

- Getting the period right

-Calculating the M value correct

-Calculating the m value correctly

-Substituting into equation to calculate (d)

-and converting the distance to meters..

I really hope that this is the layout, we will be so LUCKY!

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I did medical and EM waves and both were okay thank god! Paper 3 was a blessing after that godforsaken paper 2...

Lol Thank God for Paper 3

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But guys wait there is something I don't understand

Since it is a cepheid variable star, its is going through expansions and contractions. This means that its apperant brightness will change with the period. What I am trying to say is that if you take any point in the graph you should get the same answer because the (m) is changing with the period. I basically took one accurate point on the graph, plugged the T in the equation, found m and then plugged that value into the other equation to find the distance (because you already have the apperant magnitude on the graph and the absolute on after doing the calculations.

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But guys wait there is something I don't understand

Since it is a cepheid variable star, its is going through expansions and contractions. This means that its apperant brightness will change with the period. What I am trying to say is that if you take any point in the graph you should get the same answer because the (m) is changing with the period. I basically took one accurate point on the graph, plugged the T in the equation, found m and then plugged that value into the other equation to find the distance (because you already have the apperant magnitude on the graph and the absolute on after doing the calculations.

Well, if I'm not mistaken, the question said that the given equation can only be used to give the average absolute magnitude (M). So I guess that we need to find the average apparent magnitude (m) instead taking any point in the graph; but correct me if I am wrong!

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But guys wait there is something I don't understand

Since it is a cepheid variable star, its is going through expansions and contractions. This means that its apperant brightness will change with the period. What I am trying to say is that if you take any point in the graph you should get the same answer because the (m) is changing with the period. I basically took one accurate point on the graph, plugged the T in the equation, found m and then plugged that value into the other equation to find the distance (because you already have the apperant magnitude on the graph and the absolute on after doing the calculations.

Well, if I'm not mistaken, the question said that the given equation can only be used to give the average absolute magnitude (M). So I guess that we need to find the average apparent magnitude (m) instead taking any point in the graph; but correct me if I am wrong!

That is probably true, cant remember to be honest... I am just hoping I did the right thing or at least get two marks out of the 5

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