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Author Topic: VCE Physics Question Thread!  (Read 603286 times)  Share 

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Rod

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Re: VCE Physics Question Thread!
« Reply #720 on: November 08, 2014, 12:51:38 pm »
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Hey guys, need help with question 22.c

http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/2013physics-cpr-w.pdf

Here is my working out:

Let / = wavelength (because I don't know how to do it on this LOL)

So since it is the second brightest band, path different must be 1/ because central maxima (the first brightest band) is 0/.

Therefore 1/ = PD

1/ = 1.4X10^3X10^-9

/ = 1.4X10^3 nm

Therefore, since we have the wavelength, we can find the path difference of the first dark band

(n-1/2)/ = PD

1-1/2/ = PD

1/2 x 1.4x10^3x10^-9 = PD

PD = 7x10-7 m


-----------------------------------------

Okay so I got 2 out of the possible 3 marks for this. Where have I gone wrong? According to the assesors report my path difference was wrong for the bright band, it should have been 2/. Why 2? THanks
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Rod

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Re: VCE Physics Question Thread!
« Reply #721 on: November 08, 2014, 12:58:28 pm »
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And just another quick q sorry;

23.B

How does momentum influence the wavelength of an electon? ANd is 'fringe spacing' just referring to the diffraction pattern? Thanks
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Zealous

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Re: VCE Physics Question Thread!
« Reply #722 on: November 08, 2014, 01:10:21 pm »
+1
Hey guys, need help with question 22.c

http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/2013physics-cpr-w.pdf

Here is my working out:
....

-----------------------------------------

Okay so I got 2 out of the possible 3 marks for this. Where have I gone wrong? According to the assesors report my path difference was wrong for the bright band, it should have been 2/. Why 2? THanks

The path difference is 2 whole wavelengths. The path difference to the first bright band is 1 whole wavelength and the path difference to the second bright band is 2 whole wavelengths. Recall the formula , we are looking at the 2nd bright band so n=2, and we get: .

And just another quick q sorry;
23.B
How does momentum influence the wavelength of an electon? ANd is 'fringe spacing' just referring to the diffraction pattern? Thanks

The formula for de Broglie matter wavelength is . So the wavelength of an electron (matter) is inversely proportional to the momentum of the electron. Basically, the faster an electron is travelling (greater momentum), the smaller it's de Broglie wavelength and conversely, the slower an electron (and lower momentum), the larger it's de Broglie wavelength.

You're right with "fringe spacing" - the bands in the circular diffraction pattern (or rings) are refered to as fringes, and their spacing depends on the wavelength of electrons/light. It's a bit like taking Young's Double Slit interference pattern and converting it into a circle so you've got bright and dark sections from the inside to outside.
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Rod

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Re: VCE Physics Question Thread!
« Reply #723 on: November 08, 2014, 01:58:38 pm »
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The path difference is 2 whole wavelengths. The path difference to the first bright band is 1 whole wavelength and the path difference to the second bright band is 2 whole wavelengths. Recall the formula , we are looking at the 2nd bright band so n=2, and we get: .

The formula for de Broglie matter wavelength is . So the wavelength of an electron (matter) is inversely proportional to the momentum of the electron. Basically, the faster an electron is travelling (greater momentum), the smaller it's de Broglie wavelength and conversely, the slower an electron (and lower momentum), the larger it's de Broglie wavelength.

You're right with "fringe spacing" - the bands in the circular diffraction pattern (or rings) are refered to as fringes, and their spacing depends on the wavelength of electrons/light. It's a bit like taking Young's Double Slit interference pattern and converting it into a circle so you've got bright and dark sections from the inside to outside.
Oh I see! So the central maxima does not count as a bright band? Thanks!

And I get 23.b now. Thanks! Awesome explanations ....
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Re: VCE Physics Question Thread!
« Reply #724 on: November 08, 2014, 02:13:54 pm »
+1
Oh I see! So the central maxima does not count as a bright band? Thanks!
Well, the central maximum is still a bright band, but you don't use it in the PD formula because there isn't any path difference. The central band is exactly half way between the two slits, so waves from both slits would've travelled the exact same distance, there's no difference in the paths they've taken so they'll interfere constructively.

It becomes a central maximum (brighter than the other bands) because the intensity of light is inversely proportional to the square of the distance (), so the brightest band would be in the center (where the waves have travelled the least) and then the pattern will get dimmer as you go outwards (and the light travels further).
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Rod

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Re: VCE Physics Question Thread!
« Reply #725 on: November 08, 2014, 02:26:21 pm »
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Well, the central maximum is still a bright band, but you don't use it in the PD formula because there isn't any path difference. The central band is exactly half way between the two slits, so waves from both slits would've travelled the exact same distance, there's no difference in the paths they've taken so they'll interfere constructively.

It becomes a central maximum (brighter than the other bands) because the intensity of light is inversely proportional to the square of the distance (), so the brightest band would be in the center (where the waves have travelled the least) and then the pattern will get dimmer as you go outwards (and the light travels further).
Thanks again

Sorry for bothering but would it be okay if you please explaind two more

2b and 6c 

http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf

Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?
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Re: VCE Physics Question Thread!
« Reply #726 on: November 08, 2014, 02:45:42 pm »
+1
2b and 6c 

http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf

Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?
2b:
Look at the two components of the system separately.

So for m1, we've got the gravitational force working in one direction and tension in the string working in the other:



For m2, the only force acting on m2 is the tension in the string which is pulling it to the left:



The tension in a string is always the same, so we can sub in m2a=T into the first equation to get:

   Now we can sub a back into the second equation:



So this is the simultaneous equation method, where you look at the forces acting on each block - alternatively you can look at the whole entire system as the whole, imagine both blocks are attached together and there's only a force of gravity acting on it, find the acceleration then look at individual blocks (the method in the examiner report).

6c:
So they've assumed that when the mass is at position Q, that this will be a point of 0 spring potential which is incorrect. Because the spring has already been stretched by 0.5m, it is incorrect to say there's no energy there. So to prove that it's not 0, do a calculation and show that - so there is in fact 1.25J stored in the spring, not 0. Furthermore you could show that to properly calculate the spring potential, you'd need to do .
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Yacoubb

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Re: VCE Physics Question Thread!
« Reply #727 on: November 08, 2014, 02:49:17 pm »
+2
Thanks again

Sorry for bothering but would it be okay if you please explaind two more

2b and 6c 

http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf

Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?

2b. To calculate tension, we firstly need to calculate the acceleration of m2. So:

W = mg = 2*10 = 20N.
Fnet = ma.
20 = (6+2)a
a = 2.5 m/s^2

T = ma = 2.5 * 6 = 15N.

______________________________________________________________________________________________

6c. The spring potential is not 0 because the spring is being stretched 1m if you can see in the second diagram (so it has been stretched by 0.5m). Thus the strain potential energy is not equal to 0, and this is the mistake made in the collated data.


Edit: just realised Zealous beat me to the punch :)

Rishi97

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Re: VCE Physics Question Thread!
« Reply #728 on: November 08, 2014, 03:59:10 pm »
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What is an oscilloscope?
Thanks  :)
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myanacondadont

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Re: VCE Physics Question Thread!
« Reply #729 on: November 08, 2014, 05:48:52 pm »
+1
What is an oscilloscope?
Thanks  :)

When used in VCE physics it refers to a voltmeter/ammeter thingo in the sense it reads the voltage/current. But it produces a graph of the voltage/current aswell on a screen. (That's all I know). Someone can probably offer a better explanation.

Is this correct to say about transformers?: The current in the primary coil creates a magnetic field which is strengthened by the iron core, therefore threading the secondary coil and inducing a current proportional to the amount of turns in the coil.

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Re: VCE Physics Question Thread!
« Reply #730 on: November 08, 2014, 05:52:55 pm »
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Is this correct to say about transformers?: The current in the primary coil creates a magnetic field which is strengthened by the iron core, therefore threading the secondary coil and inducing a current proportional to the amount of turns in the coil.
I don't even think you need to mention an iron core unless they've given it in the question. I'd probably say "The AC current through the primary coil creates a rapidly changing flux through the secondary coil. This results in an EMF induced in the secondary coil, as dictated by Faraday's Law. The magnitude of the EMF induced is directly proportional to the amount of turns in the secondary coil." or something like that. But there's nothing wrong with your definition, just saying it in different ways.
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myanacondadont

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Re: VCE Physics Question Thread!
« Reply #731 on: November 08, 2014, 05:58:16 pm »
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I don't even think you need to mention an iron core unless they've given it in the question. I'd probably say "The AC current through the primary coil creates a rapidly changing flux through the secondary coil. This results in an EMF induced in the secondary coil, as dictated by Faraday's Law. The magnitude of the EMF induced is directly proportional to the amount of turns in the secondary coil." or something like that. But there's nothing wrong with your definition, just saying it in different ways.

Thankyou! Another question too :D

When a coil is placed in an external magnetic field and connected with slip rings, does the coil oscillate 90 degrees back and forth? Give or take an extra bit because of the momentum of the coil.

And when it's connected with slip rings to a DC power source, does it just align itself with the external field?

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Re: VCE Physics Question Thread!
« Reply #732 on: November 08, 2014, 06:04:19 pm »
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When a coil is placed in an external magnetic field and connected with slip rings, does the coil oscillate 90 degrees back and forth? Give or take an extra bit because of the momentum of the coil.

I'm not sure if you're talking about generators or motors. If you're talking about AC power generation, the coil will still rotate in the same direction (not going back and forth), but the current induced will create a sinosidual curve (sine curve) because the change in flux is positive and negative at different parts in the rotation. The momentum of the coil in rotation will completely depend on what you've got rotating the coil to generate the power.

And when it's connected with slip rings to a DC power source, does it just align itself with the external field?
Connecting a coil to a DC power source through slip rings is just like connecting it directly to the circuit. Once current goes through it, it will rotate to a position of maximum flux and get stuck at that position.
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Re: VCE Physics Question Thread!
« Reply #733 on: November 09, 2014, 11:53:23 am »
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I'm not sure if you're talking about generators or motors. If you're talking about AC power generation, the coil will still rotate in the same direction (not going back and forth), but the current induced will create a sinosidual curve (sine curve) because the change in flux is positive and negative at different parts in the rotation. The momentum of the coil in rotation will completely depend on what you've got rotating the coil to generate the power.
Connecting a coil to a DC power source through slip rings is just like connecting it directly to the circuit. Once current goes through it, it will rotate to a position of maximum flux and get stuck at that position.

Thankyou - I think I have some understanding wrong. What's the scenario when a coil will oscillate 90 degrees back and forth?

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Re: VCE Physics Question Thread!
« Reply #734 on: November 09, 2014, 02:11:57 pm »
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Thanks Zeal and yacoub :)

http://www.vcaa.vic.edu.au/Documents/exams/physics/2012/physics_assessrep_12.pdf

Having trouble understanding 2d.

Agree:

Old -> 2\
New -> 1\

Don't understand:

How can we conclude that the path differences are the same?

Thanks
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Currently: Physiotherapist working at a musculoskeletal clinic. Back pain, sore neck, headaches or any other pain limiting your study? Give me a PM (although please do see your personal health professional first!)

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