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HSC Physics Marathon 2013-2015 Archive (2 Viewers)

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Zlatman

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re: HSC Physics Marathon Archive

The simplest way I can explain why the speed relative to the earth doesn't change is with a head-on slingshot effect, where the spaceship initially moves in the opposite direction to the earth, and ends up moving in the same direction as the earth (relative to the sun).

F = V + 2U (where F is the final velocity of the spaceship, V is the initial velocity of the spaceship and U is the velocity of the earth; all relative to the sun)
NOTE: this is derived from both the conservation of kinetic energy and conservation of momentum.


Initially, the spaceship moves at a velocity V, while the earth is travelling with a velocity U in the opposite direction (both relative to the sun). Relative to the Earth, the spaceship is moving at a speed of V + U.
[V - (-U) = V + U] (imagine or use vector diagram - considering the initial direction of the spaceship as the positive direction)

After the slingshot effect, the spaceship is travelling with a velocity V + 2U in the same direction as the Earth, which is still moving at a speed of U (since the change in velocity is pretty much negligible). However, relative to the earth, the spaceship will still be travelling with a speed of V + U.
[-(V + 2U) - (- U) = - V - U = -(V + U)] (same speed, opposite direction)

Thus, the spaceship's velocity relative to the sun increases by 2U, but relative to the earth, its speed does not change.



EDIT: (This is how the Jacaranda textbook explains it)

Jupiter has an orbital velocity of about 13000 m/s relative to the Sun, and let us assume that the spacecraft has a velocity of 15000 m/s relative to the Sun.

In order to visualise how this effect occurs, it is helpful to consider the relative velocities. If we consider the velocity of the spacecraft relative to Jupiter then the situation is as if Jupiter were standing still. When a small object collides elastically with a very large, massive object, the small object will rebound without loss of speed — consider a ball bouncing against a wall. Similarly, our spacecraft approaches Jupiter at (13000 + 15000) or 28000 m/s relative to it, swings around behind it and then slingshots back out in front at 28000 m/s relative to Jupiter. Look now at what has happened to the velocity of the spacecraft relative to the Sun. If Jupiter’s velocity is 13000 m/s and the spacecraft is moving ahead 28000 m/s faster than it, the spacecraft is now travelling at 41000 m/s relative to the Sun.
 
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re: HSC Physics Marathon Archive

The simplest way I can explain why the speed relative to the earth doesn't change is with a head-on slingshot effect, where the spaceship initially moves in the opposite direction to the earth, and ends up moving in the same direction as the earth (relative to the sun).

F = V + 2U (where F is the final velocity of the spaceship, V is the initial velocity of the spaceship and U is the velocity of the earth; all relative to the sun)
NOTE: this is derived from both the conservation of kinetic energy and conservation of momentum.


Initially, the spaceship moves at a velocity V, while the earth is travelling with a velocity U in the opposite direction (both relative to the sun). Relative to the Earth, the spaceship is moving at a speed of V + U.
[V - (-U) = V + U] (imagine or use vector diagram - considering the initial direction of the spaceship as the positive direction)

After the slingshot effect, the spaceship is travelling with a velocity V + 2U in the same direction as the Earth, which is still moving at a speed of U (since the change in velocity is pretty much negligible). However, relative to the earth, the spaceship will still be travelling with a speed of V + U.
[-(V + 2U) - (- U) = - V - U = -(V + U)] (same speed, opposite direction)

Thus, the spaceship's velocity relative to the sun increases by 2U, but relative to the earth, its speed does not change.



EDIT: (This is how the Jacaranda textbook explains it)

Jupiter has an orbital velocity of about 13000 m/s relative to the Sun, and let us assume that the spacecraft has a velocity of 15000 m/s relative to the Sun.

In order to visualise how this effect occurs, it is helpful to consider the relative velocities. If we consider the velocity of the spacecraft relative to Jupiter then the situation is as if Jupiter were standing still. When a small object collides elastically with a very large, massive object, the small object will rebound without loss of speed — consider a ball bouncing against a wall. Similarly, our spacecraft approaches Jupiter at (13000 + 15000) or 28000 m/s relative to it, swings around behind it and then slingshots back out in front at 28000 m/s relative to Jupiter. Look now at what has happened to the velocity of the spacecraft relative to the Sun. If Jupiter’s velocity is 13000 m/s and the spacecraft is moving ahead 28000 m/s faster than it, the spacecraft is now travelling at 41000 m/s relative to the Sun.
Ahhh thank you, so this is assuming that it will always gain U - 13000 m/s in the example - worth of speed?
 

Zlatman

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re: HSC Physics Marathon Archive

Ahhh thank you, so this is assuming that it will always gain U - 13000 m/s in the example - worth of speed?
No worries!

I think it's assuming it gains 2U, since relative to the Sun it starts at 15000m/s and ends at 41000m/s in the example, but I'm not sure how it works for cases when it's not front-on. I'd imagine it is still the same concept, but the calculations won't be as straight-forward.
 

Mr_Kap

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Can i report CSSA for copyright??


In their physics paper this year they copied a whole question (part a b and c) from the 2001 HSC paper. Or are people allowed to copy HSC questions and pass them as your own and make profit???
 

TQuadded

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Can i report CSSA for copyright??


In their physics paper this year they copied a whole question (part a b and c) from the 2001 HSC paper. Or are people allowed to copy HSC questions and pass them as your own and make profit???
Do you mean last year?
 

astroman

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Q: Discuss how energy savings can be made in the field of transportation and electricity generation/transmission through the use of superconductors.
 
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Q: Discuss how energy savings can be made in the field of transportation and electricity generation/transmission through the use of superconductors.
* Transportation can use superconducting coils in applications like Maglev trains. This is far more efficient that current usage of electromagnets in places like Shanghai and Japan (No resistance and heat loss in wiring). This is also because of the Meissner effect which allows for effective levitation due to the exclusion of external magnetic fields (bottom of the train opposing superconducting track of vice versa). This means that energy savings will occur due to the absence of loss through heat and friction.

* Electrical generation will be more efficient and require smaller generators -not really sure about this part
* Electrical transmission will result in less power loss (from heat due to resistance [P=VI]), as superconductors below the critical temperature have no resistance, saving energy in long distance power transmissions.
 

atargainz

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re: HSC Physics Marathon Archive

* Transportation can use superconducting coils in applications like Maglev trains. This is far more efficient that current usage of electromagnets in places like Shanghai and Japan (No resistance and heat loss in wiring). This is also because of the Meissner effect which allows for effective levitation due to the exclusion of external magnetic fields (bottom of the train opposing superconducting track of vice versa). This means that energy savings will occur due to the absence of loss through heat and friction.

* Electrical generation will be more efficient and require smaller generators -not really sure about this part
* Electrical transmission will result in less power loss (from heat due to resistance [P=VI]), as superconductors below the critical temperature have no resistance, saving energy in long distance power transmissions.
Since it's a discuss question, the only disadvantage I can think of at the moment is the technological and infrastructural costs required to cool and maintain the superconductors below critical temp.
 

atargainz

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re: HSC Physics Marathon Archive

The only thing affecting it is g (always constant hence the straight line), whereby it will accelerate uniformly downwards until it hits the ground and acceleration abruptly stops. B
 

atargainz

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A cathode ray oscilloscope is a modified version of the cathode ray tube. Justify this statement, with reference to the function of a cathode ray oscilloscope and the features of a cathode ray tube. [4 marks]
 

malcolm21

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^ Why isnt it a horizontal line but negative for that mc?
 

InteGrand

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^ Why isnt it a horizontal line but negative for that mc?
The choice of positive direction is arbitrary.

(And out of the given options, only (B) has a graph with constant non-zero acceleration, so (B) is the answer.)
 
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A magnet is hovering above a superconducting disk. Explain why the magnet is able to hover above the superconductor. (3)

Is Lenz's law relevant here. My teacher and my friends (tutoring) have always told me that the Meissner effect is a completely separate thing to Lenz's. Is this so, and how would you answer?
 

astroman

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u have to use meisner affect and explain the balance between the weight force and the force of magnetic repulsion resulting from the interaction
of the fields of the induced current and hovering magnet.
 
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