HSC Physics Marathon 2013-2015 Archive (5 Viewers)

Mr_Kap · Oct 8, 2015

re: HSC Physics Marathon Archive

leehuan said:
That question seems rather dodgy...

Cathode rays are essentially beams of electrons (negatively charged particles with q=-1.602*10^-19C, m=9.109*10^-31kg) that are directed from an unjoined cathode and anode in a very low pressure tube. Normally, cathode rays seek to travel in a straight line, and with the use of a fluorescent display screen this pathway can be clearly observed. We can manipulate charged particles through observing the fluorescent display screen when we include an electric field or magnetic field which deflects it, or by inserting a paddle wheel to observe how they carry momentum.

When are they going to change momentum to thermionic effect...
_______________________________________________
NEXT QUESTION:
a) Explain the forces involved when a spacecraft is being launched into orbit. (3)
b) Explain the forces involved when a spacecraft is in orbit AND during reentry. (5)
c) Discuss the necessity of having a spacecraft reenter at a specific range of angles. (4)

a) When a spacecraft is being launched, the combustion of the fuel from the bottom of the rocket provides the force necessary for the rocket to leave earth and overcome the force provided by gravity. This explosion of fuel gives fuel particles a high velocity, where the momentum of these particles flying out is equal and opposite to the momentum of the rocket by the conservation of momentum. Because P = mv, and 'P' is constant, as the mass of the fuel decreased (from being combusted), the velocity of the rocket is hence increased. The force provided from the combustion is constant and is called thrust. Because of this, and F = ma, as mass decreases the acceleration increases, meaning that the velocity of the rocket is increasing at an increasing rate. To be launched into orbit, the rocket must be launched from the equator from a west to easterly direction, where it experiences a force from earth's rotational orbit.

No clue if what I wrote even answers the question, lol.

Mr_Kap · Oct 8, 2015

re: HSC Physics Marathon Archive

BUMP someone answer this question / or tell me what is wrong with my crap answer.

atargainz · Oct 8, 2015

re: HSC Physics Marathon Archive

leehuan said:
That question seems rather dodgy...

Cathode rays are essentially beams of electrons (negatively charged particles with q=-1.602*10^-19C, m=9.109*10^-31kg) that are directed from an unjoined cathode and anode in a very low pressure tube. Normally, cathode rays seek to travel in a straight line, and with the use of a fluorescent display screen this pathway can be clearly observed. We can manipulate charged particles through observing the fluorescent display screen when we include an electric field or magnetic field which deflects it, or by inserting a paddle wheel to observe how they carry momentum.

When are they going to change momentum to thermionic effect...
_______________________________________________
NEXT QUESTION:
a) Explain the forces involved when a spacecraft is being launched into orbit. (3)
b) Explain the forces involved when a spacecraft is in orbit AND during reentry. (5)
c) Discuss the necessity of having a spacecraft reenter at a specific range of angles. (4)

c) Upon re-entry a spacecraft's ideal angle for optimal re-entry is between 5.2-7.2 degrees. If the angle of re-entry is >7.2 ie too steep, the spacecraft will accumulate increased speeds and experience extreme heat build up which may eventually cause it to burn up and vaporise. More so this heightened velocity will result in the astronaut experiencing dangerous levels of g-forces, making the re-entry unsafe. If the angle is <5.2 ie too shallow, the spacecraft will bounce off the atmosphere of the earth, causing it to launch back into space. Two possibilities arise with this, if the angle is extremely shallow the spacecraft will not lose enough speed, causing it to fly off into space forever. In the second scenario, the spacecraft will enter an elliptical rather than hyperbolic trajectory and rise back into space. At this point, the spacecraft will be de-orbiting as it keeps on hitting the atmosphere over and over again, losing speed each time.

feedback pls, also can't think of enough points to write 5 marks for part b soz

Kaido · Oct 8, 2015

re: HSC Physics Marathon Archive

Mr_Kap said:
a) When a spacecraft is being launched, the combustion of the fuel from the bottom of the rocket provides the force necessary for the rocket to leave earth and overcome the force provided by gravity. This explosion of fuel gives fuel particles a high velocity, where the momentum of these particles flying out is equal and opposite to the momentum of the rocket by the conservation of momentum. Because P = mv, and 'P' is constant, as the mass of the fuel decreased (from being combusted), the velocity of the rocket is hence increased. The force provided from the combustion is constant and is called thrust. Because of this, and F = ma, as mass decreases the acceleration increases, meaning that the velocity of the rocket is increasing at an increasing rate. To be launched into orbit, the rocket must be launched from the equator from a west to easterly direction, where it experiences a force from earth's rotational orbit.

No clue if what I wrote even answers the question, lol.

It seems like you're shooting ten arrows in hopes of hitting the bullseye, when in reality you only have to shoot three. Your answer is not focused on the question at hand, consider revising it and narrowing it down (since you most definitely will not be given that much space to write a 3marker)

Kaido · Oct 8, 2015

re: HSC Physics Marathon Archive

atargainz said:
c) Upon re-entry a spacecraft's ideal angle for optimal re-entry is between 5.2-7.2 degrees. If the angle of re-entry is >7.2 ie too steep, the spacecraft will accumulate increased speeds and experience extreme heat build up which may eventually cause it to burn up and vaporise. More so this heightened velocity will result in the astronaut experiencing dangerous levels of g-forces, making the re-entry unsafe. If the angle is <5.2 ie too shallow, the spacecraft will bounce off the atmosphere of the earth, causing it to launch back into space. Two possibilities arise with this, if the angle is extremely shallow the spacecraft will not lose enough speed, causing it to fly off into space forever. In the second scenario, the spacecraft will enter an elliptical rather than hyperbolic trajectory and rise back into space. At this point, the spacecraft will be de-orbiting as it keeps on hitting the atmosphere over and over again, losing speed each time.

feedback pls, also can't think of enough points to write 5 marks for part b soz

Yup, strong response for a 4 marker. Try to compress your answer as you won't have that much space to write in the exams
(Perhaps mention what happens if the correct angle of re-entry is achieved)

InteGrand · Oct 8, 2015

re: HSC Physics Marathon Archive

$Incidentally, $F=ma=m\frac{\mathrm{d}v}{\mathrm{d}t}$ is actually wrong for an object that is changing mass. The proper equation is $F=m\frac{\mathrm{d}v}{\mathrm{d}t}+v\frac{\mathrm{d}m}{\mathrm{d}t}$. As you can see, in the case where mass is constant, $\frac{\mathrm{d}m}{\mathrm{d}t}=0$ and we have our standard $F=ma$. This formula for general mass which may be non-constant comes about from the definition of force in physics: $\vec{F}=\frac{\mathrm{d} \vec{p}}{\mathrm{d}t}$, where $\vec{p}$ is the momentum vector. Since $\vec{p}=m\vec{v}$, it follows that $\vec{F}=\frac{\mathrm{d}(m \vec{v} )}{\mathrm{d}t}=m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}+\vec{v}\frac{\mathrm{d}m}{\mathrm{d}t}$, by the product rule.$

Kaido · Oct 8, 2015

re: HSC Physics Marathon Archive

InteGrand said:
$Incidentally, $F=ma=m\frac{\mathrm{d}v}{\mathrm{d}t}$ is actually wrong for an object that is changing mass. The proper equation is $F=m\frac{\mathrm{d}v}{\mathrm{d}t}+v\frac{\mathrm{d}m}{\mathrm{d}t}$. As you can see, in the case where mass is constant, $\frac{\mathrm{d}m}{\mathrm{d}t}=0$ and we have our standard $F=ma$. This formula for general mass which may be non-constant comes about from the definition of force in physics: $\vec{F}=\frac{\mathrm{d} \vec{p}}{\mathrm{d}t}$, where $\vec{p}$ is the momentum vector. Since $\vec{p}=m\vec{v}$, it follows that $\vec{F}=\frac{\mathrm{d}(m \vec{v} )}{\mathrm{d}t}=m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}+\vec{v}\frac{\mathrm{d}m}{\mathrm{d}t}$, by the chain rule.$

Awesome, I'll write it in the exams

iforgotmyname · Oct 8, 2015

re: HSC Physics Marathon Archive

porcupinetree said:
I talked about this in an exam once for Physics and my teacher said that it was mainly unnecessary, especially when the question is only worth 3 marks. However, you do raise a good point, and personally I reckon it'd be useful to mention the 2dsintheta stuff in an exam, especially if it's worth more than 3 marks

Ah, i did it when the question was worth 7 so lol

iforgotmyname · Oct 8, 2015

re: HSC Physics Marathon Archive

InteGrand said:
$Incidentally, $F=ma=m\frac{\mathrm{d}v}{\mathrm{d}t}$ is actually wrong for an object that is changing mass. The proper equation is $F=m\frac{\mathrm{d}v}{\mathrm{d}t}+v\frac{\mathrm{d}m}{\mathrm{d}t}$. As you can see, in the case where mass is constant, $\frac{\mathrm{d}m}{\mathrm{d}t}=0$ and we have our standard $F=ma$. This formula for general mass which may be non-constant comes about from the definition of force in physics: $\vec{F}=\frac{\mathrm{d} \vec{p}}{\mathrm{d}t}$, where $\vec{p}$ is the momentum vector. Since $\vec{p}=m\vec{v}$, it follows that $\vec{F}=\frac{\mathrm{d}(m \vec{v} )}{\mathrm{d}t}=m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}+\vec{v}\frac{\mathrm{d}m}{\mathrm{d}t}$, by the chain rule.$

This guy's Physic is Mint

InteGrand · Oct 8, 2015

re: HSC Physics Marathon Archive

Kaido said:
Awesome, I'll write it in the exams

I was tempted to, but resisted the temptation. (Would result in loss of marks most likely haha.)

Kaido · Oct 8, 2015

re: HSC Physics Marathon Archive

InteGrand said:
I was tempted to, but resisted the temptation. (Would result in loss of marks most likely haha.)

Giving away some info as to when you did your hsc

InteGrand · Oct 8, 2015

re: HSC Physics Marathon Archive

Kaido said:
Giving away some info as to when you did your hsc

Was referring to in my trials.

Mr_Kap · Oct 8, 2015

re: HSC Physics Marathon Archive

Kaido said:
It seems like you're shooting ten arrows in hopes of hitting the bullseye, when in reality you only have to shoot three. Your answer is not focused on the question at hand, consider revising it and narrowing it down (since you most definitely will not be given that much space to write a 3marker)

can someone tell me what 'three arrows' I need to shoot. basically, what is unecessary and what is necessary.

Drsoccerball · Oct 8, 2015

re: HSC Physics Marathon Archive

Mr_Kap said:
can someone tell me what 'three arrows' I need to shoot. basically, what is unecessary and what is necessary.

Theres nothing wrong with your answer except what integrand said.

leehuan · Oct 9, 2015

re: HSC Physics Marathon Archive

Mr_Kap said:
a) When a spacecraft is being launched, the combustion of the fuel from the bottom of the rocket provides the force necessary for the rocket to leave earth and overcome the force provided by gravity. This explosion of fuel gives fuel particles a high velocity, where the momentum of these particles flying out is equal and opposite to the momentum of the rocket by the conservation of momentum. Because P = mv, and 'P' is constant, as the mass of the fuel decreased (from being combusted), the velocity of the rocket is hence increased. The force provided from the combustion is constant and is called thrust. Because of this, and F = ma, as mass decreases the acceleration increases, meaning that the velocity of the rocket is increasing at an increasing rate. To be launched into orbit, the rocket must be launched from the equator from a west to easterly direction, where it experiences a force from earth's rotational orbit.

No clue if what I wrote even answers the question, lol.

The forces were thrust and gravity. Any kind of analysis of this was enough and a brief relation to acceleration. Launching the rocket in an easterly direction was unnecessary.
But yeah, you can't get 4/3, you can only get 3/3.

leehuan · Oct 9, 2015

re: HSC Physics Marathon Archive

atargainz said:
c) Upon re-entry a spacecraft's ideal angle for optimal re-entry is between 5.2-7.2 degrees. If the angle of re-entry is >7.2 ie too steep, the spacecraft will accumulate increased speeds and experience extreme heat build up which may eventually cause it to burn up and vaporise. More so this heightened velocity will result in the astronaut experiencing dangerous levels of g-forces, making the re-entry unsafe. If the angle is <5.2 ie too shallow, the spacecraft will bounce off the atmosphere of the earth, causing it to launch back into space. Two possibilities arise with this, if the angle is extremely shallow the spacecraft will not lose enough speed, causing it to fly off into space forever. In the second scenario, the spacecraft will enter an elliptical rather than hyperbolic trajectory and rise back into space. At this point, the spacecraft will be de-orbiting as it keeps on hitting the atmosphere over and over again, losing speed each time.

feedback pls, also can't think of enough points to write 5 marks for part b soz

It was out of 4. So that is fine.

Though, if my memory serves me right, the accumulation of an increased velocity results in the resistive force from friction being too powerful, and it is that which causes excess heat build-up. I reckon for a 4 marker analysing both shallow and deep you have enough, but maybe also a brief reference as to how inbuilt heating systems will not be able to withstand it.

I must admit however, although your second scenario does make sense with being bounced into elliptical orbit, the idea of rebouncing I haven't heard about.

leehuan · Oct 9, 2015

re: HSC Physics Marathon Archive

iforgotmyname said:
Ah, i did it when the question was worth 7 so lol

7 marks on Braggs? That is harsh.

InteGrand said:
$Incidentally, $F=ma=m\frac{\mathrm{d}v}{\mathrm{d}t}$ is actually wrong for an object that is changing mass. The proper equation is $F=m\frac{\mathrm{d}v}{\mathrm{d}t}+v\frac{\mathrm{d}m}{\mathrm{d}t}$. As you can see, in the case where mass is constant, $\frac{\mathrm{d}m}{\mathrm{d}t}=0$ and we have our standard $F=ma$. This formula for general mass which may be non-constant comes about from the definition of force in physics: $\vec{F}=\frac{\mathrm{d} \vec{p}}{\mathrm{d}t}$, where $\vec{p}$ is the momentum vector. Since $\vec{p}=m\vec{v}$, it follows that $\vec{F}=\frac{\mathrm{d}(m \vec{v} )}{\mathrm{d}t}=m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}+\vec{v}\frac{\mathrm{d}m}{\mathrm{d}t}$, by the chain rule.$

Product rule?

But yeah, I'll admit because I love my maths this is beautiful. Too bad the NSW HSC course is non-calculus though and overly simplified.

InteGrand · Oct 9, 2015

re: HSC Physics Marathon Archive

Yep, meant product rule.

mrpotatoed · Oct 9, 2015

re: HSC Physics Marathon Archive

leehuan said:
That question seems rather dodgy...

It's a syllabus dot point, lol

also you forgot mention the high potential difference

leehuan · Oct 9, 2015

re: HSC Physics Marathon Archive

I did indeed. But for two marks would the full functionality be necessary? If I think about it even mentioning the charge and mass would've been unnecessary in this case.

HSC Physics Marathon 2013-2015 Archive (5 Viewers)

Well-Known Member

Well-Known Member

Active Member

be.

be.

Well-Known Member

be.

Metallic Oxide

Metallic Oxide

Well-Known Member

be.

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 5)