# Physics Exam 2019 Questions Thread (1 Viewer)

#### Arrowshaft

##### Well-Known Member
Yeah, I’d also describe the job of linear accelerators such as the ones in SLAC, the discontinuation of cyclotrons and the favoration of synchrotrons. Also, the role of multi component detectors and calorimeters in determining the particles that are released. Otherwise, seems like a solid response!
idk half of what you said here .... shit
Haha, dw I’m sure as long as you know the basic outline of them you’re all good to go. I’m just really into this stuff so I know a lot about it. They also most likely won’t ask this question as a 9 marker since they’ve already asked it in the sample q’s

#### akkjen

##### Active Member
Haha, dw I’m sure as long as you know the basic outline of them you’re all good to go. I’m just really into this stuff so I know a lot about it. They also most likely won’t ask this question as a 9 marker since they’ve already asked it in the sample q’s
i was panicking .... lol

#### akkjen

##### Active Member
Outline the difference between the Proton-Proton chain and CNO cycle for the process of fusion in stars.

#### Arrowshaft

##### Well-Known Member
Outline the difference between the Proton-Proton chain and CNO cycle for the process of fusion in stars.
Proton proton chain fuses 4 hydrogen to produce the net products after a series of transmutations of a helium nuclei, 2 electron neutrinos 2 positrons and 2 gamma rays iirc. The proton proton chain uses the principle of nuclear fusion and mass defect to produce energy which fuels the Star. The proton proton chain occurs in smaller and less heavier stars. The CNO cycle is a series of nuclear fusion reactions catalysed by the transmutation of carbon into the intermediates of nitrogen and oxygen and their isotopes in order to fuse 4 hydrogen into a helium nucleus, 2 electron neutrinos, 2 positrons and 3 gamma rays, releasing more energy than p and p. The CNO cycle occurs in heavier and hotter stars which can fuse elements passed carbon, stars like the Sun are not capable of relying on the CNO cycle as a primary production of energy hence only accounting for 2% of energy production

#### Arrowshaft

##### Well-Known Member
Sorry if it’s clunky, didn’t re read it. Also I might have misquoted some stuff

#### TheOnePheeph

##### Active Member
I really hate long answer questions in physics, seem more about memorization than actual understanding, but here is one anyway, a 9 marker from my trial:

"The Orbits of Satellites such as GPS must be known very precisely. With reference to the total energy of a satellite and Kepler's Laws, discuss how satellites' orbits can be predicted and maintained"

#### Arrowshaft

##### Well-Known Member
Ooh that’s a good question, I'm going to prepare a somewhat actual answer for this.

The Orbits of Satellites such as GPS must be known very precisely. With reference to the total energy of a satellite and Kepler's Laws, discuss how satellites' orbits can be predicted and maintained.

The orbits of satellites are dependent upon the empirical discoveries of Kepler's Laws of planetary motion and transformations of its total energies, thus by examining these factors, satellite orbits can be predicted and maintained.

For an orbiting satellite, the gravitational potential energy of a satellite is given by
$\bg_white U=-\dfrac{GMm}{r}$
To determine the total energy, the kinetic energy must also be determined. For a satellite in stable orbit, the centripetal acceleration of a circular orbit is provided by the gravitational force exerted by the planet such that,
$\bg_white F_c=F_g$

$\bg_white \dfrac{mv^2}{r}=\dfrac{GMm}{r^2}$
$\bg_white v^2=\dfrac{GM}{r}$
$\bg_white v_{orb}=\sqrt{\dfrac{GM}{r}}$
Since the kinetic energy of the satellite is
$\bg_white K=\dfrac{1}{2}mv_{orb}^2=\dfrac{GMm}{2r}$
Thus the total energy is
$\bg_white E=U+K=-\dfrac{GMm}{r}-\dfrac{GMm}{2r}=-\dfrac{GMm}{2r}$
Hence, for a satellite undergoing uniform circular motion around the Earth, the total energy is constant, so is its kinetic and potential energy.

GPS satellites are geostationary above the equator at an altitude of 36 000 km. Geostationary satellites are launched above the equator with the tangential horizontal component of Earth's rotational motion coupled with the radial gravitational attraction serving to keep the satellite in orbit above the same point on the equator and no work required to be done to keep the satellite in orbital motion (as the velocity vector is perpendicular to the force vector). The resultant orbital motion is circular, as described by Kepler's First Law (eccentricity of o). Since the change in work is zero, this means there is no change in total energy as
$\bg_white W=\Delta E=0$

By using Kepler's Second Law of planetary motion, it is observable that a satellite undergoing orbital motion will sweep equal areas across equal times, that is; its areal speed will be constant. Hence, when the satellite gains kinetic energy, it moves faster, by abiding Kepler's Second Law it moves closer to the planet to sweep across an equal area. Since the total energy is constant, this means that the potential energy of the satellite decreases as it moves faster along its orbit. This effect is heightened as the eccentricity of the satellite's orbit increases, widening the ellipse and allowing for extreme positions of high and low kinetic and potential energies. This allows scientists to monitor energies and hence positions and velocities depending on the position of the satellite in its orbit. However, for GPS satellites which are geostationary, due to the circular nature of the orbit, it travels at the same angular and tangential velocities; thus allowing easier monitoring due to its constancy. Since geostationary satellites also stay above the same point over the equator during orbit, it shares the same rotational velocity and period of the Earth's orbit, thus monitoring any variations in period or position relative to the Earth's surface can shed information on any possible orbital turmoil such as collisions with space debris.

By applying Kepler's Third Law, which states that the cube of the radius is proportional to the square of the period, this means that the radius of the motion can be calculated by analysing the satellite's period, thus being able to determine any fluctuations in the satellite's orbit which can indicate loss of energy due to collisions resulting in its spiralling toward Earth. By monitoring its radius to within an acceptable margin of error, any potential damage to the satellite can be analysed and hence prevented or minimised by using rockets or boosters to stabilise its orbit if needed. For other satellites in Low Earth Orbits (LEO's), this can be very crucial due to the presence of atmospheric gases and their high velocities and hence periods, which may cause increased effects of friction that could result in the loss of the satellite's kinetic energy causing it to spiral inward, hence monitoring is done frequently and intensely to determine when the satellite would be required to boosted up in its orbit, this is especially significant for manned satellites such as the ISS where the safety of the astronauts is prioritised.

We can calculate the orbital velocity of a geostationary satellite, since the altitude is constant at 36 000 km,
$\bg_white v_{orb}=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{36000\times10^3+6.378\times10^6}}\approx3073~ms^{-1}$

The period of a geostationary satellite can be evaluated by using Kepler's Third Law
$\bg_white \dfrac{T^2}{r^3}=\dfrac{4\pi^2}{GM}\implies T=\sqrt{\dfrac{4\pi^2\times(36 000\times10^3+6.378\times10^6)}{6.67\times10^{-11}\times6\times10^{24}}}=86646.70 s\approx24\text{ hours}$

Thus, by evaluating the total energies by using Kepler's Laws, their positions can be predicted and they can be maintained.

Last edited:

#### Arrowshaft

##### Well-Known Member
How much would you give that? Sorry, couldn't be bothered to check over it lol, just whipped it out. Its also a bit clunky and unorthodox cause I didn't plan it out

#### Drdusk

##### π
Moderator
How much would you give that? Sorry, couldn't be bothered to check over it lol, just whipped it out. Its also a bit clunky and unorthodox cause I didn't plan it out
Bro that's like the length of a whole published paper

#### Arrowshaft

##### Well-Known Member
Bro that's like the length of a whole published paper
It’s trash tho just a buncha non-calculus waffle

#### akkjen

##### Active Member
Ooh that’s a good question, I'm going to prepare a somewhat actual answer for this.

The Orbits of Satellites such as GPS must be known very precisely. With reference to the total energy of a satellite and Kepler's Laws, discuss how satellites' orbits can be predicted and maintained.

The orbits of satellites are dependent upon the empirical discoveries of Kepler's Laws of planetary motion and transformations of its total energies, thus by examining these factors, satellite orbits can be predicted and maintained.

For an orbiting satellite, the gravitational potential energy of a satellite is given by
$\bg_white U=-\dfrac{GMm}{r}$
To determine the total energy, the kinetic energy must also be determined. For a satellite in stable orbit, the centripetal acceleration of a circular orbit is provided by the gravitational force exerted by the planet such that,
$\bg_white F_c=F_g$

$\bg_white \dfrac{mv^2}{r}=\dfrac{GMm}{r^2}$
$\bg_white v^2=\dfrac{GM}{r}$
$\bg_white v_{orb}=\sqrt{\dfrac{GM}{r}}$
Since the kinetic energy of the satellite is
$\bg_white K=\dfrac{1}{2}mv_{orb}^2=\dfrac{GMm}{2r}$
Thus the total energy is
$\bg_white E=U+K=-\dfrac{GMm}{r}-\dfrac{GMm}{2r}=-\dfrac{GMm}{2r}$
Hence, for a satellite undergoing uniform circular motion around the Earth, the total energy is constant, so is its kinetic and potential energy.

GPS satellites are geostationary above the equator at an altitude of 36 000 km. Geostationary satellites are launched above the equator with the tangential horizontal component of Earth's rotational motion coupled with the radial gravitational attraction serving to keep the satellite in orbit above the same point on the equator and no work required to be done to keep the satellite in orbital motion (as the velocity vector is perpendicular to the force vector). The resultant orbital motion is circular, as described by Kepler's First Law (eccentricity of o). Since the change in work is zero, this means there is no change in total energy as
$\bg_white W=\Delta E=0$

By using Kepler's Second Law of planetary motion, it is observable that a satellite undergoing orbital motion will sweep equal areas across equal times, that is; its areal speed will be constant. Hence, when the satellite gains kinetic energy, it moves faster, by abiding Kepler's Second Law it moves closer to the planet to sweep across an equal area. Since the total energy is constant, this means that the potential energy of the satellite decreases as it moves faster along its orbit. This effect is heightened as the eccentricity of the satellite's orbit increases, widening the ellipse and allowing for extreme positions of high and low kinetic and potential energies. This allows scientists to monitor energies and hence positions and velocities depending on the position of the satellite in its orbit. However, for GPS satellites which are geostationary, due to the circular nature of the orbit, it travels at the same angular and tangential velocities; thus allowing easier monitoring due to its constancy. Since geostationary satellites also stay above the same point over the equator during orbit, it shares the same rotational velocity and period of the Earth's orbit, thus monitoring any variations in period or position relative to the Earth's surface can shed information on any possible orbital turmoil such as collisions with space debris.

By applying Kepler's Third Law, which states that the cube of the radius is proportional to the square of the period, this means that the radius of the motion can be calculated by analysing the satellite's period, thus being able to determine any fluctuations in the satellite's orbit which can indicate loss of energy due to collisions resulting in its spiralling toward Earth. By monitoring its radius to within an acceptable margin of error, any potential damage to the satellite can be analysed and hence prevented or minimised by using rockets or boosters to stabilise its orbit if needed. For other satellites in Low Earth Orbits (LEO's), this can be very crucial due to the presence of atmospheric gases and their high velocities and hence periods, which may cause increased effects of friction that could result in the loss of the satellite's kinetic energy causing it to spiral inward, hence monitoring is done frequently and intensely to determine when the satellite would be required to boosted up in its orbit, this is especially significant for manned satellites such as the ISS where the safety of the astronauts is prioritised.

We can calculate the orbital velocity of a geostationary satellite, since the altitude is constant at 36 000 km,
$\bg_white v_{orb}=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{36000\times10^3+6.378\times10^6}}\approx3073~ms^{-1}$

The period of a geostationary satellite can be evaluated by using Kepler's Third Law
$\bg_white \dfrac{T^2}{r^3}=\dfrac{4\pi^2}{GM}\implies T=\sqrt{\dfrac{4\pi^2\times(36 000\times10^3+6.378\times10^6)}{6.67\times10^{-11}\times6\times10^{24}}}=86646.70 s\approx24\text{ hours}$

Thus, by evaluating the total energies by using Kepler's Laws, their positions can be predicted and they can be maintained.
this makes me feel like i aint going to get a B1

#### Arrowshaft

##### Well-Known Member
this makes me feel like i aint going to get a B1
May have gone a bit extra lol

#### TheOnePheeph

##### Active Member
It’s trash tho just a buncha non-calculus waffle
Trust me, I got 7/9 and completely forgot what kepler's 2nd law was in my trial. You would easily get 9, but definitely don't write a response that long in hsc. I think all you needed for 9 marks was:

-Kepler's first law states that objects move in elliptical orbit, while his 2nd law states that the objects sweep out equal areas in ewual times, meaning the velocity is lower further away from the focus, spends more time further away

-Due to the conservation of energy, potential energy will convert to kinetic as the orbit approaches the apogee, while kinetic will convert to potential as the orbit approaches the perigee

-With Kepler's 3rd law, the period of the orbit can be predicted at a fixed radius, assuming a near circular orbit, mentioning equation for this

-For maintaining orbits - to move up orbits, burning in the direction of the orbit at the perigee allows for a positive work done, allowing for increase in potential energy and therefore radius, mentioning equation.

-For moving down, burning against the direction of orbit at the apogee is the most efficient way to decrease, with work done being negative. Maybe mention orbital decay for circular orbits?

While by all means there should be calculus in the hsc physics course, non calculus waffle in the course we've got is fine lol.

#### Arrowshaft

##### Well-Known Member
Trust me, I got 7/9 and completely forgot what kepler's 2nd law was in my trial. You would easily get 9, but definitely don't write a response that long in hsc. I think all you needed for 9 marks was:

-Kepler's first law states that objects move in elliptical orbit, while his 2nd law states that the objects sweep out equal areas in ewual times, meaning the velocity is lower further away from the focus, spends more time further away

-Due to the conservation of energy, potential energy will convert to kinetic as the orbit approaches the apogee, while kinetic will convert to potential as the orbit approaches the perigee

-With Kepler's 3rd law, the period of the orbit can be predicted at a fixed radius, assuming a near circular orbit, mentioning equation for this

-For maintaining orbits - to move up orbits, burning in the direction of the orbit at the perigee allows for a positive work done, allowing for increase in potential energy and therefore radius, mentioning equation.

-For moving down, burning against the direction of orbit at the apogee is the most efficient way to decrease, with work done being negative. Maybe mention orbital decay for circular orbits?

While by all means there should be calculus in the hsc physics course, non calculus waffle in the course we've got is fine lol.
Oh damn, I didn’t even know what a perigee and apogee was haha . I was thinking of discussing Holman transfer orbits as well, but thought that’s more old syllabus and wouldn’t be necessary.

#### TheOnePheeph

##### Active Member
Oh damn, I didn’t even know what a perigee and apogee was haha . I was thinking of discussing Holman transfer orbits as well, but thought that’s more old syllabus and wouldn’t be necessary.
Yeah it should be fine - Im gonna mention them personally if they ask a question about how you would efficiently move up/down an orbit but I doubt they are required

#### blyatman

##### Well-Known Member
While by all means there should be calculus in the hsc physics course, non calculus waffle in the course we've got is fine lol.
Damn I thought the new syllabus was meant to be more math based and less waffle like the old one. Doesn't look like its changed all that much haha.

Interesting fact: In reality, predicting an orbit is significantly much more complicated than simply using Kepler's laws.

#### Drdusk

##### π
Moderator
Damn I thought the new syllabus was meant to be more math based and less waffle like the old one. Doesn't look like its changed all that much haha.

Interesting fact: In reality, predicting an orbit is significantly much more complicated than simply using Kepler's laws.
I mean really all NESA did was take out a bunch of that society and effects stuff and put in 50 more formulas. Adding formulas doesn't make the course more mathematical if its just plugging into a calculator lol.

Without Calculus or even a degree of problem solving, the essence of Physics is lost.

#### blyatman

##### Well-Known Member
I mean really all NESA did was take out a bunch of that society and effects stuff and put in 50 more formulas. Adding formulas doesn't make the course more mathematical if its just plugging into a calculator lol.

Without Calculus or even a degree of problem solving, the essence of Physics is lost.
Yeh true, but that society and effects stuff was bad, like really bad. Questions like "Assess the impacts of generators on society" were absolute garbage. The fact that there's more calculations, even if it's simple, is a small, but nonetheless welcome change. Hopefully they'll keep on improving it in the future.

#### Arrowshaft

##### Well-Known Member
I mean really all NESA did was take out a bunch of that society and effects stuff and put in 50 more formulas. Adding formulas doesn't make the course more mathematical if its just plugging into a calculator lol.

Without Calculus or even a degree of problem solving, the essence of Physics is lost.
I gave up on NESA the moment the moment they used suvat smh...
Simple integration: am I a joke to you?