coolcat6778
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- HSC
- 2025
"reference to experimental evidence"Q32 is not a history question are you retarded
hmm I wonder which experiments? oh yes, the ones we were forced to rote learn
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"reference to experimental evidence"Q32 is not a history question are you retarded
YOU DO NOT KNOW WHAT ACTUAL PHYSICS QUESTIONS ARE BECAUSE YOU ARE A LOW BAND 5 HIGHSCHOOL PHYSICS STUDENT YOU HAVE NO IDEA WHAT HAPPENS AT A UNIVERSITY OR POST UNIVERSITY LEVEL YOU ARE JUST ATTEMPTING TO JUSTIFY YOUR STUPIDITYseriously, insulting me doesn't change the fact that these aren't actual physics questions...
shhh. mid band 5. thanks assessment markYOU DO NOT KNOW WHAT ACTUAL PHYSICS QUESTIONS ARE BECAUSE YOU ARE A LOW BAND 5 HIGHSCHOOL PHYSICS STUDENT YOU HAVE NO IDEA WHAT HAPPENS AT A UNIVERSITY OR POST UNIVERSITY LEVEL YOU ARE JUST ATTEMPTING TO JUSTIFY YOUR STUPIDITY
the same opinions from anyone else would be fine you're just a deluded retard thoughLol did you author this exam, is that why you're so triggered?
i don't get why you think that answer is bad?Does anyone feel the very last (non history) question is a load of bullshit?
what the fuck is this answer even doing?
"Analyse the subsequent motion" is a fucking horrible way to phrase an apparently quantitative question
sorry, where exactly did I do this lmaothey blame the world around them for their stupidity

sighan actual physics question isn't one that makes you write a 30 liner essay (x3). with how stupid I am I still know what physics is and what humanities is.
2019 dudeahhh yes the 50% HSC physics and maths student making claims about what happens at a university level
after your results come out it's made based on inputs from different students that bought their raw marks2019 dude
do u know when the raw marks -aligned thing for 2025 officially comes out?
Basically some people buy the raw marks results after the results come out and give them to the raw marks database for our benefit. So anytime dec 18th onwards2019 dude
do u know when the raw marks -aligned thing for 2025 officially comes out?
Because the was no indication we had to calculate anything (it only said to mention to relevant laws and formulas, not use).i don't get why you think that answer is bad?
i think 'analyse' is just nesa's way of saying 'tell us everything you think is relevant and back it up with calcs where possible'



They won’t necessarily mark down a qualitative answer, these aren’t the solutions given by NESA.Because the was no indication we had to calculate anything (it only said to mention to relevant laws and formulas, not use).
Hence, I don't believe they should reduce marks for a full qualitative response which mentions the laws
Want another crazy (and wrong sample solution)?
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It will clearly follow a elliptical orbit (not escape like what the retarded official solutions suggest)
Along with that, this is clearly an "Explain" question. So I don't fucking know why the official solutions even put calculations (that don't even prove anything) cause clearly their answer is wrong.
nah this all sounds right and is what i gotDoes anyone feel the very last (non history) question is a load of bullshit?
The satellite is in a geostationary orbit thus the velocity and radius can be determined as it is stationary relative to a point on Earth (T = 24 hours).
Applying Kepler’s 3rd Law to find orbital radius from known period:
r3 / T2 = GM / 4π2
r = 3√(GMT2 / 4π2)
r = 3√{6.67 x 10-11 x 6.0 x 1024 x (24 x 60 x 60)2 / 4π2}
r = 4.2297… x 107 m
r = 4.2 x 107 m (2 s.f.) from the centre of Earth
Determining orbital velocity by applying formula:
Fc = Fg → mv2 / r = GMm / r2
vorb = √(GM / r)
vorb = √(6.67 x 10-11 x 6.0 x 1024 / 4.2297… x 107)
vorb = 3075.9627… ms-1
vorb = 3100 ms-1 (2 s.f.)
An explosion can be analysed using the law of conservation of momentum. Whilst kinetic energy is not necessarily conserved, momentum is always conserved in any closed system interaction. Hence:
Σpi = Σpf
pa initial + pb initial = pa final + pb final
As their mass is identical, it will be denoted with m:
mv + mv = m(2v) + mva
2mv = 2mv + mvb
mvb = 0
Thus, vb = 0. If ma changes from v to 2v = 2 x 3075.9627… = 6151.925… = 6200 ms-1 (2 s.f.), then it carries all of the initial momentum and mb is left with no momentum and thus no velocity after the explosion.
Considering the initial and final kinetic energies:
ΣKi = ½mv2 + ½mv2 = mv2
ΣKf = ½m(2v)2 + 0 = 2mv2
Hence, it can be determined that the satellite pieces experienced an increase in kinetic energy of mv2. By the law of conservation of energy, this extra energy likely came from chemical potential energy in the explosion.
Analysing the subsequent motion of ma:
ma was originally at orbital velocity v given by v = √(GM / r). Doubling the velocity results in its velocity being v = 2√(GM / r) = √(4GM / r).
Comparing this to escape velocity which can be derived by:
Ei = Ef
Ki + Ui = Kf + Uf
Escape velocity is where the mass has no kinetic energy remaining once it reaches r = infinity where U = 0:
Ki + Ui = 0 (Kf = 0, Uf = 0)
½mv2 + (-GMm / r) = 0
v = √(2GM / r)
v = √(2 x 6.67 x 10-11 x 6.0 x 1024 / (4.2297… x 107))
v = 4350.095… ms-1
v = 4400 ms-1 (2 s.f.)
Hence, the new velocity of ma being 2v = 6200 ms-1 is larger than escape velocity of √(2GM / r) = 4400 ms-1. This means that ma will travel in a hyperbolic path away from the Earth and never return. As it travels away its total mechanical energy will remain constant as per the law of conservation of energy: E = K + U. However, as it travels further away, its potential energy increases by U = -GMm / r and thus its kinetic energy decreases to conserve E. As kinetic energy is given by K = ½mv2, its speed will decrease as it travels further away. Because potential energy asymptotes to zero as ma approaches an infinite distance away, the kinetic energy and thus speed will asymptote to a positive, non-zero value as its initial energy exceeds what is needed for escape velocity. This can be shown as follows:
Ki + Ui = Kf (Uf = 0)
½mv2 = ½mu2 + (-GMm / r)
½mv2 = 2GMm / r - GMm / r = GMm / r
v = √(2GM / r) = 4350.095… ms-1 = 4400 ms-1 (2s.f.)
Hence, ma will slow down asymptoting to a speed of 4400 ms-1 as it approaches a point infinitely far from the Earth, ignoring the influence of other gravitational fields in the universe.
Analysing the subsequent motion of mb:
mb has a velocity of 0 after the explosion as derived earlier. The only force acting on it is the gravitational force F = GMm / r2 from the Earth which will be vertically downwards towards the centre of the Earth. Hence, the acceleration is in the same direction. Given that its velocity is zero, it will simply accelerate downwards towards the centre of the Earth crashing into the surface.
Applying the law of conservation of energy, mb will have its total mechanical energy E = K + U remain constant. As it accelerates downwards U will decrease by U = -GMm / r, thus, kinetic energy will increase proportionally. By K = ½mv2, mb will continuously increase speed travelling in a straight line until it impacts the ground.
Ei = Ef
Ui + Ki = Uf + Kf
-GMm / ri + 0 = -GMm / rf + ½mv2
-GM / ri = -GM / rf + ½v2
v = √2(GM / rf - GM / ri) = √2GM(1 / rf - 1 / ri)
v = √2 x 6.67 x 10-11 x 6.0 x 1024 x (1 / (6.371 x 106) - 1 / (4.23 x 107))
v = 10330 ms-1 = 1.0 x 104 ms-1 at impact
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what the fuck is this answer even doing?
"Analyse the subsequent motion" is a fucking horrible way to phrase an apparently quantitative question
nah Ma has velocity that is greater than escape velocity so it wont stay in orbitThe key details are:
- geostationary (you can find orbital velocity from there)
- conservation of momentum, so the mb just falls back down to earth while the ma mass starts undergoing an elliptical orbit with the earth at one focus
- kinetic energy not conserved (though I have to reread the question to make sure whether that’s in there or not)
