it'll scale the same as 2024 most likely (NESA will not deem the 20 marks of history questions as band 6 questions, even with the time pressure). The only novel question in this exam is the very last one.
sorry, I think it will be better than 2022. So b6 cutoff will be lower
yea it will be maybe even high 80s because it was very easy
rawmarks.info
Tell me the hardest question in this 2025 exam and then tell me the hardest question in the 2024 exam
ahhh yes the 50% HSC physics and maths student making claims about what happens at a university level
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 --- |
| what the fuck is this answer even doing? "Analyse the subsequent motion" is a fucking horrible way to phrase an apparently quantitative question |
u are a band 4 student stop questioning the solutions and instead start learning from them
I'm definitely glad I didn't wait a year later now lol
5. why do u keep forgeting what i got
much better bro!!!!! good job. my point stands.
did you even look at the paper? it's history based rather than theory let alone calculation based (20 marks).
your existence just bewilders me so much i think half of my messages on this website have been spent pointing out how retarded you are. How can someone be so so so dumb and deluded to the point they blame the world around them for their stupidity resulting in them thinking they r actually smart and capable of giving advice/having a valid opinion.
Q32 is not a history question are you retarded
you are so fucking delusional it hurts whenever you type I cant refrain myself from pointing out how retarded you arelol what is this insult
tell me with a straight face these are physics questions
View attachment 50538
