# Thread: HSC Physics Marathon 2017

1. ## Re: HSC Physics Marathon 2017

Originally Posted by Bluee
Just apply Keplers 3rd Law.
That's pretty late into the Space topic so I doubt that many, if not any, people know it yet.

But the formula is something like:

T^2 = (4 x pi^2 x r^3)/GM

2. ## Re: HSC Physics Marathon 2017

Originally Posted by pikachu975
That's pretty late into the Space topic so I doubt that many, if not any, people know it yet.

But the formula is something like:

T^2 = (4 x pi^2 x r^3)/G*M

r^3/T^2 = G*M/4*Pi^2
My yearlies ended weeks ago so been studying since then. So quite ahead on Space.

Edit: Whats the point answering now that you have the formula.

3. ## Re: HSC Physics Marathon 2017

Originally Posted by Bluee
r^3/T^2 = G*M/4*Pi^2
My yearlies ended weeks ago so been studying since then. So quite ahead on Space.
That's the same formula that I said. By the way what is the 'r' component?

4. ## Re: HSC Physics Marathon 2017

Originally Posted by pikachu975
That's the same formula that I said. By the way what is the 'r' component?

5. ## Re: HSC Physics Marathon 2017

Originally Posted by pikachu975
That's the same formula that I said. By the way what is the 'r' component?
I know. 'r' component is orbital radius as intergrand already pointed out. Not sure if we have to go into a lot of the theory behind calling it orbital radius.

6. ## Re: HSC Physics Marathon 2017

What do I need to know from the prelim course for the HSC course?

7. ## Re: HSC Physics Marathon 2017

Originally Posted by dragon658
What do I need to know from the prelim course for the HSC course?

8. ## Re: HSC Physics Marathon 2017

Originally Posted by dragon658
What do I need to know from the prelim course for the HSC course?
And about the link that eyeseeyou posted, remember that if you do the option Astrophysics, you need to know most content from the cosmic engine.

10. ## Re: HSC Physics Marathon 2017

A crate of supplies for a scientific expedition to Greenland is being dropped by plane. When the supplies are dropped, the plane is travelling at 40 m/s horizontally at a height of 50 m. Sadly the parachute fails to open and the package falls to the ground at 9.8 m/s^2. Find the horizontal distance travelled by the package as it falls given that it hits the ground right at the feet of the scientific party.

11. ## Re: HSC Physics Marathon 2017

Originally Posted by eyeseeyou
A crate of supplies for a scientific expedition to Greenland is being dropped by plane. When the supplies are dropped, the plane is travelling at 40 m/s horizontally at a height of 50 m. Sadly the parachute fails to open and the package falls to the ground at 9.8 m/s^2. Find the horizontal distance travelled by the package as it falls given that it hits the ground right at the feet of the scientific party.
^BUMP

Also Rathin do you do Quanta, Medical physics or Astro as an option?

12. ## Re: HSC Physics Marathon 2017

Originally Posted by eyeseeyou
A crate of supplies for a scientific expedition to Greenland is being dropped by plane. When the supplies are dropped, the plane is travelling at 40 m/s horizontally at a height of 50 m. Sadly the parachute fails to open and the package falls to the ground at 9.8 m/s^2. Find the horizontal distance travelled by the package as it falls given that it hits the ground right at the feet of the scientific party.
127.78m

13. ## Re: HSC Physics Marathon 2017

$\noindent Alice drops a stone into a deep well and hears the sound of the stone hitting the bottom after a time T. Given that the speed of sound is v_s and gravity is g, how deep is the well?$

14. ## Re: HSC Physics Marathon 2017

Originally Posted by trecex1
127.78m
How did you get this answer?

15. ## Re: HSC Physics Marathon 2017

Originally Posted by InteGrand
$\noindent Alice drops a stone into a deep well and hears the sound of the stone hitting the bottom after a time T. Given that the speed of sound is v_s and gravity is g, how deep is the well?$
$\text{Let the distance be } d \\ T-\frac{d}{v_s} = \text{Time taken for it to reach the bottom}\\ d=\frac{g}{2}(T-\frac{d}{v_s})^2 \\ T^2-\frac{2d}{v_s}+\frac{d^2}{(v_s)^2}=\frac{2d}{g} \\ gd^2-2dv_s-2dgv_s+T^2v_s^2g=0\\ d=\frac{2v_s+2gv_s \pm \sqrt{(2v_s+2gv_s)^2-4g^2T^2v_s^2}}{2g} \\ \text{I'm guessing you take the positive one only for d}>0\\ \text{Not sure if you can simplify that any further though}$

Originally Posted by eyeseeyou
How did you get this answer?
Quote the formulas of motion (should be obvious) set y as -50 to get t --> sub into the x-motion.

Edit:just realised I missed a T in expanding it..oops.

16. ## Re: HSC Physics Marathon 2017

Originally Posted by trecex1
$\text{Let the distance be } d \\ T-\frac{d}{v_s} = \text{Time taken for it to reach the bottom}\\ d=\frac{g}{2}(T-\frac{d}{v_s})^2 \\ T^2-\frac{2d}{v_s}+\frac{d^2}{(v_s)^2}=\frac{2d}{g} \\ gd^2-2dv_s-2dgv_s+T^2v_s^2g=0\\ d=\frac{2v_s+2gv_s \pm \sqrt{(2v_s+2gv_s)^2-4g^2T^2v_s^2}}{2g} \\ \text{I'm guessing you take the positive one only for d}>0\\ \text{Not sure if you can simplify that any further though}$

Quote the formulas of motion (should be obvious) set y as -50 to get t --> sub into the x-motion.
Note that in your answer both solutions are positive, so you would need to decide (with justification!) which solution is right.

17. ## Re: HSC Physics Marathon 2017

How do you guys know all this stuff? You haven't even started learning HSC content yet!

18. ## Re: HSC Physics Marathon 2017

Originally Posted by WildestDreams
How do you guys know all this stuff? You haven't even started learning HSC content yet!
We self learn it

19. ## Re: HSC Physics Marathon 2017

$\text{Continuining from the second line}\\ T^2 -2\frac{dT}{v_s}+ \frac{d^2}{v_s^2}=\frac{2d}{g} \\ gd^2-2dTv_sg-2dv_s^2 +T^2v_s^2g =0 \\ d=\frac{2Tv_sg+2v_s^2 + \sqrt{(2Tv_sg+2v_s^2)^2-4g^2T^2V_s^2}}{2g} \\ or d=\frac{2Tv_sg+2v_s^2 - \sqrt{(2Tv_sg+2v_s^2)^2-4g^2T^2V_s^2}}{2g}$
Originally Posted by InteGrand
Note that in your answer both solutions are positive, so you would need to decide (with justification!) which solution is right.
Revised answer if I didn't make any more algebraic mistakes.
Edit: The value with the + (first one) should be omitted as the value of T will be negative. I.e the sound will take longer to reach her from the bottom of the well than the time it takes for the stone to fall and the sound to reach her(?!), which is impossible.

20. ## Re: HSC Physics Marathon 2017

A stone is thrown vertically upwards with a velocity of 29.4 m/s from the edge of a cliff 78.4 m high. The stone falls so that it just misses the edge of the cliff and falls to the ground at the foot of the cliff. Determine the time taken by the stone to reach the ground. Assume that the acceleration due to gravity is 9.8 ms-2

21. ## Re: HSC Physics Marathon 2017

Originally Posted by eyeseeyou
A stone is thrown vertically upwards with a velocity of 29.4 m/s from the edge of a cliff 78.4 m high. The stone falls so that it just misses the edge of the cliff and falls to the ground at the foot of the cliff. Determine the time taken by the stone to reach the ground. Assume that the acceleration due to gravity is 9.8 ms-2
https://postimg.org/image/d0zyhtyp9/

22. ## Re: HSC Physics Marathon 2017

Discuss the g-forces astronauts experience during the launch of a multi stage rocket and assess measures taken to minimize them. (7 marks)

23. ## Re: HSC Physics Marathon 2017

Originally Posted by jathu123
$\noindent a=\frac{(v_{y}-u_{u})}{t} \\ \Rightarrow t=\frac{(v_{y}-u_{u})}{a} \\ = \frac{0-400\sin 30}{-9.8} \\ \therefore time to reach max point = 20.4s \\ \Rightarrow time to reach the other end = 2 \cdot 20.4 = 40.8s\\ \therefore x=v_{x}t \\ = 400\cos 30 \cdot 40.8\\ \approx 14139m$
Jathu u nerd
it's Uy NOT Uu

24. ## Re: HSC Physics Marathon 2017

Originally Posted by pikachu975
And about the link that eyeseeyou posted, remember that if you do the option Astrophysics, you need to know most content from the cosmic engine.
Actualyyyy, I didn't bother remembering anything from Prelim and I'm doing great in Astro.
I guess it's iseful to remember some concepts like H-R Diagram and stars. But majority of it is new content. That's just my 2 cents.

25. ## Re: HSC Physics Marathon 2017

Determine the gravitational force between a 77.5 kg astronaut and Earth when he is standing on the ground. Calculate the change in gravitational force when the astronaut is inside a satellite orbiting the Earth at an altitude of 685 km. The radius of the Earth is 6380 km and the mass of the earth is 5.97*10^24 kg. [4 marks]

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