# Thread: HSC Physics Marathon 2017

1. ## HSC Physics Marathon 2017

Welcome to the HSC physics marathon
This thread can be used to ask for help or test others!
Just some simple rules:
- Answer the question asked previously, before posting a new question.
- After you have answered a question, post another question to keep the thread alive.

I will start off:
A cannon is fired at a velocity of 400.0ms^-1, 30.0° above horizontal. Determine the vertical and horizontal components of this initial velocity.

2. ## Re: HSC Physics Marathon 2017

I'm not gonna do the question, just asking how is projectile motion different in physics to 3U? I've heard it's different, but don't know how.

Edit: might as well add something to the question. Assuming I am at the same height as the cannon, how far away from me must you place the cannon if you want to hit me?

3. ## Re: HSC Physics Marathon 2017

Originally Posted by calamebe
Edit: might as well add something to the question. Assuming I am at the same height as the cannon, how far away from me must you place the cannon if you want to hit me?
$\noindent a=\frac{(v_{y}-u_{u})}{t} \\ \Rightarrow t=\frac{(v_{y}-u_{u})}{a} \\ = \frac{0-400\sin 30}{-9.8} \\ \therefore time to reach max point = 20.4s \\ \Rightarrow time to reach the other end = 2 \cdot 20.4 = 40.8s\\ \therefore x=v_{x}t \\ = 400\cos 30 \cdot 40.8\\ \approx 14139m$

4. ## Re: HSC Physics Marathon 2017

Originally Posted by calamebe
I'm not gonna do the question, just asking how is projectile motion different in physics to 3U? I've heard it's different, but don't know how.

Edit: might as well add something to the question. Assuming I am at the same height as the cannon, how far away from me must you place the cannon if you want to hit me?
There are several differences between HSC Physics projectile motion and HSC 3U. In HSC Physics, you are allowed to quote formulas like equations of motion and other relevant equations (at least ones on the Formula Sheet), whereas for 3U you need to derive them generally. For 3U, projectile motion Q's are generally going to require calculus and thus can have a much broader range of types of things that can be asked, whereas in HSC Physics there's no calculus and the Q's are mainly (or at least often) substituting values into formulas.

5. ## Re: HSC Physics Marathon 2017

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

6. ## Re: HSC Physics Marathon 2017

Originally Posted by InteGrand
There are several differences between HSC Physics projectile motion and HSC 3U. In HSC Physics, you are allowed to quote formulas like equations of motion and other relevant equations (at least ones on the Formula Sheet), whereas for 3U you need to derive them generally. For 3U, projectile motion Q's are generally going to require calculus and thus can have a much broader range of types of things that can be asked, whereas in HSC Physics there's no calculus and the Q's are mainly (or at least often) substituting values into formulas.
Oh ok, that's a little disappointing. Thanks for the reply though!

7. ## Re: HSC Physics Marathon 2017

Calculate the GPE of a 2000 kg satellite which orbits the Earth at an altitude of 35 000 km. The radius of Earth is 6378km.

8. ## Re: HSC Physics Marathon 2017

Originally Posted by eyeseeyou
Calculate the GPE of a 2000 kg satellite which orbits the Earth at an altitude of 35 000 km. The radius of Earth is 6378km.
GPE=-(Gm1m2)/r
=-(6.67x10^-11*2000*6x10^24)/(6378x10^3+35000x10^3)
=-1.93^10 Joules

9. ## Re: HSC Physics Marathon 2017

Calculate Work done when an object with mass 5x10^2kg is moved from surface of the Earth to the altitude 300km. The radius of the Earth is 6370km.

10. ## Re: HSC Physics Marathon 2017

7.5*10^-22

is that right?

11. ## Re: HSC Physics Marathon 2017

Originally Posted by JackPatel
7.5*10^-22

is that right?
Nope, show me your working out.

12. ## Re: HSC Physics Marathon 2017

I am not sure that this is right but let me try. Please don't laugh if I did this completely wrong

I think the formula for gravitational potential energy is
W= mgy

and since you would need the same amount of energy to potential energy to move the object upwards(assuming constant velocity)

then it would be

5.0*10^2*9.8*300*10^3 = 1.47*10^9 J

13. ## Re: HSC Physics Marathon 2017

Originally Posted by jiujiu1123
I am not sure that this is right but let me try. Please don't laugh if I did this completely wrong

I think the formula for gravitational potential energy is
W= mgy

and since you would need the same amount of energy to potential energy to move the object upwards(assuming constant velocity)

then it would be

5.0*10^2*9.8*300*10^3 = 1.47*10^9 J
By using E(p)=mgh we assume that g is constant at any altitude which is incorrect and this formula only works for low altitudes as we make the surface the point where E(p)=0. So a more accurate way of measuring E(P) is by -(Gm1m2)/r as it takes in consideration that g is now always constant do to many factors such as variation in the thickness of the Earth's crust and lithosphere, earth not being a perfect sphere so having flattened poles thus the value of 'g' is greater near the poles, and the spin of Earth creates a centrifugal effect which affects the 'g' near the equator. However by making E(p) a point a very large distance away thus getting the formula E(p)= -(Gm1m2)/r it assumes that GPE is the only thing changing, but rockets can loose fuel or gain/loose velocity which increases E(k). However you would still get 1 out of 2 marks for it but the correct way to do it is using E(p)= -(Gm1m2)/r.

14. ## Re: HSC Physics Marathon 2017

Work done = Gm1m2 (1/r,initial - 1/r,final)
= (6.67x10^-11)(5.972x10^24)(5x10^2)(1/(6370x1000) - 1/(6370x1000+300x1000))
= 1.4 x 10^9 Joules

To get the work done formula:
WD = Ep,final - Ep, initial
= -Gm1m2/r,final - (-Gm1m2/r,initial)
= Gm1m2/r,initial - Gm1m1/r,final
= Gm1m2 (1/r,initial - 1/r,final)

15. ## Re: HSC Physics Marathon 2017

BTW what options are you all doing?

Quanta, Astro or med?

16. ## Re: HSC Physics Marathon 2017

Originally Posted by Rathin
By using E(p)=mgh we assume that g is constant at any altitude which is incorrect and this formula only works for low altitudes as we make the surface the point where E(p)=0. So a more accurate way of measuring E(P) is by -(Gm1m2)/r as it takes in consideration that g is now always constant do to many factors such as variation in the thickness of the Earth's crust and lithosphere, earth not being a perfect sphere so having flattened poles thus the value of 'g' is greater near the poles, and the spin of Earth creates a centrifugal effect which affects the 'g' near the equator. However by making E(p) being a point a very large distance away thus getting the formula E(p)= -(Gm1m2)/r it assumes that GPE is the only thing changing, but rockets can loose fuel or gain/loose velocity which increases EPE. However you would still get 1 out of 2 marks for it but the correct way to do it is using E(p)= -(Gm1m2)/r.
You are such a nerd! I'm really looking forward to working with you and Jathu!!

17. ## Re: HSC Physics Marathon 2017

Gosh, Thank you so much pikachu975 and Rathin.

18. ## Re: HSC Physics Marathon 2017

Originally Posted by Rathin
By using E(p)=mgh we assume that g is constant at any altitude which is incorrect and this formula only works for low altitudes as we make the surface the point where E(p)=0. So a more accurate way of measuring E(P) is by -(Gm1m2)/r as it takes in consideration that g is now always constant do to many factors such as variation in the thickness of the Earth's crust and lithosphere, earth not being a perfect sphere so having flattened poles thus the value of 'g' is greater near the poles, and the spin of Earth creates a centrifugal effect which affects the 'g' near the equator. However by making E(p) being a point a very large distance away thus getting the formula E(p)= -(Gm1m2)/r it assumes that GPE is the only thing changing, but rockets can loose fuel or gain/loose velocity which increases EPE. However you would still get 1 out of 2 marks for it but the correct way to do it is using E(p)= -(Gm1m2)/r.
Did you get this info from page 22 of the booklet xD

19. ## Re: HSC Physics Marathon 2017

just out of curiosity, are you guys currently in year 11 or year 12? I do not remember learning this in the prelim course but you guys seem to be highly proficient at it.

20. ## Re: HSC Physics Marathon 2017

Originally Posted by jiujiu1123
just out of curiosity, are you guys currently in year 11 or year 12? I do not remember learning this in the prelim course but you guys seem to be highly proficient at it.
Starting year 12 now, but already learnt some content.

21. ## Re: HSC Physics Marathon 2017

Originally Posted by eyeseeyou
BTW what options are you all doing?

Quanta, Astro or med?
Most likely Med

22. ## Re: HSC Physics Marathon 2017

How come there are no question left for me to answer!? Btw I like the idea of this thread and it would have been great if some people could have moved their discussions somewhere else as you are kind of spamming.

Heres my question and this should be really easy. So like calculate the period of a satellite orbiting the earth at an altitude of say 700,000 km. Okay so the radius of the earth is 6.38 × 10^6 m and the earth’s mass is 5.97 × 102^4 kg. Work it out people and show your full working.

I will look at the solutions tomorrow morning maybe.

23. ## Re: HSC Physics Marathon 2017

Originally Posted by Bluee
How come there are no question left for me to answer!? Btw I like the idea of this thread and it would have been great if some people could have moved their discussions somewhere else as you are kind of spamming.

Heres my question and this should be really easy. So like calculate the period of a satellite orbiting the earth at an altitude of say 700,000 km. Okay so the radius of the earth is 6.38 × 106 m and the earth’s mass is 5.97 × 1024 kg. Work it out people and show your full working.

I will look at the solutions tomorrow morning maybe.
I think you forgot to put the ^ symbols for the radius and mass of the earth, but I'll try solve it now.

Edit:
Is orbital period one of Kepler's laws?

24. ## Re: HSC Physics Marathon 2017

Originally Posted by pikachu975
I think you forgot to put the ^ symbols for the radius and mass of the earth, but I'll try solve it now.
Oops thats so stupid of me how could I have not realised that!

25. ## Re: HSC Physics Marathon 2017

Originally Posted by pikachu975
Edit:
Is orbital period one of Kepler's laws?
Just apply Keplers 3rd Law.

Yes it is.

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