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HSC Physics Marathon 2017 (1 Viewer)

Green Yoda

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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.
 

calamebe

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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?
 
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InteGrand

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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.
 

calamebe

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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!
 

eyeseeyou

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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.
 

Green Yoda

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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
 
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Green Yoda

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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.
 

jiujiu1123

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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
 

Green Yoda

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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.
 
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pikachu975

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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)
 

eyeseeyou

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BTW what options are you all doing?

Quanta, Astro or med?
 

Fizzy_Cyst

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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!! :D
 

pikachu975

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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
 

jiujiu1123

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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.
 

pikachu975

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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.
 

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