HSC Physics Marathon 2013-2015 Archive (6 Viewers)

[strawberrye]

Re: HSC Physics Marathon 2014

@Fizzy_Cyst this is normally the quietest time of year (end of HSC to start of T1). Not many students do much during their holidays so it's no surprise, really.

Or alternatively, you could interpret it as some of the super, studious students appear to not be doing much in the holidays, but they are actually doing a lot by not going on BOS forums as frequently and silently studying very hard while the unsuspicious students are still relaxing, feeling complacent that there is still plenty of time left to enjoy themselves

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

Or alternatively, you could interpret it as some of the super, studious students appear to not be doing much in the holidays, but they are actually doing a lot by not going on BOS forums as frequently and silently studying very hard while the unsuspicious students are still relaxing, feeling complacent that there is still plenty of time left to enjoy themselves

Yeah this. I only had around 200 posts on BoS before HSC.

Somehow in the past 2 months I've posted over 1300 times.

 

[Carrotsticks]

Re: HSC Physics Marathon 2014

Yeah this. I only had around 200 posts on BoS before HSC.

Somehow in the past 2 months I've posted over 1300 times.

Excluding posts in NS too.

 

[anthony_wheels]

Re: HSC Physics Marathon 2014

Sorry! I forgot to post another question! Here is a pretty simple projectile motion question:

A projectile is fired at a velocity of 50m/s at an angle of 30 degrees to the horizontal. Determine the range of the projectile. 4

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

Excluding posts in NS too.

Off-topic, but over 500 in p00n.

 

[Zeref]

Re: HSC Physics Marathon 2014

Sorry! I forgot to post another question! Here is a pretty simple projectile motion question:

A projectile is fired at a velocity of 50m/s at an angle of 30 degrees to the horizontal. Determine the range of the projectile. 4

Find ux and uy.

ux=43.3
uy=25

use vy=uy+gt to find time where vy=0
the time given will be the time from launch to the peak of the parabola so you double it for total trip time

Substitute into delta x =uxt

(43.3 X 5.1)=220.83

 

[anthony_wheels]

Re: HSC Physics Marathon 2014

Find ux and uy.

ux=43.3
uy=25

use vy=uy+gt to find time where vy=0
the time given will be the time from launch to the peak of the parabola so you double it for total trip time

Substitute into delta x =uxt

(43.3 X 5.1)=220.83

Yep, good job!

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

New Question:

Determine the speed at which a clock must be moving relative to an observer, if every hour, 1 second more passes on the stationary clock compared to the moving clock

 

[Zeref]

Re: HSC Physics Marathon 2014

New Question:

Determine the speed at which a clock must be moving relative to an observer, if every hour, 1 second more passes on the stationary clock compared to the moving clock

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

Good stuff!

New Question:
Determine the initial velocity of a projectile launched at 30degrees above horizontal, if the range of projectile was 200m.

 

[Zeref]

Re: HSC Physics Marathon 2014

Good stuff!

New Question:
Determine the initial velocity of a projectile launched at 30degrees above horizontal, if the range of projectile was 200m.

I got 47.57 ms^-1

is that correct?

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

It says velocity, so you would need to add in the angle

Make sure that you only use physics equations to get the answer and don't use mathematics equations (I.e., don't use Range = (u^2sin(2@))/g)

New Question:

Determine the centripetal force experienced by the Earth during its orbit of the Sun.
Mass of the Sun = 2x10^30kg
Radius of Earths orbit around the Sun = 1.5x10^8km

 

[iStudent]

Re: HSC Physics Marathon 2014

Make sure that you only use physics equations to get the answer and don't use mathematics equations (I.e., don't use Range = (u^2sin(2@))/g)

Why? Do we lose marks if we do that?
I heard from someone who said that you can use 3u formulas in physics but not physics formulas in 3u

 

[QZP]

Re: HSC Physics Marathon 2014

Motors and generators please

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

I would dare say that you would lose marks, as st00pid as it may be.

HSC marking schemes often state 'substitute values into appropriate equations' -- appropriate equations refer to those on your formula sheet or those derived from equations in the formula sheet.

There was a 4 mark projectile question in 2006 (I think) HSC. If you used the range equation, it was solved in one step, but iirc marking scheme said you needed to use appropriate equations, plural. Meaning to get full marks, you needed to substitute values into more than one equation

Motors and generators please

Will do after my current question is answered correctly


New Question:

Determine the centripetal force experienced by the Earth during its orbit of the Sun.
Mass of the Sun = 2x10^30kg
Radius of Earths orbit around the Sun = 1.5x10^8km

 

[Zeref]

Re: HSC Physics Marathon 2014

It says velocity, so you would need to add in the angle

Make sure that you only use physics equations to get the answer and don't use mathematics equations (I.e., don't use Range = (u^2sin(2@))/g)

New Question:

Determine the centripetal force experienced by the Earth during its orbit of the Sun.
Mass of the Sun = 2x10^30kg
Radius of Earths orbit around the Sun = 1.5x10^8km

So 30 degrees from the horizontal?

We are missing the velocity component required for centripetal force so we gotta find that out. Subbing it into the orbital velocity formula gives me 29821.7

Subbing into the centripetal force formula gives me an answer of 3.556 X10^22 N

that doesn't look right HAHAHAHAHA

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

So 30 degrees from the horizontal?

We are missing the velocity component required for centripetal force so we gotta find that out. Subbing it into the orbital velocity formula gives me 29821.7

Subbing into the centripetal force formula gives me an answer of 3.556 X10^22 N

that doesn't look right HAHAHAHAHA

Say 30degrees above horizontal as 'to/from the horizontal' could mean either above or below.

Your value seems correct!

The quicker way of doing it would be to realise that the centripetal force is being provided by gravity, so you can use Newtons Law of Universal Gravitation and solve straight away! You would just need to make sure in the beginning you let the marker know that you are using Fc = Fg

Also, ensure that you give a direction, in this case the acceleration were can approximate as 'towards the Sun'

New Question (M&G as requested!):

A wire of length 50cm is placed on top of an electronic balance with a current of 25A flowing to the right of the page. The balance gives a reading of 8.154g. A second wire with a current of 20A flowing to the left of the page is placed parallel to the first and is brought to within 0.1cm of the wire. Find the new reading on the balance.

 

[Zeref]

Re: HSC Physics Marathon 2014

Say 30degrees above horizontal as 'to/from the horizontal' could mean either above or below.

Your value seems correct!

The quicker way of doing it would be to realise that the centripetal force is being provided by gravity, so you can use Newtons Law of Universal Gravitation and solve straight away! You would just need to make sure in the beginning you let the marker know that you are using Fc = Fg

Also, ensure that you give a direction, in this case the acceleration were can approximate as 'towards the Sun'

Oh hey I never really took notice of that although I do remember deriving orbital velocity through Fc=Fg

awwww haven't started and I don't plan to learn ahead for M and G yet :/

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

Oh hey I never really took notice of that although I do remember deriving orbital velocity through Fc=Fg

awwww haven't started and I don't plan to learn ahead for M and G yet :/

Yeah, can save you heaps of time Fw=Fg=Fc=Fnet in orbit!

I'm glad you haven't done M&G, it will give someone else a chance to answer a question, you tank! Hahaha

 

[QZP]

Re: HSC Physics Marathon 2014

Assuming length of second wire = length of first wire,
F = 0.05N (force exerted on first wire perpendicularly towards balance)

Now, old weight w1 = mg = (8.154)x 9.8 --- assuming g = 9.8
Therefore new weight w2 = w1 + 0.05N

Thus, new reading = w2/9.8 ~~ 8.159 g

Tell me how I did *my answer relies on the fact that scales take weight as an input then divide by 9.8 to give mass an output.