# Thread: HSC Physics Marathon 2013-2015 Archive

1. ## re: HSC Physics Marathon Archive

Originally Posted by Sy123
$A charge has charge-mass ratio \ \ 1 \times 10^{5}$

$The charge travels in a magnetic field of strength \ \ 10 T \ \ perpendicular to its trajectory, the charge moves in a circle$

$Calculate the time the charge takes to complete a full revolution of the circle$
q/m = 1 x10^5
mv^2 /r = qvB (theta = 90degrees)
q/m = v/rB

v/r = q/m x B
v/r =1 x 10^5 x 10
r/v = 1 x 10^-6
distance/speed = time
therefore time = 1x10^-6 seconds

maybe?

2. ## Re: 2013 HSC physics marathon

Originally Posted by Immortalp00n
The slingshot effect, defined as the increase in velocity given to a spacecraft because it enters the gravitational field of a planet as it passes by, is a technque used by Space agencies in accordance with physicists in order to take full advantage of a planets rotational motion in order to increase a particle's ( spacecraft) final speed.
The slingshot effect is used by many spacecraft which travel within and beyond the solar system.
A spacecraft passes close to a a planet such that its gravity pulls the spacecraft into an arc or circular motion.It leaves with the same speed relative to the planet, but its speed is increased when viewed from an alternate frame of reference such as the Sun.
The acquired speed is large enough so that the spacecraft can travel away from the planet.
Comparatively, a ball thrown from a person who is stationary would have speeds less than a ball thrown from a person who is running at an increasing or stationary speed. Thus the slingshot effect, also known as gravity assist trajectory is used in order to attain a significant change in speed and direction despite the very little expenditure of fuel, improving flight efficiency.
Should possibly consider writing this as "increase in velocity, relative to the Sun", because relative to the planet it passes, no change in velocity is experienced.

3. ## re: HSC Physics Marathon Archive

Originally Posted by yasminee96
q/m = 1 x10^5
mv^2 /r = qvB (theta = 90degrees)
q/m = v/rB

v/r = q/m x B
v/r =1 x 10^5 x 10
r/v = 1 x 10^-6
distance/speed = time
therefore time = 1x10^-6 seconds

maybe?
That looks correct to me, nice work.

4. ## re: HSC Physics Marathon Archive

Almost a correct solution

However, for one complete revolution the distance is r2pi.
It's a minor error, probs only lose 2 marks
Although in the future try not to just use formulae without thinking

5. ## re: HSC Physics Marathon Archive

Originally Posted by yasminee96
q/m = 1 x10^5
mv^2 /r = qvB (theta = 90degrees)
q/m = v/rB

v/r = q/m x B
v/r =1 x 10^5 x 10
r/v = 1 x 10^-6
distance/speed = time
therefore time = 1x10^-6 seconds

maybe?
isn't it :

t=2(r)(pi)/v

t= 2(10^-4)(pi) = 6.2831*10^-4 seconds

??

6. ## re: HSC Physics Marathon Archive

Describe three applications of the CRT and all features and characteristics of each application. (8 marks)

bump

8. ## re: HSC Physics Marathon Archive

$By using the fact that work is the change in kinetic energy$

$Prove that for a projectile in projectile motion$

$(v_y)^2 = (v_0 \sin \theta)^2 -2gy$

$Where \ \ v_y \ \ is the vertical component of velocity$
$\theta \ \ is the angle of projection$
$v_0 \ \ is the initial velocity$
$g \ \ is the acceleration due to gravity$
$y \ \ is the vertical displacement$

9. ## re: HSC Physics Marathon Archive

Originally Posted by Sy123
$By using the fact that work is the change in kinetic energy$

$Prove that for a projectile in projectile motion$

$(v_y)^2 = (v_0 \sin \theta)^2 -2gy$

$Where \ \ v_y \ \ is the vertical component of velocity$
$\theta \ \ is the angle of projection$
$v_0 \ \ is the initial velocity$
$g \ \ is the acceleration due to gravity$
$y \ \ is the vertical displacement$

$W = change in Ek.$

$W = Ek(final) - Ek(initial)$

$Ek(final)y = -mgh + Ek(initial)y$ (F=mg, W= Fr (Tungsten = Francium )) (BUT REMEMBER THAT GPE IS ALWAYS NEGATIVE!!!)

$0.5m v(final)^2 = -mgh + 0.5m v(initial)^2$

$v(final)^2 = -2gh + v(initial)^2$ (1)

$sin(theta) = V(initial)y/ V(initial)o$ (Draw a triangle of the object at rest at the starting position)

$V(initial)y = Vosin(theta)$ (2)

Therefore, substituting equation (2) into (1) gives us:

$Vy^2 = (VoSin(theta))^2 - 2gy$

Now someone have a go at my last question before I forget the band 6 response!!! (8 marks)

Describe three applications of the CRT and all features and characteristics of each application. (8 marks)

10. ## re: HSC Physics Marathon Archive

Originally Posted by anomalousdecay
$W = change in Ek.$

$W = Ek(final) - Ek(initial)$

$Ek(final)y = -mgh + Ek(initial)y$ (F=mg, W= Fr (Tungsten = Francium )) (BUT REMEMBER THAT GPE IS ALWAYS NEGATIVE!!!)

$0.5m v(final)^2 = -mgh + 0.5m v(initial)^2$

$v(final)^2 = -2gh + v(initial)^2$ (1)

$sin(theta) = V(initial)y/ V(initial)o$ (Draw a triangle of the object at rest at the starting position)

$V(initial)y = Vosin(theta)$ (2)

Therefore, substituting equation (2) into (1) gives us:

$Vy^2 = (VoSin(theta))^2 - 2gy$

Now someone have a go at my last question before I forget the band 6 response!!! (8 marks)

Describe three applications of the CRT and all features and characteristics of each application. (8 marks)
I can't read it well but if you showed it, you showed it, nice work.

11. ## re: HSC Physics Marathon Archive

Originally Posted by Sy123
$A 5T magnetic field directed out of the page changes to 0T in 0.1 seconds.$

$Determine the emf generated in a wire loop of \ \ 2m^2 \ \ in area. Find the magnitude and direction of the current if the resistance of the wire loop is 20 Ohms$
5A anticlockwise through the loop?

12. ## re: HSC Physics Marathon Archive

Originally Posted by Sy123
I can't read it well but if you showed it, you showed it, nice work.
Thanks.

What code do you use for the font you have? I tried to use the same one but it didn't work.
Also, sub- and super-scripts don't work here the same as word, so yeah I had troubles with that.

13. ## re: HSC Physics Marathon Archive

Originally Posted by anomalousdecay
Thanks.

What code do you use for the font you have? I tried to use the same one but it didn't work.
Also, sub- and super-scripts don't work here the same as word, so yeah I had troubles with that.
It's Latex: http://www.codecogs.com/latex/eqneditor.php

Question: Explain why the motor of an electric drill is more likely to overheat when the drill is experiencing a load.

14. ## re: HSC Physics Marathon Archive

Originally Posted by iBibah
It's Latex: http://www.codecogs.com/latex/eqneditor.php

Question: Explain why the motor of an electric drill is more likely to overheat when the drill is experiencing a load.
When the motor of an electric drill is experiencing a load, its rotational speed lowers which in turn lowers the back EMF of the motor. Net EMF = Supply EMF - Back EMF. If the back EMF is decreased, the net EMF increases and so does the current flowing through the coils of the motor. The greater the current flowing through the coils the greater the electrical power lost to heat by power loss=I^2*R making the motor more likely to overheat.

15. Originally Posted by bleakarcher
When the motor of an electric drill is experiencing a load, its rotational speed lowers which in turn lowers the back EMF of the motor. Net EMF = Supply EMF - Back EMF. If the back EMF is decreased, the net EMF increases and so does the current flowing through the coils of the motor. The greater the current flowing through the coils the greater the electrical power lost to heat by power loss=I^2*R making the motor more likely to overheat.
Good answer. Post a question if you like.

16. ## re: HSC Physics Marathon Archive

Originally Posted by anomalousdecay
Thanks.

What code do you use for the font you have? I tried to use the same one but it didn't work.
Also, sub- and super-scripts don't work here the same as word, so yeah I had troubles with that.
If you couldn't be bothered learning LaTeX, then you can just use the equation system on Google Docs (https://drive.google.com/), then use http://puush.me/ to take screenshots of it.

Insert -> equation gets you started and there's a toolbar for a bunch of mathematical symbols and what not.

It's really, really easy to use. ^ is superscript, and _ is subscript, etc.

17. ## re: HSC Physics Marathon Archive

Questions:

1. Explain why Michelson and Morley expected a change in the interference pattern received by the detector of the interferometer.
2. A circular disc of conducting material is rotating clockwise. The bottom part of the disc is in a magnetic field going into of the page. Explain what happens to the disc and why.

18. ## re: HSC Physics Marathon Archive

Explain the purpose of using a radial magnet in a galvanometer.

19. ## re: HSC Physics Marathon Archive

Originally Posted by RealiseNothing
Explain the purpose of using a radial magnet in a galvanometer.
According to the equation F=BIL sinx , where x is the angle at which a conductor is orientated to an uniform magnetic field, once the galvanometer needle is moved, the force induced will vary as the needle moves. Radial magnetic fields are uniform within a consistent radius from the centre of the field, at the tip of the needle. This means that as the needle tilts, its angle to the radius of the circle of particular strength will be 90 degrees. This occurs because the needle and magnetic field form a radius and tangent.
Hence, a radial magnetic field provides equal force on the needle, no matter how much it is tilted. Since a galvanometer detects tiny currents, it is necessary to have a uniform magnetic field as the needle is tilted, hence the use of the radial magnetic field.

Someone do my question if they are bothered . 3 posts, but still failed to be answered.

Describe three applications of the CRT and all features and characteristics of each application. (8 marks)

20. ## re: HSC Physics Marathon Archive

$A particle initally travels with constant acceleration with a velocity \ \ u \ \ after time \ \ t \ \ it has a velocity of \ \ v$

$The distance travelled after time \ \ t \ \ is given by \ \ x$

$i) Prove that \ \ v=u+at$

$ii) Prove that \ \ x=ut+ \frac{1}{2}at^2$

$iii) Hence prove that \ \ v^2=u^2+2ax$

$iv) The Work done on the particle is defined by the change in kinetic energy, using the equation from part (iii), prove that the work \ \ W \ \ is given by$

$W=Fx$

$Where \ \ F \ \ is the force applied to the particle$

21. ## re: HSC Physics Marathon Archive

$If it were possible to squash the Earth into a smaller sphere without losing any mass$

$How much would you have to squash the Earth, so that the Earth becomes a black hole?$

22. ## re: HSC Physics Marathon Archive

$i) How much energy does a 500g energy drink contain?$

$ii) It is impractical to extract this energy from the mass, the only way to convert energy from mass is a collision of matter and anti-matter$

$e^{+} + e^{- } \rightarrow 2\gamma$

$Is the collision of an electron and anti-electron, determine the energy released on collision$

23. ## re: HSC Physics Marathon Archive

Sy are you doing q2q? I might do Age of Silicon or q2q.

24. ## re: HSC Physics Marathon Archive

Originally Posted by anomalousdecay
Sy are you doing q2q? I might do Age of Silicon or q2q.
I'm doing Astrophyiscs.

25. ## re: HSC Physics Marathon Archive

Originally Posted by Sy123
$If it were possible to squash the Earth into a smaller sphere without losing any mass$

$How much would you have to squash the Earth, so that the Earth becomes a black hole?$
$A black hole is a body that light cannot escape$

$Hence its a body where the escape velocity is greater than the speed of light$

$Knowing that escape velocity is given by$

$v_e = \sqrt{\frac{2GM}{r}}$

$We have to find \ \ r \ \ such that \ \ v_e = c$

$\therefore \ \ r = \frac{2GM_e}{c^2}$

Hence we need to shrink the Earth to a radius r, whereby, $M_e \ = \ mass of earth$

$G \ = \ gravitation constant$

$c \ = \ speed of light in vacuum$

========

$If the Earth was compressed to produce a black hole, predict what would happen with this compressed body of mass of radius \ \ r$

It isn't directly related to the Physics syllabus but I think its rather interesting, but of course can be answered with HSC physics knowledge, just a bit of imagination.

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