HSC Physics Marathon 2013-2015 Archive (1 Viewer)

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Fizzy_Cyst

Well-Known Member
re: HSC Physics Marathon Archive

Yeah, when both of their 'F' is equal on the probe, it'll experience 0 net force but I don't know how to work it out :angry:

ps i thought the answer was infinite because i never knew he meant the probe being in between the two planets
If the probe is a distance of 'r' fron Saturn, think about an expression for the distance away from Jupiter

If ur really stuck, I will put up working

superSAIyan2

Member
re: HSC Physics Marathon Archive

This ones' from my trials: The distance to Alpha Centauri is 4.367 light years as measured from earth. using relevant calculation explain how a rocket could complete the journey from earth to alpha centauri in exactly 3.28 years.

Try not to use trial and error; my teachers still gave full marks to those who solved it with trial and error, but it takes the fun out of it

psychotropic

Member
re: HSC Physics Marathon Archive

This ones' from my trials: The distance to Alpha Centauri is 4.367 light years as measured from earth. using relevant calculation explain how a rocket could complete the journey from earth to alpha centauri in exactly 3.28 years.

Try not to use trial and error; my teachers still gave full marks to those who solved it with trial and error, but it takes the fun out of it
Due to the effect of length contraction, the journey will appear less than 4.367 light years for those within the spacecraft, which means that they may be able to journey in 3.28 years. Quantitatively,

d = 4.367*c(1-v^2/v^2)^0.5
t = d/v

Using these equations, with t = 3.28 and the contracted length, we get that:
v^2 = (4.37*c)^2/(4.37^2 + 3.28^2)
v = 2.399x10^8 m/s

Hence, if the spacecraft were to be travelling at 2.399x10^8 m/s, it would be able to travel to alpha centauri in exactly 3.28 years

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superSAIyan2

Member
re: HSC Physics Marathon Archive

thats not the right answer. but you are pretty close.
You might have made an error in squaring ? Also i dont see how the denominator is (4.37^2 + 3.28)

psychotropic

Member
re: HSC Physics Marathon Archive

thats not the right answer. but you are pretty close.
You might have made an error in squaring ? Also i dont see how the denominator is (4.37^2 + 3.28)
changed?

i think i typed something in the calculator incorrectly. As to your qn:

t = d/v

3.28 = 4.37c(1-v^2/c^2)^0.5/v
squaring both sides and rearranging:
v^2 (4.37^2 + 3.28^2) = (4.37c)^2
which gives the answer. I forgot to square the 3.28 when i squared both sides, which I have now corrected

ocatal

Active Member
re: HSC Physics Marathon Archive

$\bg_white 3.28 = \frac{4.367c\sqrt{1 - \frac{v^2}{c^2}}}{v}$

$\bg_white 3.28^2v^2 = 4.367^2c^2 - 4.367^2c^2 \times \frac{v^2}{c^2}$

$\bg_white 3.28^2v^2 + 4.367^2v^2 = 4.367^2c^2$

$\bg_white v = \sqrt{\frac{4.367^2c^2}{3.28^2 + 4.367^2}}$

$\bg_white v = 2.40 \times 10^8 \ m \ s^-^1$

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Menomaths

Exaı̸̸̸̸̸̸̸̸lted Member
re: HSC Physics Marathon Archive

If the probe is a distance of 'r' fron Saturn, think about an expression for the distance away from Jupiter

If ur really stuck, I will put up working
I give up =(

psychotropic

Member
re: HSC Physics Marathon Archive

$\bg_white 3.28 = 4.367\sqrt{1 - \frac{v^2}{c^2}}$

$\bg_white 0.564 = 1 - \frac{v^2}{c^2}}$

$\bg_white 0.660 = \frac{v}{c}$

$\bg_white \therefore v = 1.98 \times 10^8 \ m \ s^-^1$

Is this correct?
I believe the RHS should be on v because t = d/v

superSAIyan2

Member
re: HSC Physics Marathon Archive

changed?

i think i typed something in the calculator incorrectly. As to your qn:

t = d/v

3.28 = 4.37c(1-v^2/c^2)^0.5/v
squaring both sides and rearranging:
v^2 (4.37^2 + 3.28^2) = (4.37c)^2
which gives the answer. I forgot to square the 3.28 when i squared both sides, which I have now corrected
yep, you got the right answer

ocatal

Active Member
re: HSC Physics Marathon Archive

I believe the RHS should be on v because t = d/v
Ohhh I see what I did wrong. Misread the question thinking 3.28 was given as a distance (i.e. light years).

Fizzy_Cyst

Well-Known Member
re: HSC Physics Marathon Archive

Challenge? Question:

A space cowboy is standing on the surface of a spherical asteroid of mass 1000kg and volume 1ML and is moving through space at 1000ms-1. He launches a 100g rock directly upwards and it reaches a maximum height of 1125m, it then falls down and hits the cowboy on the head.

Determine the speed at which the rock hit the cowboys head.

MUAHAHAHAHAHAHAHA

psychotropic

Member
re: HSC Physics Marathon Archive

Challenge? Question:

A space cowboy is standing on the surface of a spherical asteroid of mass 1000kg and volume 1ML and is moving through space at 1000ms-1. He launches a 100g rock directly upwards and it reaches a maximum height of 1125m, it then falls down and hits the cowboy on the head.

Determine the speed at which the rock hit the cowboys head.

MUAHAHAHAHAHAHAHA
the speed is 1.97 x 10^-4 m/s

superSAIyan2

Member
re: HSC Physics Marathon Archive

also got 1.974 x 10^-4 m/s.

edit : is planets velocity irrelevant?

Fizzy_Cyst

Well-Known Member
re: HSC Physics Marathon Archive

I get a different answer, closer to 1.5x10-4.

Planets velocity is a distractor.

Can u guys show working?

psychotropic

Member
re: HSC Physics Marathon Archive

using V = 4pir^3/3
We find r = 62.035m
then we use g = GM/r^2
and we get g = 1.7332 x 10^-11
Then we use v^2 = u^2 + 2 ar, with v = o and r = 1125m.
This yields u = 1.97 x 10^-4 m/s
Using the symmetry of the parabola, the velocity which it hits his head with is 1.97 x 10^-4 m/s

Fizzy_Cyst

Well-Known Member
re: HSC Physics Marathon Archive

You forgot to change volume into appropriate units! Should be in m^3, not L

SATNAV101

New Member
Re: 2013 HSC physics marathon

Explain how the slingshot effect can increase the speed of a rocked without using fuel or violating the law of conservation of energy. 4 marks

skillstriker

Member
re: HSC Physics Marathon Archive

When a spacecraft approaches a planet, its incoming speed and outgoing speed are the same relative to the planet but there is a change in direction. However relative to the sun the speed increases because the planet acquires angular momentum from planet and the planet loses a similar amount of angular momentum (law of conservation of momentum). Thus, the sling-shot effect allows the rocket to increase its speed without using fuel.

SATNAV101

New Member
re: HSC Physics Marathon Archive

Nice.. maybe include a diagram for 4 marks with the earth and rocket system and then the earth moving relative to the sun

iBibah

Well-Known Member
re: HSC Physics Marathon Archive

Question:

Bob bought a new drill. He attempts to drill through concrete, but as the concrete is hard, the drill slows right down. However Chris holds the trigger, waiting for it to drill though.

Explain why Chris is an idiot. (3 marks).

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