HSC Physics Marathon 2013-2015 Archive (1 Viewer)

[Zeref]

Re: HSC Physics Marathon 2014

Zeref, that is because when you consider it from the perspective of delta y, you are considering the entire time (the time taken to reach ground) however, when you consider it from the perspective of delta x, you only consider the time taken for the ball to be directly above the net (at exactly 12.2m) that's why if you work it out from delta y, I got a time of 0.357973 sec, which is longer than the 0.27286s time achieved by using delta x. Does that make sense?

And so by using the delta x equation, you can get the time taken to get directly above the net and use this in your calculations to find the delta y at this exact point. Does that make sense?

(At least this is what I think)

Oh I see.......

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

^^^Do this for Space.

As for Motors and Generators:

How many of you here have done the first quarter of MandG?

I think maybe only QZP has done it so far.

 

[BERRY]

Re: HSC Physics Marathon 2014

I've started a bit and I think I might know how to do part a-
Area = 0.05^2
=0.0025
n=1
B=2
Since magnetic field is linear, cos90=1
Current=10A

Therefore, torque=nBIAcosQ
=10 X 0.0025 X 2
= 0.05 Nm torque

I haven't yet done this in school, so I'm hoping its correct...

 

[QZP]

Re: HSC Physics Marathon 2014

I don't know how to deal with gears mainly because I don't know how they work My first idea was that torque on coil = torque on big gear but that didn't make as much sense as force on coil = force on big gear. So yeah idk my approach because I've never learnt anything about gears. Please clarify.

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

I've started a bit and I think I might know how to do part a-
Area = 0.05^2
=0.0025
n=1
B=2
Since magnetic field is linear, cos90=1
Current=10A

Therefore, torque=nBIAcosQ
=10 X 0.0025 X 2
= 0.05 Nm torque

I haven't yet done this in school, so I'm hoping its correct...

It is a bit of a tricky question. You missed one vital point which QZP pointed out below. Don't worry its a difficult question that you won't see in HSC.

Also, just make sure to label your units on everything when doing calculations.

And also, in exams write a full sentence answer like:

The maximum torque on the bigger gear is .......

I don't know how to deal with gears mainly because I don't know how they work My first idea was that torque on coil = torque on big gear but that didn't make as much sense as force on coil = force on big gear. So yeah idk my approach because I've never learnt anything about gears. Please clarify.

Correct. The force is the same.

The trick is that you have to know which formula to use for torque according to the turning effect of the gears NOT the coil.

Just think of gears as the same as a spanner, with a locus which only subtends from one point (if you form a circle with a coil, the coil will form two semi-circles, whereas a gear forms it as one full circle from the same point).

Hint:

For a) find the maximum force on the coil, which is the same as the maximum force applied on the gear, then find the maximum torque.

 

[Fizzy_Cyst]

Re: HSC Physics Marathon 2014

What would you guys say if I told you the ball does hit the net xD

Sorry to the person who PM'd me their answer and I said it looked good! I just made up the numbers and hadnt checked the answer, just checked it now and it does hit the net! There is a fundamental flaw in the working. I am not going to point it out directly, but has something to do with the angle

 

[QZP]

Re: HSC Physics Marathon 2014

a) Maximum (total) force on coil F = 2BIl = 2(2)(10)(0.05) = 2 N
and F on coil = F on big gear
Therefore Maximum total torque T on big gear = Fd = (2)(0.1) = 0.2 Nm

b) Torque on small gear T: Fd = 10 Nm
Therefore F = 10/d = 10/0.05 = 200N
Hence force on big gear = force on coil = 200N

So for coil: F = BIl sin 45
200 = (2)(I)(2x0.05) sin 45

Therefore I = 1000root(2) A lol.

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

a) Maximum (total) force on coil F = 2BIl = 2(2)(10)(0.05) = 2 N
and F on coil = F on big gear
Therefore Maximum total torque T on big gear = Fd = (2)(0.1) = 0.2 Nm

b) Torque on small gear T: Fd = 10 Nm
Therefore F = 10/d = 10/0.05 = 200N
Hence force on big gear = force on coil = 200N

So for coil: F = BIl sin 45
200 = (2)(I)(2x0.05) sin 45

Therefore I = 1000root(2) A lol.

Well done. Everything is correct. I would say both are worth two marks, so 4/4 for you.

Yeah that coil would have fried way before it got to that Current (unless it was one metre thick haha)

New Question for Motors and Generators:

I may add more questions later to this graph whenever I think of something a little more interesting (had a potentially great idea but stuffed up and was wrong).

 

[QZP]

Re: HSC Physics Marathon 2014

If the motor is connected to a supply of 240V, then wouldn't the voltage drop across the motor be 240V?

 

[anomalousdecay]

Re: HSC Physics Marathon 2014

If the motor is connected to a supply of 240V, then wouldn't the voltage drop across the motor be 240V?

Not necessarily.

There is a form of inertia/friction. That's why the graph is there.

Without the graph, you can assume it is an ideal motor and that there is no inertia/friction/resistance on it.

However, in this situation there is, due to the normal load.

However I worded it slightly off with "constant pace". I have now changed it to "normal load" as this is more correct. Sorry about the confusion there.

 

[nexusbrah]

Re: HSC Physics Marathon 2014

Ok, couldn't wait til tomorrow. Not heaps hard, but tricky.

New Question:

Whilst playing tennis, John serves a ball with a velocity of 45ms-1 at an angle of 6.5degrees below the horizontal. The height of the ball as it leaves his racquet is 2.45m above the ground. The net is 12.2m away and is of height 0.92m.
Determine whether the ball clears the net.

I am going to solve this, before i sleep.

 

[nexusbrah]

Re: HSC Physics Marathon 2014

Yay i got it

In July 1969 the Apollo 11 Command Module with Michael Collins on board orbited the Moon waiting for the Ascent Module to return from the Moon’s surface. The mass of the Command Module was 9.98×103 kg, its period was 119 minutes, and the radius of its orbit from the Moon’s centre was 1.85×106 metres.

a) Assuming the Command Module was in circular orbit, calculate
i) Mass of the moon
ii)The magnitude of the orbital velocity of the command module

 

[timeflies]

Re: HSC Physics Marathon 2014

In July 1969 the Apollo 11 Command Module with Michael Collins on board orbited the Moon waiting for the Ascent Module to return from the Moon’s surface. The mass of the Command Module was 9.98×103 kg, its period was 119 minutes, and the radius of its orbit from the Moon’s centre was 1.85×106 metres.

a) Assuming the Command Module was in circular orbit, calculate
i) Mass of the moon
ii)The magnitude of the orbital velocity of the command module

I assume 1.85×106 kg means 1.85×10^6 kg? If so:

New Question: A satellite is propelled from a certain Orbit 1 to Orbit 2. The radius of Orbit 2 is four times that of Orbit 1. What is the ratio of the new orbital period to the original period?

 

[nexusbrah]

I assume 1.85×106 kg means 1.85×10^6 kg? If so:

New Question: A satellite is propelled from a certain Orbit 1 to Orbit 2. The radius of Orbit 2 is four times that of Orbit 1. What is the ratio of the new orbital period to the original period?

Oh yes sorry about that. The solution is correct. Ill have a crack at your one later in the day

 

[Zeref]

Re: HSC Physics Marathon 2014

I assume 1.85×106 kg means 1.85×10^6 kg? If so:

New Question: A satellite is propelled from a certain Orbit 1 to Orbit 2. The radius of Orbit 2 is four times that of Orbit 1. What is the ratio of the new orbital period to the original period?

So 1:8?

 

[nexusbrah]

So 1:8?

That quality.

Im on my phone can barely see tje working out

 

[Zeref]

Re: HSC Physics Marathon 2014

A car comes to a bridge during a storm and finds out the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side of the road the car is on is 21.3m above the river, while the opposite side is a mere 1.8m above the river. The river itself is a raging torrent 61.0m wide.

a) How fast should the car be traveling at the time it leaves the road in order to just clear the river and land safely on the opposite side?
b) What is the speed of the car just before it lands on the other side?

 

[Zeref]

Re: HSC Physics Marathon 2014

That quality.

Im on my phone can barely see tje working out

How can you get bad quality? It's screenshotted LOL

 

[nexusbrah]

How can you get bad quality? It's screenshotted LOL

What it looks like



Also ill try out your question as soon as I get home! Looks interesting

 

[Zeref]

What it looks like

View attachment 29838

Also ill try out your question as soon as I get home! Looks interesting