VCE Physics Question Thread!

[Saikyo]

My hatred for the jacaranda physics textbook is at my boiling point here...

In an elastic collision between two objects of mass m1 and m2, show that the speed of approach (u2-u1) is equal to the speed of seperation (v2+v1). The symbols u1, u2, v1 and v2 each represent speeds, not velocities.

there is NO WAY that this is a vcaa type question right guys??

[sorry for the bad drawing i made from the photo for the diagram in the question!]

[lzxnl]

This is just proving the conservation of momentum (if initial total masses are equal to final total masses, then total initial speeds are equal to total final speeds)

VCAA won't give anything this vague, it will always have actual figures for you to plug into equations

It's not even proving the conservation of momentum; that is a principle you'd have to use to prove it

I wish VCAA gave things without figures.

[Brunette15]

Hi everyone ,
Can someone please help me out with this question for light:
Estimate the size of the smallest object that can be clearly imaged by a microscope that uses visible light. Explain this limitation.

[PB]

Hi everyone ,
Can someone please help me out with this question for light:
Estimate the size of the smallest object that can be clearly imaged by a microscope that uses visible light. Explain this limitation.

A microscope works by shining light through a slide and the specimen absorbs some frequencies of light, producing an image that enters the lens and into your eye (or something like that).
Now, we all know that light whose wavelength is greater than the width of a gap will diffract significantly. The same thing goes if the wavelength is greater than the width of an object! Fortunately, this doesn't affect our day to day lives because every-day objects are definitely not smaller than the wavelengths of the visible light spectrum. If not, there will be crazy diffractions going everywhere, and nobody will be able to see anything.
 However when we are dealing with extremely tiny objects, whose width is actually smaller than the wavelength of visible light, significant diffraction starts to occur, and (as you can imagine) starts to create blurry images.
So there you go. Now go and find the smallest wavelength of visible light, and the object of interest should be no smaller than that in order to be clearly imaged by a light microscope.

[Brunette15]

A microscope works by shining light through a slide and the specimen absorbs some frequencies of light, producing an image that enters the lens and into your eye (or something like that).
Now, we all know that light whose wavelength is greater than the width of a gap will diffract significantly. The same thing goes if the wavelength is greater than the width of an object! Fortunately, this doesn't affect our day to day lives because every-day objects are definitely not smaller than the wavelengths of the visible light spectrum. If not, there will be crazy diffractions going everywhere, and nobody will be able to see anything.
 However when we are dealing with extremely tiny objects, whose width is actually smaller than the wavelength of visible light, significant diffraction starts to occur, and (as you can imagine) starts to create blurry images.
So there you go. Now go and find the smallest wavelength of visible light, and the object of interest should be no smaller than that in order to be clearly imaged by a light microscope.

Thanks so much! 

[hyunah]

a ball is thrown vertically upward from a platform 16m above the ground and has a initial velocity of 24.5m/s. Find:
a) max. height the ball reaches
b) time taken to reach max height
c) time taken to reach ground from max height
d) total time the ball is in the air
e) speed of ball when it returns to the platform
f) speed of ball when it hits the ground

thanks you and please

[RKTR]

a ball is thrown vertically upward from a platform 16m above the ground and has a initial velocity of 24.5m/s. Find:
a) max. height the ball reaches
b) time taken to reach max height
c) time taken to reach ground from max height
d) total time the ball is in the air
e) speed of ball when it returns to the platform
f) speed of ball when it hits the ground

thanks you and please

(a) v^2=u^2+2ax
      at max height,v=0
    0=24.5^2+2(-10)x
x=30m
but this is the displacement from the platform therefore max height=30+16=46m
(b) v=u+at
      0=24.5-10t
        t=2.45s
(c) x=ut+1/2 a t^2
   from max height to ground x=-46
  -46=1/2(-10)(t^2)
  t=3.03s
(d) total time = ans of (b) + ans of (c)
                     =2.45+3.03
                      =5.48s
(e)v^2=u^2+2ax
     v^2=2(-10)(-30)
    v^2=600
      speed=24.5 m/s
(f)v^2=u^2+2ax
    v^2=2(10)(46)
   v^2=920
speed=30.3 m/s

[hyunah]

thank you RKTR

someone please help me here too!

A bus travels 60 metres in 10 seconds and the next 60 metres in 15 seconds. If the acceleration is constant, find:
i) how much further it will travel before ocming to rest
ii) how many more seconds it takes before coming to rest

thanks and please

[speedy]

When using the RHG rule for a solenoid, the side that interacts as north is at the tip of the thumb, and the side that interacts as south is at the base of the thumb...? Right? My teacher says the opposite, which is true for inside the solenoid, but when trying to determine the direction of current flow using lenz's law, for example a bar magnet approaching a solenoid, wouldn't you want to focus on the interactions between the poles outside of the solenoid.

Take this image -

Spoiler

- my teacher would say that the poles should be switched, and we use those to determine how solenoids will interact.

[lzxnl]

Ask your teacher why he/she thinks that. The magnetic field lines are leaving at the marked N terminal, so I see nothing wrong with the diagram given.

[speedy]

Ask your teacher why he/she thinks that. The magnetic field lines are leaving at the marked N terminal, so I see nothing wrong with the diagram given.

What he says is that any point is north relative to a point further down the line (in the direction of the field), which is a south. Basically that labelling a north and south is relative, so INSIDE the solenoid, the field starts at a north and goes to a south, thus doing the opposite to the diagram. He argues that you should never label the poles of a solenoid as they are relative (he's a VCAA assesor, and took a mark off my friend in a test for labelling the poles). However when I asked him a question on lenz's law, with a bar magnet going into a loop, he said to determine the direction of current, you need a north end produced by the coil that opposes the north coming towards it - to do this he used the north and south relative to inside the solenoid. Which, when looking over some questions, I found to be wrong.

[lzxnl]

What he says is that any point is north relative to a point further down the line (in the direction of the field), which is a south. Basically that labelling a north and south is relative, so INSIDE the solenoid, the field starts at a north and goes to a south, thus doing the opposite to the diagram. He argues that you should never label the poles of a solenoid as they are relative (he's a VCAA assesor, and took a mark off my friend in a test for labelling the poles). However when I asked him a question on lenz's law, with a bar magnet going into a loop, he said to determine the direction of current, you need a north end produced by the coil that opposes the north coming towards it - to do this he used the north and south relative to inside the solenoid. Which, when looking over some questions, I found to be wrong.

The problem is, you can't say that north and south are relative terms. They're not. Magnetic field lines form closed loops, so by his logic you can keep going more and more and more north, which is ludicrous.
Inside the solenoid, you go from a south pole to a north pole, not the other way around, exactly like a bar magnet. I don't know what he's thinking, but honestly I don't follow his logic.

As for Lenz's law and a bar magnet going into the loop, if it's the north pole of the bar magnet facing the loop, you have a stronger magnetic field going into the loop so the resulting current should produce a magnetic field which opposes the original magnetic field. And vice versa.

[speedy]

The problem is, you can't say that north and south are relative terms. They're not. Magnetic field lines form closed loops, so by his logic you can keep going more and more and more north, which is ludicrous.
Inside the solenoid, you go from a south pole to a north pole, not the other way around, exactly like a bar magnet. I don't know what he's thinking, but honestly I don't follow his logic.

As for Lenz's law and a bar magnet going into the loop, if it's the north pole of the bar magnet facing the loop, you have a stronger magnetic field going into the loop so the resulting current should produce a magnetic field which opposes the original magnetic field. And vice versa.

Yeah I completely agree. This is really annoying because he's a very good teacher and does know his stuff. I'll bring it up with him tomorrow, tell him what you said, show him the textbook and a few questions. Thank you

Also with that, for a magnet approaching a loop from the left, would you say that the field is increasing to the left, thus the induced current must produce a field increasing to the right, which would be coming out of the loop in the same direction the magnet was moving?

Also, just another question I came across, how do you find the average voltage produced by a coil that has been turned one complete cycle? You use faraday's law with the time for one quarter of the cycle, right?

[Brunette15]

Can someone please help me with this question:
What accelerating potential difference would be required to give an alpha particle a de Broglie wavelength of 2.0nm?

mass of alpha is 6.67x10^-27kg

I attached the solution and am just unsure why they doubled the coulomb of charge. The way i tried to work it out is find the velocity using de broglie's wavelength formula which gave me 49.7m/s. Then use this to find the kinetic energy which was 8.24x10^-24J. Once i did this i used the formula W=qV and subbed in 8.24x10^-24=(1.6x10^-19)V which gave me the answer of 5.1x10^-5V required.

[hyunah]

can someone please help with this question:

what is the importance of frequency in determining the non-ideal behaviour of transformers. are there any other parameters?

thank you