HSC Physics Question Thread

[jamonwindeyer]

An object that is travelling at a speed of 0.9c relative to an observer has a mass of 100kg measured by the observer. Calculate the mass of the object measured by the same observer when it is at rest relative to the observer?

Why is the answer to this question not 100kg? I know the formula and how to get the answer, i just don't under stand why we still have to use the formula.

Thanks guys.

Hey! Okay, so it sounds like you know the idea here is mass dilation. The mass of an object (as measured by a stationary observer) changes based on its velocity due to this principle (the idea being that mass and energy are equivalent, and increasing kinetic energy also results in slight increase in mass). Since the measurement of 100kg is made when travelling close to the speed of light, the actual mass of the object must be much less. It is only be speeding the object to 0.9c that the mass dilates and increases to the value of 100kg. Hence, we use the formula to find the initial mass. Does that help? 

Edit: Combine Rui's comment about frames of reference with this for a pretty good explanation. We pretty much said the same thing 

[MysteryMarker]

Can't explain it carefully anymore but the idea is to keep in mind that all frames of references are relative. Mass dilation says that moving objects become heavier. If the object is moving at 0.9c, it is going to appear way heavier than if it were at rest (relatively speaking).

Oh so when it says 'at rest relative to the observer, would that mean that the observer is also travelling at 0.9c?

Thanks.

[jamonwindeyer]

Oh so when it says 'at rest relative to the observer, would that mean that the observer is also travelling at 0.9c?

Thanks.

It would mean that the two are travelling with the same velocity. Possibly 0.9c, possibly stationary, but that makes no difference either way (relativity) 

[RuiAce]

Oh so when it says 'at rest relative to the observer, would that mean that the observer is also travelling at 0.9c?

Thanks.

Yeah they either travel at 0.9c relative to each other, or they're at rest relative to each other. Always remember no absolute.

[MysteryMarker]

Damn that just changed the whole game. *Mindblows*

Thanks guys.

Another question --> A plane flies high up in the atmosphere and then plunges towards the earth to produce weightlessness for the occupants inside. Comment on the use of the term weightlessness in this context.

The answer says that the would not be weightless as they are still under the influence of gravity, but they would 'feel' weightless because there is no contact force. Could someone expand on this and explain how they are not weightlessness but are at the same time?

Cheers.

[Swagadaktal]

Swag's got you covered, it's definitely Faraday, my favourite Physics hoe 

Faraday's Law goes like this:



In words, the induced EMF/voltage is given by the rate of change of magnetic flux with respect to time (there are more complex versions of this formula out there, but this is the principle).

Note that electromotive force (EMF) is interchangeable with the term "voltage" at this level, they are really one in the same. So, the voltage we generate is given by how quickly we are changing the magnetic field. The fact that the negative sign is there reflects Lenz's Law, the induced voltage should OPPOSE the changing field that created it, so essentially:



The two cancel each other out (as we'd expect), if they didn't that would violate Lenz's Law.

Anyway, in terms of the question, the formula itself is enough to answer mathematically. A sine curve is shown, so the negative derivative is a negative cosine curve, answer D. However, a Physics student isn't expected to know how to do this, so let's do it the longer way.

The rate of change of the magnetic field can be considered properly by taking the derivative of the sine curve, or just simply notice that the EMF should be at a peak when the magnetic field is changing rapidly, and when the magnetic field is at a peak (not changing), the induced EMF should be zero. Remember we only have induced EMF when magnetic flux is changing. That eliminates options A and C, because they don't have peaks/troughs in the correct places. We want either B or D.

Now, the next one is where we bring in Faraday's Law for ease of use. The negative sign makes selecting D easier to see. However, Faraday's Law isn't explicitly in the syllabus in the mathematical form, so instead, you could use a Lenz's Law explanation. The EMF must be opposite in sign to the rate of change of magnetic field, because the induced EMF must oppose the change that created it. So, as the rate of change of magnetic flux is positive (EG - at the start of the top graph), we expect a negative peak (trough) for the EMF. D matches this response 

Does this make sense? Definitely a tough question this one, a tad beyond the syllabus IMO, but definitely within realms of reason, where did you find it? 

Yeah see what I mean their explanations are on fleek

Btw can you teach me how to input formulas? - How you get that fancy writing :3

[jakesilove]

Yeah see what I mean their explanations are on fleek

Btw can you teach me how to input formulas? - How you get that fancy writing :3

Hey! Give this epic guide a read to learn how LaTeX works!

[Skidous]

Damn that just changed the whole game. *Mindblows*

Thanks guys.

Another question --> A plane flies high up in the atmosphere and then plunges towards the earth to produce weightlessness for the occupants inside. Comment on the use of the term weightlessness in this context.

The answer says that the would not be weightless as they are still under the influence of gravity, but they would 'feel' weightless because there is no contact force. Could someone expand on this and explain how they are not weightlessness but are at the same time?

Cheers.

Hi Mystery Marker

The reason as to why they are not weightless is due to the fact they are still within the gravitational field of earth and thus are experiencing their own weight force as described by W=mg
The reason why they experience this feeling of weightlessness is because they are experiencing no g-forces or 0g. This means that the force of gravity they feel exerted on them is not their regular weight (or g-forces<1), similar to that of going down in an elevator and feeling slightly less heavier (this is an actual thing that happens and an experiment you may want to try yourself). This gives them to sense that there is no gravity by they are just free falling in the craft.
There is also the contrast to this weightlessness as the ascent back up to that height also produces more g-forces (g-forces>1) and makes people feel heavier than they are.
I hope this answers your question

Skidous

[Happy Physics Land]

Hi Mystery Marker

The reason as to why they are not weightless is due to the fact they are still within the gravitational field of earth and thus are experiencing their own weight force as described by W=mg
The reason why they experience this feeling of weightlessness is because they are experiencing no g-forces or 0g. This means that the force of gravity they feel exerted on them is not their regular weight (or g-forces<1), similar to that of going down in an elevator and feeling slightly less heavier (this is an actual thing that happens and an experiment you may want to try yourself). This gives them to sense that there is no gravity by they are just free falling in the craft.
There is also the contrast to this weightlessness as the ascent back up to that height also produces more g-forces (g-forces>1) and makes people feel heavier than they are.
I hope this answers your question

Skidous

Josh made a really good point explaining weightlessness in terms of g-force. Good job on his part, definitely would include it in my answer.
I would just like to drop in my 2 cents as well. Weight always exists, unless you arrive at a point in the universe where there is 0 gravitational field (hypothetically) you can't not have weight. Now we experience the idea of "weight" in terms of normal force - something that we feel. We feel heavy because the ground pushes against us with greater magnitude if we exert too much weight force onto the ground (Newton's 3rd Law of Motion). The scale doesnt measure our "weight" but instead measures the amount of reaction force it has to exert on us when we stand onto the scale. Keeping that in mind, when the answer mentions contact force, they are really talking about NORMAL FORCE. When the plane plunges down into Earth, our inertia makes us want to remain in the same prior position (Newton's 1st law) and hence momentarily we would not be able to "feel" our weight because we are not exerting a force on anything (i.e. we arent really have physical contact with anything". So this is why they say "We still have weight, but we just FEEL like we are weightless".

Just to go a step further (you dont necessarily need this for you answer). You can perhaps use the word "free fall", but lm not going to risk that because free fall usually refers to when you and your frame of reference are accelerating in the same magnitude. E.g. You feel weightless in space because your spaceship and you are both experiencing 9.8 m/s² towards Earth. Technically speaking, weight is a numerical concept but the FEELING of weight is really a relative concept. You can mention this in your response as well but what l wrote in the first paragraph combined with what Josh wrote was be sufficient enough.

[MysteryMarker]

For ideas to implementation,

Is the grid in a CRT connected to a voltage source? If so, is it positive or negative? And, how does it control both the velocity of the electrons and the number of the electrons?

i guess what I'm trying to ask is, how does a grid in a cathode ray tube work?

Cheers.

[wyzard]

For ideas to implementation,

Is the grid in a CRT connected to a voltage source? If so, is it positive or negative? And, how does it control both the velocity of the electrons and the number of the electrons?

i guess what I'm trying to ask is, how does a grid in a cathode ray tube work?

Cheers.

I've attached a simple schematic of a CRT from wikipedia. I'm guessing the grid you're talking about refers to the deflecting coils to control the electrons.

A CRT basically works by having an electron gun, the cathode and the heater, firing electrons to a fluorescent screen which lights up when an electron lands on it. The focus and deflecting coil will then control the direction of the fired electron. The coils are connected to a voltage source to produce an electric field, and they're usually oppositely charged. If the above is positively charged, the bottom part will be negatively charged which will deflect the electron upwards. Their polarity depends on where you want the electron to land on the screen.

The use of electric field though is the old design; more modern designs uses a varying magnetic field to deflect the electrons, which also requires voltage to drive current through the coils.

Last but not least, the deflecting coil usually does not affect the velocity, it just changes their direction; the velocity of electrons depends on the speed they are fired from the cathode.

[RuiAce]

Yeah basically.

This is copied out of Physics in Focus

Second, there exists another electrode in between the cathode and anode, which is named the grid. Making the grid more positive or negative with respect to the cathode controls the number of electrons reaching the anodes and hence striking the display screen per unit time, which consequently controls the intensity of the cathode ray and thus the brightness of the display.

[jamonwindeyer]

For ideas to implementation,

Is the grid in a CRT connected to a voltage source? If so, is it positive or negative? And, how does it control both the velocity of the electrons and the number of the electrons?

i guess what I'm trying to ask is, how does a grid in a cathode ray tube work?

Cheers.

Hey!! Okay, so this question has the potential to be very simple or extremely complicated! You'll never need the level of detail I'm about to give you in the HSC 

Before I start, the grid you are referring to is just another electrode like the cathode/anode. It is therefore, definitely connected to a voltage source, which will cause it to be positively or negatively charged 

Basically, Cathode Ray Tubes can contain a number of grids between the cathode and anode, which fulfil different functions. When you refer to the velocity/number of electrons, we can actually control those with separate grids. By placing a negatively charged grid immediately in front of the cathode, we can deflect a certain number of electrons based on how negatively charged the grid is. This way we can control the number of electrons. Placing the grid halfway between the cathode and anode won't change the number of electrons too much. Instead, it will just alter their velocity, because by this point they are already moving fast enough that they are almost definitely just going to continue to the anode

Pretty much, we can fulfil different functions based on where we place the grid and how it is placed. That can get quite complex (I simplified it a tad above, and it could be slightly off). Of course the simple version of this answer is just that the grid (or grids) is placed halfway between the cathode and anode, charged positively or negatively, to change the number of electrons striking the screen. That is the version Rui grabbed from Physics in Focus, and all that you'll need in any exam 

[Happy Physics Land]

100% agree. Who is josh, by the way? Is he an ATARNotes lecturer?

We all love the work you have done for economics and business!!! Extremely happy to see you involved and your work has been fabulous mate!!! (Is that what you want me to say???)

[MysteryMarker]

Outline Einsteins explanation for the photoelectric effect. (6marks)

How would i go about this question to get the maximum? The marking criteria says 'an explanation of at least three observations', what does this mean?

Thanks guys.