Physics Questions

[#1procrastinator]

1) Is stress a vector or scalar? Apparently it's a tensor but for the purposes of VCE, should I just treat it as scalar or vector?

2)  2 people complete a 100m race in 10.2 seconds, Person A accelerated uniformly for the first 2 seconds, Person B accelerated uniformly for the first 3 seconds. How do I determine their constant velocity throughout the rest of the race?

Thanks

[DisaFear]

1) Is stress a vector or scalar? Apparently it's a tensor but for the purposes of VCE, should I just treat it as scalar or vector?

The Physics study design thingy says

Investigations of the shape and composition of the material will be carried out to
determine its behaviour under stress up to the point of its destruction.

...calculate the stress and strain resulting from the application of compressive and tensile forces and
loads to materials in structures...

I'm thinking just treat it as scalar, just something that you calculate.

Still thinking on how to tackle that motion problem. Thinking it'll need simultaneous equations, maybe

[mark_alec]

Draw trapezia for velocity vs time, you know the area of it (100 m), and you know length of the top and bottom edges.

[DisaFear]

I found a very similar problem in my uni physics textbook (it was in the challenge problems!)

Initial velocity of Person A is 0 m/s, as he starts from rest
Let the distance that Person A traveled during the acceleration phase of 2 seconds be 's'

We know from kinematic formulas:




(equation 1)

Velocity of the sprinter after his 2 seconds of acceleration can be calculated by:



(equation 2)

If we fiddle around with both equation 1 and equation 2, we come up with:

(equation 3)

So, this Person A uses this constant 'v' velocity to complete his race. The distance he covers is 100-s, where 's' is the distance that guy accelerated for. We can use kinematic formulas to work out the part of constant velocity

We know that:


WE ARE NOW WORKING OUT THE CONSTANT VELOCITY PHASE WITH NO ACCELERATION, conditions have been changed, 's' will be 100-s, the distance he has to travel after his fit of acceleration, and his initial velocity, 'u', will be the velocity he obtained from acceleration in the previous part, which is 'v' (I should be using proper physics notation, but it'd probably make it more confusing to you)
The time in this part will be 10.2-2 as well! No acceleration, keep that in mind



This is where we bring equation 3 back into play, substitute 'v' in for the 's'





And that's that. Work it out for the other guy. The place where I was going wrong, I thought both A and B were connected somehow. Seems not. Answer has been confirmed by graph method, as worked out by laseredd. Graph method is way easier, as may be proved below

[Lasercookie]

Answer has been confirmed by graph method, as worked out by laseredd. Graph method is way easier, as may be proved below

heh, might as well post it up now then

Draw trapezia for velocity vs time, you know the area of it (100 m), and you know length of the top and bottom edges.

The 'graph method' that DisaFear referred to was was just following this method of working out which mark_alec outlined

It should be fairly easy to draw the graphs. The area under the curve (the integral) of velocity-time graphs is displacement (this point should be fairly clear too)

Area of trapezium = Area of Triangle + Area of Rectangle = 100m



(where t_a is the time spent accelerating)

So for person A:






Person B:

[nina_rox]

Question - what do we need to know about AM/FM?

[#1procrastinator]

Wow, awesome, thanks a lot guys!

[#1procrastinator]

The lifetime of a muon is about 2.2 microseconds in the laboratory, at rest to an observer. When they are created in the atmosphere, they're moving super fast and from our frame of reference on Earth, they appear to have a lifetime of around 64 microseconds. Estimate the speed that the muons are traveling at in the atmosphere as a percentage of c.

A) 99.92%
B) 99.94%
C) 99.96%
D) 99.98%

Book got the value of gamma = 29 and somehow got B. I got D. They divided 64/2.2 but I did 2.2/64...shouldn't 2.2 microseconds the lifetime of the muon in its reference frame and 64 microseconds be from our frame?

[DisaFear]

The lifetime of a muon is about 2.2 microseconds in the laboratory, at rest to an observer. When they are created in the atmosphere, they're moving super fast and from our frame of reference on Earth, they appear to have a lifetime of around 64 microseconds. Estimate the speed that the muons are traveling at in the atmosphere as a percentage of c.

A) 99.92%
B) 99.94%
C) 99.96%
D) 99.98%

Book got the value of gamma = 29 and somehow got B. I got D. They divided 64/2.2 but I did 2.2/64...shouldn't 2.2 microseconds the lifetime of the muon in its reference frame and 64 microseconds be from our frame?

You probably don't need an answer to this anymore, but still XD

Afaik, proper time is measured in the frame where both events happen at the same place. That muon in the atmosphere, if you hold it as stationary, then the Earth is traveling towards it. It is sitting there, biding its time, as Event A (passing atmosphere) and Event B (reaching Earth), occur. (I think!) That would probably give their answer.

But...I always thought the proper time was the shortest interval measured...

Yea, I don't really know, maybe my musings can lead you to the right path. Feel free to correct me, I hate these questions.

[#1procrastinator]

^ Eh, I'm still interested :p

I only skimmed over the textbook's definition of proper time because it seemed like it result was the same as how I'd been thinking of it, like you said, proper time is the shortest time interval. I think of the left hand side of the time dilation equation as the time of the guy that's watching some object moving at a relativistic velocity. So the object, say the muon moves so that it appears to us that it's ageing/decaying slower than it should (64 microseconds), that would be the time of the left hand side and so its time would be 2.2 microseconds, which should never change in its reference frame, it doesn't feel as though its taking slower to age.

So if you plug in the numbers and do the algebra, you get 2.2/64 = gamma which is how I got what I got.

The solution does 64/2.2 which suggests that the proper time is 64 microseconds (frame of the muon) and that the time we see it last is 2.2 microseconds which seems to go against what I think I know about SR.

I think I rambled a bit there...

[Lasercookie]

The lifetime of a muon is about 2.2 microseconds in the laboratory, at rest to an observer. When they are created in the atmosphere, they're moving super fast and from our frame of reference on Earth, they appear to have a lifetime of around 64 microseconds. Estimate the speed that the muons are traveling at in the atmosphere as a percentage of c.

Irrelevant: The experiment that the book is talking about is this: http://en.wikipedia.org/wiki/Time_dilation_of_moving_particles#Rossi.E2.80.93Hall_experiment

The idea was that if you didn't consider time dilation, then we should observe barely any muons reaching the Earth.
I always thought of it like this: proper = property = the observer's own clock --> not some other observer's clock.

The time on the muon's clock is the value of 2.2 microseconds.
This means that what we observe is the dilated time. This is the value of 64 microseconds.

Muon frame of reference = proper time: 2.2 microseconds
Earth frame of reference = relativistic time: 64 microseconds

(the error seems to be due to algebra for this step here? - or have I just forgotten something crucial?)



Rearrange for v:










then figure out what percentage of the speed of light that is:





Therefore B

edit: fixed LaTeX (forgot to escape the sqrt's!)

[#1procrastinator]

^ Thanks laseredd

yeah, it was an algebraic mistake. i had but I multiplied both sides by the denominator and then divided both sides by

[#1procrastinator]

What would the implication of a non-massless rope be in a pulley problem (e.g. two masses attached to either end of a rope which is suspended over a pulley)? Also, can there be nonuniform tension in a rope?

[DisaFear]

The tension at one end would be higher than the other end, this difference in tension is needed to accelerate the string (this is, of course, for accelerating strings)

This equation is in my textbook regarding ropes with mass



Where A and B are blocks, S is the string and x is the motion in the x-plane (Sorry, I can't replicate the diagram in the book)

Hope that helps

[#1procrastinator]

^ thanks...I'm trying to visualise the difference between an accelerating system and one where the rope is accelerating