Post by: #1procrastinator on May 11, 2012, 11:07:26 am
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
Post by: DisaFear on May 11, 2012, 06:00:22 pm
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
Quote
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.
Quote
...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
Post by: mark_alec on May 11, 2012, 06:27:16 pm
Post by: DisaFear on May 11, 2012, 09:12:44 pm
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:
Velocity of the sprinter after his 2 seconds of acceleration can be calculated by:
If we fiddle around with both equation 1 and equation 2, we come up with:
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
Post by: Lasercookie on May 11, 2012, 09:26:29 pm
Answer has been confirmed by graph method, as worked out by laseredd. Graph method is way easier, as may be proved belowheh, might as well post it up now then :P
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
So for person A:
Person B:
Post by: nina_rox on May 13, 2012, 01:21:35 pm
Post by: #1procrastinator on May 15, 2012, 01:16:23 pm
Post by: #1procrastinator on June 12, 2012, 02:10:34 pm
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?
Post by: DisaFear on June 19, 2012, 09:51:41 pm
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.
Post by: #1procrastinator on June 24, 2012, 08:10:18 am
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...
Post by: Lasercookie on June 24, 2012, 01:02:20 pm
Quote
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
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!)
Post by: #1procrastinator on June 24, 2012, 02:27:14 pm
yeah, it was an algebraic mistake. i had
Post by: #1procrastinator on June 26, 2012, 09:24:26 pm
Post by: DisaFear on June 26, 2012, 09:29:41 pm
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
Post by: #1procrastinator on June 26, 2012, 10:23:49 pm
Post by: paulsterio on June 26, 2012, 10:42:22 pm
Question - what do we need to know about AM/FM?
No, just AM.
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?
Basically what DisaFear said.
But there are many other situations to consider as well, a pulley with mass (angular acceleration of the pulley, assuming that the pulley does not slip), you can also consider the situation of elastic strings (take into account strain potential energy) and other situations like that, but they're all well beyond the VCE course, so best not to worry really.
Post by: #1procrastinator on June 26, 2012, 11:37:38 pm
but yeah, i don't want to delve that deep into yet. i was just curious cause in every problem the strings/ropes are massless and wondered what would happen if they weren't