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Phy 122
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Speed of Light
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Problem set 7 number 24 states:
A spacecraft moving at constant velocity .7 * c relative to a nonaccelerating observer keeps time by reflecting a light pulse back and forth between two mirrors oriented at a right angle to the direction of motion. The mirrors are 70 meters apart (big spaceship) and are mounted within a sealed vertical vacuum tube (vertical being perpendicular to the direction of motion) so that the light pulse will travel at its vaccuum speed.
If we assume that the laws of physics are the same in the reference frame of the spaceship as in the frame of the observer, it follows that the speed of light in a vacuum will be identical in both reference frames.
Under this assumption:
How long does it take a pulse to travel from the lower mirror to the higher, as measured in the spaceship?
How long does it take the pulse to travel from the lower mirror to the higher as measured by the observer outside the spaceship?
If the pulse rate of an inhabitant of the spaceship is 64 beats/minute, as measured by that individual, then what pulse rate would be measured by the outside observer, assuming some accurate means of making the observation?
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I can see from the very long solution that using symbols is clearly the way to go!
I can calculate the how much greater the time is from the observer's standpoint, but I wanted to run my process by you and see if I got it right.
From the spacecraft the light takes 70 meters/c
this is 70m/(3*10^8m/s) = 2.33 *10^-7 sec
From the outside observer's standpoint, we calculate that
time is multiplied by a factor of 1/`sqrt(1-(v^2/c^2)
1/`sqrt(1- (.7^2))= 1.4
Thus all outside observed times will be 1.4 times that observed from the spacecraft.
So since the light takes 2.33* 10^-7 sec when observed from within, it would take 2.33* 10^-7 sec * 1.4 or 3.26*10^-7 secs.
For the pulse, I waffled a bit but finally settled on the concepts that the beats would stay the same but the time would change. Thus 64beats/minute would be observed from outside as 64 beats/1.4 minutes which would be about 45.7 beats per minute.
??? Am I approaching this the right way???
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Conceptually the key is that the path of the light is longer as observed from the observer's frame, where it moves in both the horizontal and vertical directions, than in the rocket frame where it moves only in the vertical direction.
The expression 1 / sqrt( 1 - v^2 / c^2 ) can be obtained directly from analysis of the resulting triangles.
You do end up with a time interval factor of 1.4 and (64 beats / minute) / 1.4.
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Phy 122
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Alpha Particle
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An alpha particle consists of two protons and two neutrons. The neutrons are slightly more massive than the protons, but the average mass of all the particles is reasonably close to 1.6 * 10^-27 kg. The fundamental charge is 1.6 * 10^-19 Coulomb. If a beam of alpha particles moving at 4.5 * 10^6 m/s enters a magnetic field of .005 Tesla, with the velocity and field perpendicular to one another, what will be the radius of the circular path of the beam in the magnetic field?
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This is how I would approach this:
F = qvB
F = 1.6 * 10^-19 C * 4.5 * 10^6 m/s * .005 Tesla
F = 3.6 * 10^-15 N
We calculate the total mass of the particle as 1.6*10^-27 *4 = 6.4*10^-27
(this is the average mass of each particle times the 4 particles in the neutron)
F= ma
a= F/m
a= 5.63*10^11 m/s^2
this will be the centA
centA = v^2/r
r= v^2/centA
r=(4.5*10^6 m/s)^2/5.63*10^11 m/s^2
r=( 2.025*10^13m^2/s^2 )/ 5.63*10^11 m/s^2
r= 36.0 meters
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???#1 is that correct reasoning???
???#2 If I have 2 protons in this alpha particle, am I right in assuming it still only have a charge of 1.6*10^-19 (and not twice that amount)???
Thanks for your help
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Good, but each proton carries a charge of 1.6 * 10^-19 Coulombs.
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This will change your final result, but the calculations will be easy enough to adjust.
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