#$&*
course Phy 202
submitted 02/10/12
PH2 Class 120118
Bottlecap and tube experiments:
Put some water into a bottle and screw the bottlecap on it, being sure the end of the tube extends down into the water. Support the upper end of the tube so that it’s more or less vertical.
Squeeze the tube until water is 90% of the way to the top of the tube. The force you have to exert to achieve this will designated as ‘10’ on the squeeze scale.
Release and squeeze the tube until water is only halfway to the 90% height. We’ll call this squeeze a ‘5’ on the squeeze scale.
We’ll think of dividing the first 90% of the tube into 10 equal parts, with a numbered height scale running from 1 to 10.
Close your eyes and try to repeat the ‘5’ squeeze, then look and see how you did. You should be able to estimate the actual height you reached on your 1-10 scale. How close did you get to the ‘5’ height?
Spend about a minute practicing this. Then try again, four times, to repeat the ‘5’ squeeze, without looking. On each trial, look and see how you did and write down how high the water column really rose, as you estimate it on the 1-10 scale.
*********************************************
Question: A BB of mass .1 grams, moving at 5 m/s perpendicular to a wall, collides elastically with the wall 60 times per second. What is its momentum change during a single collision?
Answer: The momentum of the BB is p = m v = .1 g * 5 m/s = .5 g m / s = .5 (.001 kg) m / s = .0005 kg m/s, or 5 * 10^-4 kg m/s.
On collision the ball comes to rest, so that the magnitude of its momentum change when coming to rest is .0005 kg m/s. The momentum was originally toward the wall, so the direction of the change in momentum is away from the wall.
It then rebounds, gaining momentum equal in magnitude to its original momentum, so there is another change of magnitude .0005 kg m/s. The direction of this change in momentum is also away from the wall.
So the total change in momentum is the sum of .0005 kg m/s directed away from the wall, and .0005 kg m/s directed away from the wall, i.e., .001 kg m/s.
Another way to get this: Let the direction of motion toward the wall be positive. Momentum then goes from +.0005 kg m/s to -.0005 kg m/s, a change of -.001 kg m/s. The negative sign means that the change in momentum is in the direction opposite the positive direction, so the change is .001 kg m/s away from the wall.
*********************************************
Question: What is the average force exerted by the wall on the BB?
The momentum of the BB changes by .001 kg m/s every 1/60 second, so by the impulse-momentum theorem
F_ave = .001 kg m/s / (1/60 second) = .06 kg m/s^2 = ,06 Newton.
*********************************************
Question: If a .1 gram BB bounces to the right and left, back and forth along the axis of a cylinder of length 20 cm, colliding elastically with the ends of the cylinder while moving at 10 meters / second, then what average force does it exert on the left end of the cylinder?
We will again use the impulse-momentum theorem, dividing the change in momentum by the associated time interval.
The momentum of the BB has magnitude .1 gram * 10 m/s = .001 kg m/s, so on collision is change in the momentum has magnitude 2 * .001 kg m/s = .002 kg m/s.
Between collisions with the left end of the cylinder the BB must move to the other end and back, a distance of 40 cm. At 10 m/s this requires .004 second.
So the average force has magnitude
F_ave = `dp / `dt = .002 kg m/s / (.004 s) = .5 Newton.
*********************************************
Question: If there are 9 identical BBs all moving at 10 m/s, and we manage to keep them all moving parallel to the axis, what is the average force on the left end?
Pick any BB. In the .004 s interval between collisions of that BB with the left end, all of the other 9 BB’s will also make collisions, so the total momentum change in .004 s has magnitude 9 * .002 kg m/s = .018 kg m/s. The average force thus has magnitude
F_ave = `dp / `dt = .018 kg m/s / (.004 s) = 4.5 Newtons, which as expected is97 times as great as the average force exerted by one BB.
Observation: If you put 9 BB’s into the cylinder it will be difficult to keep them all moving axially. Should a collision occur, two of the BB’s will cease moving along the axis and will begin crossing the paths of the others, which will quickly lead to more collisions. Some BB’s will be sped up, some slowed down by collisions. If collisions are elastic the total energy of the system will remain constant, but the directions will become randoms and speeds of the BB’s will become nonuniform. The collisions of the BB’s with the ends of the container will no longer occur at right angles to the ends, so less momentum change will occur for each collision. The velocities of the BB’s will no longer take them directly from one end to the other, so there will also be fewer collisions.
This randomization of direction from 1 dimension to 3 dimensions ends up reducing the average force by a factor of 3.
*********************************************
Question: Once velocities have been randomized, how much average force will the 9 BB’s therefore exert on the left end?
The 4.5 Newtons we would expect if the velocity of the BB’s remained perpendicular to the ends is reduced, once the velocities have become randomized in direction, by a factor of 3. The resulting average force is therefore
F_ave = 1/3 * 4.5 Newtons = 1.5 Newtons, approx..
*********************************************
Question: If the number of BB’s in a cylinder of length L is N, with each BB moving with speed v perpendicular to the ends, and if each BB has mass m, then what average force is exerted on the left end of the cylinder?
Using the same reasoning as above:
The momentum of a single BB is m v, and the magnitude of the momentum change as a BB collides with the left end is 2 m v.
A given BB must move distance 2 L between collisions with the left end. So the time between its collisions is 2 L / v.
In the time between two collisions of any selected BB with the left end, all N of the BB’s in the container will collide with that end.
Thus the total momentum change at the left end, during time interval `dt = 2 L / v, is N * (2 m v), and the average force exerted on that end is
F_ave = `dp / `dt = 2 N m v / (2 L / v) = 2 N m v * (v / (2 L) ) = N m v^2 / L.
*********************************************
Question: If the cross-sectional area of the cylinder is A then what is the average pressure on the left end?
Pressure is force / area.
The left end experiences average force N m v^2 / L, and the area of the end is the same as that of the cross-section, so the pressure on this end is
Pressure on end, assuming aligned BB velocities: P_ave = F_ave / A = (N m v^2 / L) / A = N m v^2 / (L * A) = N m v^2 / V,
where V = L * A is the volume of the cylinder.
Note also that m v^2 is just 2 * (1/2 m v^2) = 2 * KE, so the average pressure is
Pressure on end, assuming aligned BB velocities: P_ave = N * (2 KE) / V.
*********************************************
Question: Realistically, the velocities of the particles will randomize in direction, while the total KE of the particles remains the same. What will this do to the expression for the pressure?
As before, going from aligned 1-dimensional motion to directions randomized in 3-dimensional space, the total force will become 1/3 as great, making the pressure 1/3 as great. So we have
P_ave = 1/3 N * (2 KE) / V
or
P_ave = 2/3 N * KE / V,
where KE is the average KE of a particle and V the volume of the container.
*********************************************
Question: The Ideal Gas Law says that P V = n R T, where n is the number of moles of the gas and R is the gas constant 8.31 J / (particle Kelvin), n is the number of moles of gas and T is the Kelvin temperature. How does this compare with the result P = 2/3 N * KE / V?
Multiplying the latter expression by V we get
P V = 2/3 N * KE
The right-hand sides of our two expressions for PV are therefore equal, so
n R T = 2/3 N * KE
where we recall that KE is the average kinetic energy ½ m v^2 of a single particle.
*********************************************
Question: What therefore is the expression for the average KE of a particle?
Solving n R T = 2/3 N * KE we get
KE = 3/2 n / N * R T.
*********************************************
Question: What is the meaning of n / N?
N is the number of particles and n the number of moles. The number of particles is equal to the number of moles, multiplied by Avagodro’s Number, which we will denote N_A. Thus N = n * N_A, so that n / N = n / (n * N_A) = 1 / N_A.
Note that n / N is a number that does not depend on either n or N. n / N is just the constant number 1 / N_A.
So we can write the expression for the KE of a particle as
KE = 3/2 ( 1 / N_A ) * R T = 3/2 * (R / N_A) * T, or
KE = 3/2 k T, where k = R / N_A.
*********************************************
Question: What is the meaning of k, and what is its value?
R is the gas constant, in Joules per (mole Kelvin). When we divide by N_A, we’re dividing by the number of particles in a mole, so we get Joules per (particle Kelvin).
So k T has units of J / (particle Kelvin) * Kelvin = J / particle.
3/2 k T has the same unit. Thus the calculation
KE = 3/2 k T
gives us the number of Joules per particle, at temperature T.
The value of k is
k = R / N_A = 8.31 J / (mole Kelvin) / (6.02 * 10^23 particles / mole) = 1.38 * 10^-23 Joules / (particle Kelvin).
`q001. Phy 201 students only: Report the data you took in class on the dimensions of the dominoes. Be sure to use an organized table to report your measurements.
****
Paper Meter Stick
L W H L W H
A 12.2 6.1 2.0 5.2 2.5 0.9
B 12.2 6.0 2.2 5.2 2.5 0.9
C 12.3 6.0 2.3 5.2 2.6 1.0
D 12.1 6.0 2.4 5.1 2.6 1.0
E 12.2 6.2 2.3 5.2 2.6 1.0
F 12.3 6.1 1.8 5.2 2.6 0.8
#$&*
Indicate the uncertainty in your measurements for the meter stick, and for the paper rulers. Explain why you don’t think your uncertainty is much lower than you estimate, and why you don’t think it’s much higher.
****
You calculate uncertainty by taking the mean of the calculations and then dividing the deviation of the measurement to the mean by the mean then multiplying the answer by 100.
ex. The mean of the length of the paper ruler is 12.2.
The deviation of domino A from the mean is zero. so, 0/12.2 = 0 * 100 = 0% uncertainty
Percent uncertainty
Paper Meter Stick
L W H L W H
A 0% 0% 9.1% 0% 3.8% 0%
B 0 1.6 0 0 3.8 0
C 0.8 1.6 4.5 0 0 11.1
D 0.8 1.6 9.1 1.9 0 11.1
E 0 1.6 4.5 0 0 11.1
F 0.8 0 18.2 0 0 11.1
The uncertainty was lower than expected because the domino was big enough to be accurately measured on the millimeter scale.
#$&*
@&
What you have indicated here is the percent deviation of each measurement form the mean.
However this is not the same thing as uncertainty. You can't have an uncertainty of 0 in any measurement of a quantity whose actual value varies continuously. The meter stick's smallest measurement is a millimeter, and the human eye is not capable of resolving anything as small as .01 millimeter. So the physiological limit of your uncertainty in the width, for example, is around +-0.1%. Practically, it's unlikely that you can estimate positions on the meter stick to within +- .1 millimeter; and the act of positioning and lining up your eye with the domino involves additional uncertainties.
*@
What is your percent uncertainty for the measurement of the length of a domino, for each ruler?
****
To get uncertainty of the length you add all the percent uncertainties together.
ex. paper ruler uncertanties 0%+0+0.8+0.8+0+0.8 = 2.4% uncertainty
Meter stick is 1.9% uncertainty
#$&*
What is your percent uncertainty for the measurement of the width of a domino, for each ruler?
****
Again add all the % uncertainties together.
ex. paper ruler uncertainties 0+1.6+1.6+1.6+1.6+0 = 6.4% uncertainty
Meter stick is 7.6% uncertainty
#$&*
What is your percent uncertainty for the measurement of the thickness of a domino, for each ruler?
****
Paper ruler uncertainties 9.1+0+4.5+9.1+4.5+18.2 = 36.3% uncertainty
Meter stick is 44.4%
#$&*
Calculate the volume of each domino, in the units of each of your rulers.
****
The volume is calculated by multiplying the length by the width and the height.
ex. domino A paper ruler was 12.2 long X 6.1 wide X 2.0 high = 148.84
Volume
Paper Meter Stick
A 148.84 11.7 cm
B 161.04 11.7
C 169.74 13.52
D 174.24 13.26
E 173.97 13.52
F 135.05 10.82
#$&*
What do you think is the percent uncertainty in your calculation of the volume, for each ruler?
****
add the uncertainties together to get the total uncertainty of each ruler
ex. paper ruler volume 7.3%+0.3+5.8+8.6+8.4+15.8 = 46.2% uncertainty
Meter stick volume uncertainty is 48.7%
#$&*
@&
You need to calculation 'in the units of each of your rulers'.
You've calculated for the meter stick but haven't yet done so for the paper ruler.
*@
`q002. 500 BB’s each of mass 0.15 gram are place in a cylinder 50 cm long and 20 cm in diameter, and the cylinder is shaken rapidly and randomly, giving the BB’s an average velocity of 8 m/s. Assume that their directions of motion are randomized over 3 dimensions of space.
What average pressure is exerted by the BB’s on the ends of the container?
****
first, .15 g bb (8m/s) = 1.2 g m/s (.001 kg) = .0012 kg m/s per bb
second, 50 cm length convert to .5 m in lenght divided by 8 m/s = .0625 sec to hit each end
third, .0012 kg m/s / .0625 s = 0.0192 N
fourth, surface area of cylinder to get pressure by, 2 pi .1m^2 + 2 pi (.1m)(.5m) = .3769 m^2
fifth, .0192 N / .3769 m^2 = 0.0509 Pascals per bb so .0509 * 500 = 25.47 Pascals for all 500 bb's
@&
Good sequence of reasoning, and mostly good calculations.
However the BB has to make a round trip between collisions with a given end, so the time interval is longer.
Also
.15 g bb (8m/s) is not equal to.0012 kg m/s per bb
so you have misused the '=' sign in your calculation
.15 g bb (8m/s) = 1.2 g m/s (.001 kg) = .0012 kg m/s per bb
'=' means 'equal to', not 'and the next thing we calculate is ... '
The sequence of calculations is fine; the incorrect use of the '=' sign is potentially confusing and should be avoided.
*@
#$&*
What is the average force exerted by the BB’s on one of the ends of the container?
****
Average force is .0012 kg m/s / 0.125 (doubling the .0625 for each side) = 0.0096 N then multiplied by 500 bb's = 4.8 N for one end of cylinder.
then because it is in 3 dimensions it will be 1/3 * 4.8 N = 1.6 N
#$&*
Atmospheric pressure is about 10^5 Newtons / meter^2. What percent of atmospheric pressure is achieved by this system?
****
101300 pascals equals one atm so (25.47 / 101300) *100 = 24.7%
#$&*
@&
25 is .00025 of 100 000, or .025%.
25% of 100 000 would be 25 000.
*@
What is the total KE of all the BB’s?
****
1/2 m v^2 = 1/2 (.15 g)(8 m/s)^2 = 4.8 N * 500 bb's = 2400 N
#$&*
@&
Again your thinking is clear, but '=' means 'equal to' and that is its only meaning.
*@
To what Kelvin temperature is this situation equivalent?
****
Room temp is approx 20 degrees celsius so 293.15 K
#$&*
@&
The average KE per particle is 3/2 k T.
This gives an impossibly large result for T, but that's what we get.
*@
`q003. Suppose now that you push on one end of the cylinder, moving it 1 cm closer to the other end, so that the cylinder’s length decreases to 49 cm.
You previously figured out how much average force is exerted on an end of the container by the BB’s. Assuming that the end you are pushing on moves without frictional resistance, how much work must you do to move it through its 1 cm displacement?
****
.49 / 8 m/s = .06125 so 0.0012 / .06125 = 0.01959 N
So then subtract 0.0192 from 0.01959 = 0.00039 N * (.01 m) = 3.9 x 10^-6 Joules of work
#$&*
@&
You could calculate the average force exerted if the length is reduced to 49 cm.
However the result won't be a whole lot different than is was for 50 cm (it will differ by about 2% from that value).
So you're pretty safe just assuming that the force doesn't change significantly.
To get the work you multiply force by displacement.
*@
You just did work on the system. What happened to that energy? Assume no energy loss to friction, and assume that all collisions are perfectly elastic, which means that no kinetic energy is lost in any collision.
****
That energy is then moved to potential energy.
#$&*
@&
Nothing moves higher or lower, so there's no change in gravitational PE.
There are other forms of PE, but they don't change either.
If none of your work goes into PE, and if the collisions are elastic so no energy is dissipated, what are you left with (according to the work-energy theorem)?
*@
`q004. Two pendulums are released simultaneously. One oscillates at 60 cycles / minute, the other at 56 cycles / minute. So most of the time the oscillations of the pendulums are not synchronized, returning to the starting point at different times.
How many times per minute will the two pendulums become synchronized, in the sense that both return to their starting points at the same time?
****
the first one will reach its starting point once every second and the second pendulum reaches it starting point every 0.933 seconds.
Which means they will become synchronized four times a minute.
#$&*
`q005. To the left of the y axis, a line runs parallel to the x axis, 1 centimeter higher than the x axis. At the point (0, 1), the line changes direction so that it passes through the x axis at the point (5 cm, 0). This broken line will constitute path 1.
Another line, which defines path 2, runs below the x axis at a constant distance of ½ cm from the axis.
At what point do these two paths intersect?
****
they will cross at (7.5, -0.5) becuase the slope is -1/5
#$&*
A third path consists of a straight line passing through the origin and the point at which the first two paths intersect. At what point to the left of the origin does this third path intersect the first?
****
they will intersect at (-15, 1) becuase the slope is -5/7.5
#$&*
@&
RIght idea, but the slope is -0.5 / 7.5.
*@
`q006. How far is the point (0, 4 cm) from the point (-20 cm, 0)?
****
need distance of thrid line by pathagoreon therom so 4^2 + 20^2 = Z^2 = SQRT 416 = Z = 20.39 cm distance from both points
#$&*
How much further from the point (-20 cm, 0) is the point (0, 5 cm)?
****
20^2 + 5^2 = Z^2 = SQRT 425 = 20.62 cm from these two points so these points are 20.39 - 20.62 = .23 cm further than the previous points.
#$&*
How far would you have to move along the y axis, starting at (0, 4 cm), in order to move 1 millimeter further from the point (-20 cm, 0)?
****
Y^2 = 20.49^2 (adding one millimeter to 20.39) - 20^2
Y = 4.45 cm
#$&*
How far would you have to move along the y axis, starting at (0, 4 cm), in order to move 1 millimeter closer to the point (-20 cm, 0)?
****
X^2 = 4^2 - 20.29^2
X = 19.89 cm
#$&*
@&
If you move 19.89 cm along the y axis from (0, 4 cm) your distance from (-20, 0) is going to change by way more than 1 millimeter. And you'll be further from, not closer to that point.
*@
At what points along the y axis would your distance from the point (-20 cm, 0) differ from the distance between this point and (0, 4) by 2 millimeters?
****
I am sorry, I cannot figure out what you are trying to ask. Is there any way you could rephrase it for me.
#$&*
@&
I transposed the statement, making the question unanswerable.
What I meant to type was
'At what points along the y axis would your distance from the point (0, 4 cm) differ from the distance between this point and (-20 cm, 0) by 2 millimeters?'
*@
Take a good look at your results. Do you see any patterns?
****
I see that since the x axis is larger, only moving the point by a tenth will increase the length of the line by a millimeter but to increase the line a millimeter on the y axis you have to move almost 4.5 mm.
#$&*
@&
Good.
*@
`q007. A well-insulated container is divided into two compartments.
Suppose the compartments are of equal volume and contain equal amounts of air, with the air in one compartment at 80 degree Fahrenheit and the other at 50 degrees Fahrenheit. If the divider is removed and the air allowed to mix throughly, what would you expect to be the final temperature?
****
They should mix equally so the final temperature would be 65 degrees Fahrenheit.
#$&*
Suppose the compartment at 80 Fahrenheit has double the volume of the other and contains twice as much air. If the divider is removed and the air allowed to mix throughly, what would you expect to be the final temperature?
****
the difference is 30 degrees between 80 and 50. So, if equal the temperature should be in the middle at 65 degrees. because the higher temper compartment has double volume and double air add one-fourth of 30 = 7.5 to 65 to equal 72.5 degrees Fahrenheit once they are mixed equally.
#$&*
@&
If it was double the volume as well as double the density, you would end up at this temperature.
However you have only twice as much air in the 'double-volume' side, not 4 times as much.
*@
The meaning of ‘equal amounts’ is a little vague. ‘Equal amounts’ could possibly mean equal volumes, or equal masses. If the volumes of the compartments are the same and the masses of air are the same, then one of the compartments will be at higher pressure than the other. Which will this be?
****
It would be the one with the higher temperature.
#$&*
If the two compartments have the same volume and are at the same pressure, will the masses of air be equal? If not, which will be greater?
****
yes becuase they should contain the same amount of particles and moles.
#$&*"
end document
@&
Very good overall.
You do need another set of domino volumes. My note will clarify.
In addition you could use a few revisions, which however shouldn't take you long.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.
Spend a reasonable amount of time on your revision, but don't let yourself get too bogged down. After a reasonable amount of time, if you don't have at least a reasonable attempt at a solution, insert the best questions you can showing me what you do and do not understand, and I'll attempt to clarify further. &#
*@