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course phy 201
1/31/12 7:50 pm
PH2 Class 120118Bottlecap 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.
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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.
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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.
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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.
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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.
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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..
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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Domino ruler length width thickness
E paper 7.7 cm 3.8 cm 1.4 cm
Meter stick 5 cm 2.5 cm .9 cm
F paper 7.7 cm 3.8 cm 1 cm
Meter stick 5 cm 2.5 cm .7 cm
D paper 7.8 cm 3.8 cm 1.4 cm
Meter stick 5.05 cm 2.5 cm .9 cm
C paper 7.7 cm 3.7 cm 1.4 cm
Meter stick 5 cm 2.4 cm .9 cm
A paper 7.7 cm 3.7 cm 1 cm
Meter stick 5 cm 2.4 cm .8 cm
B paper 7.8 cm 3.8 cm 1.3 cm
Meter stick 5.05 cm 2.5 cm .875 cm
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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.
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I would say that my uncertainty is about 1 mm. I used my finger to keep the domino in line with the end of the meter stick and then used a piece of paper to measure across the end of the domino to the marks on the meter stick. The paper ruler is more like 2 mm. Not only were the spacings between the marks closer together, but it was also a bit more difficult to line the domino up with the end of the paper.
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What is your percent uncertainty for the measurement of the length of a domino, for each ruler?
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Paper ruler: 2 mm / 77 mm = 2.6%
Meter stick: 1 mm / 50 mm = 2%
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What is your percent uncertainty for the measurement of the width of a domino, for each ruler?
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Paper ruler: 2 mm / 38 mm = 5.3%
Meter stick: 1 mm / 25 mm = 4%
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What is your percent uncertainty for the measurement of the thickness of a domino, for each ruler?
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Paper ruler: 2 mm / 12 mm = 16.7%
Meter stick: 1 mm / 8 mm = 12.5%
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Calculate the volume of each domino, in the units of each of your rulers.
Paper ruler: 7.7cm * 3.8cm * 1.2cm= 35.11 cm^3
Meter stick: 5cm * 2.5cm *.8 cm= 10cm^3
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You're doing the calculation correctly, but you need a volume calculation for each domino.
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What do you think is the percent uncertainty in your calculation of the volume, for each ruler?
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Paper ruler: 8.2%
Meter stick: 6.17%
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If there's a 12% uncertainty in the thickness, for example, there's more than a 12% uncertainty in the volume.
You are multiplying three quantities. How do the uncertainties in those quantities combine to give you the uncertainty in the result?
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`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?
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P ave= 500* 2 * .0015 kg * 8 m/s= 12 kg m/s
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This would be the total momentum of the BB's. But it's not the average pressure.
You need to first find the average force, then the average pressure.
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What is the average force exerted by the BB’s on one of the ends of the container?
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F ave= 12 kg m/s)/ (1 m/8 m/s)= 98 kg m/s^2
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Atmospheric pressure is about 10^5 Newtons / meter^2. What percent of atmospheric pressure is achieved by this system?
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???
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What is the total KE of all the BB’s?
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Ke= 1/2mv^2
=1/2(500*.15g)(8m/s)^2
=.0375kg*64m^2/s^2
=2.4J
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To what Kelvin temperature is this situation equivalent?
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12 kgm/s* .016m^3=500 *.0821 *T
=.005deg K
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KE_ave = 3/2 k T.
What therefore is T?
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`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?
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You have to exert enough force to overcome the force exerted by the BB's. How much force is that, how far are you pushing the end, and how much work is done?
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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.
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`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?
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4
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`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?
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(7.5, -.5)
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Right. Good.
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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?
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(1, -15)
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Also right. Very good.
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`q006. How far is the point (0, 4 cm) from the point (-20 cm, 0)?
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-20cm on the x-axis
4 cm on the y-axis
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There is only one answer to 'how far' one point is from another.
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How much further from the point (-20 cm, 0) is the point (0, 5 cm)?
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20cm on the x-axis
5cm on the y-axis
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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)?
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-4.1cm
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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)?
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-3.9cm
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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?
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.1 and -.1
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Take a good look at your results. Do you see any patterns?
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`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?
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65 deg F
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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?
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70 deg F
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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?
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The one with the higher temperature
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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?
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No, the one with the higher temperature will have less mass.
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end document
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