ic_class_100125

these last two should be in terms of m, v, v' etc.; if you substitute your expression m v - m v ' for `dp they would be

This is good reasoning, though grams * m/s^2 will not give you Joules.

However you were asked to plug this into your expression. Your expression was

F_ave = ‘dP / ‘dt ; should have been

F_ave = (m v - m v ') / ‘dt so you could plug in your information directly.

1000 Pa supports a 10 cm column; you would need 5000 Pa to support a 50 cm column

right for 1 mole; but the bottle contains only about .02 moles of gas

dividing mass by Celsius temperature has no meaning, so this reasoning won't work

Some of your solutions are very good.

Most need at least some refinement, and some are off the mark.

I've made notes on a few problems; the rest you can compare with the solutions given below.

Check out the document below, compare with your solutions, and let me know if you have questions.

100125

We'll worry about the energy questions and the optics this

Impulse-Momentum:

The impulse of a force F acting during time interval `dt

The impulse of the net force F_net acting on a system

• F_net * `dt = `d ( m * v ).

Today's Class:

If you bounce a ball off an analog bathroom scale you can

We bounced a ball off the scale. 

• We observed that
  after striking the scale, the ball rose to a height of 3 or 4 meters (or would
  have had the ceiling not intervened.  This is consistent with a rebound
  velocity near 10 m/s (if vf = 0, `ds = 4.9 m and a = -9.8 m/s^2 we get a 10 m/s
  velocity). 
• A ball dropped on the scale rebounds to about 1/4 its drop
  height, indicating that rebound velocity is about half of the final drop
  velocity ( | `dPE | is about 1/4 that of the drop so | `dKE | is about 1/4, so
  since KE = 1/2 m v^2 we conclude that the magnitude of rebound velocity is about
  half that of drop velocity).  So we estimate a downward speed of about 20
  m/s (double the 10 m/s rebound velocity) prior to impact.  These speeds are probably overestimates, by perhaps
  20-30% or so. 
• Estimating the mass of the ball to be 50 grams, its downward
  momentum is 50 g * 20 m/s = 1000 g m / s = 1.0 kg m/s, and its rebound momentum
  is 50 g * 10 m/s = 500 g m/s = .5 kg m/s in the upward direction.  Choosing
  upward to be positive the momentum just before impact is -1.0 kg m/s and the
  momentum just after impact is +.5 kg m/s, so the change in momentum at the scale
  is

`dp = p_f - p_0 = +.5 kg m/s - (-1.0 kg m/s) = +1.5 kg m/s.

If we had 10 people each throwing one ball downward at the

When the scale is struck once with the ball its reading can be

• The impact of the
  ball lasts much less than .1 second (more like .01 second, as we can infer from
  the speed of the ball and its dimensions), so the force exerted at impact is
  much more than the +15 N we estimated previously (if impact lasts .01 second,
  the 1.5 kg m/s momentum change implies a force of 1.5 kg m/s / (.01 s) = 150
  N, around 30 lb). 
• The impact gives the spring kinetic energy,
  and the scale system then undergoes damped SHM, which lasts for a few cycles. 

If we were to continue bombarding the scale with 10 impacts every second, each

`q001. If the mass of the ball is m, the before-collision

• What is the
  before-collision momentum, and what is the after-collision momentum of the ball? 
  (Be sure to pick a positive direction and use the correct signs in your
  quantities.)

****

We need to begin by choosing a positive direction. 

Let the positive direction be that of the ball after collision. 

Then the before-collision momentum is -m v and the

after-collision momentum is + m v '.

#$&*

• What therefore is the change in the ball's momentum? 

****

The change in the ball's momentum is

`dp = p_f - p_0 = m v ' - (-m v) = m ( v ' + v ).

#$&*

• If a number of such balls are thrown at the scale, with
  the average time interval `dt between impacts, then what is the average force
  exerted by the scale?  What is the average force exerted on the scale?

****

The average force on the ball, by the impulse-momentum

theorem, is

F_ave = `dp / `dt = m ( v ' + v ) / `dt.

#$&*

• If the area of the scale is about A, then what is the
  average force per unit area exerted by the balls?

****

We need just divide the average force by the area:

• F_ave / A = (m ( v ' + v) / `dt) / A.

Force / area is average pressure, so we can say that

• P_ave = F_ave / A = (m ( v ' + v) / `dt) / A.

#$&*

• Plug m = 50 grams, v = 20 m/s, v ' = 10 m/s and `dt = .1
  second into your expression for the average force.  What is your result?

****

F_ave = 50 grams * (10 m/s + 20 m/s) / (.1 sec) = .050

kg * 30 m/s / (.1 s) = 15 kg m / s^2 = 15 N.

#$&*

• Plug the same numbers, along with area A = .07 m^2, into
  your expression for force per unit area.  What do you get?

****

P_ave = F_ave / A = (m ( v ' + v) / `dt ) / A = (.050

kg * (10 m/s + 20 m/s) / (.1 s) ) / (.07 m^2) = 15 N / (.07 m^2) = 214 N /

m^2, approx.

This can also be expressed as 214 Pa.

#$&*

Molecules in a gas, striking a surface, behave in a manner similar

`q002.  If the directions of the balls striking the

****

P_ave = F_ave / A = (m ( v ' + v) / `dt) / A.

#$&*

Consider the piece of tubing, containing some water,

This can be explained by a model similar to that of the

• If one side was higher than the other, there
  would be more weight on one side than the other and the net force on the water
  in the tube would be nonzero, tending to accelerate the water toward the 'lower'
  side. 
• So the water would quickly settle into a state where the level is
  the same.

However if you use your mouth to 'pull' water up, as in a

• This
  results in greater average distance between the molecules, which in turn results
  in fewer collisions, during a given time interval, with the water surface on
  that side (there are fewer molecules within 'striking distance' if the spacing
  increases). 
• As a result the average force on that side is less, so that
  the force on the other side of the tube now exceeds that on your side and the water in the
  tube is accelerated toward your side, raising the level on your side and
  decreasing it on the other. 
• This results in greater
  gravitational force on your side. 
• When the extra gravitational force on
  your side balances the now-greater pressure on the other, the system will again
  be in equilibrium, this time with the water higher on your side.  (Again
  this model neglects some of the finer details of the situation; however in this
  case these details have an insignificant effect).

`q003.  If you were to block one end of the tube with

• How will this affect the forces exerted on the water
  surfaces on the two sides of the tube?

****

The side which remains open to the atmosphere will

experience the same force as before.

However if the length of the air column trapped

between thumb and water surface increases (which increases the volume of the

air), the molecules will have to travel further between collisions with the

water surface and the average force they exert will decrease.

More correctly we can say that since the volume

increases, fewer molecules are available to hit the water surface within a

given time interval `dt, which reduces the force they exert.

#$&*

• How will this affect the relative water level on the two
  sides?

****

The water in the tube is in equilibrium.  More

force is exerted on it by the pressure on the open side than on the closed

side.  To remain in equilibrium, gravity must therefore exert more

force on the closed side.  So the water level on the closed side will

be higher than the level on the open side.

#$&*

Suppose you have a long tube, open to the atmosphere, with

• The top 10 cm forms a column with cross sectional area
  5 cm^2. 
• This column is supported by the water directly beneath it, and by
  nothing else (that water might ultimately be supported by something else, but
  the only thing the 10 cm column is actually in contact with is the water
  directly below it--and of course the walls of the container, which exert no
  force in the upward direction). 
• The volume of the water in the 10 cm
  column is clearly 10 cm * 5 cm^2 = 50 cm^3, so its mass is 50 grams or .05 kg. 
  The weight of the water is therefore .49 N, which for now we will round to .5 N. 
  This weight exerts a downward force on the water column.
• It follows that, if everything is in equilibrium, the water directly below our
  column exerts an upward force of .5 N. 
• The cross-sectional area of the
  tube is 5 cm^2, so the force per unit area exerted by the water directly below
  our column is .5 N / (5 cm^2) = .1 N / cm^2.  Since 1 m^2 = (100 cm)^2 = 10
  000 cm^2, our result .1 N / cm^2 is the same as 1000 N / m^2 (10 000 cm^2 each
  with force .1 N results in force 1000 N).

Pressure is force / area.  So we see that the

`q004.  We generalize the above results, using

• If a water column has cross-sectional area A and
  height h, what is the expression for its volume? 

****

The volume of a column with uniform cross-sectional

area A and altitude h is the product of the cross-sectional area and the

height:

V = A * h.

#$&*

• If the density of the water is rho (i.e., if rho
  represents the mass of water per unit of volume), then what is the mass of
  the water in the column? 

****

The mass is the product of the volume and the density:

m = rho * V = rho * A * h.

#$&*

• What therefore is the weight of the water (hint: 
  to get the weight of something you multiply its mass by g, the acceleration
  of gravity)? 

****

The weight of mass m is just m g, where g is the

acceleration of gravity.

So the weight of the mass m = rho * A * h is

wt = rho * g * A * h.#$&*

• What therefore is the pressure at the bottom of the
  column (remember pressure = force / area)?

****

The bottom of the column must exert an upward force

equal to the weight of the column, which is rho g A h.

The pressure is the force divided by the area over

which that force is applied.

Thus the pressure is

P = rho g A h / A = rho g h.

#$&*

When we brought the bottle in from the cold, we estimated

You are expected to be able to convert Fahrenheit to

Celsius, and to know that the freezing point of water at atmospheric

pressure is about 272 Celsius degrees above absolute zero.  The F to C

conversion requires that you know only two things:  Water freezes at 0

Celsius and 32 Fahrenheit, and a Celsius degree is 9/5 of a Fahrenheit

degree. 

For example, 45 Fahrenheit is 13 F above freezing, so

it's 5/9 * 13 = 7 Celsius degrees above freezing, which is 7 Celsius. 

Similarly, 68 F is 36 F above freezing so its 5/9 * 36 = 20 Celsius degrees

above freezing.  Thus the air inside the bottle warmed from 7 C to 20

C, an increase of about 13 Celsius degrees. 

Applying the gas law P V = n R T to this situation: 

• The bottle was sealed (air had no way to move from
  inside the bottle to outside, or vice versa) so the number n of moles of air
  remained constant. 
• R is a physical constant. 
• The bottle might have expanded a little, increasing V
  slightly, but this would have a minimal effect.   A little water
  was displaced into the tube, also slightly increasing V, but again the
  effect would be slight.  So we can assume that V remains about
  constant.

Rearranging the equation so that the constant terms are

• Since P / T is constant, the percent increase in P
  must be the same as the percent increase in T. 
• We can calculate the percent increase in T, based on
  our temperature estimates.  Remember that T is the absolute
  temperature, so that when the temperature goes from 7 C to 20 C, the value
  of T goes from 273 K + 7 K = 280 K to 273 K + 20 K = 293 K. 

• The 13 K increase is 13/280 = .05, or about 5%, of
  the original temperature.   Thus the pressure goes up by factor of
  about .05, to about 1.05 times atmospheric pressure. 

We estimated that this resulted in a water column in the

• Recalling that a 10 cm column requires pressure 1000
  Pa = 1000 N / m^2 to support it, we conclude that the pressure in the bottle
  is about 3 times this, or about 3000 N / m^2 = 3000 Pa. 
• If this is 5% of atmospheric pressure, then
  atmospheric pressure must be about 3000 Pa, then atmospheric pressure must
  be 3000 Pa / (.05) = 60 000 Pa. 

In fact, atmospheric pressure is about 70% higher than

`q005.  In a careful experiment we determine that a

• What pressure, in Pa or N / m^2, is required to
  support a column 50 cm high? 

****

As we saw previously the pressure required to support

a column of height h is

P = rho g h.

The density of water is 1000 kg / m^3, so the required

pressure is

P = 1000 kg / m^3 * 9.8 m/s^2 *.50 m = 4900 N / m^2 =

4900 Pa.

#$&*

• What do we conclude is the atmospheric pressure?

****

If 4900 Pa is .06 of the atmospheric pressure, then

using P_atm for atmospheric pressure we have

.06 * P_atm = 4900 Pa so that

P_atm = 4900 Pa / (.06) = 82 000 Pa, approx..

This isn't a bad estimate of the accepted value, which

is about 100 000 Pa.

#$&*

`q006.  If the pressure of a gas is 100 000 Pa,

****

P V = n R T, so

V = n R T / P = 1 moles * 8.31 Joules / (mol Kelvin) *

273 Kelvin / (100 000 N / m^2) = .022 m^3,

which is equal to about 22 liters.

This is the volume of a mole of gas at STP (standard

temperature and pressure), to two significant figures.

Using a more accurate expression for atmospheric

pressure we find that the 3-significant-figure volume of a mole at STP is

22.4 liters.

#$&*

• What would be the volume of a mole of gas at 100 000
  Pa and 293 K (about room temperature)? 

****

The same calculation as before, but with the 293 K

temperature, would give us a volume of about 23 liters.

#$&*

• The bottles we used had a little water at the bottom,
  and had gas volume about .5 liter.  How many moles of gas did each
  bottle contain? 

****

Room temperature was around 290 Kelvin, pressure was

about 100 000 Pa.  .5 liters is .5 * .001 m^3 = .0005 m^3.  So 

the number of moles would have been

n = P V / (R T) = 100 000 Pa * .0005 m^3 / (8.31 J /

(mol K * 290 K) = .02 moles, approximately.

NOTE COMMON ERROR:

100 000 Pa (.5 liters) = n (8.31 J/mole*K) (293K)

20.5353 moles = n

INSTRUCTOR COMMENT:  Good solution, but this last calculation doesn't

come out in moles.

The units of the given calculation are N / m^2 * liters / ( J / (mole

Kelvin) * Kelvin).

The unit 'liter' isn't directly compatible with the SI units of the other

quantities:

A liter is .001 m^3, and a Joule is a N * m. So

our units are

N / m^2 * (.001 m^3) / ( N * m / (mole Kelvin) * Kelvin) =

N / m^2 * (.001 m^3) / ( N * m / mole) =

.001 N * m^3 / (N * m * m^2 * mole) =

.001 mole so your result would be

20.5353 (.001 mole) = .0205 mole.

It would probably have been easier to express the

liters in moles in the first place, using 10^-3 m^3 instead of liters:

• 100 000 Pa (.5 * 10^-3 m^3) = n (8.31 J/mole*K)
  (293K)

The numbers and the units would then have worked out

as they did in the given solution.

#$&*

`q007.  A mole of a diatomic gas has a specific heat

• How much energy do you think were required to raise
  the temperature of the gas in the bottle by 13 Kelvin degrees?

****

With .02 moles of gas in the bottle, the energy

required to raise temperature of the gas, per degree, is (.02 moles * 31 J /

(mole Kelvin) ) = 0.62 J / Kelvin. 

Thus the system requires .62 Joules for every Kelvin

degree increase in temperature.

To raise the temperature of the system by 13 Kelvin

degrees requires energy (0.62 J / Kelvin) * 13 Kelvin = 8 Joules, approx. .

#$&*

`q008.  A gram of water requires about 4 Joules of

• If you have 200 grams of water at 25 Celsius, and after adding snow you
  have 240 grams of water at 10 Celsius, then how many Joules of energy did
  the original water lose to the snow?

****

The original water has mass 200 grams.  So the

energy to raise this water, per Celsius degree, is 200 grams * 4 Joules /

(gram Celsius) = 800 Joules / Celsius.

The temperature of the original water decreased by 15

Celsius, so the energy change of the water is -15 Celsius * 800 J / Celsius

= -12 000 J.

The original water therefore lost about -12 000 J of

energy.

#$&*

• If the process happened quickly, so that minimal energy was exchanged
  with the room, then how many Joules of energy did the snow gain?

****

If the system was indeed isolated so that no energy

could be exchanged with the surroundings, the loss of 12 000 J by one part

of the system is balanced by a gain of 12 000 J by the other part of the

system.

So the snow gained 12 000 Joules of energy.

#$&*

• How much of this energy was gained after melting, assuming that the snow
  remained at 0 Celsius until melted? How much energy was gained by the snow
  before the melting was complete?

****

After melting the snow will be in the form of water,

and every gram will gain 4 Joules for every Celsius degree change in

temperature.

The temperature of the melted snow increased from 0 C

to 10 C, a change of +10 C.

40 grams of water will require 40 grams * 4 J / (gram

* C) = 160 J / C, i.e., 160 Joules for every Celsius degree increase in

temperature.  The 10 C increase will therefore require ( 160 J / C ) *

10 C = 1600 Joules of energy.

#$&*

• How much energy per gram was gained by the snow before the melting was
  complete?

****

The snow gained 12 000 J of energy, only 1600 J after

melting.  So it gained the difference 12 000 J - 1600 J = 10 400 J in

the melting process.

This is the energy gained by 40 grams of snow. 

We conclude that the energy per gram is 10 400 J / (40 grams) = 260 J /

gram.

This differs by about 25% from the accepted value,

which is about 330 J / gram.

#$&*

You observed in class the effect of snow melted in

water.  Give your data, and repeat the analysis outlined in the

preceding questions for your data.

****

Your reasoning should be equivalent to that in the

above series of questions.

#$&*

`q009.  The bottle you observed contained

• What is the volume of the water in the 20 cm section
  of tubing?
• What is this volume as a percent of the volume of the air in the bottle?

****

A cross section of the water in the tube is a circle whose

The cross-sectional area of the tube is pi r^2 = pi * (4.5

A 20 cm section of tubing therefore contains volume 20 cm

The air in the bottle has a volume of about 500 cm^3. 

#$&*

`q010.  When the bottle was brought in from

• By how much did the air expand? (hint: why is it that the air was able
  to expand a bit, other than the slight expansion of the
  container?)

****

The air expanded by an amount equal to the volume of

water in the tube, about 3 cm^3.  This is less than a 1% expansion.

#$&*

• By how much did the pressure increase? Use PV = n R T, or just P V / T =
  constant (go ahead and calculate P V / T; get used to using absolute
  temperature).

****

We estimated that the temperature in the bottle

increased from about 273 K to about 293 K.

The of the air volume increased from about 500 cm^3 to

about 503 cm^3.

The original pressure of the air was about 1

atmosphere.

So in its original state we had

• P_1 V_1 / T_1 = 1 atmosphere * 500 cm^3 /
  (273 K).= 1.82 atm cm^3 / K.

If P_2, V_2 and T_2 denote the final values of

pressure, volume and  temperature then, since P V / T is constant,

we have

P_2 * V_2 / T_2 = 1.82 atm cm^3 / K.  We

easily conclude that

• P_2 = 1.82 atm cm^3 / K * (T_2 / V_2) = 1.82
  atm cm^3 / K * (293 K) / (503 cm^3) = 1.067 atm.

This means that the pressure inside the bottle

exceeded that of the atmosphere by a factor of .067.

Compare this to the result we obtain if we assume

volume to be constant.  If V and n are constant we conclude that P / T

is constant, so

P_2 / T_2 = P_1 / T_1 so that

P_2 = P_1 * T_2 / T_1 = 1 atm * (293 K) / (273 K)

= 1 atm * 1.073 = 1.073 atm.

By ignoring the volume change we would estimate the

pressure to change by a factor of .073, rather than .067.  So

considering the volume change due to water entering the tube did make a

difference, but not a great difference, in our result.

We can independently determine the pressure change:

A 20 cm column of water requires excess pressure

equal to (1000 kg / m^3) * 9.8 m/s^2 * 0.20 m = 2000 N / m^2, or 2000 Pa

of excess pressure.

From this we would conclude that the .067

atmosphere increase in pressure was equivalent to a 2000 Pa increase in

pressure, so that 1 atmosphere would be about 2000 Pa / (.067 atm) = 300

000 Pa.

Since the accepted value of atmospheric pressure is

about 100 000 Pa, we are led to suspect either the accuracy of our data or

our ability to control extraneous variables in our setup.

#$&*

`q011.  Assume that a domino is 1 cm thick and

• By how much does the PE of the system increase?

****

One domino has mass 20 grams = .020 kg, so its weight

is .020 kg * 9.8 m/s^2 = .2 N, approx..

If we raise a domino by 1 cm the gravitational force,

which acts downward, does work `dW_dom_cm = -.2 N * .01 m = .002 Joules on

it.  So its gravitational PE increases by .002 Joules.

If we raise a domino by 2 cm it's easy to see that the

gravitational force does twice as much work, .004 Joules.

To build the stack we have to raise the first domino 1

cm, the second 2 cm, the third 3 cm, etc..   It's easy to see,

then, that the PE of the system must in the process of building the stack

increase by

• `dPE = .002 J + .004 J + .006 J + .008 J + .010 J
  + .012 J + .014 J + .016 J + .018 J = .090 J.

A shorter and more efficient calculation is also

possible. 

The 10 dominoes lying on the tabletop all had

centers of mass which were half their thickness, or .5 cm, above the

tabletop.  When stacked their center of mass was 5 cm above the

tabletop.  So the center of mass of the 10-domino system increased

in vertical position from .5 cm to 5 cm, an increase of 4.5 cm.

The gravitational PE of the stack therefore

increased by .2 N * .045 m = .090 J.

#$&*

• Does it matter if we stack the dice one at a time, or make three stacks
  of 3 and stack these stacks, one at a time,
  starting with the first stack on the first domino?

****

We could calculate the PE increase in making the 3

stacks, then the increase when each stack is added to the preceding stack. 

The result would be the same.

Since gravity is a conservative force, it doesn't

matter how we get from the initial system to the final, the PE will be the

same in any case.

#$&*

• What has greater mechanical energy, a stack of 10 or a stack of 9 with
  one domino on the table scooting off at 2 meters / second?

****

The original stack had a PE relative to the original

configuration of .090 J, and no KE.  Its total mechanical energy was

therefore KE + PE  = .090 J.

A stack of 9 dominoes has PE of about .002 J + .004 J

+ ... + .016 J = .072 J.  Alternative the center of mass is at 4.5 cm,

which is 4.0 cm above that of the original configuration, from which we

conclude that the PE relative to the original system is (9 * .020 kg) *

(.040 m) * 9.8 m/s^2 =  .072 J.

A single domino moving at 2 m/s has KE of (1/2 * .020

kg * (2 m/s)^2 ) = .040 Kg m^2 / s^2.  If that domino is on the

tabletop its PE relative to the original configuration is zero.

The total mechanical energy of the second

configuration is therefore .072 J + .040 J = .112 J, which is greater than

that of the 10 domino stack.

#$&*

• Compare both of the above with a stack of 9 dominoes,
  plus one additional domino at a height of 6
  cm moving off at 1 m/s.

****

The KE of the moving domino is (1/2 * .020 kg * (1

m/s)^2 ) = .010 J.  The PE of this domino, relative to its original

position on the tabletop, is easily found to be .012 J (it wasn't specified

so we assume the 6 cm height is relative to its previous position on the

tabletop).

So the system as currently configured has PE, relative

to the original position, of .072 J + .012 J = .084 J.  This is less

than the .090 J potential energy of the 10-domino stack.  However it

also has KE of .010 J, raising its total mechanical energy to .094 J,

greater than that of the original stack.

An alternative way to make the comparison:

The difference in PE is due to the change in vertical

position #$&*

`q012.  Now consider a cylinder with water to

• What has greater mechanical energy, the cylinder with water to the 30 cm
  mark, or the cylinder filled to the 29 cm mark with a 10-gram puddle of water lying
  on the tabletop?

****

The center of mass of the first system is higher than

that of the second, so its PE is greater.  Nothing is moving in either

system so both have zero KE.  The first system therefore has the

greater mechanical energy.

#$&*

• Make the same comparison if the 10 grams of water lost from the top of
  the cylinder are at the 5 cm height, flowing out at 2 m/s.

****

If 10 grams of water are removed at the 30 cm height

and replaced by water at the 5 cm height, then we have a PE change of -(.010

kg) * ( 9.8 m/s^2) * (.25 m) = -.025 Joules.

A 10 gram mass of water moving at 2 m/s has KE = (1/2

* .010 kg * (2 m/s)^2 ) = .020 Joules.

Relative to the original system, we have a PE loss of

-.025 J and a KE gain of .020 J.  The total mechanical energy therefore

has changed by `dKE + `dPE = .020 J + (-.025 J) = -.005 J.

#$&*

• What is the speed of outflow at the  5 cm height, if the mechanical
  energy of the system is to be the same as the original mechanical energy?

****

As we have seen the potential energy has decreased by

.025 J.  If the mechanical energy is to remain the same, then the PE

loss must be matched by a KE gain of equal magnitude.  Thus the 10 gram

mass of water, which was originally at rest and therefore had zero KE, now

has kinetic energy .025 J.

If we set KE = 1/2 m v^2 and solve for v we obtain v =

+-sqrt( 2 KE / m) so to preserve total mechanical energy the water must have

velocity

v = +-sqrt( 2 KE / m) = +- sqrt( 2 * .025 J / (.010

kg) ) = +- sqrt( 5 m^2 / s^2) = 2.2 m/s, approx..

#$&*

`q013.  A plastic bead of mass .4 grams falls

• What is its momentum just before hitting the
  floor, and what is its momentum just after hitting the floor?  Give your
  answers relative to your chosen positive direction.  You may choose the
  positive direction to be either up or down; be sure to state which direction
  you choose as positive.

****

Choosing up as the positive direction the momentum of

the bead changes from .4 g * (-5 m/s) = -20 g m/s = -.020 kg m/s, to .4 g *

4.5 m/s = 18 g m/s = .018 kg m/s

#$&*

• What therefore is the change in the bead's momentum?  Your answer will
  be positive or negative, and this will depend on which direction you chose
  as positive.

****

The momentum change is therefore

`dp = p_f - p_0 = .018 kg m/s - (-.020 kg m/s) = .038

kg m/s.

#$&*

`q014.  If a particle of mass 3 milligrams

****

Choosing upward as the positive direction, the particles

The momentum change is therefore

`dp = p_f - p_0 = .0006 kg m/s - (-.0006 kg m/s) = .0012

#$&*

`q015.  A particle of mass 5 milligrams was

• How much time would elapse between collisions
  with one of the walls?

****

From the time the particle collides with the wall,

until it again collides with the wall, it has to travel to the other wall

and back.  The distance is double the distance between the wall, or 20

cm.

At 300 m/s the time required to move 20 cm is `ds / `dt

= .20 m / (300 m/s) = .0007 sec, approx..

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• How much would the momentum of the particle change in a collision?

****

Let the direction toward the other wall be positive. 

Then the momentum of the particle approaching the wall is 5 milligrams *

(-300 m/s) = -1500 milligram * m/s, or -.000 0015 kg m/s, or in scientific

notation -1.5 * 10^-6 kg m/s.  The momentum of the particle after

bouncing off the wall is +1.5 * 10^-6 kg m/s. 

The change in momentum is therefore

`dp = p_f - p_0 = 3 * 10^-6 kg m/s.

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• What therefore is the average force exerted on that wall?
  ****
  The 3 * 10^-6 kg m/s momentum change takes place once
  every .0007 sec, approx., so by the impulse-momentum theorem the approximate
  average force is
  F_ave = `dp / `dt = 3 * 10^-6 kg m/s / (.0007 s) = 4 *
  10^-3 kg m^2 / s^2 = .004 Newtons.

  #$&*

`q016.  Using P V = n R T, rearrange the equation so

• V and T remain constant.

****

If V and T are constant then we rearrange the equation

with V, T and R (the last always being constant) on one side:

P V = n R T.  Divide both sides by V to get

P = n R T / V.  All constant terms are now on the

right-hand side, along with the non-constant quantity n.  Dividing both

sides by n we get

P / n = R T / V.

Thus the ratio P / n of pressure to number of moles is

constant.  If one quantity goes up or down, the other must go up or

down, respectively, by the same proportion.

This makes sense.  If you keep volume and

temperature constant and pump in more gas, the pressure should go up in the

same proportion as the amount of gas.

#$&*

• P and V remain constant.

****

If P and V are constant then we rearrange the equation

with P, V and R (the last always being constant) on one side:

P V = n R T.  Divide both sides by R to get

P V / R = n * T.  All constant terms are now on

the left-hand side.

Thus the product n * T of temperature and number of

moles is constant.  If one of these quantities goes up, the other goes

down and vice versa. 

This makes sense.  If you add gas to a

constant-volume container, the only way you can keep pressure constant is to

decrease the temperature.

#$&*

• n and P remain constant.

****

If P and n are constant then we rearrange the equation

with P, n and R (the last always being constant) on one side:

P V = n R T.  Divide both sides by P to get

V = n R T / P.  All constant terms are now on the

right-hand side, along with the non-constant quantity T.  Dividing both

sides by T we get

V / T = n R / P.

Thus the ratio V / T of volume to temperature is

constant.  If one quantity goes up or down, the other must go up or

down, respectively, by the same proportion.

This makes sense.  The only way to increase the

temperature while keeping the pressure and number of moles constant is to

increase the volume.

#$&*

• V and n remain constant.

****

If V and n are constant then we rearrange the equation

with V, n and R (the last always being constant) on one side:

P V = n R T.  Divide both sides by V to get

P = n R T / V.  All constant terms are now on the

right-hand side, along with the non-constant quantity T.  Dividing both

sides by T we get

P / T = n R / V.

Thus the ratio P / T of pressure to temperature is

constant.  If one quantity goes up or down, the other must go up or

down, respectively, by the same proportion.

This makes sense.  If we increase the temperature

of a fixed quantity of gas, while keeping the volume of the container the

same, the pressure goes up.

#$&*