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Newton Laws

VIEWS: 8 PAGES: 35

									          Friday Dec 10th

Friday – Dec 10th
1)Test
2)Asst: (a) Notes on work, power, and
  energy pp: 256-265, pp 284-301 (write
  the page number, as always)
        Monday Dec 13th
1. Turn in all dittoes, organize
   notebooks
2. Lecture notes part 1 – W, P, E
3. Asst: parts A and B on asst sheet
           Work, Energy, Power
Note – this unit looks VERY similar to the last unit:
• In that unit you had 2 basics equations:
• (1) J = F  t = p, or if the impulse was zero (as it is in a
  closed & isolated system), then:
• (2) p initial total = p final total
• See in later slides that, just like the impulse-momentum
  theorem above, we now have the work-energy theorem:
  (3) W = F  d = E
• And see in later slides that, just like we had the
  conservation of momentum in a closed & isolated system
  above, we now have the conservation of energy when the
  net work done is zero: (4) E initial total = E final total
• So, you will have both of those types of math-problems to
  solve on the test, as well as many conceptual problems
  again.
• (But note things are slightly more complicated this time, as
  the F  d, written & pronounced “F dot d” is a special type of
  multiplication, and there is more than one type of energy!)
                W = F  d = E
• Let’s talk about just the middle part of that equation…
• It is NOT normal multiplication;
• It is said as “Work equals F dot d”, where d is
  displacement.
• The “dot” means that only that component of the force
  that is in the direction of the displacement can be used
  when calculating work.
• This means that 90% of the time, W = F  d = F d cos .
  But in a few problems, if a weird angle is given to you, it
  MIGHT be a sine, so be careful!
• Similarly, if the force and displacement are in opposite
  directions, the work will be negative.
• If the force and displacement are perpendicular, the
  work will be zero.
• examples: slide a book along the floor & carry a book ...
• The unit for work is the Joule [J].
              P=W/t =Fv
• In the high school textbook, it is most common to use
  P = W / t, but there are a few problems that require
  P=Fv
• (The third part comes from making W = F  d and then
  dividing by time.)
• It is said as “Power is F dot v”, where v is velocity.
• The “dot” means that only that component of the force
  that is in the direction of the velocity can be used when
  calculating power.
• Similarly, if the force and velocity are in opposite
  directions, the power will be negative.
• If the force and velocity are perpendicular, the power
  will be zero.
• The unit for power is the Watt [W].
            Tuesday Dec 14
1.   Questions on last night’s HW / turn in
2.   And W, E, P notes
3.   Lecture notes
4.   Asst: parts C and D (but cross out #69,
     answer to #75b = 89 m) on asst sheet
        Work, Energy, Power
           continued...
• KE = ½mv2
• PE = mgh
• Kinetic energy has the symbol KE
• Gravitational potential energy has the
  symbol PE
• Both are also measured in Joules [J].
    There are other types of energy
   than those listed in the last slide:
• KE and PE are both types of what we call “mechanical energy”, which
  is the only type of energy we physicists care about.
• There is also the potential energy of a spring, which is a type of
  mechanical energy too, but that high school textbooks don’t talk about
  much.
• There are other types of non-mechanical energies, such as
  chemical energy, heat or thermal energy, light energy, sound energy,
  etc.
• Fact: The TOTAL amount of energy in the universe is a constant.
• Confusion: We say in the W-E equation, that Energy could either
  increase or decrease.
• Understand: Physicists speak only about mechanical energy.
  When they say “energy is not conserved (because work was
  done)” what they really mean, but are too lazy to say, is
  “mechanical energy is not conserved (because work was done),
  and it was changed into or came from other types such as
  chemical, thermal, light or sound energy.” (You see how much
  harder that is to say? Physicists are inherently lazy!)
         W = F  d = E …revisited:
 • NOTE – the textbook is WRONG here. It says W equals
   the change in kinetic energy, but it is really the change in
   over-all energy.
 • High school problems are sometimes:
       W = F  d = KE = KEF – KEI = ½mvF2  ½mvI2
 • BUT, sometimes those problems are:
        W = F  d = PE = PEF – PEI = mghF  mghI
Work done:         Why?            Energy change?                      Simple Examples:
                                                     (a) Pull or push something (even at an angle) …
    +          F & d same dir          increase          increases KE
                                                     (b) Raise or lift something … increases PE
                                                     (a) Friction … decreases KE
              F & d opp dir          decrease
                                                     (b) Lower something … decreases PE
             F & d are perpend;
                                                     (a) You carry something … energy stays constant
              or no net force by        none
    0                                                (b) Something falls … energy is conserved: PE → KE
             anything other than     overall/total
                                                     (c) Something rises … energy is cons: KE → PE
                   gravity
  Let’s talk about throwing or dropping things: YOU do no work after the release!
       HERE ARE THE EQUATIONS
       YOU NEED ONE MORE TIME:

W = F  d = E   E initial total = E final total   P=W/ t = F v   ( PE = mgh & KE = ½ mv2 )




                                Specifically:
                                (a)
                                W = F  d = ½mvF2  ½mvI2
                                Or (b)
                                W = F  d = mghF  mghI
             Wed Dec 15th
1. Questions on last nights HW? #75??
2. Lecture notes on Collisions
3. Asst: Elastic collisions 1-6 TWO
    different ways ...
(b) as stated – they are perfectly elastic, and
(a) NOT as stated, but the two things STICK
    together after the collision, called
    “inelastic”. (but only for #’s 1&2, 5&6)
         NOTE - #2 has wrong answer!
              Old way = -2.4 m/s
       New way = -10.## m/s and 0.## m/s
    Please make that change on your sheet 
    Elastic & inelastic collisions:
• **ALL collisions – elastic & inelastic – conserve momentum.
• **Elastic collisions also conserve (mechanical) energy.
• **Inelastic collisions don’t conserve (mechanical) energy. (Energy
  goes in through chemicals or springs; energy comes out through
  heat and sound)
• IT is a FACT that the total amount of energy in universe stays same.
  (Conservation of Energy you learned about in Chem class.)
• So, for elastic & inelastic, we physicists actually mean MECHANICAL
  energy - just PE + KE. (That’s why I put the parentheses around
  “mechanical” up above, because its true, but physicists are often lazy
  and don’t say it specifically.)
• In inelastic collisions the chemicals, springs, heat & sound are just
  OTHER forms of energy. So energy DOES stay constant in the whole
  universe total.
• So even inelastic collisions conserve energy, just not the mechanical
  types we are talking about / care about in this unit, and are really the
  most important types in Physics.
• **Elastic collisions are usually “bouncing” collisions, inelastic are
  usually explosions or “sticking” collisions.
Example: A 3.2-kg block is traveling EAST with a speed
of 11 m/s. It collides with a 4.8-kg block traveling WEST
at 5.0 m/s. What will be the speeds of these two blocks
after the collision if it is perfectly/completely (in)/elastic?

• First, let’s do this the old way (stick/inelastic)
  with chapter 9 conservation of momentum stuff
  only, where most collisions stuck together:
            • (**old way from last chapter)
• p I total = p F total: m1v1I + m2v2I = (m1 + m2)vF
• (3.2 kg * 11 m/s) + (4.8 kg * 5.0 m/s) = (3.2 + 4.8 kg)*vfinal
• Solve for vfinal = ...
• .... = 1.4 m/s
 Example: A 3.2-kg block is traveling EAST with a speed of 11 m/s. It
collides with a 4.8-kg block traveling WEST at 5.0 m/s. What will be the
  speeds of these two blocks after the collision if it is perfectly elastic?

• Now, we do it with the new way where they bounce and the
   collision is “completely elastic” as the new instructions say:
                               (**new way part 1)
             pI total = pF total: m1v1I + m2v2I = m1v1F + m2v2F
 (3.2 kg * 11 m/s) + (4.8 kg *5.0 m/s) = (3.2 kg * v1F) + (4.8 kg * v2F)


• “elastic” means energy is also conserved. But remember, when a
   physicist says energy, they usually mean “mechanical energy”, so
   we mean PE + KE. But, collisions usually always all happen on
   the same level/height, so the PE all stays at the same level,
   which we’ll deem our reference level, which we can just call zero.
   Thus, what we really mean for an elastic collision is:
                             (**new way part 2)
     KEI total = KEF total: ½m1v1I2 + ½m2v2I2 = ½m1v1F2 + ½m2v2F2
 (½ * 3.2 * 112) + (½ * 4.8 *(-5.0)2) = (½ * 3.2 * v1F2) + (½ * 4.8 * v2F2)
     Of the 3 algebraic methods we talked about to solve eqs,
 substitution is messy (very!) and we can’t use matrices because
    these are not both equations of lines. So, we must use the
   graphing method. Thus, let x = v1F and y = v2F and solve both
equations for y with as few parentheses as you can get away with :
                               pI total = pF total1 pt
                                                                         4 pts
     (3.2 kg * 11 m/s) + (4.8 kg *5.0 m/s) = (3.2 kg * x) + (4.8 kg * y)
                                becomes…
                                                                    4 pts
              y1 = ( 3.2*11 +        4.8*5.0      3.2x )/4.8
    (only use 1 parenthesis when doing this,
            and simplify NOTHING!!!)
                                                    1
                              KE I total = KE F total pt
                                                                            4 pts
      (½ * 3.2 * 112) + (½ * 4.8 *(-5.0)2) = (½ * 3.2 * x2) + (½ * 4.8 * y2)
                                 becomes…
                                                                             4 pts
       y2 = ((    .5*3.2*112 +       .5*4.8*5.02        .5*3.2x2 )/(
                                    .5*4.8))
    (only use 3 parentheses when doing this,
    Example: A 3.2-kg block is traveling EAST with a speed of 11 m/s. It
   collides with a 4.8-kg block traveling WEST at 5.0 m/s. What will be the
     speeds of these two blocks after the collision if it is perfectly elastic?
• y1 = ( 3.2*11 + 4.8*5.0  3.2x ) / 4.8
• y2 = (( .5*3.2*112 + .5*4.8*5.02  .5*3.2x2 )/( .5*4.8 ))
• y3 = y2 (do not type in y2 all over again, use the y2 “button”)
• I put the subscripts y1 and y2 there because that’s how your TI will
  show them. But, you’ll also need to do eq y3, because TI’s won’t
  calculate the bottom of the circle/ellipse without this little help.
  (You can find y2 in vars / y-vars / function.)
• When you graph it, do a -20 to 20 window for both x and y.
• Find the intersect of the line and circle/ellipse, by doing...
• 2nd Calc, Intersect, 1st curve, 2nd curve, guess.
• There are actually 2 intersections. You should find them both and
  get:
• 8.2 m/s & 7.8 m/s; and 11 m/s & 5.0 m/s
• It turns out the second intersection (of the two I have written above)
  is the exact same answers as the originally velocities, and thus not
  physically possible, so the other intersection is the correct answer.
• Note: you only have to do # 1 and 2 on the worksheet the “old
  way”. See hints and stuff on the assignment ditto itself.
Let’s do HW #1 together too…




    Elastic collisions 1-5 TWO different ways ...
    (b) as stated – they are perfectly elastic, and
    (a) NOT as stated, but the two things STICK together after the
          collision, called “inelastic”. (but only for #’s 1 & 2, 5 & 6)
    NOTE - #2 redo MAY have the wrong answer!
#1 :
           OLD WAY = INELASTIC = ONLY MOMENTUM CONSERVED
                           (usually stick) …..

                                      pI total = pF total
               (1.6 kg * 5.5 m/s) + (2.4 kg * 2.5 m/s) = (1.6 kg + 2.4) vfinal

       ---------------------------------------------------------------------------------------------

        NEW WAY = ELASTIC = BOTH MOMENTUM & KE CONSERVED
                         (usually bounce) …..

                                     pI total = pF total
          (1.6 kg * 5.5 m/s) + (2.4 kg * 2.5 m/s) = (1.6 kg * x) + (2.4 kg * y)
                                      becomes…
                         y1 = ( 1.6*5.5 + 2.4*2.5  1.6x )/2.4

                                   KE I total = KE F total
            (½ * 1.6 * 5.52) + (½ * 2.4 * 2.52) = (½ * 1.6 * x2) + (½ * 2.4 * y2)
                                       becomes…
                 y2 = (( .5*1.6*5.52 + .5*2.4*2.52  .5*1.6x2 )/( .5*2.4))
#2 :
           OLD WAY = INELASTIC = ONLY MOMENTUM CONSERVED
                           (usually stick) …..

                                      pI total = pF total
              (1.6 kg * 5.5 m/s) + (2.4 kg * 2.5 m/s) = (1.6 kg + 2.4) vfinal

       ---------------------------------------------------------------------------------------------

        NEW WAY = ELASTIC = BOTH MOMENTUM & KE CONSERVED
                         (usually bounce) …..

                                     pI total = pF total
          (1.6 kg * 5.5 m/s) + (2.4 kg * 2.5 m/s) = (1.6 kg * x) + (2.4 kg * y)
                                      becomes…
                         y1 = ( 1.6*5.5 + 2.4*2.5  1.6x )/2.4

                                   KE I total = KE F total
           (½ * 1.6 * 5.52) + (½ * 2.4 * 2.52) = (½ * 1.6 * x2) + (½ * 2.4 * y2)
                                       becomes…
                 y2 = (( .5*1.6*5.52 + .5*2.4*2.52  .5*1.6x2 )/( .5*2.4))
#3 :
           OLD WAY = INELASTIC = ONLY MOMENTUM CONSERVED
                           (usually stick) …..

               DON’T DO THIS ONE THIS WAY! (They won’t “stick”)

       ---------------------------------------------------------------------------------------------

        NEW WAY = ELASTIC = BOTH MOMENTUM & KE CONSERVED
                         (usually bounce) …..

                                    pI total = pF total
            (220 kg * 45 m/s) + (46 kg * 0 m/s) = (220 kg * x) + (46 kg * y)
                                     becomes…
                          y1 = ( 220*45 + 46*0  220x )/46

                                  KE I total = KE F total
             (½ * 220 * 452) + (½ * 46 * 02) = (½ * 220 * x2) + (½ * 46 * y2)
                                      becomes…
                  y2 = (( .5*220*452 + .5*46*02  .5*220x2 )/( .5*46))
#4 :
           OLD WAY = INELASTIC = ONLY MOMENTUM CONSERVED
                           (usually stick) …..

              CAN’T DO THIS ONE THIS WAY! (They explode apart!)

       ---------------------------------------------------------------------------------------------

        NEW WAY = ELASTIC = BOTH MOMENTUM & KE CONSERVED
                         (usually bounce) …..

                                       pI total = pF total
                  (1 kg * 0 m/s) + (2 kg * 0 m/s) = (1 kg * x) + (2 kg * y)
                                        becomes…
                                  y1 = (1*0 + 2*0 1x)/2

                       KE I total + 60 Joulesgiven to system= KE F total
                (½ * 1 * 02) + (½ * 2 * 02) + 60 = (½ * 1 * x2) + (½ * 2 * y2)
                                        becomes…
                     y2 = (( .5*1*02 + .5*2*02 + 60  .5*1x2 )/( .5*2))
  #5-8 Don’t forget to do these too!
• (5 & 6 you will do inelastic also, 7&8 you will not.)
• I STRONGLY encourage you to make sure you can use
  your calculator! Don’t just copy the answers off the sheet,
  but make sure YOU can get them yourself given a TI
  calculator! Many students do NOT know how to find the
  intersection correctly!
• Some answers:

             #1 REDO (sticking/inelasti/old way answer = 7.4 m/s
             & new "elastic" way answers = ?? m/s and 12.2 m/s)
       #2 REDO- (sticking/inelastic/old way answer = NEGATIVE ?? m/s
           & new "elastic" way answers = -10.## m/s and 0.## m/s)
#3 REDO - only do REDO as stated on ditto. Try bigger window of 90 x 90 when
     finding REDO intersection (and your answer is really close to there!)
#4 REDO - only do REDO as stated on ditto. REDO answers = 11.5 m/s and ???
                           m/s in opposite directions
Thurs Dec 16th
1. Hand in: Cons. Of Mom write-up
2. Experiment: Work & Energy
3. Each person goes into Excel and
   starts Data
4. Asst a: Make sure collisions done
5. Asst b: HAND-Written Introduction for
   experiment write-up (see website for
   what should be included!)
         Fri Dec 17th (min day)

1.   Conservation of energy problems (E -
     next slide)
2.   “15-story problem” (F - 2nd next slide)
3.   Computers: Work-Energy data – SIG
     FIGS!
4.   ASST: parts E and F on asst sheet
5.   ALSO: Work-Energy write-ups
     Purpose, Materials, Procedure
6.   AND FINALLY… Enjoy the time with
     your family 
               E initial total = E final total
• This is a statement of the conservation of energy.
• If net Work done = 0, then by the equations in previous
  slides, Etotal also equals zero.
• Since Etotal = E final total  E initial total = 0, then we will write
  E initial total = E final total on top of all of our HW problems that
  use the conservation of energy.
• For high-school problems, that means we have:
                  (PEinitial + KEinitial) = (PEfinal + KEfinal)
• Usually in high-school problems either
    – the KEinitial & PEfinal are both zero (when an object is falling), or
    – the PEinitial & KEfinal are both zero (when an object is being
      thrown up)
• So you can usually set up problems as:
    – PEinitial + 0 = 0 + KEfinal, i.e.: PEinitial = KEfinal when falling, or
    – 0 + KEinitial = PEfinal + 0, i.e.: KEinitial = PEfinal when rising
  A 6-kg squirrel is dropped from the top of a 15-
  story building. At the top of each floor, find the
following. (Assume each story is 3.5 meters high.)
  Floor Height       PE   TOTAL     KE      vel
         (m)         (J) Energy (J)  (J)   (m/s)
           floor x = mgh energy is T - PE KE = ½
             3.5         conserved          mv2
   15
   14
   13
   .....
    1
    0
               Mon Jan 3rd
1. Bill Nye – (desks cleared, no working on asst
   during video!)
2. Start in class ….
3. Asst: Concepts due tomorrow (Tues):
(See hints posted on website!)
•   Page 265 #17, 21-23
•   Page 278 #36-38**, 44-46**, 89*****
•   Page 292 #12, 14a(no bar graph)&b:
•   Page 301 #22:
•   Page 305 #30&32** together, 33-35**,
    36, 38, 39**&46, 51**, 52, 88
                   Tues Jan 4th
1.   Go over asst F (squirrel) and E (cons of energy); turn
     in
2.   Do an elastic/inelastic problem .... (& turn in 8 others)

3.   Go over Concepts **(23, 38, 89; 22, 30&32, 34, 35,
     51); turn in
4. Asst:
(a) PRINT OFF STUDY GUIDE for class tomorrow!
(b) Work on ch 10/11 review packet (due Thursday, so you
     have 2 nights)
Note – first 6 parts of the experiment write-up due Friday
Wed Jan 5th
 1. Continue yesterday:
   •   Do another elastic/inelastic problem
   •   Concepts…..
 2. Work on review packet (due tomorrow)
 3. Asst:
   •   Complete review packet (due TOMORROW)
   •   DOWNLOAD THE STUDY GUIDE IF YOU
       HAVEN’T!! (and all the other ones too!)
Concepts Key
                  Thursday Jan 6th
1.       Go over review packet
2.       Look at printed off Study Guide
3.       Do another elastic/inelastic if time
     •     Note about quick checks:
     •     (a) is the “other” answer the starting velocities?
     •     (b) IF YOU HAVE TIME (not so quick check), and you DO
           simplify the y= equations, do you get the same answers?
4.       Asst: for the W-E lab - which is on your test …
         You must bring the INTRODUCTION (as specified
         on the template on the webpage) through the
         DATA section (done in Excel) to class tomorrow

TEST MONDAY!
Next week: Study for comprehensive final (Finals: Jan
    19-21)
Friday Jan 7th
 1.   Stamp EXPERIMENT write-ups…
 2.   Look at Study Guide if didn't yesterday
 3.   Practice test
 4.   Do Fill-in “Study Guide” together if time????
 5.   Asst: Study for test Monday!

 Next week: study for comprehensive final
       (Finals: Jan 19-21)
Practice Test:
1.  Chart on work/energy for ... (a) friction (b) carrying bks (c) lifting a book
    (d) something drops
 Work done, Why?, E change?, What type E?
2.  Bow exerts 100.-N of force over 0.50 meters. If it exerts this force on a
    0.20 kilogram arrow, what a) speed will the arrow obtain? b) height will it
    obtain if shot straight up in the air?
3.  What is the speed of a cat that falls 10. meters?
4.  A 1.00-kg object falls from a two-story bldg where each story is 1.00
    meter high. Make a chart with Height, PE, Total E, KE, Vel for floors 2-0.
5.  A 350-N force pulls a suitcase at an angle of 20 degrees. The suitcase
    moves 15 meters. What is the work done?
6.  Discuss energy changes in a pendulum and 2 ways it can be used to
    show energy is conserved.
7.  What data was collected in the lab, what calculations were made from
    that, what was the point, why did the cart have to be moving at a constant
    speed?
8.  A 2.4 kg block traveling EAST at 4.5 m/s, collides with a 1.8 kg block
    traveling WEST at 2.0 m/s. (a) If the collision is completely inelastic, what
    is the final velocity of the blocks? (b) If the collision is completely elastic,
    find the final velocity of both the blocks?
THIS IS NOT THE COMPLETE TEST! THERE IS MORE THAN THIS ON
    MONDAY’S TEST. LOOK A THE ENTIRE STUDY GUIDE!
                Mon Jan 10th
1.   Test (make sure your calc is in degrees!)
2. REVIEW HW:
(a) kinematics (ditto A – you may select any 3
     problems; ditto D – do #5, 7, 9, 11)
& (b) projectiles (do either 4 or 5 from BOTH
     types)
AND …do the projectiles questions at the bottom:
• Angles give maximum range, same range?
• Range & time on Jupiter vs moon?
• vel and accel on way up, at top, on way down
     in both x and y

								
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