Physics 121C Mechanics - PowerPoint by fjwuxn

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									Physics 121B - Mechanics
Lecture 12 (T&M: 7.1-3)
Mass-Energy & Momentum
          April 28, 2010


         John G. Cramer
   Professor Emeritus of Physics
            B451 PAB
         jcramer@uw.edu
                 Announcements
 Homework Assignment #5 should be submitted
on the Tycho system by 11:59 PM on Friday, April 30.
 On Friday, May 7, we will have Exam 2 covering
T&M Chapters 5-8. Sections will be Lecture
multiple-choice (55 pts), Laboratory (25 pts), and
Tutorial (20 pts). There will be assigned seating.
Send E-mail if you have a new seating preference.

 As of 9:10 AM today, 153/175 Physics 121B
students have registered their clickers. If you have
not already done so, register your clicker at:
http://courses.washington.edu/p121bs10/clicker.htm
April 28, 2010      Physics 121B - Lecture 12        2/28
            Lecture Schedule (Part 2)
                               Physics 121B - Prof. John G. Cramer - 11:30-12:20 PM MWF - A118 PAB
          Textbook: Physics for Scientists and Engineers, 6th Edition, Paul A. Tipler and Gene Mosca, W. H. Freeman & Co. (2008)


Week     Date     L#           Lecture Topic            Text Reading   Pages Slides   Hwk           Tutorial                  Lab

                                                                                                     Forces          Free Fall & Projectiles
        16-Apr-10 E1 EXAM 1 - Chapters 1, 2, 3, and 4                                 HW#3

        19-Apr-10 8 Friction, Circular Motion         Ch. 5.1,2,3       5      29

 4      21-Apr-10 9 Center of Mass; Work              Ch. 5.5, 6.1-2    19     32              Newton's 2nd & 3rd        1-D Dynamics

        23-Apr-10 10 Energy;Potential Energy          Ch. 6.3-5, 7.1    10     19     HW#4

        26-Apr-10 11 Conservation of Energy           Ch. 7.2 to 7.3    12     32

 5      28-Apr-10 12 Mass-Energy; Momentum            Ch. 7.4 to 8.1    26     28                 Work and KE       Newton's Laws, Tension

        30-Apr-10 13 Kinetic En; Impulse & Colliison Chs. 8.2 to 8.5    16     24     HW#5

        3-May-10 14 Explosions & Rockets              Ch. 8.4 to 8.5    19     21
                                                                                                      We are here!
 6      5-May-10 15 Review 2                                                   22                Cons. of Energy        Work & Energy

        7-May-10 E2 EXAM 2 - Chapters 5, 6, 7, and 8                                  HW#6

                                                                                               Cons. of Momentum    Momentum & Collisions
       10-May-10 16 Rotational Kinematics & Energy Ch. 9.1,2,3          6


     April 28, 2010                                  Physics 121B - Lecture 12                                                     3/28
                     Potential
                 Energy and Force




                              dU ( x)
                   F ( x)  
                               dx

April 28, 2010       Physics 121B - Lecture 12   4/28
                 Clicker Question 1

   A particle moves along the x-axis
with the potential energy shown.

   What is the x component of the
force on the particle when it is at
x=4 m ?




    (a) 2 N       (b) 1 N         (c) 4 N              (d) 2 N   (e) 1 N


April 28, 2010              Physics 121B - Lecture 12                     5/28
  Stable and Unstable Equilibrium
Condition for Stable Equilibrium: In stable
equilibrium, a small displacement in any direction
results in a restoring force that accelerates the
particle back to its equilibrium position. d2U/dx2 > 0.



Condition for Unstable Equilibrium: In unstable
equilibrium, a small displacement in any direction
results in a force that accelerates the particle
away from its equilibrium position. d2U/dx2 < 0.



Condition for Neutral Equilibrium: In neutral
equilibrium, a small displacement in any direction
results in a zero force and the particle remains in
equilibrium. d2U/dx2 = 0.
 April 28, 2010                Physics 121B - Lecture 12   6/28
          Example: Force and
     the Potential-Energy Function
  In the region –a < x < a the force on a
   particle is represented by the potential
   energy function: U(x) = -b[1/(a+x) +1/(a-x)],
   where a and b are positive constants.
(a) Find the force Fx in the region –a < x < a.
(b) At what value of x is the force 0?
(c) At that location, is the equilibrium stable or
    unstable?
          dU    d   1           1            1           1 
 Fx          b                    b                  2 
          dx    dx   (a  x) (a  x)         (a  x ) (a  x ) 
                                                          2

At x  0, Fx  0.
d 2U      d2   1              1             1            1 
       2  b                      2b                   3 
dx 2      dx   (a  x) (a  x)              (a  x ) (a  x ) 
                                                         3


            d 2U
At x  0,         4b / a  0, so the equilibrium is unstable.
            dx 2
 April 28, 2010                       Physics 121B - Lecture 12          7/23
     Suspension and Equilibrium
                                                         By
                                                      choosing
                                                      several
                                                      pivot points,
                                                      one can
                                                      locate the
                                                      CM as the
                                                      cross point
                                                      of the
                                                      several
   An object hangs so that the CM is                  verticals
below the pivot point.                                from the
                                                      pivot points.




April 28, 2010            Physics 121B - Lecture 12          8/28
   The Conservation of Energy
        The total energy of the universe is constant. Energy can be
     converted from one form to another or transmitted from one region
     to another, but energy can never be created or destroyed.


                       Ein  Eout  Esys
                 Esys  Emech  Etherm  Echem  Eother


                     The Work-Energy Theorem
      Wext  Esys  Emech  Etherm  Echem  Eother

April 28, 2010             Physics 121B - Lecture 12                     9/28
                 Clicker Question 2
                                                      4.0 m/s




         A spring-loaded gun shoots a plastic ball with a speed of 4.0 m/s.

         If the spring is compressed twice as far, what is the ball’s speed?



      a) 2.0 m/s    b) 4.0 m/s     c) 8.0 m/s      d) 16.0 m/s   e) 32.0 m/s




April 28, 2010                Physics 121B - Lecture 12                        10/28
                  Example: Falling Clay
   A ball of modeling clay with mass m is released from rest from a height h
and falls to the perfectly rigid floor (thud). Discuss the application of the law
of conservation of energy to
         (a) the system consisting of the clay ball alone, and

 Wext  mgh; Emech  0; Wext  Emech  Etherm
 Etherm  mgh
          (b) the system consisting of Earth, the floor, and the clay ball.

 Wext  0; Wext  Emech  Etherm  0
  Emechi  mgh; Emech f  0
  Emech  0  mgh  mgh
  Etherm  Emech  mgh
 April 28, 2010                Physics 121B - Lecture 12                      11/28
                      Problems Involving
                       Kinetic Friction

   Wext  0  Emech  Etherm

   Emech  Kblock  Kboard              1
                                             2
                                                 mv 2  1 mvi2 
                                                    f   2         1
                                                                    2
                                                                        MV f2  0   
  fk  max ;  fk x  max x; 2ax x  v2  vi2
                                          f


 f k x  1 m  v2  vi2   1 mv2  1 mvi2
           2      f           2   f   2                f k X  MAx X  1 M V f2  Vi 2   1 MV f2  0
                                                                         2                    2


 fk  x  X   1 mv2  1 mvi2  1 MVf2
                   2   f   2        2
                                                             f k srel  Emech ;   f k srel  Etherm


                  Wext  Emech  Etherm  Emech  f k srel

 April 28, 2010                      Physics 121B - Lecture 12                                      12/28
                     Example: Pushing a Box
     A 4.0-kg box is initially at rest on a horizontal
tabletop. You push the box a distance of 3.0 m
along the tabletop with a horizontal force of 25
N. The coefficient of kinetic friction between the
box and tabletop is 0.35. Find
(a) the external work done on the block–table
system,
(b) the energy dissipated by friction,
(c) the final kinetic energy of the box, and
(d) the speed of the box.
   Wext  Wby you on block  Wby gravity on block  Wby gravity on table  Wby floor on table
            Fpush x  0  0  0   25 N  3.0 m   75 J

  Etherm  f k x  k Fn x  k mg x   0.35 4.0 kg  9.81 m/s2  3.0 m   41 J


  K f  Wext  Etherm   75 J    41 J   34 J               v f  2K f / m  2  34 J  /  4.0 kg   4.1 m/s

   April 28, 2010                              Physics 121B - Lecture 12                                   13/28
        Example: A Playground Slide
   A child of mass 40 kg goes down an 8.0-m-
long slide inclined at 30° with the horizontal.
The coefficient of kinetic friction between the
child and the slide is 0.35. If the child starts
from rest at the top of the slide, how fast is
he traveling when he reaches the bottom?

 Wext  Emech  Etherm  (U  K )  f k srel

K  K f  0  1 mv2 ; Wext  0; U  mgh
               2


Fn  mg cos  0;      fk  k Fn  k mg cos ;       h  s sin 

0  mgh  1 mv2  fk s  mgs sin   1 mv2  k mg cos s
           2   f                       2   f


v f  2 gs  sin   k cos    2  9.81 (4.0)  sin 30  0.35cos30   5.6 m/s

 April 28, 2010                  Physics 121B - Lecture 12                   14/28
   Friction and Chemical Energy




                   Fk srel  Etherm

Wext  Emech  Etherm  Emech  Fk srel                 Echem  (mgh  Etherm )



  April 28, 2010               Physics 121B - Lecture 12                       15/28
                 Mass & Energy


   E  mc2

     E
 M  2
      c




April 28, 2010     Physics 121B - Lecture 12   16/28
 Example: Nuclear Binding Energy
   A hydrogen atom consisting of a proton and an electron has a binding
energy of 13.6 eV. By what percentage is the mass of a proton plus the
mass of an electron greater than that of the hydrogen atom?




 April 28, 2010             Physics 121B - Lecture 12                     17/28
                  Example: Nuclear Fusion
   In a typical nuclear fusion reaction, a triton (t) and a deuteron (d) fuse
together to form an alpha particle (a) plus a neutron (n). The reaction is
written d + t → a + n.
   How much energy is released per deuteron produced for this fusion
reaction?




 April 28, 2010               Physics 121B - Lecture 12                    18/28
     Nonrelativistic (Newtonian)
      Mechanics and Relativity
    As the speed of a particle approaches a significant fraction of the speed
of light, Newton’s second law breaks down, and we must modify Newtonian
mechanics according to Einstein’s theory of relativity. The criterion for the
validity of Newtonian mechanics can also be stated in terms of the energy
of a particle.
    In nonrelativistic (Newtonian) mechanics, the kinetic energy of a particle
moving with speed v is:


where E0 = mc2 is the rest energy of the particle. Solving for v/c gives:




   Nonrelativistic mechanics is valid if the speed v of the particle is much
less than c, the speed of light, or alternatively, if the kinetic energy K of a
particle is much less than its rest energy E0.
 April 28, 2010               Physics 121B - Lecture 12                     19/28
            Quantization of Energy



   In microscopic mechanical systems (e.g.,
molecules), because of quantum mechanics, the
system energy cannot change continuously, but
rather can change only in discrete steps to well
defined “state” values. The “ground state” E0 is the
lowest of these energy values.

En  (n  )hf , n  0,1, 2,3,
            1
                                         E0  1 hf
                                              2
            2

h  Planck's constant  6.636 10-34 J s =4.136 10-15 eV s
Erad  E f  Ei      Ephoton  hf
 April 28, 2010                 Physics 121B - Lecture 12     20/28
Momentum
Momentum:

 p  mv
                       Conservation
                   of Linear Momentum
            Momentum:              p  mv
dp d (mv )    dv                                 dp
          m     ma                     Fnet 
dt    dt      dt                                 dt

P   mi vi   pi  Mvcm
 sys
            i             i

                          dPsys  dvcm
F
 i
      ext    Fnet ext   
                           dt
                              M
                                  dt
                                       Macm

If  Fext  0, then P   mi vi  Mvcm  constant             Conservation
                     sys
                                                              of Momentum

April 28, 2010                    Physics 121B - Lecture 12           22/28
              Example: A Space Repair
   During repair of the Hubble Space Telescope, an
astronaut replaces a damaged solar panel during a
spacewalk. Pushing the detached panel away into space,
she is propelled in the opposite direction. The astronaut’s
mass is 60 kg, and the panel’s mass is 80 kg. Both
astronaut and panel are initially at rest relative to the
telescope, until the astronaut gives the panel a shove,
giving it a velocity of 0.30 m/s relative to the telescope.
   Assuming her tether is slack, what is her velocity
relative to the telescope?
                   dPsys
 F    ext   0
                    dt
                           , so Psys  constant

  mP vPf  mAv Af  mP vPi  mAv Ai  0
                                                                mP vPf  mAv Af
             mP         (80 kg)            ˆ             ˆ
  v Af        vPf           (0.30 m/s)i  (0.40 m/s)i
             mA         (60 kg)
 April 28, 2010                     Physics 121B - Lecture 12                 23/28
   Kinetic Energy of a System
    K   K i   1 mi vi2   1 mi (vi  vi )
                  2            2
                                                              vi  vCM  ui
             i            i           i


    K   1 mi (vCM  ui )  (vCM  ui )   1 mi (v CM  u i2  2vCM  ui )
          2                                  2
                                                     2

             i                                        i


    K   1 mi v CM   1 mu i2  vCM   mi ui
          2
             i
                 2
                        2
                              i                   i
                                                              mu
                                                               i
                                                                   i i   0


                 K   1 mi vCM   1 mi ui2  1 MvCM  K rel
                       2
                             2
                                    2          2
                                                   2

                      i                   i

    In an isolated system, only the relative kinetic energy can change
 due to internal forces. I. e., no “space drives”.
April 28, 2010                    Physics 121B - Lecture 12                   24/28
            Example: Hubble Repair
   During repair of the Hubble Space Telescope, an
astronaut replaces a damaged solar panel during a
spacewalk. Pushing the detached panel away into
space, she is propelled in the opposite direction. The
astronaut’s mass is 60 kg and the panel’s mass is 80
kg. Both the astronaut and the panel initially are at
rest relative to the telescope. The astronaut then
gives the panel a shove. After the shove it is moving
at 0.30 m/s relative to the Hubble Telescope.
    What is her subsequent velocity relative to the
telescope? (During this operation the astronaut is
tethered to the ship; for our calculations assume
that the tether remains slack.)




  April 28, 2010              Physics 121B - Lecture 12   25/28
                Example:
          A Runaway Railroad Car
    A runaway 14,000-kg railroad car is rolling
horizontally at 4.00 m/s toward a switchyard. As
it passes by a grain elevator, 2000 kg of grain
suddenly drops into the car.
    How long does it take the car to cover the
500-m distance from the elevator to the
switchyard? Assume that the grain falls straight
down and that slowing due to rolling friction or air
drag is negligible.




 April 28, 2010               Physics 121B - Lecture 12   26/28
 Example: Radioactive Decay
   A thorium-227 nucleus (mass 227 u)
at rest decays into a radium-223 nucleus
(mass 223 u) by emitting an alpha particle
(mass 4.00 u) . The kinetic energy of the a particle is measured to be 6.00
MeV. What is the kinetic energy of the recoiling radium nucleus?




 April 28, 2010              Physics 121B - Lecture 12                  27/28
       End of Lecture 12
 For the next lecture, read T&M Chapter 8.2-3.

 Homework Assignment #5 should be submitted
on the Tycho system by 11:59 PM on Friday, April 30.

 As of 9:10 AM today, 152/176 Physics 121B
students have registered their clickers. If you have
not already done so, register your clicker at:
http://courses.washington.edu/p121bs10/clicker.htm
                      Isolation
                 Emech  Eth  Esys  Wext




April 28, 2010         Physics 121B - Lecture 12   29/28
       Problem Solving Strategy
     Picture: Determe that the net external force Fext (or Fext x) on
     the system is negligible for some time interval. (If the net force is
     NOT determined to be negligible, do not proceed.)
     Solve:
     1. Draw a sketch showing the system before and after the time
     interval. Include coordinate axes and label the initial and final
     velocity vectors.
     2. Equate the initial momentum to the final momentum and express
     this as a vector equation (or one or more scalar equations involving x,
     y, and z components.)
     3. Substitute the given information into the equation(s) and solve
     for the quantity or quantities of interest.
     Check: Make sure you include any minus signs that accompany
     velocity components, because momentum can have either sign.


April 28, 2010               Physics 121B - Lecture 12                    30/28
                  Example: Moving a Sled


  A sled is coasting on a horizontal snow-covered surface with an initial
speed of 4.0 m/s. If the coefficient of friction between the sled and the
snow is 0.14, how far will the sled travel before coming to rest?

Wext  Emech  Etherm  (U  K )  f k srel ;        f k  k Fn  k mg

0  0  K  k mgsrel


srel 
       1
       2
           mv 2
                 
                   v2      4.0 m/s        5.8 m
       k mg 2k g 2  0.14   9.81 m/s 
                                        2




 April 28, 2010              Physics 121B - Lecture 12                         31/28
                   Example:
             A Skateboard Workout
    A 40.0-kg skateboarder on a
3.00-kg board is training with two
5.00-kg weights. Beginning from rest,
she throws the weights horizontally,
one at a time, from her board
The speed of each weight is 7.00 m/s relative to her
after it is thrown. Assume the board rolls without friction.
(a) How fast is she moving in the opposite
    direction after throwing the first weight?
(b) After throwing the second weight?




  April 28, 2010              Physics 121B - Lecture 12        32/28

								
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