Chapter 7: Momentum

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Chapter 7: Momentum Powered By Docstoc
					Chapter 7
Linear Momentum


          老師:熊京民
          教室:ME206
          日期:12/4
Content

   7.1 Momentum…………………………………………………3
   7.2 Impulse……………………………………………………..7
   7.3 Conservation of Momentum…………………………….12
   7.4 Center of Mass…………………………………………...16
        例題7.3、7.5、7.7……………………………………….23
   7.5 Motion of the Center of Mass…………………………..29
   7.6 & 7.7 Collisions in One and Two Dimension………….33
        例題7.8、7.10、7.11…………………………………….41




                                                      1/46
Objectives

   Definition of Momentum
   Impulse
    Impulse
   Conservation of Momentum
   Center of Mass
   Motion of the Center of Mass
   Collisions (1d, 2d; elastic, inelastic)




                                              2/46
 §7.1 Momentum

Consider two interacting bodies:




                      F21       F12

                                                      Fnet
If we know the net force on each body then v  at       t
                                                       m

The velocity change for each mass will be different if the
masses are different.
                                                             3/46
Rewrite the previous result for each body as:


                  m1v1  F21t

                  m2 v 2  F12 t  F21t



Combine the two results:   m1v1  m2 v 2
                           m1 v1 f  v1i   m2 v 2 f  v 2i 



                                                                    4/46
The quantity (mv) is called momentum (p).
                                      p
p = mv and is a vector
     v          vector.



The unit of momentum is kg m/s; there is no derived unit for
momentum.




                                                          5/46
 From slide (3):   m1v1  m2 v 2
                   p1  p 2


The change in momentum of the two bodies is “equal and
                                                equal
opposite”.
opposite Total momentum is conserved during the
interaction; the momentum lost by one body is gained by
the other.




                                                      6/46
§7.2 Impulse


Definition of impulse:   p  Ft



One can also define an average impulse when the force
is variable.




                                                        7/46
Example (text conceptual question 7.2): A force of 30 N is
applied for 5 sec to each of two bodies of different masses.
                                            30 N
     Take m1<m2
                                m1 or m2


   (a) Which mass has the greater momentum change?


                   Since the same force is applied to
    p  Ft       each mass for the same interval, p
                   is the same for both masses.



                                                               8/46
Example continued:

   (b) Which mass has the greatest velocity change?


          p         Since both masses have the same p,
     v             the smaller mass (mass 1) will have the
          m          larger change in velocity.


   (c) Which mass has the greatest acceleration?

         v          Since av the mass with the greater
      a             velocity change will have the greatest
         t
                     acceleration.

                                                               9/46
Example (text problem 7.9): An object of mass 3.0 kg is
 Example
allowed to fall from rest under the force of gravity for 3.4
seconds. What is the change in momentum? Ignore air
resistance.
      Want p =mv.



             v  at
             v   gt  33.3 m/sec
              p  mv  100 kg m/s (downward)



                                                               10/46
Example (text problem 7.10): What average force is
necessary to bring a 50.0-kg sled from rest to 3.0 m/s in a
period of 20.0 seconds? Assume frictionless ice.



       p  Fav t
              p mv
       Fav      
              t     t
      Fav 
            50.0 kg3.0 m/s  7.5 N      The force will be
                  20.0 s                    in the direction
                                            of motion.



                                                              11/46
§7.3 Conservation of Momentum

            v1i
                         v2i                    m1>m2


          m1               m2


A short time later the
masses collide.

                                   m1   m2

                                What happens?
                                                        12/46
During the interaction:
                                                           y
                         N1           N2
                                                                 x
              F21                           F12

                         w1           w2

      F  y    N1  w1  0            F   y    N 2  w2  0
      F  x     F21  m1a1           F   x    F12  m2 a2


                               There is no net
                               external force on
                               either mass.                          13/46
The forces F12 and F21 are internal forces. This means that:


                p1  p 2
                p1 f  p1i  p 2 f  p 2i 
                p1i  p 2i  p1 f  p 2 f


 In other words, pi = pf. That is, momentum is conserved.
 This statement is valid during the interaction only.



                                                            14/46
Example (text problem 7.18): A rifle has a mass of 4.5 kg
and it fires a bullet of 10.0 grams at a muzzle speed of 820
m/s. What is the recoil speed of the rifle as the bullet leaves
the barrel?

    As long as the rifle is horizontal, there will be no net
    external force acting on the rifle-bullet system and
    momentum will be conserved.


   pi  p f
   0  mb vb  mr vr
           mb        0.01 kg 
   vr             4.5 kg 820 m/s  1.82 m/s
              vb           
           mr                
                                                               15/46
§7.4 Center of Mass


The center of mass (CM) is the point representing the mean
                     CM
(average) position of the matter in a body. This point need
 average
not be located within the body.




                                                          16/46
The center of mass (of a two body system) is found from:
                                  system

                             m1 x1  m2 x2
                       xcm 
                              m1  m2

This is a “weighted” average of the positions of the particles
that compose a body. (A larger mass is more important.)




                                                             17/46
                                m r        i i
                                                            where mtotal   mi
                                i
In 3-dimensions write   rcm
                                m  i
                                             i                                   i




The components of rcm are:


          m x    i i           m y             i       i                   m z    i i

         i             ycm        i                                zcm    i

                                                                             m
  xcm
          m       i            m      i
                                                     i
                                                                                 i
                                                                                      i
              i




                                                                                           18/46
Example (text problem 7.27): Particle A is at the origin and
has a mass of 30.0 grams. Particle B has a mass of
10.0 grams. Where must particle B be located so that the
center of mass (marked with a red x) is located at the point
(2.0 cm, 5.0 cm)?

  y
         x                         ma xa  mb xb    mb xb
                             xcm                
                                    ma  mb        ma  mb


                                      ma ya  mb yb    mb yb
                              ycm                  
  A                x                    ma  mb       ma  mb



                                                                19/46
Example continued:




            10 g xb            ycm 
                                         10 g yb    5 cm
    xcm                 2 cm
          10 g  30g                  30 g  10 g
     xb  8 cm                   yb  20 cm




                                                              20/46
Example (text problem 7.30): The positions of three particles
are (4.0 m, 0.0 m), (2.0 m, 4.0 m), and (-1.0 m, -2.0 m). The
masses are 4.0 kg, 6.0 kg, and 3.0 kg respectively. What is
the location of the center of mass?

                       y        2




                                     1   x


                   3



                                                           21/46
Example continued:


           m1 x1  m2 x2  m3 x3
     xcm 
              m1  m2  m3

         
           4 kg4 m   6 kg2 m   3 kg 1 m 
                          4  6  3 kg
          1.92 m


           m1 y1  m2 y2  m3 y3
     ycm 
               m1  m2  m3

         
           4 kg0 m  6 kg4 m  3 kg 2 m
                         4  6  3 kg
          1.38 m
                                                          22/46
例題
   7.3
   7.5
   7.7




          23/46
24/46
25/46
26/46
27/46
Example 7.07b




                28/46
§7.5 Motion of the Center of Mass



For an extended body, it can be shown that p = mvcm.




From this it follows that Fext = macm.




                                                       29/46
Example (text problem 7.35): Body A has a mass of 3.0 kg
and vx = +14.0 m/s. Body B has a mass of 4.0 kg and has vy
= -7.0 m/s. What is the velocity of the center of mass of the
two bodies?


 Consider a body made up of many different masses
 each with a mass mi.


 The position of each mass is ri and the displacement of
 each mass is ri = vit.




                                                           30/46
Example continued:
                                                                     m r              i        i
For the center of mass:                    rcm  v cm         t          i

                                                                     m         i
                                                                                            i




Solving for the velocity of the center of mass:
                                               ri
                                          mi  t
                                                       m v     i       i
                                 v cm    i           i

                                             mi
                                            i
                                                       m  i
                                                                    i




 Or in component form:

                       m v    i i,x                                          m v              i i, y
            v cm,x    i
                                                               v cm,y              i

                       m  i
                                 i                                            m        i
                                                                                                  i
                                                                                                         31/46
Example continued:


Applying the previous formulas to the example,

                       ma va , x  mb vb , x
            vcm, x 
                           ma  mb

                  
                    3 kg 14 m/s  4 kg 0 m/s  6 m/s
                               3  4 kg

                       ma va , y  mb vb , y
            vcm, y 
                            ma  mb

                  
                    3 kg 0 m/s  4 kg  7 m/s  4 m/s
                               3  4 kg
                                                                 32/46
§§ 7.6 & 7.7 Collisions in One and Two
Dimension


When there are no external forces present, the momentum
of a system will remain unchanged. (pi = pf)



If the kinetic energy before and after an interaction is the
same, the “collision” is said to be perfectly elastic If the
                                              elastic.
                                         inelastic.
kinetic energy changes, the collision is inelastic



                                                               33/46
                                               together,
If, after a collision, the bodies remain stuck together the
                              maximum.
loss of kinetic energy is a maximum This type of collision is
called perfectly inelastic.
                    inelastic




                                                          34/46
Example (text problem 7.41): In a railroad freight yard, an
empty freight car of mass m rolls along a straight level track
at 1.0 m/s and collides with an initially stationary, fully loaded,
boxcar of mass 4.0m. The two cars couple together upon
collision.
 (a) What is the speed of the two cars after the collision?

                       pi  p f
                p1i  p2i  p1 f  p2 f
                m1v1  0  m1v  m2 v  m1  m2 v
                               m1 
                               m  m v1  0.2 m/s
                         v          
                               1    2 



                                                                35/46
Example continued:


(b) Suppose instead that both cars are at rest after the
collision. With what speed was the loaded boxcar moving
before the collision if the empty one had v1i = 1.0 m/s.


                            pi  p f
                     p1i  p2i  p1 f  p2 f
              m1v1i  m2 v2i  0  0
                                        m1 
                               v2i   v1i  0.25 m/s
                                       m 
                                        2


                                                            36/46
Example (text problem 7.49): A projectile of 1.0 kg mass
approaches a stationary body of 5.0 kg mass at 10.0 m/s and,
after colliding, rebounds in the reverse direction along the
same line with a speed of 5.0 m/s. What is the speed of the
5.0 kg mass after the collision?

              pi  p f
       p1i  p2i  p1 f  p2 f
       m1v1i  0  m1v1 f  m2 v2 f

            v2 f   
                     m1
                        v1i  v1 f 
                     m2

                   
                     1.0 kg
                            10 m/s   5.0 m/s  3.0 m/s
                     5.0 kg
                                                               37/46
Example (text problem 7.58): Body A of mass M has an
original velocity of 6.0 m/s in the +x-direction toward a
stationary body (B) of the same mass. After the collision,
body A has vx=+1.0 m/s and vy=+2.0 m/s. What is the
magnitude of body B’s velocity after the collision?


                                 Final
    Initial                                      A

    A    vAi

               B
                                                B



                                                             38/46
Example continued:

  x momentum:                                   y momentum:
           pix  p fx                                    piy  p fy
 p1ix  p2ix  p1 fx  p2 fx                   p1iy  p2iy  p1 fy  p2 fy
 m1v1ix  0  m1v1 fx  m2 v2 fx                    0  0  m1v1 fy  m2 v2 fy

 Solve for v2fx:                               Solve for v2fy:
           m1v1ix  m1v1 fx                              m1v1 fy
 v2 fx                                       v2 fy                 v1 fy
                 m2                                       m2
       v1ix  v1 fx                           2.00 m/s
       5.00 m/s

  The mag. of v2 is           v2 f  v 2 2 fy  v 2 2 fx  5.40 m/s              39/46
Summary

•Definition of Momentum
•Impulse
•Center of Mass
•Conservation of Momentum




                            40/46
例題
   7.8
   7.10
   7.11




           41/46
42/46
Example 7.08b




                43/46
Example 7.10




               44/46
Example 7.11a




                45/46
Example 7.11b




                46/46
第七章完

				
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