Momentum by yaoyufang

VIEWS: 12 PAGES: 66

									               Momentum
• Do the Yankees or the Phillies have Big Mo
  (momentum) with them tonight?
• If you have ever been in an auto accident
  physics may have saved your life or at least
  prevented more serious injuries.
• Conservation of Momentum as a forensic tool.
                Momentum
              Inertia in motion
• Inertia is a measure of the resistance of an
  object to a change in motion.
• We have a sense that it is harder to stop a
  truck than a car if both are moving at the
  same speed.
• It is more difficult to stop a car moving quickly
  than one that is moving slowly.
                Momentum
• Momentum = mass x velocity
•              =mv
• Momentum is a vector
• We often write momentum as p,
                     p =mv
• The unit of momentum is kg m/s; there is no
  derived unit for momentum.
                    Momentum
• An object can have a large momentum if

• m is large,   mv , or if
                             v
• It has a large speed, m , or
• Both

• Can a skateboard have more
momentum than a cement
truck?
A) Yes    B) No
                  Impulse
• To change momentum we must change either
  mass or velocity (or both).
• If m is constant to change p we must change v
  (accelerates)
•                Δp = mΔv
• What causes a? A force F.
• But there is another variable when changing
  momentum, time!
                         Impulse
• If you are pushing a car without gas the final
  velocity depends on both how hard and how
  long you push.
• The final momentum thus depends on how large a
  force and on how long you apply the force.
• Definition of impulse



• Can also define an average impulse when force is variable
          Impulse-examples


Increase momentum (increase Δt)
• Follow through- golf, baseball
• Barrel of rifle
Decrease momentum
 Increase Δt to decrease F
                     Impulse
• The concept of impulse leads to a more general
  form of Newton’s 2nd law.
•             FΔt = Δp
•             F = Δp/Δt
• In this form we can handle
problems where the mass changes

• If m is constant Δp = mΔv
•            F = m Δv/Δt = ma
Example (text problem 7.9): An object of mass 3.0 kg is
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.




                                                       12
Example : 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 each
                  mass for the same interval, p is the
                  same for both masses.


                                                            13
Example continued:

    (b) Which mass has the greatest velocity change?


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


    (c) Which   mass has the greatest acceleration?

                      Since av the mass with the greater
                      velocity change will have the greatest
                      acceleration (mass 1).

                                                                   14
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.




                                      The force will be in
                                      the direction of
                                      motion.



                                                        15
                 Question
As a linebacker on the football team, which
  player would be the easiest for you to stop
  when you tackle him?
  A) The 350-lb lineman who can cover 10
       yards in 2.5 seconds.
  B) The 160-lb running back, who can make
       it down the field in ten seconds.
  C) The 240-lb halfback, who can make it
       down the field in twenty seconds.
  D) They are all equally hard to stop.
                   Question
As a kid playing on the playground, you would bend
  your knees when you landed after jumping off
  the monkey bars to reduce the "sting" in your
  feet. This worked because
  A) bending your knees gave you upward
       momentum which partly canceled the
  downward momentum.
  B) bending your knees lowered your center of
       gravity reducing the force of your fall.
  C) bending your knees increased the time of
       contact for the ground to bring you to rest.
  D) you didn't do it on purpose, your knees just
       buckled.
                    Momentum
Consider two interacting bodies with m2 > m1:


                   F21       F12
             m1                     m2

If we know the net force on each body then



The velocity change for each mass will be different if
the masses are different.                                18
Rewrite the previous result for each body as:




                                            Newton’s 3rd
                                            Law

Combine the two results:




                                                       19
From slide (3):




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




                                                     20
     Conservation of Momentum
• If the net external force acting on a system is
  zero, then the momentum of the system does
  not change.
• When a quantity in physics does not change
  we say it is conserved. Momentum is
  conserved.
• We have to be careful to define the system.
• It is only forces external to the system that can
  change momentum
    Conservation of Momentum
• Internal forces do not change the momentum
  of the system.

Momentum is a vector. Each component is
  conserved separately.
Vectors can cancel each other out.
                                   0
-M v + m V = 0
     Conservation of Momentum
Crash! What happens in collisions
Energy and Momentum in collisions.
     Conservation of Momentum
• If the net external force acting on a system is
  zero, then the momentum of the system does
  not change.
• We have to be careful to define the system.
• It is only forces external to the system that can
  change momentum.
• If there are no external forces to the system
  the internal changes in momentum cancel
                   Question
You peddle frantically to get your bicycle up to a
  speed of 15 m/s. On level ground, you relax
  and start to coast. Your speed
  A) stays the same, because momentum is
  conserved.
  B) increases, now that you are not expending
  energy turning the pedals.
  C) decreases, as there are still forces acting on
  the bicycle.
          Conservation of Momentum
             v1i

                         v2i                        m1>m2


           m1                  m2



A short time later the
masses collide.
                                       m1   m2


                                    What happens?
                                                            29
During the interaction:

                          N1           N2
                                                   y

           F21                               F12
                                                       x
                          w1           w2




                               There is no net
                               external force on
                               either mass.
                                                           30
The forces F12 and F21 are internal forces. This means
that:




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

                                                         31
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?




                                                            32
               Question
You paddle your own canoe forward by
  pushing back on the water. If you can
  change the velocity of 7.3 kg of water by
  3.0 m/s with each stroke, your speed
  changes by (myou + mcanoe = 93 kg)
  A) 226 m/s
  B) 38 m/s
  C) 4.2 m/s
  D) 0.24 m/s
                       Collisions

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
kinetic energy changes, the collision is inelastic.


                                                           34
Elastic collisions
Inelastic collisions
If, after a collision, the bodies remain stuck together, the loss of
kinetic energy is a maximum, but not necessarily a 100% loss of
kinetic energy. This type of collision is called perfectly inelastic.




                                                                    37
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?




                                                                      39
(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.




                                                                      40
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?




                                                                  41
                   Question
Which is an example of a perfectly elastic
 collision?
 A) A wet tissue thrown at the wall.
 B) The cue ball striking the 8-ball in a billiards
 game.
 C) A rubber ball bouncing off the floor.
Fig. 07.18
   The New York Times, Jan.13, 1920, p.
                   12
…its flight would be neither accelerated nor maintained
by the explosion of the charges… To claim that it would
be, is to deny a fundamental law of dynamics… That
Professor Goddard, with his ‘chair’ in Clark College and
the countenancing of the Smithsonian Institution, does
not know the relation of action to reaction, and of the
need to have something better than a vacuum against
which to react – to say that would be absurd.



PHY211 Fall 2009
Lecture 10-1
      The New York Times, July 17, 1969,
                    p. 34
   …further investigation and experimentation have
   confirmed the findings of Isaac Newton in the 17th
   century, and it is now definitely established that a rocket
   can function in a vacuum as well as in an atmosphere.
   The Times regrets the error.
   …an editorial feature of the New York Times dismissed
   the notion that a rocket could function in a vacuum and
   commented on the ideas of Robert H. Goddard.



PHY211 Fall 2009
Lecture 10-1
                 Question
A boxcar with mass m moving to the right with
  speed v collides with a second boxcar of the
  same mass which is at rest. After the
  collision, both boxcars move off to the right
  with speed v/2. In this collision
  A) momentum is conserved.
  B) energy is conserved.
  C) momentum and energy are conserved.
  D) neither momentum nor energy is
      conserved.
              Center of Mass
• Why is your belly button where it is?

• Something simple about complicated motion

• Why can I safely walk out over the edge of the
  table?
There is something special about this point
Page 250
          §7.5 Center of Mass


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



                                               54
The center of mass (of a two body system) is found
from:




This is a “weighted” average of the positions of the
particles that compose a body.

A larger mass is more important.
If m1 = 100 m2 then xcm ~ x1
                                                       55
In 3-dimensions write        where




The components of rcm are:




                                     56
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


                                                 M     r
                                                 4 (4,0)
                              1    x             6 (2,4)
          3
                                                 3 (-1,-2)

                                                                       57
Example continued:




                     58
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




 A                x




                                                         59
    Where is the CM       of a Donut?



A                     B




         C



D) It could be A or C or anywhere on the donut.
     Motion of the Center of Mass


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

From this it follows that Fext = macm.



                                          62
Fig. 07.11
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.

                                                        64
Example continued:

 For the center of mass:


Solving for the velocity of the center of mass:




 Or in component form:




                                                  65
Example continued:


Applying the previous formulas to the example,




                                                 66

								
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