Momentum

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Momentum
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Friday, November 17, 2006

• Momentum and Momentum Change

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Momentum

• Momentum is a measure of how hard it is to stop
  or turn a moving object.
• Momentum is related to both mass and velocity.
• Momentum is possessed by all moving objects.

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Calculating Momentum

• For one particle
• p mv
• For a system of multiple particles
• P ?pi ?mivi
• Momentum is a vector with the same direction as
  the velocity vector.
• The unit of momentum is
• kg m/s or Ns

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Which has the most momentum?
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Sample Problem

• Calculate the momentum of a 65-kg sprinter
  running east at 10 m/s.

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Sample Problem

• Calculate the momentum of a system composed of a
  65-kg sprinter running east at 10 m/s and a 75-kg
  sprinter running north at 9.5 m/s.

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Change in momentum

• Like any change, change in momentum is calculated
  by looking at final and initial momentums.
• Dp pf pi
• Dp change in momentum
• pf final momentum
• pi initial momentum

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Momentum change demonstration

• Using only a meter stick, find the momentum
  change of each ball when it strikes the desk from
  a height of exactly one meter.
• Which ball, Bouncy or Lazy, has the greatest
  change in momentum?

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Wording dilemma

• In which case is the magnitude of the momentum
  change greatest?
• In which case is the change in the magnitude of
  the momentum greatest?

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Monday, November 20, 2006

• Impulse

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Announcements

• Exam corrections
• Today at lunch
• Monday and Tuesday (AM and Lunch)
• Tomorrow
• Geometric optics exam
• CW packet for optics is due
• Wednesday
• HW due (Momentum 1 and 2)

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Impulse (J)

• Impulse is the product of an external force and
  time, which results in a change in momentum of a
  particle or system.
• J F t
• J ?P
• Units N s or kg m/s (same as momentum)

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Impulsive Forces

• Usually high magnitude, short duration.
• Suppose the ball hits the bat at 90 mph and
  leaves the bat at 90 mph, what is the magnitude
  of the momentum change?
• What is the change in the magnitude of the
  momentum?

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Impulse (J) on a graph
F(N)
3000
2000
area under curve
1000
0
0
1
2
3
4
t (ms)
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Sample Problem

• Suppose a 1.5-kg brick is dropped on a glass
  table top from a height of 20 cm.
• What is the magnitude and direction of the
  impulse necessary to stop the brick?
• If the table top doesnt shatter, and stops the
  brick in 0.01 s, what is the average force it
  exerts on the brick?
• What is the average force that the brick exerts
  on the table top during this period?

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Solution a)

• Find the velocity of the brick when it strikes
  the table using conservation of energy.
• mgh ½ mv2
• v (2gh)1/2 (29.8 m/s20.20 m) 1/2 2.0 m/s
• Calculate the bricks momentum when it strikes
  the table.
• p mv (1.5 kg)(2.0 m/s) 3.0 kg m/s (down)
• The impulse necessary to stop the brick is the
  impulse necessary to change to momentum to zero.
• J Dp pf pi 0 3.0 kg m/s -3.0 kg m/s
• or 3.0 kg m/s (up)

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Solution b) and c)

• b) Find the force using the other equation for
  impulse.
• J Ft
• 3.0 N s F (0.01 s)
• F 300 N (upward in the same direction as
  impulse)
• c) According the Newtons 3rd law, the brick
  exerts an average force of 300 N downward on the
  table.

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Sample Problem
F(N)
2,000
1,000
0.20
0.40
0.60
0.80
t(s)

• This force acts on a 1.2 kg object moving at
  120.0 m/s. The direction of the force is aligned
  with the velocity. What is the new velocity of
  the object?

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Solution

• Find the impulse from the area under the curve.
• A ½ base height ½ (.1 s)(2500 N) 125 Ns
• J 125 N s
• Since impulse is equal to change in momentum and
  it is in the same direction as the existing
  momentum, the momentum increases by 125 kg m/s.
• Dp 125 kg m/s
• Dp pf - pi mvf - mvi
• mvf mvi Dp
• (1.2 kg)(120 m/s) 125 kg m/s 269 kg
  m/s
• vf (269 kg m /s) / (1.2 kg) 224 m/s

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Wednesday, November 22, 2006

• Law of Conservation of Momentum

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Announcements

• Energy exam corrections
• Monday and Tuesday (AM and Lunch)
• Due today
• HW due -- Momentum 1 and 2
• Due next Wednesday
• Momentum 3 and 4
• Have a happy Thanksgiving!!!

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Law of Conservation of Momentum

• If the resultant external force on a system is
  zero, then the vector sum of the momentums of the
  objects will remain constant.
• SPbefore SPafter

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Sample problem

• A 75-kg man sits in the back of a 120-kg canoe
  that is at rest in a still pond. If the man
  begins to move forward in the canoe at 0.50 m/s
  relative to the shore, what happens to the canoe?

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Solution

• The momentum before the man moves is equal to the
  momentum after the man moves.
• Spb Spa
• 0 mmvm mcvc
• 0 (75 kg)(0.50 m/s) (120 kg)v
• v - (75 kg)(0.50 m/s)/(120 kg)
• v -0.31 m/s
• The canoe slips backward in the water at -0.31 m/s

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External versus internal forces

• External forces forces coming from outside the
  system of particles whose momentum is being
  considered.
• External forces change the momentum of the
  system.
• Internal forces forces arising from interaction
  of particles within a system.
• Internal forces cannot change momentum of the
  system.

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An external force in golf

• Consider the collision between the club head and
  the golf ball in the sport of golf.
• The club head exerts an external impulsive force
  on the ball and changes its momentum.

• The acceleration of the ball is greater because
  its mass is smaller.

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An internal force in pool

• Consider the collision between two balls in pool.
  
• The forces they exert on each other are internal
  and do not change the momentum of the system.

• Since the balls have equal masses, the magnitude
  of their accelerations is equal.

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Explosions

• When an object separates suddenly, as in an
  explosion, all forces are internal.
• Momentum is therefore conserved in an explosion.
• There is also an increase in kinetic energy in an
  explosion. This comes from a potential energy
  decrease due to chemical combustion.

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Recoil

• Guns and cannons recoil when fired.
• This means the gun or cannon must move backward
  as it propels the projectile forward.
• The recoil is the result of action-reaction force
  pairs, and is entirely due to internal forces.
• As the gases from the gunpowder explosion expand,
  they push the projectile forwards and the gun or
  cannon backwards.

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Sample problem

• Suppose a 5.0-kg projectile launcher shoots a 209
  gram projectile at 350 m/s. What is the recoil
  velocity of the projectile launcher?

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Solution

• Momentum conservation is used to calculate recoil
  speed.
• Spb Spa
• 0 mpvp mlvl
• 0 (0.209 kg)(350 m/s) (5.0 kg)v
• v - (0.209 kg)(350 m/s)/(5.0 kg)
• v - 14.6 m/s

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Sample Problem

• An exploding object breaks into three fragments.
  A 2.0 kg fragment travels north at 200 m/s. A 4.0
  kg fragment travels east at 100 m/s. The third
  fragment has mass 3.0 kg. What is the magnitude
  and direction of its velocity?

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Solution

• The momentum before is zero, so the momentum
  after is zero.
• This is a vector addition problem. Each fragment
  has a momentum magnitude of 400 kg m/s according
  to the formula p mv.

v p/m 566/3 189 m/s due SW
(4002 4002)1/2 566 kg m/s due southwest
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Monday, November27, 2006

• Collisions

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Announcements

• Exam corrections
• Today and Tuesday (AM and Lunch)
• Tonights assignment
• Momentum 4

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Sample problem

• An exploding object breaks into three fragments.
  A 2.0 kg fragment travels north at 200 m/s. A 4.0
  kg fragment travels east at 100 m/s. The third
  fragment has mass 3.0 kg. What is the magnitude
  and direction of its velocity?

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Today

• Bumper cars demo lab.
• You will need to be clever and quick as you
  attempt to simulate various types of collisions
  using the carts and cart track.
• For each collision type, you will be given a few
  minutes to create the simulation and answer the
  associated questions.
• You are not to harm the carts or each other in
  your simulations.

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Simulation 1

• Your mission Create a situation in which an
  impulse changes the momentum of a moving cart.
  You express (a) the magnitude of the momentum
  change, and (b) the change in the magnitude of
  the momentum in terms of m and v. You must
  identify the force causing the impulse.
• Rule 1 Use just one cart.
• Rule 2 No kinetic energy may be lost or gained
  by the cart in this collision.

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Simulation 2

• Your mission Create a situation in which an
  impulse changes the momentum of a moving cart.
  You express (a) the magnitude of the momentum
  change, and (b) the change in the magnitude of
  the momentum in terms of m and v. You must
  identify the force causing the impulse.
• Rule 1 Use just one cart.
• Rule 2 All kinetic energy must be lost by the
  cart in this collision.

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Simulation 3

• Your mission Create a collision between two
  carts in which momentum and kinetic energy (of
  the system of two carts) are both conserved. You
  must be able to express momentum and kinetic
  energy of each cart before and after the
  collision.
• Rule 1 One cart must lose ALL of its momentum
  and kinetic energy in this simulation.

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Simulation 4

• Your mission Create a collision between two
  carts in which the following happens
• Rule 1 The two carts become one cart.
• Rule 2 The system of two carts loses all of its
  kinetic energy.
• Be prepared to identify the initial and final
  momentum and kinetic energy of the carts in terms
  of m and v.

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Simulation 5

• Your mission Create simulation of an explosion
  creating two equal mass fragments. You must
  express (a) the initial and final momentum of the
  system and (b) the final momentum of each of the
  fragments.
• Rule 1 Kinetic energy must be zero initially.
• Rule 2 You must identify the source of the
  kinetic energy using Conservation of Energy
  thinking.

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Collisions

• When two moving objects make contact with each
  other, they undergo a collision.
• Conservation of momentum is used to analyze all
  collisions.
• Newtons Third Law is also useful. It tells us
  that the force exerted by body A on body B in a
  collision is equal and opposite to the force
  exerted on body B by body A.

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Collisions

• During a collision, external forces are ignored.
• The time frame of the collision is very short.
• The forces are impulsive forces (high force,
  short duration).

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Collision Types

• Elastic collisions
• Also called hard collisions
• No deformation occurs, no kinetic energy lost
• Inelastic collisions
• Deformation occurs, kinetic energy is lost
• Perfectly Inelastic (stick together)
• Objects stick together and become one object
• Deformation occurs, kinetic energy is lost

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(Perfectly) Inelastic Collisions

• Simplest type of collisions.
• After the collision, there is only one velocity,
  since there is only one object.
• Kinetic energy is lost.
• Explosions are the reverse of perfectly inelastic
  collisions in which kinetic energy is gained!

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Sample Problem

• An 80-kg roller skating grandma collides
  inelastically with a 40-kg kid. What is their
  velocity after the collision?

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Tuesday, November 28, 2006

• More on Collisions

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Announcements

• Exam corrections
• Tuesday (AM and Lunch)
• Tonights assignment
• Momentum 5
• Lunch Bunch tomorrow

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Sample Problem

• A train of mass 4m moving 5 km/hr couples with a
  flatcar of mass m at rest. What is the velocity
  of the cars after they couple?

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Sample Problem

• A fish moving at 2 m/s swallows a stationary fish
  which is 1/3 its mass. What is the velocity of
  the big fish and after dinner?

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Elastic Collision

• After the collision, there are still two objects,
  with two separate velocities
• Kinetic energy remains constant before and after
  the collision.
• Therefore, two basic equations must hold for all
  elastic collisions
• Spb Spa (momentum conservation)
• SKb SKa (kinetic energy conservation)

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Sample Problem

• A pool ball traveling at speed v strikes a second
  pool ball at rest such that the first pool ball
  stops completely. Show that the second pool ball
  now must have speed v. Assume that the collision
  is elastic.

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Sample Problem

• A 500-g cart on an air track strikes a 1,000-g
  cart at rest. What are the resulting velocities
  of the two carts? (Assume the collision is
  elastic, and the first cart is moving at 2.0 m/s
  when the collision occurs.)

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Solution

• before after
• m1v1 m1v1 m2v2
• 1.0 0.50v1 v2
• ½ m1 v12 ½ m1v12 ½
  m2v22
• 2.0 0.50v12 v22
• Solve simultaneously
• v1 -0.67 m/s
• v2 1.33 m/s

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Sample Problem

• Suppose three equally strong, equally massive
  astronauts decide to play a game as follows The
  first astronaut throws the second astronaut
  towards the third astronaut and the game begins.
  Describe the motion of the astronauts as the game
  proceeds. Assume each toss results from the
  same-sized "push." How long will the game last?

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Thursday, November 30

• Lab

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Announcements

• Friday
• HW 3-5 will be checked
• Lab report (partial) due
• Monday
• HW quiz
• Second two free response from packet are due at
  the beginning of the period
• Tuesday
• Momentum exam

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Lab turn in data, calculations, and results only

• Analyze collisions of the carts in terms of
  momentum and energy conservation.
• 3 or 4 Trials
• Perfectly inelastic collision equal masses
• Perfectly inelastic collision unequal masses
• Elastic collision equal masses
• Elastic collision unequal masses (BONUS!!)
• Clearly show all data collected (mass, width,
  time) for each trial
• Clearly show a comparison of the momentum before
  and after collision.
• Clearly show a comparison of kinetic energy
  before and after collision.

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Friday, December 1

• 2D Collisions

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Announcements

• Lab write-up due pass forward
• Put boot homework in folder.
• Coming up
• HW quiz on Monday
• Free response due Monday
• Exam on Momentum next Tuesday. Makeup exams for
  English Essay people are at lunch the same day.
  Field trip people take it the day before at
  lunch.
• Momentum Jeopardy

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2D-Collisions

• Momentum in the x-direction is conserved.
• SPx (before) SPx (after)
• Momentum in the y-direction is conserved.
• SPy (before) SPy (after)
• Treat x and y coordinates independently.
• Ignore x when calculating y
• Ignore y when calculating x
• Lets look at a simulation
• http//surendranath.tripod.com/Applets.html

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Sample problem

• Calculate velocity of 8-kg ball after the
  collision.

2 m/s
y
2 kg
y
3 m/s
50o
x
x
2 kg
8 kg
0 m/s
8 kg
v
After
Before