Linear momentum and Collisions

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Linear momentum and Collisions

• Chapter 9

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• Linear Momentum and its Conservation
• Impulse and Momentum
• Collisions in One Dimension
• Collisions in Two Dimensions
• V. The Center of Mass
• Motion of a System of Particles
• Deformable Systems
• Rocket Propulsion

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Linear Momentum

• The linear momentum of a particle, or an object
  that can be modeled as a particle of mass m
  moving with a velocity , is defined to be the
  product of the mass and velocity
• The terms momentum and linear momentum will be
  used interchangeably in the text

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I. Linear momentum
The linear momentum of a particle is a vector p
defined as
Momentum is a vector with magnitude equal mv and
has direction of v .
The dimensions of momentum are ML/T
SI unit of the momentum is kg-meter/second
Momentum can be expressed in component form px
m vx py m vy pz m vz
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Newton and Momentum

• Newtons Second Law can be used to relate the
  momentum of a particle to the resultant force
  acting on it
• with constant mass

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Newton called the product mv the quantity of
motion of the particle
Newton II law in terms of momentum
The time rate of change of the momentum of a
particle is equal to the net force acting on the
particle and is in the direction of the force.
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System of particles

The linear momentum of a system of particles is
equal to the product of the total mass M of the
system and the velocity of the center of mass.
Net external force acting on the system.
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Conservation
If no external force acts on a closed, isolated
system of particles, the total linear momentum P
of the system cannot change.
Closed no matter passes through the system
boundary in any direction.
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Conservation of Linear Momentum
If no net external force acts on the system of
particles the total linear momentum P of the
system cannot change. Each component of the
linear momentum is conserved separately if the
corresponding component of the net external force
is zero.
If the component of the net external force
on a closed system is zero along an axis ?
component of the linear momentum along that axis
cannot change.
The momentum is constant if no external
forces act on a closed particle system. Internal
forces can change the linear momentum of portions
of the system, but they cannot change the total
linear momentum of the entire system.
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Conservation of Linear Momentum

• Whenever two or more particles in an isolated
  system interact, the total momentum of the system
  remains constant
• The momentum of the system is conserved, not
  necessarily the momentum of an individual
  particle
• This also tells us that the total momentum of an
  isolated system equals its initial momentum

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Conservation of Momentum

• Conservation of momentum can be expressed
  mathematically in various ways
• In component form, the total momenta in each
  direction are independently conserved
• pix pfx piy pfy piz pfz
• Conservation of momentum can be applied to
  systems with any number of particles
• This law is the mathematical representation of
  the momentum version of the isolated system model

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Conservation of Momentum, Archer Example

• The archer is standing on a frictionless surface
  (ice)
• Approaches
• Newtons Second Law no, no information about F
  or a
• Energy approach no,
• no information about work or energy
• Momentum yes

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Archer Example

• Conceptualize
• The arrow is fired one way and the archer recoils
  in the opposite direction
• Categorize
• Momentum
• Let the system be the archer with bow (particle
  1) and the arrow (particle 2)
• There are no external forces in the x-direction,
  so it is isolated in terms of momentum in the
  x-direction
• Analyze
• Total momentum before releasing the arrow is 0

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Archer Example

• Analyze.
• The total momentum after releasing the arrow is
• Finalize
• The final velocity of the archer is negative
• Indicates he moves in a direction opposite the
  arrow
• Archer has much higher mass than arrow, so
  velocity is much lower

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Impulse and Momentum

• From Newtons Second Law,
• Solving for gives
• Integrating to find the change in momentum over
  some time interval
• The integral is called the impulse, , of the
  force acting on an object over ?t

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IV. Collision and impulse
Collision isolated event in which two or more
bodies exert relatively strong forces on each
other for a relatively short time.
Impulse
Measures the strength and
duration of the collision force
Third law force pair
FR - FL
Single collision
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Impulse-linear momentum theorem
The change in the linear momentum of a body in a
collision is equal to the impulse that acts on
that body.
Units
kg m/s
Favg such that Area under F(t) vs ?t curve
Area under Favg vs t
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• An estimated force-time curve for a baseball
  struck by a bat is shown in Figure. From this
  curve, determine (a) the impulse delivered to the
  ball, (b) the average force exerted on the ball,
  and (c) the peak force exerted on the ball.

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More About Impulse

• Impulse is a vector quantity
• The magnitude of the impulse is equal to the area
  under the force-time curve
• The force may vary with time
• Dimensions of impulse are M L / T
• Impulse is not a property of the particle, but a
  measure of the change in momentum of the particle

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Impulse

• The impulse can also be found by using the time
  averaged force
• This would give the same impulse as the
  time-varying force does

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Impulse Approximation

• In many cases, one force acting on a particle
  acts for a short time, but is much greater than
  any other force present
• When using the Impulse Approximation, we will
  assume this is true
• Especially useful in analyzing collisions
• The force will be called the impulsive force
• The particle is assumed to move very little
  during the collision
• represent the momenta immediately
  before and after the collision

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Impulse-Momentum Crash Test Example

• Categorize
• Assume force exerted by wall is large compared
  with other forces
• Gravitational and normal forces are perpendicular
  and so do not effect the horizontal momentum
• Can apply impulse approximation

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Crash Test Example

• Analyze
• The momenta before and after the collision
  between the car and the wall can be determined
• Find
• Initial momentum
• Final momentum
• Impulse
• Average force
• Finalize
• Check signs on velocities to be sure they are
  reasonable

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Collisions Characteristics

• We use the term collision to represent an event
  during which two particles come close to each
  other and interact by means of forces
• May involve physical contact, but must be
  generalized to include cases with interaction
  without physical contact
• The time interval during which the velocity
  changes from its initial to final values is
  assumed to be short
• The interaction forces are assumed to be much
  greater than any external forces present
• This means the impulse approximation can be used

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Collisions Example 1

• Collisions may be the result of direct contact
• The impulsive forces may vary in time in
  complicated ways
• This force is internal to the system
• Momentum is conserved

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Collisions Example 2

• The collision need not include physical contact
  between the objects
• There are still forces between the particles
• This type of collision can be analyzed in the
  same way as those that include physical contact

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Types of Collisions

• In an elastic collision, momentum and kinetic
  energy are conserved
• Perfectly elastic collisions occur on a
  microscopic level
• In macroscopic collisions, only approximately
  elastic collisions actually occur
• Generally some energy is lost to deformation,
  sound, etc.
• In an inelastic collision, kinetic energy is not
  conserved, although momentum is still conserved
• If the objects stick together after the
  collision, it is a perfectly inelastic collision

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Collisions

• In an inelastic collision, some kinetic energy is
  lost, but the objects do not stick together
• Elastic and perfectly inelastic collisions are
  limiting cases, most actual collisions fall in
  between these two types
• Momentum is conserved in all collisions

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Perfectly Inelastic Collisions

• Since the objects stick together, they share the
  same velocity after the collision

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Elastic Collisions

• Both momentum and kinetic energy are conserved

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Elastic Collisions

• Typically, there are two unknowns to solve for
  and so you need two equations
• The kinetic energy equation can be difficult to
  use
• With some algebraic manipulation, a different
  equation can be used
• v1i v2i v1f v2f
• This equation, along with conservation of
  momentum, can be used to solve for the two
  unknowns
• It can only be used with a one-dimensional,
  elastic collision between two objects

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Elastic Collisions

• Example of some special cases
• m1 m2 the particles exchange velocities
• When a very heavy particle collides head-on with
  a very light one initially at rest, the heavy
  particle continues in motion unaltered and the
  light particle rebounds with a speed of about
  twice the initial speed of the heavy particle
• When a very light particle collides head-on with
  a very heavy particle initially at rest, the
  light particle has its velocity reversed and the
  heavy particle remains approximately at rest

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Problem-Solving Strategy One-Dimensional
Collisions

• Conceptualize
• Image the collision occurring in your mind
• Draw simple diagrams of the particles before and
  after the collision
• Include appropriate velocity vectors
• Categorize
• Is the system of particles isolated?
• Is the collision elastic, inelastic or perfectly
  inelastic?

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Problem-Solving Strategy One-Dimensional
Collisions

• Analyze
• Set up the mathematical representation of the
  problem
• Solve for the unknown(s)
• Finalize
• Check to see if the answers are consistent with
  the mental and pictorial representations
• Check to be sure your results are realistic

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Example Stress Reliever

• Conceptualize
• Imagine one ball coming in from the left and two
  balls exiting from the right
• Is this possible?
• Categorize
• Due to shortness of time, the impulse
  approximation can be used
• Isolated system
• Elastic collisions

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Example Stress Reliever

• Analyze
• Check to see if momentum is conserved
• It is
• Check to see if kinetic energy is conserved
• It is not
• Therefore, the collision couldnt be elastic
• Finalize
• Having two balls exit was not possible if only
  one ball is released

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Example Stress Reliever

• What collision is possible
• Need to conserve both momentum and kinetic energy
• Only way to do so is with equal numbers of balls
  released and exiting

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Collision Example Ballistic Pendulum

• Conceptualize
• Observe diagram
• Categorize
• Isolated system of projectile and block
• Perfectly inelastic collision the bullet is
  embedded in the block of wood
• Momentum equation will have two unknowns
• Use conservation of energy from the pendulum to
  find the velocity just after the collision
• Then you can find the speed of the bullet

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Ballistic Pendulum

• A multi-flash photograph of a ballistic pendulum
• Analyze
• Solve resulting system of equations
• Finalize
• Note different systems involved
• Some energy was transferred during the perfectly
  inelastic collision

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V. Momentum and kinetic energy in collisions
Assumptions Closed systems (no mass enters or
leaves them)
Isolated systems (no external forces act on
the bodies within the system)
Elastic collision
If the total kinetic energy of the system of two
colliding bodies is unchanged (conserved) by the
collision.
Example Ball into hard floor.
Inelastic collision
The kinetic energy of the system is not conserved
? some goes into thermal energy, sound, etc.
After the collision the bodies lose energy and
stick together.
Completely inelastic collision
Example Ball of wet putty into floor
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VII. Elastic collisions in 1D
In an elastic collision, the kinetic energy of
each colliding body may change, but the total
kinetic energy of the system does not change.
Stationary target
Closed, isolated system ?
Linear momentum
Kinetic energy
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Stationary target
v2f gt0 always v1f gt0 if m1gtm2 ? forward mov. v1f
lt0 if m1ltm2 ? rebounds
Equal masses m1m2 ? v1f0 and v2f v1i ?
In head-on collisions bodies of equal masses
simply exchange velocities.
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Massive target m2gtgtm1 ? v1f -v1i and
v2f (2m1/m2)v1i ? Body 1 bounces back with
approximately same speed. Body 2 moves forward at
low speed.
Massive projectile m1gtgtm2 ? v1f v1i and
v2f 2v1i ? Body 1 keeps on going scarcely
lowed by the collision. Body 2 charges ahead at
twice the initial speed of the projectile.
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VII. Elastic collisions in 1D
Moving target
Closed, isolated system ?
Linear momentum
Kinetic energy
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VIII. Collisions in 2D
Closed, isolated system ?
Linear momentum conserved
Elastic collision ?
Kinetic energy conserved
Example
If the collision is elastic ?
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• Two blocks of masses M and 3M are placed on a
  horizontal, frictionless surface. A light spring
  is attached to one of them, and the blocks are
  pushed together with the spring between them. A
  cord initially holding the blocks together is
  burned after this, the block of mass 3M moves to
  the right with a speed of 2.00 m/s. (a) What is
  the speed of the block of mass M? (b) Find the
  original elastic potential energy in the spring
  if M 0.350 kg.

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• A tennis player receives a shot with the ball
  (0.0600 kg) traveling horizontally at 50.0 m/s
  and returns the shot with the ball traveling
  horizontally at 40.0 m/s in the opposite
  direction. (a) What is the impulse delivered to
  the ball by the racquet? (b) What work does the
  racquet do on the ball?

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• Two blocks are free to slide along the
  frictionless wooden track ABC shown in Figure. A
  block of mass m1 5.00 kg is released from A.
  Protruding from its front end is the north pole
  of a strong magnet, repelling the north pole of
  an identical magnet embedded in the back end of
  the block of mass m2 10.0 kg, initially
  at rest. The two blocks never touch. Calculate
  the maximum height to which m1 rises after the
  elastic collision.

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• As shown in Figure, a bullet of mass m and speed
  v passes completely through a pendulum bob of
  mass M. The bullet emerges with a speed of v/2.
  The pendulum bob is suspended by a stiff rod of
  length l and negligible mass. What is the
  minimum value of v such that the pendulum bob
  will barely swing through a complete vertical
  circle?

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• A small block of mass m1 0.500 kg is released
  from rest at the top of a curve-shaped
  frictionless wedge of mass m2 3.00
  kg, which sits on a frictionless horizontal
  surface as in Figure (a). When the block leaves
  the wedge, its velocity is measured to be 4.00
  m/s to the right, as in Figure (b). (a)
  What is the velocity of the wedge after the block
  reaches the horizontal surface? (b) What is the
  height h of the wedge?

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Two-Dimensional Collisions

• The momentum is conserved in all directions
• Use subscripts for
• Identifying the object
• Indicating initial or final values
• The velocity components
• If the collision is elastic, use conservation of
  kinetic energy as a second equation
• Remember, the simpler equation can only be used
  for one-dimensional situations

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Two-Dimensional Collision, example

• Particle 1 is moving at velocity and
  particle 2 is at rest
• In the x-direction, the initial momentum is m1v1i
• In the y-direction, the initial momentum is 0

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Two-Dimensional Collision, example

• After the collision, the momentum in the
  x-direction is m1v1f cos ? m2v2f cos f
• After the collision, the momentum in the
  y-direction is m1v1f sin ? m2v2f sin f
• If the collision is elastic, apply the kinetic
  energy equation
• This is an example of a glancing collision

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Problem-Solving Strategies Two-Dimensional
Collisions

• Conceptualize
• Imagine the collision
• Predict approximate directions the particles will
  move after the collision
• Set up a coordinate system and define your
  velocities with respect to that system
• It is usually convenient to have the x-axis
  coincide with one of the initial velocities
• In your sketch of the coordinate system, draw and
  label all velocity vectors and include all the
  given information

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Problem-Solving Strategies Two-Dimensional
Collisions

• Categorize
• Is the system isolated?
• If so, categorize the collision as elastic,
  inelastic or perfectly inelastic
• Analyze
• Write expressions for the x- and y-components of
  the momentum of each object before and after the
  collision
• Remember to include the appropriate signs for the
  components of the velocity vectors
• Write expressions for the total momentum of the
  system in the x-direction before and after the
  collision and equate the two. Repeat for the
  total momentum in the y-direction.

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Problem-Solving Strategies Two-Dimensional
Collisions

• If the collision is inelastic, kinetic energy of
  the system is not conserved, and additional
  information is probably needed
• If the collision is perfectly inelastic, the
  final velocities of the two objects are equal.
  Solve the momentum equations for the unknowns.
• If the collision is elastic, the kinetic energy
  of the system is conserved
• Equate the total kinetic energy before the
  collision to the total kinetic energy after the
  collision to obtain more information on the
  relationship between the velocities

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Problem-Solving Strategies Two-Dimensional
Collisions

• Finalize
• Check to see if your answers are consistent with
  the mental and pictorial representations
• Check to be sure your results are realistic

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Two-Dimensional Collision Example

• Conceptualize
• See picture
• Choose East to be the positive x-direction and
  North to be the positive y-direction
• Categorize
• Ignore friction
• Model the cars as particles
• The collision is perfectly inelastic
• The cars stick together

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Two dimensional collision

• m1 1500.0kg
• m2 2500.0 kg
• Find vf .

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• An unstable atomic nucleus of mass 17.0 ? 1027
  kg initially at rest disintegrates into three
  particles. One of the particles, of mass
  5.00 ? 1027 kg, moves along the y-axis with a
  speed of 6.00 ? 106 m/s. Another particle, of
  mass 8.40 ? 1027 kg, moves along the x-axis with
  a speed of 4.00 ? 106 m/s. Find (a) the velocity
  of the third particle and (b) the total kinetic
  energy increase in the process.

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The Center of Mass

• There is a special point in a system or object,
  called the center of mass, that moves as if all
  of the mass of the system is concentrated at that
  point
• The system will move as if an external force were
  applied to a single particle of mass M located at
  the center of mass.
• M is the total mass of the system

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II. Newtons second law for a system of particles
Motion of the center of mass
Center of the mass of the system moves as a
particle whose mass is equal to the total mass of
the system.
Fnet is the net of all external forces that act
on the system. Internal forces (from one part of
the system to another are not included). The
system is closed no mass enters or leaves the
system during the movement. (Mtotal system
mass). acom is the acceleration of the
systems center of mass.
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I. Center of mass
The center of mass of a body or a system of
bodies is the point that moves as though all the
mass were concentrated there and all external
forces were applied there.
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System of particles
Two particles of masses m1 and m2 separated by a
distance d
Origin of reference system coincides with m1
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System of particles
Choice of the reference origin is arbitrary ?
Shift of the coordinate system but center of
mass is still at the same distance from each
particle.
The center of mass lies somewhere between the two
particles.
General
M total mass of the system
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System of particles
We can extend this equation to a general
situation for n particles that strung along
x-axis. The total mass of the system
Mm1m2m3mn The location of center of the
mass
3D
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System of particles
3D The vector form Position of the particle
Position COM
M total mass of the object
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Center of Mass, Extended Object

• Similar analysis can be done for an extended
  object
• Consider the extended object as a system
  containing a large number of particles
• Since particle separation is very small, it can
  be considered to have a constant mass distribution

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Center of Mass, position

• The center of mass in three dimensions can be
  located by its position vector,
• For a system of particles,
• is the position of the ith particle, defined by
• For an extended object,

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Solid bodies
Continuous distribution of matter. Particles dm
(differential mass elements).
3D
M mass of the object
Assumption
Uniform objects ? uniform density
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Motion of a System of Particles

• Assume the total mass, M, of the system remains
  constant
• We can describe the motion of the system in terms
  of the velocity and acceleration of the center of
  mass of the system
• We can also describe the momentum of the system
  and Newtons Second Law for the system

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Velocity and Momentum of a System of Particles

• The velocity of the center of mass of a system of
  particles is
• The total momentum of the system can be expressed
  as
• The total linear momentum of the system equals
  the total mass multiplied by the velocity of the
  center of mass

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Acceleration of the Center of Mass

• The acceleration of the center of mass can be
  found by differentiating the velocity with
  respect to time

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Forces In a System of Particles

• The acceleration can be related to a force
• If we sum over all the internal forces, they
  cancel in pairs and the net force on the system
  is caused only by the external forces

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Newtons Second Law for a System of Particles

• Since the only forces are external, the net
  external force equals the total mass of the
  system multiplied by the acceleration of the
  center of mass
• The center of mass of a system of particles of
  combined mass M moves like an equivalent particle
  of mass M would move under the influence of the
  net external force on the system

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Impulse and Momentum of a System of Particles

• The impulse imparted to the system by external
  forces is
• The total linear momentum of a system of
  particles is conserved if no net external force
  is acting on the system

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The center of mass of an object with a point,
line or plane of symmetry lies on that point,
line or plane.
The center of mass of an object does not need to
lie within the object (Examples doughnut,
horseshoe )
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Finding Center of Mass, Irregularly Shaped Object

• Suspend the object from one point
• Then suspend from another point
• The intersection of the resulting lines is the
  center of mass

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Center of Gravity

• Each small mass element of an extended object is
  acted upon by the gravitational force
• The net effect of all these forces is equivalent
  to the effect of a single force acting
  through a point called the center of gravity
• If is constant over the mass distribution,
  the center of gravity coincides with the center
  of mass

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Center of Mass, Rod
Show that the center of mass of a rod of mass M
and length L lies midway between its ends,
assuming the rod has a uniform mass per unit
length.
81
Problem solving tactics

• (1) Use objects symmetry.
• (2) If possible, divide object in several parts.
  Treat each of these parts as a particle located
  at its own center of mass.
• Chose your axes wisely. Use one particle of the
  system as origin of your reference system or let
  the symmetry lines be your axis.

82

• A water molecule consists of an oxygen atom with
  two hydrogen atoms bound to it. The angle between
  the two bonds is 106?. If the bonds are 0.100 nm
  long, where is the center of mass of the molecule?

83

• A uniform piece of sheet steel is shaped as in
  Figure. Compute the x and y coordinates of the
  center of mass of the piece.

84
Rocket Propulsion
IV. Systems with varying mass

• The operation of a rocket depends upon the law
  of conservation of linear momentum as applied to
  a system of particles, where the system is the
  rocket plus its ejected fuel.

The initial mass of the rocket plus all its fuel
is M ?m at time ti and velocity v The initial
momentum of the system is pi (M ?m) v
85
Rocket Propulsion

• At some time t ?t, the rockets mass has been
  reduced to M and an amount of fuel, ?m has been
  ejected.
• The rockets speed has increased by ?v

86
Rocket Propulsion

• Because the gases are given some momentum when
  they are ejected out of the engine, the rocket
  receives a compensating momentum in the opposite
  direction
• Therefore, the rocket is accelerated as a result
  of the push from the exhaust gases
• In free space, the center of mass of the system
  (rocket plus expelled gases) moves uniformly,
  independent of the propulsion process

87
IV. Systems with varying mass
Example most of the mass of a rocket on its
launching is fuel that gets burned during the
travel.
System rocket exhaust products Closed and
isolated ? mass of this system does not change
as the rocket accelerates.
Pconst ? PiPf
After dt
Linear momentum of exhaust products released
during the interval dt
Linear momentum of rocket at the end of dt
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Velocity of rocket relative to frame (velocity
of rocket relative to products) (velocity of
products relative to frame)
RRate at which the rocket losses mass -dM/dt
rate of fuel consumption
First rocket equation
89
Second rocket equation
90

• The basic equation for rocket propulsion is
• The increase in rocket speed is proportional to
  the speed of the escape gases (ve)
• So, the exhaust speed should be very high
• The increase in rocket speed is also proportional
  to the natural log of the ratio Mi / Mf
• So, the ratio should be as high as possible,
  meaning the mass of the rocket should be as small
  as possible and it should carry as much fuel as
  possible

91
Thrust

• The thrust on the rocket is the force exerted on
  it by the ejected exhaust gases
• Thrust
• The thrust increases as the exhaust speed
  increases
• The thrust increases as the rate of change of
  mass increases
• The rate of change of the mass is called the rate
  of fuel consumption or burn rate

92

• The first stage of a Saturn V space vehicle
  consumed fuel and oxidizer at the rate of 1.50 ?
  104 kg/s, with an exhaust speed of 2.60 ? 103
  m/s. (a) Calculate the thrust produced by these
  engines. (b) Find the acceleration of the vehicle
  just as it lifted off the launch pad on the Earth
  if the vehicles initial mass was 3.00 ? 106 kg.
  Note You must include the gravitational force
  to solve part (b).

93

• Model rocket engines are sized by thrust, thrust
  duration, and total impulse, among other
  characteristics. A size C5 model rocket engine
  has an average thrust of 5.26 N, a fuel mass of
  12.7 grams, and an initial mass of 25.5 grams.
  The duration of its burn is 1.90 s. (a) What is
  the average exhaust speed of the engine? (b) If
  this engine is placed in a rocket body of mass
  53.5 grams, what is the final velocity of the
  rocket if it is fired in outer space? Assume the
  fuel burns at a constant rate.

94

• A rocket for use in deep space is to be capable
  of boosting a total load (payload plus rocket
  frame and engine) of 3.00 metric tons to a speed
  of 10 000 m/s. (a) It has an engine and fuel
  designed to produce an exhaust speed of 2 000
  m/s. How much fuel plus oxidizer is required?
  (b) If a different fuel and engine design could
  give an exhaust speed of 5 000 m/s, what amount
  of fuel and oxidizer would be required for the
  same task?

95
A 60.0-kg person running at an initial speed of
4.00 m/s jumps onto a 120-kg cart initially at
rest. The person slides on the carts top surface
and finally comes to rest relative to the cart.
The coefficient of kinetic friction between the
person and the cart is 0.400. Friction between
the cart and ground can be neglected. (a) Find
the final velocity of the person and cart
relative to the ground. (b) Find the friction
force acting on the person while he is sliding
across the top surface of the cart. (c) How long
does the friction force act on the person? (d)
Find the change in momentum of the person and the
change in momentum of the cart. (e) Determine the
displacement of the person relative to the ground
while he is sliding on the cart. (f) Determine
the displacement of the cart relative to the
ground while the person is sliding. (g) Find the
change in kinetic energy of the person. (h) Find
the change in kinetic energy of the cart. (What
kind of collision is this, and what accounts for
the loss of mechanical energy?)
96

• A 2.00-kg block situated on a frictionless
  incline is connected to a spring of negligible
  mass having a spring constant of 100 N/m. The
  pulley is frictionless. The block is released
  from rest with the spring initially unstretched.
  (a) How far does it move down the incline before
  coming to rest? (b) What is its acceleration at
  its lowest point? Is the acceleration constant?

97

• A 2.00-kg block situated on a rough incline is
  connected to a spring of negligible mass having a
  spring constant of 100 N/m. The pulley is
  frictionless. The block is released from rest
  when the spring is unstretched. The block moves
  20.0 cm down the incline before coming to rest.
  Find the coefficient of kinetic friction between
  the block and incline.

98
Example Tennis Racket

• A 50 g ball is struck by a racket. If the ball
  is initially travelling at 5 m/s up and ends up
  travelling at 5 m/s down after 0.1 s of contact
  what is the (average) force exerted by the racket
  on the ball? What is the impulse?

99
Example Bowling balls

• Two particles masses m and 3m are moving towards
  each other at the same speed, but opposite
  velocity, and collide elastically
• After collision, m moves off at right angles
• What are vmf and v3mf? At what angle is 3m
  scattered?

y
x
100
Sulfur dioxide (SO2) consist of two oxygen atoms
(each of mass 16u) and single sulfur atom (of
mass 32u). The center-to-center distance between
the atoms is 0.143nm. Angle is given on the
picture. Find the x- and y-coordinate of center
of the mass of this molecule.