Chapter 9
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Chapter 9. Linear Momentum and Collisions. Intro. Consider bowling: Bowling ball collides with initial pin Force on/Acceleration of the Pin Force on/Acceleration of the ball Momentum- simplified way to study these moving objects. 9.1 Linear Momentum and Conservation.

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Chapter 9

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Chapter 9

Chapter 9

Linear Momentum and Collisions


Intro

Intro

  • Consider bowling:

    • Bowling ball collides with initial pin

      • Force on/Acceleration of the Pin

      • Force on/Acceleration of the ball

  • Momentum- simplified way to study these moving objects.


9 1 linear momentum and conservation

9.1 Linear Momentum and Conservation

Consider two particles (isolated) m1 and m2 moving at v1 and v2

From Newton’s 3rd Law


Chapter 9 2838603

9.1

  • And if the the masses are constant

  • If the derivative of a function is 0 it is constant (conserved)


Chapter 9 2838603

9.1

  • Linear Momentum- the product of the mass and velocity of a moving particle.

    • Momentum is a vector quantity

    • Has dimensions MLT-1 and SI units kg.m/s

    • All momentum is conserved

    • Three Components of Momentum


Chapter 9 2838603

9.1

  • Directly related to Newton’s 2nd Law

    • Instead of Net Force equals mass times accel.

    • Can be described as Net Force equals time rate of change of momentum


Chapter 9 2838603

9.1

  • Conservation of Momentum

    • For isolated systems the time derivative of the total momentum is 0

    • The total momentum is therefore constant or conserved.

      Law of Conservation of Momentum

    • And in 3 components


Chapter 9 2838603

9.1

  • Quick Quizzes p 254-255

  • Examples 9.1-9.2


9 2 impulse and momentum

9.2 Impulse and Momentum

  • The momentum of an object changes when a net force acts on it

    or

  • Integrating this gives


Chapter 9 2838603

9.2

  • Impulse-Momentum Theorem- the impulse of the force F acting on a particle equals the change in momentum of the particle

  • Force Varies, impulse time is short impulse can generally be calculated with the average force.


Chapter 9 2838603

9.2

  • Quick Quizzes p. 258

  • Examples 9.3-9.4


9 3 collisions in 1 d

9.3 Collisions in 1-D

  • Momentum is conserved

  • Three types of Collisions

    • Inelastic

    • Perfectly Inelastic

    • Elastic


Chapter 9 2838603

9.3

  • Inelastic Collision- Momentum is conserved but kinetic energy is not.

    • Inelastic- objects collide and separate, some K is lost

    • Perfectly Inelastic- objects collide and stick together (moving as one), some K is lost


Chapter 9 2838603

9.3

  • Elastic Collisions-A collision in which no energy is lost to (surroundings / internal / potential)

    • Both momentum and kinetic energy are conserved


Chapter 9 2838603

9.3

  • By combining the Conservation of p and K equations

  • In all collision types careful attention to the direction (and sign) of velocities must be paid.


Chapter 9 2838603

9.3

  • Quick Quizzes p. 262

  • Examples 9.5 - 9.9


9 4 2 d collisions

9.4 2-D Collisions

  • Momentum is conserved on each axis

  • Examples 9.10 – 9.12


9 5 center of mass

9.5 Center of Mass

  • We can describe the overall motion of a mechanical system by tracking its center of mass

    • System could be a group of particles

    • System could be a large extended object

  • A force applied to the center of mass will cause no rotation to the system


Chapter 9 2838603

9.5

  • To find the center of mass in 3-D space for a number (i) particles

  • Or in terms of the position vector of each particle


Chapter 9 2838603

9.5

  • For extended objects that have a continuous mass distribution

  • Consider them an infinite number of closely spaced particles

  • The sum becomes an integral


Chapter 9 2838603

9.5

  • Or in terms of the position vector

  • For symmetrical objects, the center of mass lies on the axis/plane of symmetry

    • Examples: uniform rod,

      sphere,

      cube,

      donut?


Chapter 9 2838603

9.5

  • For extended objects, the force of gravity acts individually on each small piece of mass (dm)

  • The net effect of all these forces is equivalent to the single force Mg, through a point called the center of gravity.

  • If the gravitational field is uniform across all dm, the center of gravity and center of mass are one and the same.


Chapter 9 2838603

9.5

  • Quick quiz p 272

  • Examples 9.13, 9.14, 9.15


9 6 motion of a system

9.6 Motion of a System

  • If the mass of a system remains constant (no particles entering/leaving) then we can track the motion of the center of mass, rather than the individual particles.

  • Also assumes any forces on the system are internal (isolated)


Chapter 9 2838603

9.6

  • Velocity of the center of mass

  • Acceleration of the center of mass


Chapter 9 2838603

9.6

  • If there is a net force on the system, it will move equivalent to the way a single M with the same net force would move.

  • And if the net force is zero


Chapter 9 2838603

9.6

  • Quick Quizzes p. 276

  • Examples 9.17, 9.18


9 7 rocket propulsion

9.7 Rocket Propulsion

  • Most forms of vehicular motion result from action/reaction friction.

  • A rocket has nothing to push against so its motion/control depend on conservation of motion of the system.

  • The system includes the rocket body (and payload) plus the ejected fuel


Chapter 9 2838603

9.7

  • The rocket burns fuel and oxidizer creating expanding gases that are directed through the nozzle.

  • Each gas molecule has a mass (that was once part of the rockets total mass) and velocity, therefore a downward momentum.

  • The rocket receives the same compensating momentum upward.


Chapter 9 2838603

9.7

  • Looking a rocket initially with mass M + Δm, moving with velocity v…


Chapter 9 2838603

9.7

  • And some time, Δt, later...

  • The rocket now has mass, M and velocity v + Δv, compensating the momentum of the exhausted mass, Δm.


Chapter 9 2838603

9.7

  • The conservation of momentum expression for this change…

  • Can be simplified to…


Chapter 9 2838603

9.7

  • A rocket motor produces a continuous flow of exhaust gas a fairly constant speed, through the burn

  • For continually changing values…

    Δv dv

    Δm dm

    So… 


Chapter 9 2838603

9.7

  • Because the increase in exhaust mass = the decrease in rocket mass…

  • Then integrate this expression


Chapter 9 2838603

9.7

  • Discuss integral of M-1

  • Evaluating from vi to vf gives the basic expression for rocket propulsion.


Chapter 9 2838603

9.7

  • Mi is the total mass of the rocket/payload plus fuel

  • Mf is the mass of the rocket/payload

  • Mi – Mf is the mass of fuel needed to achieve a certain speed (eg. Escape speed to power down rocket)


Chapter 9 2838603

9.7

  • Thrust- the actual force on the rocket at any given time is

  • Thrust is proportional to exhaust speed and also the rate of change of mass (burn rate).

  • Examples 9.19 p. 279


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