chapter8_PC

# chapter8_PC - Chapter 8 Momentum and Collisions Linear...

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Chapter 8 Momentum and Collisions

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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:
Linear Momentum, cont Linear momentum is a vector quantity Its direction is the same as the direction of The dimensions of momentum are ML/T The SI units of momentum are kg · m / s Momentum can be expressed in component form: p x = m v x p y = m v y p z = m v z

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Newton and Momentum Newton called the product m the quantity of motion of the particle Newton’s Second Law can be used to relate the momentum of a particle to the resultant force acting on it with constant mass
Newton’s Second Law The time rate of change of the linear momentum of a particle is equal to the net force acting on the particle This is the form in which Newton presented the Second Law It is a more general form than the one we used previously This form also allows for mass changes Applications to systems of particles are particularly powerful

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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
Conservation of Momentum, 2 Conservation of momentum can be expressed mathematically in various ways In component form, the total momenta in each direction are independently conserved Conservation of momentum can be applied to systems with any number of particles

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Conservation of Momentum, Archer Example The archer is standing on a frictionless surface (ice) Approaches: Newton’s Second Law – no, no information about F or a Energy approach – no, no information about work or energy Momentum – yes
Archer Example, 2 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 Total momentum before releasing the arrow is 0 The total momentum after releasing the arrow is p 1f + p 2f = 0

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Archer Example, final The archer will move in the opposite direction of the arrow after the release Agrees with Newton’s Third Law Because the archer is much more massive than the arrow, his acceleration and velocity will be much smaller than those of the arrow
Conservation of Momentum, Kaon Example The kaon decays into a positive π and a negative π particle Total momentum before decay is zero Therefore, the total momentum after the decay must equal zero

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Impulse and Momentum From Newton’s Second Law, Solving for d p 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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