# W7.1 - Physics for Scientists &amp; Engineers 1 Spring...

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February 18, 2008 Physics for Scientists&Engineers 1 1 Physics for Scientists & Physics for Scientists & Engineers 1 Engineers 1 Spring Semester 2008 Lecture 23

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February 18, 2008 Physics for Scientists&Engineers 1 2 New Chapter: New Chapter: Momentum and Collisions Momentum and Collisions
February 18, 2008 Physics for Scientists&Engineers 1 3 Linear Momentum Linear Momentum Linear momentum is the product of mass (scalar) and velocity (vector) Momentum vector and velocity vector are parallel to each other Magnitude of the linear momentum ! p = m ! v p = mv

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February 18, 2008 Physics for Scientists&Engineers 1 4 Momentum and Force Momentum and Force Take the time derivative of the definition of the momentum If the mass is constant in time, 2 nd term is zero This Newton’s Second Law for momentum In components: d dt ! p = d dt ( m ! v ) = m d ! v dt + dm dt ! v d dt ! p = m d ! v dt = m ! a = ! F F x = dp x dt ; F y = dp y dt ; F z = dp z dt ! F = d ! p dt
February 18, 2008 Physics for Scientists&Engineers 1 5 Momentum and Kinetic Momentum and Kinetic Energy Energy We already know Use p=mv and then obtain Find relationship between momentum and kinetic energy We can reformulate all of mechanics we have studied thus far in terms of momentum instead of velocity K = 1 2 mv 2 K = mv 2 2 = m 2 v 2 2 m = p 2 2 m K = p 2 2 m

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February 18, 2008 Physics for Scientists&Engineers 1 6 Impulse Impulse Change in momentum ( f =final, i =initial) Obtain an expression for momentum change by going back to the relationship between momentum and force, and integrating both sides over time Impulse: and therefore: ! ! p " ! p f # ! p i F x dt t i t f ! = dp x dt dt t i t f ! = dp x p x , i p x , f ! = p x , f " p x , i # \$ p x ! Fdt t i t f ! = d ! p dt dt t i t f ! = d ! p ! p i ! p f ! = ! p f " ! p i # \$ ! p ! J ! ! Fdt t i t f " ! J = ! ! p
February 18, 2008 Physics for Scientists&Engineers 1 7 Impulse Impulse and and Average Force Average Force Define the average force acting over a time interval Δ t = t f - t i Then the impulse is simply: This result seems rather trivial (the integral is still in definition of average force), but is very useful for practical purposes ! F av = ! F i f ! dt dt i f ! = 1 t f " t i ! F i f ! dt = 1 # t !

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## This note was uploaded on 04/06/2008 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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W7.1 - Physics for Scientists &amp; Engineers 1 Spring...

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