PHY183-Lecture17pre

# PHY183-Lecture17pre - 1 Physics for...

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Unformatted text preview: 1 February 14, 2012 Physics for Scientists&Engineers 1 1 Physics for Scientists & Engineers 1 Physics for Scientists & Engineers 1 Spring Semester 2012 Lecture 17 Momentum and Collisions, Impulse, Conservation of Momentum February 14, 2012 Physics for Scientists&Engineers 1 2 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 = m v p = mv p = mv February 14, 2012 Physics for Scientists&Engineers 1 3 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 x = dp x dt ; F y = dp y dt ; F z = dp z dt F = d p dt F = d p dt February 14, 2012 Physics for Scientists&Engineers 1 4 Momentum and Kinetic Energy Momentum and Kinetic 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 K = p 2 2 m February 14, 2012 Physics for Scientists&Engineers 1 5 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: Δ p ≡ p f − p i Δ 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 ≡ Δ p x F dt t i t f = d p dt dt t i t f = d p p i p f = p f − p i ≡ Δ p J ≡ F dt t i t f J ≡ F dt t i t f J = Δ p J = Δ p February 14, 2012 Physics for Scientists&Engineers 1 6 Impulse and Average Force Impulse and 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 the definition of the average force), but is very useful for practical purposes F ave = F i f dt dt i f = 1 t f − t i F i f dt = 1 Δ t F i f dt J = F ave Δ t = m Δ v J = F ave Δ t = m Δ v 2 Example: A positive force and then a negative force acting on a mass Example: A positive force and then a negative force acting on a mass A 0.442 kg object is at rest at the origin of a coordinate system. A 3.38 N force in the positive x direction acts on the object for 1.57 s. What is the velocity at the end of this interval?...
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PHY183-Lecture17pre - 1 Physics for...

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