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Unformatted text preview: Chapter 9: Center of Mass and Linear Momentum
91 What Is Physics? 92 The Center of Mass 93 Newton's Second Law for a System of Particles 94 Linear Momentum 95 The Linear Momentum of a System of Particles 96 Collision and Impulse 97 Conservation of Linear Momentum 98 Momentum and Kinetic Energy in Collisions 99 Inelastic Collisions in One Dimension 910 Elastic Collisions in One Dimension 911 Collisions in Two Dimensions 912 Systems with Varying Mass: A Rocket Chapter 9 Homework 1, 3, 10, 14, 18, 23, 36, 44, 49, 53, 60, 63 Collisions in One Dimension
Total linear momentum is conserved
m1v1i + m2 v2i = m1v1 f + m2 v2 f Elastic Collision Total kinetic energy is conserved
1 2 2 2 2 2 m1v1i + 1 m2 v2i = 1 m1v1 f + 1 m2 v2 f 2 2 2 or v1i  v2i = (v1 f  v2 f ) Inelastic Collision Total kinetic energy is not conserved Completely Inelastic Collision Two bodies move together after collision
vf1 = vf2 = vf = vcm
elastic inelastic completely inelastic (m1 + m2 )vf = m1v01 + m2 v02 Elastic Collision
m1v1i + m2 v2i = m1v1 f + m2 v2 f v1i  v2i = (v1 f  v2 f ) v1 f = v2 f m1  m2 2 m2 v1i + v 2i m1 + m2 m1 + m2 v1 f Equal masses m1 = m2 v1 f = v2i and v2 f = v1i Stationary Target v2i = 0 m1  m2 2m1 = v1i and v2 f = v1i m1 + m2 m1 + m2 2m1 m  m2 = v1i + 1 v 2i m1 + m2 m1 + m2 Sample Problem 911. Elastic Collision m1 = 30 g, m2 = 75 g, h1 = 8.0 cm. Find: velocity of m1 after collision v1f Inelastic Collision
m1v1i + m2 v2i = m1v1 f + m2 v2 f
Problem 50. Inelastic Collision (bullet strikes through wooden block) m = 5.20 g, M = 700 g Before: vi = 672 m/s, Vi = 0. After: vf = 428 m/s. Find: Vf and vcom Sample Problem 99. Ballistic Pendulum M = 5.4 kg initially at rest, m = 9.5 g, h = 6.3 cm. Find: initial speed of the bullet v m1v1i + m2 v2i = (m1 + m2 )v f ...
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This note was uploaded on 04/17/2008 for the course PHY 2048 taught by Professor Chen during the Fall '08 term at UNF.
 Fall '08
 Chen
 Center Of Mass, Mass, Momentum

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