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Unformatted text preview: Momentum Forces Kinetic energy Rest energy Conservation of momentum I • “Conservation may be a sign of personal virtue, but it is not a sufficient basis for a sound, comprehensive energy policy.” Former Vice President Dick Cheney, April 30, 2001. • In science, conservation of energy is one of our most fundamental laws, and we can even sort of believe it. • But what about the conservation of momentum mv ? What’s so special about that? Why is it conserved? Momentum Forces Kinetic energy Rest energy Conservation of momentum II • Well, think of a snapshot of several balls on a pool table. We can calculate the center of mass position from vector r cm = summationdisplay vector r i m i summationdisplay m i (1) • If no external force acts on the system, then there’s no acceleration on vector r cm and its velocity remains unchanged. • If we now shift into the frame moving at the constant velocity of the position vector r cm of the center of mass position, we have made vector r cm stationary. No external force means no change in velocity; so velocity is zero! d vector r cm / dt = 0! • Taking d / dt of Eq. 1 then gives ∑ m i v i = 0, or conservation of momentum . Momentum Forces Kinetic energy Rest energy Relativistic momentum I • Conservation of momentum involves constant center of mass with no external forces, and bookkeeping on how this changes between different inertial frames. • Special relativity tells us to be more careful in shifting between different inertial frames. • Since momentum involves mass and velocity, let’s remember the velocity transforms we’ve derived before: v 2 , x = v 1 , x − v 1 − vv 1 , x / c 2 (2) v 2 , y = v 1 , y γ ( 1 − vv 1 , x / c 2 ) (3) v 2 , z = v 1 , z γ ( 1 − vv 1 , x / c 2 ) (4) Momentum Forces Kinetic energy Rest energy Relativistic momentum: vector p ⊥ vector v • Let’s pick our frame S 1 to have v 1 , x = 0 and p 1 , x = 0, and ask what observers in frame S 2 (at velocity v relative to S 1 ) see. • In the ˆ y direction (orthogonal to, or ⊥ ˆ x ), we use velocity transformation from Eq. 3 of v 2 , y = v 1 , y /γ [ 1 − v v 1 , x / c 2 ] to give p 2 , y = m v 2 , y = m v 1 , y γ ( 1 − vv 1 , x / c 2 ) = m v 1 , y γ (...
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This note was uploaded on 05/28/2011 for the course PHY 251 taught by Professor Rijssenbeek during the Fall '01 term at SUNY Stony Brook.
 Fall '01
 Rijssenbeek
 Physics, Energy, Force, Kinetic Energy, Momentum

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