# l4 - Momentum Forces Kinetic energy Rest energy...

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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?

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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)

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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 [ 1 v v 1 , x / c 2 ] to give p 2 , y = m 0 v 2 , y = m 0 v 1 , y γ ( 1 vv 1 , x / c 2 ) = m 0 v 1 , y γ ( 1 v · 0 / c 2 ) = m 0 v 1 , y γ , (5) giving p 1 , y = m 0 v 1 , y = γ p 2 , y or p 1 , y = γ m 0 v 2 , y . (6) This can be interpreted by saying that the inertial mass m 0 of the
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