l4 - Momentum Forces Kinetic energy Rest energy...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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?
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 .
Image of page 2
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)
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern