PC Chapter 01

PC Chapter 01 - 1 Relativity I 1-1 dP . Consider the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 1 Relativity I 1-1 = d F dt P . Consider the special case of constant mass. Then, this equation reduces to = AA m Fa in the stationary reference system, and =+ B A BA vvv where the subscript A indicates that the measurement is made in the laboratory frame, B the moving frame, and BA v is the velocity of B with respect to A. It is given that = BA 1 d dt v a . Therefore from differentiating the velocity equation, we have B A1 aaa . Assuming mass is invariant, and the forces are invariant as well, the Newton’s law in frame B should be = =- A B1 m mm F a aa , which is not simply B m a . So Newton’s second law = B m is invalid in frame B. However, we can rewrite it as += 1B F , which compares to B Fga . It is as if there were a universal gravitational field g acting on everything. This is the basic idea of the equivalence principle (General Relativity) where an accelerated reference frame is equivalent to a reference frame with a universal gravitation field. 1-2 IN THE REST FRAME: In an elastic collision, energy and momentum are conserved. =+= i 1 1i 2 2i f 1 1f 2 2f p mv mv p or - = -- 1 1i 1f 2 2i 2f () m vv m . The energy equation is = =+=+ 2222 i f 1 1i 2 2i 1 1f 2 2f 11 22 E E or ( 29 ( ( ( -+ = - 1 1i 1f 1i 1f 2 2i 2f 2i 2f m vvvv m . Substituting the momentum equation into the energy equation yields a very simple and general result (true even for three- dimensional collision if the velocities are replaced as vectors) ( ( +=+ 1i 1f 2i 2f or as Newton put it originally, the final relative velocity is opposite to the initial relative velocity: ( 29 ( -= 1 2 12 i i ff v v
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 CHAPTER 1 RELATIVITY I for elastic collision (and a fraction of the initial for general collisions). Now, putting in the numerical values, the momentum equation, and this relative velocity equation gives: + = + =⋅ 1 1f 2 2f 1f 2f 0.3 0.2 0.9 kg m s m v m v vv , and (29 ( 29 - - = -- 1f 2f 53 . Solving the two equations, two unknowns, we find =- 1f 1 .4 ms v and =+ 2f 6 .6 ms v . IN THE MOVING FRAME: The Galilean velocity transformations hold. ( ( ( ( ( ( ( ′′ - =-= = - = - = + = - = ×⋅ = + = -= = 1i 2i 3 i 1 1i 2 2i f f ff 3 f 20 ms 10 ms 10 ms 0 ms 10 ms 10 ms 2 000 kg 10 m s 1 500 kg 10 m s 5 10 kg m s 2 000 kg 1 500 kg 3 500 kg 10 m s , and because 11.4 m s, 5 10 kgms vvv p mv mv p v p 1-3 IN THE REST FRAME: In an elastic collision energy and momentum are conserved. ( ( ( ( = + = + -=⋅ i 1 1i 2 2i f 1 1f 2 2f 0.3 kg 5 m s 0.2 kg 3 m s 0.9 kg m s p This equation has two unknowns, therefore, apply the conservation of kinetic energy = =+=+ 2222 i f 1 1i 2 2i 1 1f 2 2f 11 22 E E and conservation of momentum one finds that 1f 1.31 m s v and = 2f 6.47 m s v or 1f 1.56 m s v and = 2f 6.38 m s v . The difference in values is due to the rounding off errors in the numerical calculations of the mathematical quantities. If these two values are averaged the values are 1f 1 v and = 2f 6 v , f 0.9 kg m s p . Thus, = if pp . IN THE MOVING FRAME: Make use of the Galilean velocity transformation equations. i 1 1i 2 2i ; where ( = - 1i 1i 5 ms 2 ms 7 ms v . Similarly, ′ =- 2i 1 ms v and i 1.9 kg m s p . To find f p use 1f 1i v and 2f 2i v because the prime system is now moving to the left. Using these results give f 1 .9 kgms p .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/05/2009 for the course PHSX 313 taught by Professor Staff during the Fall '09 term at Kansas.

Page1 / 16

PC Chapter 01 - 1 Relativity I 1-1 dP . Consider the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online