ch37-p059 - 59. (a) Before looking at our solution to part...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
59. (a) Before looking at our solution to part (a) (which uses momentum conservation), it might be advisable to look at our solution (and accompanying remarks) for part (b) (where a very different approach is used). Since momentum is a vector, its conservation involves two equations (along the original direction of alpha particle motion, the x direction, as well as along the final proton direction of motion, the y direction). The problem states that all speeds are much less than the speed of light, which allows us to use the classical formulas for kinetic energy and momentum ( Km v = 1 2 2 and G G pm v = , respectively). Along the x and y axes, momentum conservation gives (for the components of G v oxy ): oxy oxy, oxy, oxy oxy oxy, oxy, oxy 4 17 1 0. 17 xx p yp p y p p m mv m v v v v m m m v v v v m α αα =⇒ = =+ = To complete these determinations, we need values (inferred from the kinetic energies given in the problem) for the initial speed of the alpha particle ( v ) and the final speed of the proton ( v p ). One way to do this is to rewrite the classical kinetic energy expression as
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Page1 / 2

ch37-p059 - 59. (a) Before looking at our solution to part...

This preview shows document pages 1 - 2. Sign up to view the full document.

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