909_Ch4a2

909_Ch4a2 - CHAPTER 4 ANSWERS JOS E& SALETAN c& 1...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

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

Unformatted text preview: CHAPTER 4 ANSWERS, JOS ¶ E & SALETAN c & 1 Problem 1 solution. For the repulsive potential change the f in x 4.1.1 to ¡ f: Since f appears quadrat- ically everywhere except in the equation for b ( # ), even in the equation for & ( ¡ ) ; the ¡nal result will come out the same. Problem 2 solution. Write & ( # ) = f … 1 ¡ #=… (2 ¡ #=… ) 2 ( #=… ) 2 sin # : Find b ( # ): b 2 = 2 Z … # & ( # )sin #d# = 2 f … Z … # 1 ¡ #=… (2 ¡ #=… ) 2 ( #=… ) 2 sin # d# & = #=… = 2 f Z 1 #=… 1 ¡ ¡ (2 ¡ ¡ ) 2 ¡ 2 d¡: By partial fractions Z 1 ¡ ¡ (2 ¡ ¡ ) 2 ¡ 2 d¡ = 1 4 &Z ¡ d¡ (2 ¡ ¡ ) 2 + Z d¡ ¡ 2 ¶ = 1 2 ¡ ( ¡ ¡ 2) ) b 2 = f ( # ¡ … ) 2 … 2 ¡ ( # ¡ … ) 2 ) # = … & 1 § b p b 2 + f ¶ : # < … ) must choose negative sign. Then T = Z 1 y 1 ¡ b p b 2 + f p b 2 ¡ y db ‡ = b 2 = 1 2 Z 1 y 2 ‡ ¡ 1 = 2 ¡ ( ‡ + f ) ¡ 1 = 2 p ‡ ¡ y 2 d‡: This integral can be performed by using R ( x 2 ¡ c 2 ) ¡ 1 = 2 dx = ln( p x 2 ¡ c 2 + x ): Z ‡ ¡ 1 = 2 ¡ ( ‡ + f ) ¡ 1 = 2 p ‡ ¡ y 2 d‡ = ln ( p ‡ ( ‡ ¡ y 2 ) + ‡ ¡ 1 2 y 2 p ( ‡ + f )( ‡ ¡ y 2 ) + ‡ ¡ 1 2 y 2 + 1 2 f ) : Now put in the limits. As ‡ ! 1 this approaches ln(2 ‡= 2 ‡ ) = 0 : When ‡ = y 2 this becomes ¡ ln(1 + f=y 2 ) ) T ( y ) = ln p 1 + f=y 2 : CHAPTER 4 ANSWERS, JOS ¶ E & SALETAN c & 2 This and Eq.(4.18) ) r = y p 1 + f=y 2 ) y 2 = r 2 ¡ f; and then Eq.(4.19) ) V = E f r 2 : Problem 3 solution. Write ˜ q j = ! 2 q j ; j = 1 ; 2 ; where ! 2 = k=m . q j ( t ) = a j e !t + b j e ¡ !t : (a) R ( t ) · q 2 ( t ) q 1 ( t ) = a 2 e !t + b 2 e ¡ !t a 1 e !t + b 1 e ¡ !t : lim t !1 R ( t ) = a 2 =a 1 ; and lim t !¡1 R ( t ) = b 2 =b 1 ; QED. (b) Choose the axes and measure time so that the point of closest approach lies on the q 1 axis at t = 0 : Then q 1 (0) = B; q 2 (0) = _ q 1 (0) = 0 ; _ q 2 (0) · u ) q 1 ( t ) = 1 2 B ¡ e !t + e ¡ !t ¢ ; q 2 ( t ) = u 2 ! ¡ e !t ¡ e ¡ !t ¢ : To Fnd the scattering angle £ ; Fnd R · q 2 =q 1 at t = §1 : R ( 1 ) = tan & out = u B! = ¡ R ( ¡1 ) = ¡ tan & in ) & in = & out ) £ = … ¡ 2arctan ‡ u B! · : The total energy E = 1 2 mv 2 ¡ 1 2 k j q j 2 = 1 2 mu 2 ¡ 1 2 kB 2 = 1 2 m ( u 2 ¡ ! 2 B 2 ) ) £ = … ¡ 2arctan r 2 E kB 2 + 1 : Some values: E = 0 ) £ = …= 2; E = E min ( B ) = ¡ 1 2 kB 2 ) £ = … ; E ! 1 ) £ ! ; etc. (c) 3D is the same as 2D since the force is central, so the motion is planar. Problem 4 solution. ¡igure A is essentially ¡ig.4.7. Trajectories that leave above the two disks leave at & > …= 2: the solid trajectories. Trajectories leave below the two disks (with & < …= 2) CHAPTER 4 ANSWERS, JOS ¶ E & SALETAN c & 3 are the dashed ones. S labels the curve of trapped trajectories (the stable manifold), those that do not leave the scattering region. S is clearly the separatrix between the solid and dashed ones....
View Full Document

This note was uploaded on 04/22/2010 for the course PH 531 taught by Professor Clavelli,l during the Fall '08 term at Alabama.

Page1 / 18

909_Ch4a2 - CHAPTER 4 ANSWERS JOS E& SALETAN c& 1...

This preview shows document pages 1 - 4. Sign up to view the full document.

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