105Sol7 - Physics 105 Solutions to Problem Set 7 By Stephen...

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Physics 105 Solutions to Problem Set 7 By Stephen Glassman, Fall 2007 14.7 Let’s begin by calculating the solid angle Ω subtended by a cone with half- angle θ 0 . Ω = d Ω = 2 π 0 θ 0 0 sin θdθdφ = 2 π (1 - cos θ 0 ) A sphere (like the moon or sun) of radius R at a distance d occupies a conical solid angle with tan θ 0 = R/d . For the sun, R s 6 . 96 × 10 8 m and d s 1 . 50 × 10 11 m giving θ s 4 . 64 × 10 - 3 . 266 and Ω s 6 . 76 × 10 - 5 sr. For the moon, R m 1 . 74 × 10 6 m and d m 3 . 84 × 10 8 m giving θ m 4 . 53 × 10 - 3 . 260 and Ω m 6 . 45 × 10 - 5 sr. These solid angles are very close! It precisely because the sun and moon are so close in angular size that we get both solar and lunar eclipses (along with interesting effects). 14.13 Conservation of energy implies that v i = v f . Conservation of angular mo- mentum implies that mv i R sin φ i = mv f R sin φ f . Combining these, we conclude that sin φ i = sin φ f . The only possibilities are “non-scattering” with φ f = φ i and “law of reflection” scattering with φ f = π - φ i . 14.14 The analogs of equations (14.21) and (14.22) are: = db d Ω = Alternatively, we could insert factors of 2 into both equations and restrict θ to [0 , π ]. Note that a further factor of 2 will also appear later when we substitute d Ω = 2 in our integral. Dividing these equations in the manner of § 14.5, we conclude that 1
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d Ω = db (b) The hard disc will satisfy a law of reflection for the same reasons as the hard sphere. (Because the disc is “hard”, energy will be conserved. If we place the disc at the origin, then there will be no torque (relative to the origin) due to any contact force. Therefore, angular momentum will be conserved. Then by the previous problem, we conclude that the hard disc will satisfy a law of reflection.) In fact, the entire analysis of example 14.5 leading to equation (14.24) on page 573 still holds. Equation (14.25) changes however. d Ω = db = d ( R cos ( θ/ 2)) = R sin ( θ/ 2) 2 (c) Our total cross section is: σ = π - π d Ω d Ω = π - π R sin ( θ/ 2) 2 d Ω = 2 π 0 R sin ( θ/ 2) 2 = [ - 2 R cos ( θ/ 2)] π 0 = 2 R as required. 14.19 It is helpful to recall the discussion of the effective potential from § 8.4 of the text. Because the potential is assumed everywhere non-negative and radial kinetic energy is always non-negative, our projectile has positive radial mechanical energy E . Because the potential is assumed continuous and approaches infinity at the ori- gin, there will be atleast one point r such that U eff ( r ) = E . Let r max be the largest r satisfying U eff ( r ) = E . This will be the unique turn-around point for the projectile. 2
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Let’s first show that r max is a turn-around point and then that there are no other turn-around points. Now in order for ˙ r to change signs, it must be zero for a moment. Therefore the radial kinetic energy must be zero for a moment. Therefore, by energy conservation for effective potentials, U eff = E for a moment. See equation (8.34). So the projec- tile can only turn around when U eff = E . Therefore r max is certainly a promising candidate for a turn-around point.
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