CHAPTER 1. Chapter 1 Page 1.1. Problem 1.1: (a) Because the system is conservative, ΔE = 0 and ΔK = ΔU. G m 2. M kg.

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1 Chapter Page. CHAPTER Problem.: (a) Because the system is conservative, ΔE = 0 an ΔK = ΔU M kg G m newton R m kg ΔK= v e = MmG = Mm G r R r=r so, v e M G v e.84 km R sec (b) A circular orbit woul imply that r = 0 mθ r = mm but θ = r v r so v r = MG r r 50km km istance from center of earth to shuttle MG v v 7.85 time km r sec (c) Distance from earth to moon from orbit information τ 7.34hr θ r= MG r 3 π θ θ sec τ conition for a circular orbit MG r r θ 5 km Problem.: (a) x ux ( ) 4 x x 6 f( x) x ux ( )

2 Chapter Page. ux ( ) f( x) x (b) Using Newton's law, F = 4 x 3 x = 7 x t IC : t x = 0 when t=0 IC : x =.5 when t=0 Problem.3: E = m v m v ux Definitions m x m x X = M x = x x x = x x Substitute x into X m x x m x X = = M x m x m m M m = x M x or x = X m M x Substitute x into X

3 Chapter Page.3 m x x m x m m X = = x M M x or x = X M x Subsitute these equations into the expression for E E = m X m M x m m X M x Ux which simplifies to E M X = μ x Ux Problem.4: m x = rcosθ L = x y Equations for transformation to polar coorinates y = rsinθ x = rcosθ rθsinθ by ifferentiation of the transformation eqns. y = rsinθ rθcosθ x y = r cosθ sinθ r θ cosθ sinθ = r r θ after some algebra Thus, L m r = m r θ Problem.5: Note: Assume L is much greater than the pulley iameter K M x = M x = U = M g( L x) M gx x M M (see figure on next page) L = M M x M g( L x) M gx Fin erivatives for use in Langrange's equations:

4 Chapter Page.4 x L = M M x x L = M M g Thus, the equation of motion becomes M M x g M M = 0 or x = g M M M M Problem.6: The potential energy is u(r) = -C/r so the Lagrangian becomes: L m r m r θ C = see problem.4 for erivation of kinetic energy term in r polar coorinates r L r L = = mr mr θ C r θ L θ L = mr θ = 0 Fin partial erivatives to use in Lagrange's equations t ( mr ) t mr θ mr θ = 0= C = 0 () r t l Rearranging gives: mr = mr θ C rθ = rθ r Lagrange's eqns of motion () Note: l mr θ is the angular momentum. Since by this eqn, l/t = 0, angular momentum is conserve. Problem.7:

5 Chapter Page.5 V 4 πr 3 π r = h = 3 3 π r3 πr h A = 4π r πrh πr = 3πr πrh Minimize A subject to constraint on V. Thus, f = 3πr 3 πrh g = F = f λg 3 π r3 πr h V r F h F = 6πr πh λπr λπrh = 0 = 3r h λr λrh = πr λπ r = 0 Thus, r = an h = λ λ Now subsititue into constraint eqn. π 8 λ 3 Finally, r = h= 3 3V 5π V = 0 or V = 40π an λ = 3λ 3 3 5π 3V 3 Problem.8 U = Mg( D rcosθ) L M r = M r θ Mg ( D rcosθ) Constraint: r=0; i.e., r=d where D is constant r L r L = = Mr Mr θ Mg cosθ θ L θ L = Mr θ = Mgrsinθ Lagrange's eqns of motion by substituiting constraint yiels:

6 Chapter Page.6 Mr θ Mg rsinθ = 0 now replace r with D to obtain Using the constraint metho: Mr Thus, Mr θ Mg cosθ = λ But r = D an r = 0 g θ = sinθ D λ = MD θ Mg cosθ= F c Problem -9: (a) Ψ Cτ ψ Cτ ψ where we use C = C = C K L ΨKΨ x 0 L 0 ΨΨ x where h K. 8π m x To fin C, use the normalization conition: L 0 ΨΨ Simplify using L 0 ΨΨ x C L 0 τ ψ τ ψ τ ψ τ ψ τ τ n n x C L 0 ψ ψ an L x 0 L 0 ψ ψ ψ ψ x x = x = 0 by orthogonality of the basis functions

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11 Chapter Page. Problem.6: Given 6 re balls, 4 white balls an 5 blue balls, (a) 6 P r 5 P r (b) P w 5 P w (c) P b 5 P b () P 9 5 P 0.6 (e) 0 P P Probability of white or blue Probability of re or white Problem.7: Balls rawn in the orer re, white an then blue P r P w P b (a) P P r P w P b P Iniviual probabilites of rawing a re, white or blue ball Drawn with replacement (b) P P Drawn without replacement Problem.8: 4 re balls, 3 white balls an 5 blue balls There are! ways of arranging the balls, but we must ivie out equivalent arrangements. These inclue 4! ways of arranging the re balls, 3! ways of arranging the white balls an 5! ways of arranging the blue balls. N N Problem.9: (a) Put one of the bulbs in anywhere. This has a probabilty of unity an fixes the starting location. It oes not matter where this bulb is place as ientical orering is obtaine if the positions are all rotate. Thus, P 6 P 70 (b) First consier the case where the bulbs in question are ajacent. We then treat them as a single unit. Then there are 5! ways of arranging the 6 objects. However, for each of these orerings, we can switch the orer of the bulbs taken as the single unit. The number of arrangements in which the bulbs are not together is just the total number of arrangements minus the arrangements when they are together. Thus, N together 5 N total 6 N apart N total N together

12 Chapter Page. N together 40 N total 70 N apart 480 Problem.0: N = 0 C 3 0 N N 0 37 Problem.: Let tot = the total number of combinations of 5 cars taken 5 at a time; i.e., the total number of ifferent hans. tot (a) P P tot or aces - or 4 of any kin for that matter. (b) Full house: P P or tot Note there are 3 ifferent car values from which to create 3-of-a-kin an then only ifferent car values from which to create -of-a-kin (c) Two pair: P P tot or Problem.: 000! will cause an overflow on most calculators. Therefore, use Sterling's approximation. ln(000!) = 000 ln (000) Use base ten so that we can represent it as 0 to some power. log(000!) = [000 ln (000) - 000]/ ln( 000) 000 N N = 0 565

13 Chapter Page.3 Problem.3: (a) Normalization requires: 5 4 c 0 xy x y = x y Thus, c c xy x y 0 or 96 (b) Px ( ) 5 = cx y y P( x) x = 96 x 8 (c) Py ( ) 4 = cx y x P( y) y = y () x y Px ( ) P( y) = = Px ( y) Thus, they are inepenent 8 (e) Ex ( ) 4 0 x x 8 x Ex ( ).6667 E( y) 5 y y y Ey ( ) Ex ( y) = Ex ( ) E( x) = (f) Because P(x) an P(y) are inepenent, P(x y=3) must be inepenent of the value of y Thus, P(x y=3) = P(x) = x 8

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