Initials: Problem 3 - 25 points A fireworks rocket of mass M is launched with initial speed do at an angle 0 from the horizontal direction. It explodes into two unequal fragments at the highest point of its trajectory, the heavier one having three times the mass of the lighter one. The heavier fragment has speed 12 and v1? $1? direction $2 right after the explosion, as shown in the figure. You may assume that x the gravitational force has a negligible effect on the rocket only during the explosion. U2 Both the speed v1 and the angle or are unknown, but the information gathered about the motion of the bigger fragment implies that $1 Wi = 4K 40 02 - 01 WAV = At t2 - t1 U + K =E de AE = WNC and AU = -Wc w = lim At-0 At at Ug = mgh (height axis pointing upward) AW W2 - W 1 a AV Usp = = kx2 (x measured from relaxed position) At t2 - t1 Aw dw a = lim - At-0 4t dt Universal Gravitation w (t ) = wot alt - tol FG = GMm 0 (t) = 00 + wolt - to] +5 alt- to]? GM g = - 7 = wg + 2a[0 -Oo] GMm s = ro (arc length) UG = = rw atan = ra GM 2 2 Vorb = arad = - =rw2 Rotational Dynamics Mys 2173/2 T =- VGM I = Et=1mir? (point masses) Ip = ICM + Md2 (parallel axis theorem) 2GM It| = rF sino notVesc = [Text = 1a Linear Momentum p = mi P = > Pi - Quadratic Equation [ Fest = at ax2 + bx + c = 0 has the solutions X = 2a -(-bvb2 - 4ac)Derivatives d(x12) Initials: = nx -1 dx Integrals dx = X2 - X1 Lengths, areas and volumes Circumference of circle: 2nR Area of disk: TR Surface area of sphere: 4TR- Lateral area of cylinder: 2ntRh Volume of cylinder: ntR h Volume of sphere: 4TR3/3 Trigonometry sin (90) = cos (09) = - cos (1809) = 1 sin (0) = cos (90) = sin (180) = 0 cos (60) = sin (30) = 1/2 sin (60) = cos (309) = V3/2 sin (450) = cos (450) = 12/2 cos (180- 0) = - cose sin (180- 0) = sine cos (90 + 0) = - cos (90 - 0) = - sine sin (90 + 0) = sin (90 - 0) = cose cos' 0 + sin- 0 = 1Initials: Table of rotational inertias M is the mass of the system Axis Axis Axis Hoop about central axis Annular cylinder Solid cylinder (or ring) about (or disk) about central axis central axis 1=MR2 (a) 1 = *M(R, 2 + R2?) ( b ) 1 = 4MR (C) Axis Axis Axis Solid cylinder Thin rod about Solid spherea (or disk) about axis through center bout any central diameter perpendicular to diameter length 2P 1= +MR + + ML2 (P) 1= LML? (e) 1 = 3MR (D) Axis Axis Axis Thin Hoop about any Slab about spherical shella diameter perpendicular bout any axis through 2R diameter center 1 = 1MR2 (8) 1 =MR (h ) 1 = 12 M(al + b? ) (1)