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Problem 4 - 25 points Initials: A ball of mass m is traveling to the left with speed vo before it collides elastically head- on with a ball of mass 3m that is initially at y 4 rest. The heavier ball is hanging down from an ideal string of length L that is attached to the ceiling. The acceleration due 3m m to gravity is g and you may assume that the VO pendulum satisfies the small angle approximation. a- Determine the velocity vector of each one of the two balls right after the collision. Use the x-y coordinate system defined in the figure. b- Determine the initial conditions, i.e. the angular position 0 (t=0) and angular velocity 2 (t=0) of the pendulum right at the start of the oscillatory motion. Take the zero of the that: linear speed=angular speed x length. angular position when the string is vertical and take counterclockwise as positive. RememberInitials: c- Without using conservation of energy, determine the amplitude and phase shift of the function 0 (t) describing the angular position of the pendulum as a function of time. Hint: initial conditions might be helpful. consolub gilt, and a tank lace proteon -shaped tuber Bypres oras ei bluff ord to go vipolav of wiw minigys svoda bobavong nonghoesb or no bees d- Using conservation of energy, recalculate the amplitude of the oscillations. Explain any possible difference with your result from part (c).Initials: ing to the 139 ~4 Initials: Simple Marmonk Physics 8A Final Equation Sheet 7 (t ) = x(t) i + y(t)j AT = 12-11 =4xi+ 4yj Newton's 1s' Law AT B = constant = > F = 0 VAV = At - = VAU-x1 + VAD-y) B = lim 47 = vxi + vyj Newton's 2nd Law At - 0 4t [F = ma aAv = = QAV-xi + aAV-yl a = lim 40 = Newton's 3rd Law At to at = axi + ay ) FMe/You = -Fyou/Me B (t ) = vo + alt - tol Tension: r (t) = ro + volt - to] + salt - to]2 T Ax = X2 -X1 Ax X2 - X1 VAV-x At t2 - t1 Normal Force Ax dx Vx = lim - N At-0 At dt Avx Vx2 - Vx1 aAV-x - At Gravity t2 - t1 AVX dux Fo = mg ax = lim - At-0 At dt Ux (t) = Vox + axlt - tol Static Friction fs S HSN x (t) = xo + Vox[t - tol+ 5 ax[t - to]2 v2 = Vox + 2ax[x - xol Kinetic Friction Ay = y2 - y1 fx = HKN Ay _ yz - y1 VAV-y At tz - t1 Spring Force Ay dy Fap = - kx (x measured from relaxed position) Vy = lim At-0 4t dt AVy Vy2 - Vy1 Uniform Circular Motion aAv-y At t2- 1 a = Avy dvy r ay = lim At-0 At dt v = rw Vy (t) = Voy + dy[t - tol 2 TT 1 T = W y(t) = yo + Voy[t - to] + 5ay[t - to]2 vy = Voy vay + 2ayly - yolInitials: Fext = 0 = AP = 0 Work and Energy 2mz B mi - miz unit - Uzi W = F . di mi + mz my + mz 2m1 m2 - m1 W = FL cose (constant angle and magnitude) V26 mi + mz m + mz W. = -mgAh (height axis pointing upward) WSp = - =(X3 - XA) (x measured from relaxed position) K = =mv Rotational Kinematics Wnet = > Wi = 4K U + K =E e (t) = angular position AE = WNC 40 = 02 - 01 Ug = mgh (height axis pointing upward) 40 02 - 01 WAD Usp = = kx2 (x measured from relaxed position) At t2- t1 40 do w = lim - Universal Gravitation At-0 At dt AW W 2 - W 1 GMm aAV =- At t2 FG = 12 Aw dw GM a = lim g = At-0 At dt GMm w (t ) = wo+ alt- to] UG = r 0 (t) = 0 0 + wolt - to] += alt- to]2 GM W2 = wo+ 2a[0 - Oo] Vorb = s = ro (arc length) v = rw 2TY 3/2 atan = ra T = 12 2 VGM arad = -= rw2 Rotational Dynamics 2GM Vesc = 1 = Et=1 miri (point masses) Ip = ICM + Md2(parallel axis theorem) Linear Momentum It| = rF sin 0 p = mv Text = la P = P L = rp sino; L = Iw (@=angular velocity) Fext = dp W = | Tdo JO1 Krot = =102Initials: nitials: Cu. Cp and Y = Cp/Cy for an Ideal Gas Simple Harmonic Oscillator dzx 2+ w 2 x = 0 1. For a monoatomic gas, d=3 dt2 x(t) = A cos(wt + ) Cy = 38, Cp = 38 and Y = = 1.67 w = 2nf = ~ (@=angular frequency) 2. For a diatomic gas, d=5 Mass-spring: w = Cy = SR, Cp = 12 and Y = = = 1.4 3. For a triatomic or polyatomic gas, d=6 Simple pendulum: w = Cy = 3R, Cp = 4R and y = = = 1.33 Physical pendulum: w = mgLem Quadratic Equation Torsional pendulum: w = ax2 + bx + c = 0 has the solutions 20 (-b + vb2 - 4ac Fluids Derivatives P = V; P = perp Pgauge = Pabs - Patm d(xn) - = nxn-1 P = Po + pgh dx FB = PgVsub d(cos (ax + b)) -a sin(ax + b) Qm = PVA = pQv = const. dx 1' P + pgy + 5 puz = const. Lengths, areas and volumes Circumference of circle: 2nR Thermodynamics Area of disk: TR Surface area of sphere: 4TR PV = nRT = NKBT Lateral area of cylinder: 2TRh Eint = nCUT Volume of cylinder: TRZh AEint = Q - W R = NAKB; Co = AR Volume of sphere: 4TR3/3 2Cp Cp = CD + R ; Y = Cy Trigonometry e =- Wnet = 1 - Lout (efficiency) sin (90) = cos (09) = - cos (180) = 1 Qin Qin sin (09) = cos (909) = sin (180) = 0 TL eideal = 1 - TH cos (60) = sin (30) = 1/2 sin (60) = cos (309) = V3/2 sin (450) = cos (450) = 12/2 W Isobaric CPRAT PAV cos (180- 0) = - cose CVRAP P=const. Isochoric V= const. sin (180- 0) = sine Isothermal nRT In ( vo) nRT In T=const. cos (90 + 0) = - cos (90 - 0) = - sine Adiabatic 0 - # ( P, VJ - POVO) PV=const. sin (90 + 0) = sin (90 - 0) = cose cos2 0 + sin- 0 = 1Initials: Table of rotational inertias Axis Axis Axis Solid cylinder Hoop about Annular cylinder (or disk) about central axis (or ring) about central axis central axis 1=MR2 ( @ ) 1 = +M(R, 2 + R2? ) ( b ) 1 = 1MR (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 2R 1=!MR + $ ML2 (d ) 1= AML? (e) 1=3MR (D) Axis Axis Axis Thin spherical shella Hoop about any bout any R diameter Slab about 2 F perpendicular diameter axis through center 1 = AMR (8) 1= ;MR? (h ) 1 = 12 M(2 + 6? ) (1 )