1. The problem statement, all variables and given/known data Masses A and B, each having a mass of 32.2 slugs, are constrained to move in frictionless slots. They are connected by a rigid, massless rod of length L = 2 ft. Mass B is connected to two massless linear springs, each having a spring constant k = 60 lbF / ft. The springs are unstretched when the connecting rod to masses A and B is vertical. The accompanying schematic: https://drive.google.com/file/d/0B2-REaK_0ruSZ3cwV2dZS1VuNVE/view?usp=sharing If B is given a slight nudge, what are the velocities of B and A when A descends a distance of D = 3 inches? 2. Relevant equations W1->2 = ΔKE + ΔPEG + ΔPEE 3. The attempt at a solution Masses: mA = mB = m Take upward to be positive y and right as positive x. At position 1: Velocity of A: vA1 = 0 Velocity of B: vB1 = 0 Height of A: hA1 = L Height of B: hB1 = 0 Displacement of left spring: xL1 = 0 Displacement of right spring: xR1 = 0 Finding the displacement of each spring: (L - D)2 + x2= L2 x = 11.62 in. = 0.9682 ft -> symmetric for both springs At position 2: Velocity of A: vA2 = ? Velocity of B: vB2 = ? Height of A: hA2 = L - D Height of B: hB2 = 0 Displacement of left spring: xL2 = x Displacement of right spring: xR2 = x Work: W1->2 = 0 because there are no non-conservative forces acting on the system Change in kinetic energy: ΔKE = (1/2) * mA * (vA22 - vA12) + (1/2) * mB * (vB22 - vB12) ΔKE = (1/2) * m * (vA22 + vB22) Change in gravitational potential energy: g = 32.174 ft/s2 ΔPEG = mA * g * (hA2 - hA1) + mB * g * (hB2 - hB1) ΔPEG = m * g * ( - D) = -259.001 slug⋅ft2/s2 = -259.001 ft⋅lbF Change in elastic potential energy: ΔPEE = (1/2) * k * (xL22 - xL12) + (1/2) * k * (xR22 - xR12) ΔPEE = (1/2) * k * ( 2 * x2) ΔPEE = k * x2 = 56.245 ft⋅lbF Relating vA2 and vB2: Take the origin at B's initial position. x ≡ horizontal distance from origin to B y ≡ vertical distance from origin to A x2 + y2 = L2 = constant 2 * x * (dx/dt) + 2 * y * (dy/dt) = 0 2 * x * vB + 2 * y * vA = 0 x * vB + y * vA = 0 At the instant when A has moved down 3 inches: vB2 = [ - (L - D) / x] * vA2 = -1.807 * vA2 Combine all equations: W1->2 = ΔKE + ΔPEG + ΔPEE 0 = (1/2) * m * [vA22 + ( -1.807 * vA2)2] - 259.001 + 56.245 vA2 = 1.718 ft/s down vB2 = 3.105 ft/s right This was a midterm question. Physics enthusiasts, please give feedback my thought process. Much appreciated.