Download Conservation of Energy - Physics - Solved Past Exam and more Exams Physics in PDF only on Docsity! PC1141 Exam Solution - AY09/10 Page 1 of 7 Question 1 Speed of heavy ball immediately before hitting the ground is (Conservation of Energy): 1 2 Mv2b = Mgh vb = √ 2gh Note 1. You may find the speed using kinematics as well. Speed of heavy ball immediately after bouncing off the ground is va = vb = √ 2gh Speed of light ball immediately before colliding with heavy ball is (Conservation of Energy): 1 2 Mv2l = Mgh vl = √ 2gh Note 2. You may find the speed using kinematics as well. By Conservation of Momentum, and the fact that speed of approach equals speed of separation (va+vl = u− U), Mva −mvl = MU +mu = M(u− va − vl) +mu (M −m) √ 2gh = M(u− 2 √ 2gh) +mu (3M −m) √ 2gh = (M +m)u u = 3M −m M +m √ 2gh Since M m, we have u = 3 √ 2gh By Conservation of Energy, 1 2 mu2 = mgh′ h′ = u2 2g h′ = 9h Note 3. You may find the height using kinematics as well. PC1141 Exam Solution - AY09/10 Page 2 of 7 Question 2 r θ L l I = ∫ r2 dm = ∫ L 2 −L2 [l sin(θ)] 2 ( M L dl ) = M sin2(θ) L ∫ L 2 −L2 l2 dl I = ML2 sin2(θ) 12 Question 3 R′ R r √ r2 +R′2 dR′ Mass of sphere is concentrated where the dot is. By invoking the Shell Theorem, we can assume that the mass of the solid sphere is concentrated at its center, which is at a vertical distance r away from the center of the disk. The potential energy associated with the sphere and a thin ring of radius R′ is given by dU = − GM dM√ r2 +R′2 = − GM√ r2 +R′2 · M πR2 2πR′ dR′ = −2GM 2 R2 · R ′ √ r2 +R′2 dR′ PC1141 Exam Solution - AY09/10 Page 5 of 7 In the accelerated frame of the wedge, with acceleration at ẍ, we have the following forces (and resultant force) on the block (the forces are resolved along the direction of the slope (̂s) and the axis perpendicular to it): mẍ sin(θ) +mg cos(θ) = N (3a) mg sin(θ)−mẍ cos(θ) = s̈ (3b) The horizontal forces on the wedge in the inertial frame are then related by: F −N sin(θ) = Mẍ (4) We equate all the above while eliminating ẍ and N to get s̈ = g sin(θ)− F −mg sin(θ) cos(θ) M +m sin2(θ) cos(θ) where in the process we have ẍ = F −mg sin(θ) cos(θ) M +m sin2(θ) The acceleration of the block back in the inertial frame is then ax = s̈ cos(θ) + ẍ = g sin(θ) cos(θ) + F −mg sin(θ) cos(θ) M +m sin2(θ) sin2(θ) ay = s̈ sin(θ) = ( g sin(θ)− F −mg sin(θ) cos(θ) M +m sin2(θ) cos(θ) ) sin(θ) Question 7 Part A Downward force due to diver: Fdiver = mg − U = 120g − ρwaterV g ≈ 392 N Downward force due to cable: Frope(x) = mg − U = µxg − ρwaterxπ d2 4 g ≈ 7.701xN Tension at x is therefore T = (392 + 7.70x) N PC1141 Exam Solution - AY09/10 Page 6 of 7 Part B v = √ T µ dx dt = √ 392 + 7.70x 1.10∫ t 0 dt = ∫ 100 0 √ 1.10 392 + 7.70x dx t = 3.89 s Question 8 Part A In the direction of motion. Part B The length of the major-axis is: a = dSun-Earth + dSun-Mars 2 = 1.89× 1011 m Kepler’s Third Law gives T = √ 4π2a3 GM = 4.481× 107 s Time taken is half the period, thus giving t = 4.481× 107 2 s = 259 days Part C By Conservation of Angular Momentum, and defining subscripts a and p to be the aphelion and perihelion respectively, mvara = mvprp va = vp rp ra = vp a− εa a+ εa = vp 1− ε 1 + ε By Conservation of Energy, defining Ms to be the mass of the Sun and ε to be the eccentricity, and using PC1141 Exam Solution - AY09/10 Page 7 of 7 the above result, 1 2 mv2a − GMsm ra = 1 2 mv2p − GMsm rp 1 2 v2a − GMs a(1 + ε) = 1 2 v2p − GMs a(1− ε)( vp 1− ε 1 + ε )2 − 2GMs a(1 + ε) = v2p − 2GMs a(1− ε) v2p 4ε (1 + )2 = 4GMs a ε 1− ε2 v2p = GMs a 1 + ε 1− ε The eccentricity can be found easily by: ε = c a = a− dSun-Earth a = 0.26 The speed when it just leaves Earth is therefore: vp = √ GMs a 1 + ε 1− ε ≈ 3.46× 10 4 m s−1 Part D The Earth is moving through space at a speed of: ve = ωr = 2π T r = 2.989× 104 m s−1 The spacecraft has to escape Earth’s gravity, and defining Me to be the mass of Earth, hence we have: 1 2 m(ve + v) 2 − GMem Re = 1 2 mv2p (ve + v) 2 = v2p + 2GMe Re v = √ v2p + 2GMe Re − ve v = 6470 m s−1
