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Unformatted text preview: obennoskey (blo242) – hw9 – Demkov – (59910) 1 This printout should have 50 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A balloon of mass M is floating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon. The person begins to climb the ladder at a uniform speed v relative to the ground. How does the balloon move relative to the ground? 1. Down with a speed less than v correct 2. Up with speed v 3. Down with speed v 4. The balloon does not move. 5. Up with a speed less than v Explanation: Let the mass of the person be m . Total momentum is conserved (because the exterior forces on the system are balanced), especially the component in the vertical di rection. When the person begins to move, we have mv + M v M = 0 , = ⇒ v M = − m M v < v = ⇒  v M  = m M v < v, since m < M = ⇒ m M < 1 . Thus the balloon moves in the opposite direction. 002 10.0 points A uranium nucleus 238 U may stay in one piece for billions of years, but sooner or later it de cays into an α particle of mass 6 . 64 × 10 − 27 kg and 234 Th nucleus of mass 3 . 88 × 10 − 25 kg, and the decay process itself is extremely fast (it takes about 10 − 20 s). Suppose the uranium nucleus was at rest just before the decay. If the α particle is emitted at a speed of 2 . 17 × 10 7 m / s, what would be the recoil speed of the thorium nucleus? Correct answer: 3 . 71361 × 10 5 m / s. Explanation: Let : v α = 2 . 17 × 10 7 m / s , M α = 6 . 64 × 10 − 27 kg , and M Th = 3 . 88 × 10 − 25 kg . Use momentum conservation: Before the de cay, the Uranium nucleus had zero momentum (it was at rest), and hence the net momentum vector of the decay products should total to zero: vector P tot = M α vectorv α + M Th vectorv Th = 0 . This means that the Thorium nucleus recoils in the direction exactly opposite to that of the α particle with speed bardbl vectorv Th bardbl = bardbl vectorv α bardbl M α M Th = (2 . 17 × 10 7 m / s) (6 . 64 × 10 − 27 kg) 3 . 88 × 10 − 25 kg = 3 . 71361 × 10 5 m / s . 003 10.0 points A uniform flat plate of metal with a circular hole is situated in the reference frame shown in the figure below. − 10 − 8 − 6 − 4 − 2 0 2 4 6 8 10 − 10 − 8 − 6 − 4 − 2 2 4 6 8 10 obennoskey (blo242) – hw9 – Demkov – (59910) 2 Calculate the xcoordinate of the center of mass x cm of the metal plate. Correct answer: − 4 . 17668. Explanation: Basic Concept: The center of mass coor dinate is x cm ≡ ∑ x i m i ∑ m i ≡ integraldisplay xdm M , (1) where M ≡ integraldisplay dm, dm = σ y dx, and σ is the areal density parenleftBig mass area parenrightBig of the plate....
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 Spring '08
 DEMOVK
 Kinetic Energy, Mass, Momentum, m/s

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