This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This note was uploaded on 12/04/2008 for the course PHY PHY301 taught by Professor Demovk during the Spring '08 term at University of Texas at Dallas, Richardson.
 Spring '08
 DEMOVK
 Mass

Click to edit the document details