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Unformatted text preview: This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An object having an initial momentum that may be represented by the vector below strikes an object that is initially at rest. Which of the following sets of vectors may represent the momenta of the two objects af ter the collision? Note carefully: The original vector above and the following vectors are all drawn to the same length scale. 1. correct 2. 3. 4. 5. 6. 7. Explanation: There is no external force for the twoobject system, so the total momentum is a constant. From the choices, applying the vector sum mation for the momenta of the two objects, we can easily identify the correct choice. The figure below shows the sum of the x and ycomponents of the vectors which repre sent the correct answer. initial momentum The horizontal vectors add to be the same length as the vector presented in the question. The vertical vectors cancel, as expected, since there is no vertical momentum. 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 . 45 × 10 7 m / s, what would be the recoil speed of the thorium nucleus? Correct answer: 4 . 19278 × 10 5 m / s. Explanation: Let : v α = 2 . 45 × 10 7 m / s , homework 23 – Turner – (58185) 1 hinojosa (jlh3938) – homework 23 – Turner – (58185) 2 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 . 45 × 10 7 m / s) (6 . 64 × 10 − 27 kg) 3 . 88 × 10 − 25 kg = 4 . 19278 × 10 5 m / s . 003 10.0 points A(n) 1480 kg car moving with a speed of 9 . 13 m / s collides with a utility pole and is brought to rest in 0 . 344 s. Find the magnitude of the average force exerted on the car during the collision. Correct answer: 39280 . 2 N. Explanation: The impulse equals the change in momen tum, therefore F Δ t = Δ p = m v f − m v i . Thus F = m v f − m v i Δ t = − (1480 kg) (9 . 13 m / s) (0 . 344 s) = − 39280 . 2 N bardbl vector F bardbl = 39280 . 2 N , where the negative sign indicates the direction of the force is opposite to that of the car’s original motion....
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This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Momentum, Work

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