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Unformatted text preview: brown (twb493) – midterm 02 practice 2 – turner – (57340) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 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 . 35 × 10 7 m / s, what would be the recoil speed of the thorium nucleus? Correct answer: 4 . 02165 × 10 5 m / s. Explanation: Let : v α = 2 . 35 × 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 . 35 × 10 7 m / s) (6 . 64 × 10 − 27 kg) 3 . 88 × 10 − 25 kg = 4 . 02165 × 10 5 m / s . 002 (part 1 of 2) 10.0 points A pulley is massless and frictionless. The masses 4 kg, 2 kg, and 7 kg are suspended as in the figure. 3 . 6 m 23 . 7 cm ω 2 kg 4 kg 7 kg T 2 T 1 T 3 What is the tension T 1 in the string be tween the two blocks on the lefthand side of the pulley? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 42 . 2154 N. Explanation: Let : R = 23 . 7 cm , m 1 = 4 kg , m 2 = 2 kg , m 3 = 7 kg , and h = 3 . 6 m . Consider the free body diagrams 4 kg 2 kg 7 kg T 1 T 2 T 3 m 1 g T 1 m 2 g m 3 g a a For each mass in the system vector F net = mvectora . Since the string changes direction around the pulley, the forces due to the tensions T 2 and T 3 are in the same direction (up). The acceleration of the system will be down to the right ( m 3 > m 1 + m 2 ), and each mass in brown (twb493) – midterm 02 practice 2 – turner – (57340) 2 the system accelerates at the same rate (the string does not stretch). Let this acceleration rate be a and the tension over the pulley be T ≡ T 2 = T 3 . For the lower lefthand mass m 1 the accel eration is up and T 1 − m 1 g = m 1 a , for the upper lefthand mass m 2 the accelera tion is up and T − T 1 − m 2 g = m 2 a , and for the righthand mass m 3 the accelera tion is down and − T + m 3 g = m 3 a . Adding, ( m 3 − m 1 − m 2 ) g = ( m 1 + m 2 + m 3 ) a ....
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This note was uploaded on 04/13/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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