old midterm 2 - brown (twb493) midterm 02 practice 2 turner...

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Unformatted text preview: brown (twb493) midterm 02 practice 2 turner (57340) 1 This print-out should have 20 questions. Multiple-choice 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 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 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 = 2 . 35 10 7 m / s , M = 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 vectorv + M Th vectorv Th = 0 . This means that the Thorium nucleus recoils in the direction exactly opposite to that of the particle with speed bardbl vectorv Th bardbl = bardbl vectorv bardbl M 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 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 left-hand 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 left-hand mass m 1 the accel- eration is up and T 1 m 1 g = m 1 a , for the upper left-hand mass m 2 the accelera- tion is up and T T 1 m 2 g = m 2 a , and for the right-hand 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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old midterm 2 - brown (twb493) midterm 02 practice 2 turner...

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