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exam 3 phy - Version 097 Exam 3 chelikowsky(59005 This...

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Version 097 – Exam 3 – chelikowsky – (59005) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points A 29 kg gun is standing on a frictionless sur- face. The gun fires a 52 g bullet with a muzzle velocity of 302 m / s. The positive direction is that of the bullet. Calculate the momentum of the bullet im- mediately after the gun was fired. 1. 17.664 2. 16.0 3. 16.3631 4. 14.0883 5. 16.6152 6. 14.5848 7. 15.704 8. 15.15 9. 17.446 10. 14.2596 Correct answer: 15 . 704 kg · m / s. Explanation: Let : m b = 52 g and v b = 302 m / s . The momentum of the bullet is p b = m b v b = (52 g) (302 m / s) = 15 . 704 kg · m / s . 002 (part 2 of 4) 10.0 points Calculate the momentum of the gun immedi- ately after the gun was fired. 1. -17.728 2. -14.3369 3. -14.7376 4. -16.492 5. -18.7525 6. -17.112 7. -16.0868 8. -15.704 9. -14.04 10. -15.47 Correct answer: 15 . 704 kg · m / s. Explanation: By conservation of momentum, p bo + p go = p bf + p gf 0 + 0 = p b + p g p g = p b = 15 . 704 kg · m / s . 003 (part 3 of 4) 10.0 points Calculate the kinetic energy of the bullet im- mediately after the gun was fired. 1. 2153.53 2. 2625.73 3. 2397.08 4. 2474.25 5. 2286.04 6. 2125.62 7. 2560.0 8. 2431.01 9. 2371.3 10. 2512.93 Correct answer: 2371 . 3 J. Explanation: The kinetic energy of the bullet is K b = 1 2 m b v 2 b = 1 2 (52 g) (302 m / s) 2 = 2371 . 3 J . 004 (part 4 of 4) 10.0 points Calculate the kinetic energy of the gun imme- diately after the gun was fired. 1. 4.41389 2. 3.14193 3. 4.80087 4. 5.18616 5. 3.69522 6. 3.9998 7. 4.25199 8. 4.18704 9. 4.61194 10. 3.49953 Correct answer: 4 . 25199 J.
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Version 097 – Exam 3 – chelikowsky – (59005) 2 Explanation: Let : m g = 29 kg . The recoil velocity of the gun comes from its momentum: p g = m g v g v g = p g m g Thus the kinetic energy of the gun is K g = 1 2 m g v 2 g = p 2 g 2 m g = ( 15 . 704 kg · m / s) 2 2 (29 kg) = 4 . 25199 J . 005 (part 1 of 3) 10.0 points A car accelerates uniformly from rest and covers a distance of 55 m in 6 . 9 s. If the diameter of a tire is 45 cm, find the angular acceleration of the wheel. 1. 29.8587 2. 11.0222 3. 21.6049 4. 11.3786 5. 8.62411 6. 10.2686 7. 13.7392 8. 6.04335 9. 23.4874 10. 12.898 Correct answer: 10 . 2686 rad / s 2 . Explanation: Let : Δ x = 55 m , Δ t = 6 . 9 s , and R = 22 . 5 cm = 0 . 225 m . From kinematics, Δ x = 1 2 a Δ t 2 a = 2 Δ x Δ t 2 , so a = R α α = a R = 2 Δ x R t ) 2 = 2 (55 m) (0 . 225 m) (6 . 9 s) 2 = 10 . 2686 rad / s 2 . 006 (part 2 of 3) 10.0 points Find the final angular velocity of one of the car’s wheels. 1. 132.987 2. 109.501 3. 70.8535 4. 154.448 5. 90.5983 6. 86.3636 7. 93.3063 8. 67.9487 9. 69.2641 10. 83.6237 Correct answer: 70 . 8535 rad / s. Explanation: Since the acceleration is uniform, the final angular velocity is ω f = ω 0 + α t = α t = (10 . 2686 rad / s 2 ) (6 . 9 s) = 70 . 8535 rad / s . 007 (part 3 of 3) 10.0 points Find total number of rotations of the wheel. 1. 60.5816 2. 49.6874 3. 77.0105 4. 46.2996 5. 41.3803 6. 33.1302 7. 54.1987 8. 39.4427 9. 38.9045 10. 35.2173 Correct answer: 38 . 9045. Explanation:
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Version 097 – Exam 3 – chelikowsky – (59005) 3 x = n C = n (2 π r ) n = x 2 π r = 55 m 2 π (0 . 225 m) = 38 . 9045 . keywords: 008 10.0 points Block m 1 of mass 2 m and velocity v 0 is trav- eling to the right (+ x ) and makes an elastic head-on collision with block m 2 of mass m and velocity 2 v 0 ( i.e. , traveling to the left). What is the velocity v 1 of block m 1 after the collision?
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