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Unformatted text preview: baum (lmb2768) HW03 distler (56295) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 4) 10.0 points Denote the initial speed of a cannon ball fired from a battleship as v . When the initial projectile angle is 45 with respect to the horizontal, it gives a maximum range R . y x 45 R R/ 2 The time of flight t max of the cannonball for this maximum range R is given by 1. t max = 1 2 v g 2. t max = 2 v g 3. t max = 1 4 v g 4. t max = 2 v g correct 5. t max = 4 v g 6. t max = 1 2 v g 7. t max = 2 3 v g 8. t max = 1 3 v g 9. t max = 3 v g 10. t max = v g Explanation: The cannonballs time of flight is t = 2 v y g = 2 v sin 45 g = 2 v g . 002 (part 2 of 4) 10.0 points The maximum height h max of the cannonball is given by 1. h max = 1 4 v 2 g correct 2. h max = 1 2 v 2 g 3. h max = 2 v 2 g 4. h max = 1 3 v 2 g 5. h max = v 2 g 6. h max = 2 v 2 g 7. h max = 4 v 2 g 8. h max = 3 v 2 g 9. h max = 2 3 v 2 g 10. h max = 1 2 v 2 g Explanation: Use the equation v 2 y = v 2 y 2 g h . At the top of its trajectory v y = 0. Solving for h yields h = v 2 y 2 g = v 2 sin 2 45 2 g = 1 4 v 2 g . 003 (part 3 of 4) 10.0 points The speed v h max of the cannonball at its max imum height is given by 1. v h max = 2 v baum (lmb2768) HW03 distler (56295) 2 2. v h max = v 3. v h max = 1 2 v 4. v h max = 1 2 v correct 5. v h max = 2 3 v 6. v h max = 1 3 v 7. v h max = 3 v 8. v h max = 4 v 9. v h max = 1 4 v 10. v h max = 2 v Explanation: At the top of the cannonballs trajectory, v y = 0. Hence the speed is equal to v x .  v  = v x = v cos 45 = 1 2 v . 004 (part 4 of 4) 10.0 points At a new angle, , the new range is given by R = R 2 . The corresponding angle, , which is greater than 45 , is given by 1. 78 < 80 2. 72 < 74 3. 66 < 68 4. 76 < 78 5. 74 < 76 correct 6. 70 < 72 7. 64 < 66 8. 62 < 64 9. 60 < 62 10. 68 < 70 Explanation: The maximum range corresponds to when = 45 R = v 2 sin[2 (45 )] g = v 2 g . Thus for R = R 2 , we need sin[2 ] = 1 2 . There are two solutions for that satisfy the above = 15 or = 75 . 005 10.0 points A mass slides with negligible friction on a circular track of 1 m radius oriented vertically. Its speed at the position shown in the figure is 3.13 m/s. v At the position shown in the figure, which arrow best represents the direction of the ac celeration of the mass? The acceleration of gravity is 9.8 m/s 2 and no external forces act on the system....
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This note was uploaded on 11/30/2011 for the course PHYSICS 302K taught by Professor Irenepolycarpou during the Spring '09 term at University of Texas at Austin.
 Spring '09
 IRENEPOLYCARPOU
 Physics

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