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HW03-solutions

# HW03-solutions - baum(lmb2768 HW03 distler(56295 This...

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baum (lmb2768) – HW03 – distler – (56295) 1 This print-out should have 22 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 Denote the initial speed of a cannon ball fired from a battleship as v 0 . When the initial projectile angle is 45 with respect to the horizontal, it gives a maximum range R . y x θ 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 0 g 2. t max = 2 v 0 g 3. t max = 1 4 v 0 g 4. t max = 2 v 0 g correct 5. t max = 4 v 0 g 6. t max = 1 2 v 0 g 7. t max = 2 3 v 0 g 8. t max = 1 3 v 0 g 9. t max = 3 v 0 g 10. t max = v 0 g Explanation: The cannonball’s time of flight is t = 2 v 0 y g = 2 v 0 sin 45 g = 2 v 0 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 0 g correct 2. h max = 1 2 v 2 0 g 3. h max = 2 v 2 0 g 4. h max = 1 3 v 2 0 g 5. h max = v 2 0 g 6. h max = 2 v 2 0 g 7. h max = 4 v 2 0 g 8. h max = 3 v 2 0 g 9. h max = 2 3 v 2 0 g 10. h max = 1 2 v 2 0 g Explanation: Use the equation v 2 y = v 2 y 0 - 2 g h . At the top of its trajectory v y = 0. Solving for h yields h = v 2 y 0 2 g = v 2 0 sin 2 45 2 g = 1 4 v 2 0 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 0

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baum (lmb2768) – HW03 – distler – (56295) 2 2. v h max = v 0 3. v h max = 1 2 v 0 4. v h max = 1 2 v 0 correct 5. v h max = 2 3 v 0 6. v h max = 1 3 v 0 7. v h max = 3 v 0 8. v h max = 4 v 0 9. v h max = 1 4 v 0 10. v h max = 2 v 0 Explanation: At the top of the cannonball’s trajectory, v y = 0. Hence the speed is equal to v x . | v | = v x = v 0 cos 45 = 1 2 v 0 . 004 (part 4 of 4) 10.0 points At a new angle, θ , the new range is given by R = R 2 . The corresponding angle, θ , which is greater than 45 , is given by 1. 78 < θ 80 2. 72 < θ 74 3. 66 < θ 68 4. 76 < θ 78 5. 74 < θ 76 correct 6. 70 < θ 72 7. 64 < θ 66 8. 62 < θ 64 9. 60 < θ 62 10. 68 < θ 70 Explanation: The maximum range corresponds to when θ = 45 R = v 2 0 sin[2 (45 )] g = v 2 0 g . Thus for R = R 2 , we need sin[2 θ ] = 1 2 . There are two solutions for θ that satisfy the above θ = 15 or θ = 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. 1. 2. 3. 4. correct 5.
baum (lmb2768) – HW03 – distler – (56295) 3 6. The mass is traveling at a constant veloc- ity, so it has no acceleration. 7. 8. 9. Explanation: The magnitude of the centripetal accelera- tion is a r = v 2 r = (3 . 13 m / s) 2 1 m 9 . 8 m / s 2 , acting inward ( - ˆ r ) with gravity acting down- ward ( - ˆ k ). a a r g 006 (part 1 of 2) 10.0 points During World War I, the Germans had a gun called Big Bertha that was used to shell Paris.

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HW03-solutions - baum(lmb2768 HW03 distler(56295 This...

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