HW03-solutions - baum (lmb2768) HW03 distler (56295) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 . 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....
View Full Document

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.

Page1 / 8

HW03-solutions - baum (lmb2768) HW03 distler (56295) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online