Final Exam-solutions - Version 048 Final Exam chelikowsky...

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: Version 048 Final Exam chelikowsky (59005) 1 This print-out should have 50 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Given: The battleship and enemy ships A and B lie along a straight line. Neglect air friction. A battleship simultaneously fires two shells (with the same muzzle velocity) at these two enemy ships. battleship A B If the shells follow the parabolic trajectories shown in the figure, which ship gets hit first? 1. need more information 2. A 3. B correct 4. both at the same time Explanation: The time interval for the entire projectile motion is given by t trip = t rise + t fall = 2 t rise , where t rise is the rising time from 0 to the maximum height, and t fall the falling time from h to 0. In the absence of air resistance t rise = t fall , h = 1 2 g t 2 fall , or t trip = 2 radicalBigg 2 h g . So the smaller is h , the smaller is t trip . In other words, enemy ship B will get hit first. 002 10.0 points When a metal ball of unknown mass M is suspended from a spring of unknown force constant k , the springs equilibrium length increases by L e . And when the ball is out of equilibrium, it oscillates up and down with a period T . Find L e . The acceleration of gravity is 9 . 8 m / s 2 and the period is 0 . 412 s. 1. 12.303 2. 6.76406 3. 9.54223 4. 1.704 5. 12.1289 6. 4.21367 7. 8.06522 8. 4.52608 9. 15.258 10. 13.7039 Correct answer: 4 . 21367 cm. Explanation: Let : g = 9 . 8 m / s 2 , and T = 0 . 412 s . Let L be the length of the free spring with- out the ball. When a ball is suspended in equi- librium from the spring, the spring lengthens by L e to generate the tension force F S e = k L e = Mg, so (1) L e = Mg k . (2) Now consider the oscillating ball. When the ball is at height y above the equilibrium point, the spring is stretched by L ( y ) = L e y (3) and has tension F S ( y ) = k L ( y ) = k ( L e y ) = Mg k y. (4) Consequently, the net vertical force on the ball is F net y = F S Mg = k y, (5) which makes the ball oscillate with angular frequency = radicalbigg k M (6) Version 048 Final Exam chelikowsky (59005) 2 and period T = 2 = 2 radicalbigg M k . (7) We do not know the balls mass M nor the springs force constant k , but given the oscillation period T we may find the ratio M k = parenleftbigg T 2 parenrightbigg 2 , so (8) L e = Mg k = gT 2 4 2 = (9 . 8 m / s 2 )(0 . 412 s) 2 4 2 = 4 . 21367 cm . (9) 003 10.0 points Given: Two vectors vector A = A x + A y and vector B = B x + B y , where A x = 2, A y = 9, B x = 2, and B y = 10....
View Full Document

This note was uploaded on 03/26/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

Page1 / 22

Final Exam-solutions - Version 048 Final Exam chelikowsky...

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