Exam 2 - Morby Grant Quiz 2 Due Mar 8 2006 10:00 pm Inst...

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

View Full Document Right Arrow Icon
Morby, Grant – Quiz 2 – Due: Mar 8 2006, 10:00 pm – Inst: Drummond 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A(n) 5 . 53 g bullet is fired into a(n) 1 . 59 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9 . 8 m / s 2 . If the pendulum rises a vertical distance of 5 . 78 cm, calculate the initial speed of the bullet. Correct answer: 307 . 094 m / s. Explanation: Basic Concepts: Applying conservation of mechanical energy from just after the colli- sion until the end of the swing is reached, we have 1 2 ( M + m ) V 2 = ( M + m ) g h where M is the mass of the pendulum, m the mass of the bullet, h the vertical height through which the pendulum swings, and V is the velocity of the (pendulum plus bullet) immediately after the collision. The equation above reduces to V = p 2 g h . (1) Now apply conservation of momentum from just before to just after the collision. We have m v 0 = ( M + m ) V , (2) where v 0 is the velocity of the bullet just prior to collision. Thus, m v 0 = ( M + m ) p 2 g h , so v 0 = ( M + m ) m p 2 g h = (1 . 59 kg) + (0 . 00553 kg) (0 . 00553 kg) × q 2 (9 . 8 m / s 2 ) (0 . 0578 m) = 307 . 094 m / s . 002 (part 1 of 4) 10 points The following figure shows a Ferris wheel that rotates 3 times each minute and has a diame- ter of 17 m. The acceleration of gravity is 9 . 8 m / s 2 . What is the centripetal acceleration of a rider? Correct answer: 0 . 838916 m / s 2 . Explanation: The period of the Ferris wheel is T = (60 s) (3) = 20 s . The speed of the wheel is v = 2 π r T = 2 π (8 . 5 m) 20 s = 2 . 67035 m / s , so the centripetal acceleration is a = v 2 r = (2 . 67035 m / s) 2 (8 . 5 m) = 0 . 838916 m / s 2 . 003 (part 2 of 4) 10 points What force does the seat exert on a 51 kg rider at the lowest point of the ride? Correct answer: 542 . 585 N. Explanation: The force exerted by the seat balances the gravity and provides the centripetal force, so F l = m [ g + a ] = (51 kg) (9 . 8 m / s 2 + 0 . 838916 m / s 2 ) = 542 . 585 N .
Image of page 1

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

View Full Document Right Arrow Icon
Morby, Grant – Quiz 2 – Due: Mar 8 2006, 10:00 pm – Inst: Drummond 2 004 (part 3 of 4) 10 points What force does the seat exert on a 51 kg rider at the highest point of the ride? Correct answer: 457 . 015 N. Explanation: The gravity is partly balanced by the force exerted by the seat and this resultant provides the centripetal force, so F l = m [ g - a ] = (51 kg) [(9 . 8 m / s 2 ) - (0 . 838916 m / s 2 )] = 457 . 015 N . 005 (part 4 of 4) 10 points What force (magnitude) does the seat exert on a rider when the rider is halfway between top and bottom? Correct answer: 501 . 628 N. Explanation: In this case, the force exerted by the seat has two components: the vertical one balanc- ing the gravity and the horizontal one provid- ing the centripetal force. Thus we have F m = m p g 2 + a 2 = (51 kg) q (9 . 8 m / s 2 ) 2 + (0 . 838916 m / s 2 ) 2 = 501 . 628 N . 006 (part 1 of 1) 10 points Consider a frictionless roller coaster such as depicted below. The acceleration of gravity is 9 . 8 m / s 2 .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern