HS8 - Adams, Russell Homework 8 Due: Nov 3 2006, 11:00 pm...

Info iconThis preview shows pages 1–2. 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: Adams, Russell Homework 8 Due: Nov 3 2006, 11:00 pm Inst: Opyrchal, H 1 This print-out should have 9 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 4 kg steel ball strikes a wall with a speed of 11 . 9 m / s at an angle of 30 . 7 with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 1 1 . 9 m / s 4 kg 1 1 . 9 m / s 4 kg 30 . 7 30 . 7 If the ball is in contact with the wall for . 254 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 322 . 275 N. Explanation: Let : M = 4 kg , v = 11 . 9 m / s , and = 30 . 7 . The y component of the momentum is un- changed. The x component of the momentum is changed by P x =- 2 M v cos . Therefore, using impulse formula, F = P t =- 2 M v cos t =- 2(4 kg)(11 . 9 m / s) cos30 . 7 . 254 s k ~ F k = 322 . 275 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign. keywords: 002 (part 1 of 1) 10 points A superball with a mass of 25 . 3 g is dropped from a height of 2 . 9 m. It rebounds to a height of 2 . 4 m. The acceleration of gravity is 9 . 8 m / s 2 . What is the change in its linear momentum during the collision with the floor? Correct answer: 0 . 364264 kgm / s. Explanation: Given the height, the speed of the ball right before it hits the floor is v i =- p 2 g h i . Like- wise, thespeedoftheballrightafter theithits the floor is v f = q 2 g h f . Thus the change in momentum, or the impulse is I = mv f- mv i = m hq 2 g h f + p 2 g h i i = (25 . 3 g) [(7 . 53923 m / s) + (6 . 85857 m / s)] = 0 . 364264 kgm / s , where v i = p 2 g h i = q 2(9 . 8 m / s 2 )(2 . 9 m) = 7 . 53923 m / s and v f = q 2 g h f = q 2(9 . 8 m / s 2 )(2 . 4 m) = 6 . 85857 m / s and = 6 . 85857 m / s ....
View Full Document

Page1 / 5

HS8 - Adams, Russell Homework 8 Due: Nov 3 2006, 11:00 pm...

This preview shows document pages 1 - 2. Sign up to view the full document.

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