solutions 13 - Patel, Kinal Homework 13 Due: Oct 19 2007,...

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: Patel, Kinal Homework 13 Due: Oct 19 2007, 7:00 pm Inst: D Weathers 1 This print-out should have 12 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 Two ice skaters approach each other at right angles. Skater A has a mass of 38 . 8 kg and travels in the + x direction at 2 . 05 m / s. Skater B has a mass of 79 . 4 kg and is moving in the + y direction at 2 . 19 m / s. They collide and cling together. Find the final speed of the couple. Correct answer: 1 . 61772 m / s. Explanation: From conservation of momentum p = 0 m A v A + m B v B = ( m A + m B ) v f Therefore v f = p ( m A v A ) 2 + ( m B v B ) 2 m A + m B = p (79 . 54 kg m / s) 2 + (173 . 886 kg m / s) 2 38 . 8 kg + 79 . 4 kg = 1 . 61772 m / s keywords: 002 (part 1 of 2) 10 points m 1 1 m 2 before after v v v 2 v m 1 m 2 3 4 An m 2 = 1 . 5 kg can of soup is thrown upward with a velocity of v 2 = 5 . 9 m / s. It is immediately struck from the side by an m 1 = 0 . 43 kg rock traveling at v 1 = 8 . 6 m / s. The rock ricochets off at an angle of = 40 with a velocity of v 3 = 6 m / s. What is the angle of the cans motion after the collision? Correct answer: 76 . 5373 . Explanation: Basic Concepts: Conservation of Mo- mentum p before = p after . Solution: Horizontally m 1 v 1 + m 2 (0) = m 1 v 3 cos + m 2 v 4 cos , so that m 2 v 4 cos = m 1 v 1- m 1 v 3 cos (1) = (0 . 43 kg)(8 . 6 m / s)- (0 . 43 kg)(6 m / s)cos40 = 1 . 72161 kg m / s . Vertically m 1 (0) + m 2 v 2 = m 1 v 3 sin + m 2 v 4 sin , so that m 2 v 4 sin = m 2 v 2- m 1 v 3 sin (2) = (1 . 5 kg)(5 . 9 m / s)- (0 . 43 kg)(6 m / s)sin40 = 7 . 19161 kg m / s . Thus tan = m 2 v 4 sin m 2 v 4 cos = (7 . 19161 kg m / s) (1 . 72161 kg m / s) = 4 . 17727 , and = arctan(4 . 17727) = 76 . 5373 . 003 (part 2 of 2) 10 points With what speed does the can move immedi- ately after the collision? Correct answer: 4 . 92989 m / s. Explanation: Using equation (1) above, v 4 = m 1 v 1- m 1 v 3 cos m 2 cos = (0 . 43 kg)(8 . 6 m / s) (1 . 5 kg)cos76 . 5373 - (0 . 43 kg)(6 m / s)cos(40 ) (1 . 5 kg)cos(76 . 5373 ) = 4 . 92989 m / s Patel, Kinal Homework 13 Due: Oct 19 2007, 7:00 pm Inst: D Weathers 2 or using equation (2) above, v 4 = m 2 v 2- m 1 v 3 sin m 2 sin = (5 . 9 m / s) sin(76 . 5373 )- (0 . 43 kg)(6 m / s)sin(40 ) (1 . 5 kg)sin(76 . 5373 ) = 4 . 92989 m / s . keywords: 004 (part 1 of 1) 10 points Assume an elastic collision (ignoring friction and rotational motion)....
View Full Document

This note was uploaded on 10/22/2010 for the course PHYS PHYS 1710 taught by Professor Weathers during the Fall '07 term at North Texas.

Page1 / 6

solutions 13 - Patel, Kinal Homework 13 Due: Oct 19 2007,...

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