# 12 - roofner(bar784 – Homework 12 – Weathers –(17104...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: roofner (bar784) – Homework 12 – Weathers – (17104) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points m 1 1 m 2 before after v v v 2 v m 1 m 2 3 4 α β An m 2 = 1 . 4 kg can of soup is thrown upward with a velocity of v 2 = 6 m / s. It is immediately struck from the side by an m 1 = 0 . 48 kg rock traveling at v 1 = 8 . 3 m / s. The rock ricochets off at an angle of α = 51 ◦ with a velocity of v 3 = 5 . 8 m / s. What is the angle of the can’s motion after the collision? Correct answer: 70 . 3081 ◦ . 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 α + m 2 v 4 cos β , so that m 2 v 4 cos β = m 1 v 1- m 1 v 3 cos α (1) = (0 . 48 kg) (8 . 3 m / s)- (0 . 48 kg) (5 . 8 m / s) cos51 ◦ = 2 . 23197 kg m / s . Vertically m 1 (0) + m 2 v 2 = m 1 v 3 sin α + m 2 v 4 sin β , so that m 2 v 4 sin β = m 2 v 2- m 1 v 3 sin α (2) = (1 . 4 kg) (6 m / s)- (0 . 48 kg) (5 . 8 m / s) sin51 ◦ = 6 . 23642 kg m / s . Thus tan β = m 2 v 4 sin β m 2 v 4 cos β = (6 . 23642 kg m / s) (2 . 23197 kg m / s) = 2 . 79413 , and β = arctan(2 . 79413) = 70 . 3081 ◦ . 002 (part 2 of 2) 10.0 points With what speed does the can move immedi- ately after the collision? Correct answer: 4 . 7313 m / s. Explanation: Using equation (1) above, v 4 = m 1 v 1- m 1 v 3 cos α m 2 cos β = (0 . 48 kg) (8 . 3 m / s) (1 . 4 kg) cos70 . 3081 ◦- (0 . 48 kg) (5 . 8 m / s) cos(51 ◦ ) (1 . 4 kg) cos(70 . 3081 ◦ ) = 4 . 7313 m / s or using equation (2) above, v 4 = m 2 v 2- m 1 v 3 sin α m 2 sin β = (6 m / s) sin(70 . 3081 ◦ )- (0 . 48 kg) (5 . 8 m / s) sin(51 ◦ ) (1 . 4 kg) sin(70 . 3081 ◦ ) = 4 . 7313 m / s . 003 10.0 points Two ice skaters approach each other at right angles. Skater A has a mass of 74 . 4 kg and travels in the + x direction at 2 . 6 m / s. Skater B has a mass of 101 kg and is moving in the + y direction at 0 . 853 m / s. They collide and cling together. Find the final speed of the couple. Correct answer: 1 . 20729 m / s. roofner (bar784) – Homework 12 – Weathers – (17104) 2 Explanation: From conservation of momentum ∆ p = 0 m A v A ˆ ı + m B v B ˆ = ( m A + m B ) v f Therefore v f = radicalbig ( m A v A ) 2 + ( m B v B ) 2 m A + m B = radicalbig (193 . 44 kg m / s) 2 + (86 . 153 kg m / s) 2 74 . 4 kg + 101 kg = 1 . 20729 m / s 004 10.0 points A uniform flat plate of metal with a circular hole is situated in the reference frame shown in the figure below....
View Full Document

## This document was uploaded on 10/26/2011 for the course PHYS 1710 at North Texas.

### Page1 / 6

12 - roofner(bar784 – Homework 12 – Weathers –(17104...

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

View Full Document
Ask a homework question - tutors are online