Physics 114A - Mechanics Lecture 8 (Walker: 4.1&2) 2D Motion Basics January 17, 2012 John G. Cramer Professor Emeritus, Department of Physics B451 PAB

Announcements Homework Assignments #1 is due on WebAssign at 11:59 PM tonight. Assignments #2 is due on Thursday, Jan. 19. On Friday we will have Exam 1. It will have a 75 point multiple-choice part and a 25 point free-response part. It will have assigned seating, specified by seating charts posted before the exam on the entry doors. Today is your last chance to request a left-handed seat, a right-handed aisle seat, or a front row seat. My office hours are in the hour before class on Tuesdays and in the hour after class on Thursdays, both in the “114” area of the Physics Study Center on the Mezzanine floor of PAB A. January 17, 20122/21Physics 114A - Lecture 8

Physics 114A - Introduction to Mechanics - Winter-2012 Lecture: Professor John G. Cramer Textbook: Physics, (Edition: UW Vol. 1 or Complete 4th), James S. Walker WeekDateL#Lecture TopicPagesSlidesReadingHW DueLab 1 2-Jan-12H1 New Year Holiday No Lab 1st week 3-Jan-121Introduction to Physics1221Chapter 1 5-Jan-122Position & Velocity to 2-3No HW 6-Jan-123Velocity & Acceleration to Jan-124Equations of Motion to D Kinematics 10-Jan-125Vectors to Jan-126r, v & a Vectors to 3-5HW1 13-Jan-127Relative Motion Jan-12H2 MLK Birthhday Holiday Free Fall & Projectiles 17-Jan-1282D Motion Basics to Jan-1292D Examples to 4-5HW2 20-Jan-12E1 EXAM 1 - Chapters 1-4 Lecture Schedule (Part 1) We are here. January 17, 20123/21Physics 114A - Lecture 8

Two-Dimensional Kinematics January 17, 20124/21Physics 114A - Lecture 8

Constant Velocity If velocity is constant, motion is along a straight line: A turtle walks from the origin at a speed of v 0 =0.26 m/s and an angle of In a time t, the turtle moves through a straight line distance of d = v 0 t, with horizontal displacement x = d cos  and a vertical displacement y = d sin . Equivalently, the turtle’s horizontal and vertical velocity components are v x = v 0 cos  and v y = v 0 sin . January 17, 20125/21Physics 114A - Lecture 8

(a)How long does it take for the eagle to reach the water? (b)How far has the eagle traveled horizontally when it reaches the water? Example: The Eagle Descends An eagle perched on a tree limb 19.5 m above the water spots a fish swimming near the surface. He pushes off the limb and descends toward the water, maintaining a constant speed of 3.20 m/s at 20 ° below horizontal. January 17, 20126/21Physics 114A - Lecture 8

Constant Acceleration in 2D January 17, 20127/21Physics 114A - Lecture 8

Motion in Two Dimensions Motion in the x - and y -directions should be solved separately: January 17, 20128/21Physics 114A - Lecture 8

Example: A Hummer Accelerates A hummingbird is flying in such a way that it is initially moving vertically with a speed of 4.6 m/s and accelerating horizontally at a constant rate of 11 m/s 2. Find the horizontal and vertical distance through which it moves in 0.55 s. January 17, 20129/21Physics 114A - Lecture 8

Assumptions: Ignore air resistance Use g = 9.81 m/s 2, downward Ignore the Earth’s rotation If the x -axis is horizontal and the y -axis points upward, the acceleration in the x -direction is zero and the acceleration in the y-direction is  9.81 m/s 2 Projectile Motion: Basic Equations January 17, /21Physics 114A - Lecture 8

Constant Acceleration with a x = 0 and a y =  g January 17, /21Physics 114A - Lecture 8

The acceleration is independent of the direction of the velocity: Projectile Motion: Basic Equations January 17, /21Physics 114A - Lecture 8

Independence of Vertical and Horizontal Motion When you drop a ball while walking, running, or skating with constant velocity, it appears to you to drop straight down from thew point where you released it. To a person at rest, the ball follows a curved path that combines horizontal and vertical motions. January 17, /21Physics 114A - Lecture 8

A.x 1 = x 2 B.2x 1 = x 2 C.x 1 = 2x 2 D.4x 1 = x 2 E.x 1 = 4x 2 Clicker Question 1 January 17, /21Physics 114A - Lecture 8 Two divers run horizontally off the edge of a vertical cliff. Diver 2 runs with twice the speed of Diver 1. When they reach the surface of the water below, how are the horizontal distances x i that each diver covers related?

Projectile Motion in Vector Form January 17, /21Physics 114A - Lecture 8

Graphs of x-y and v-t t (s)x (m)y (m)v (m/s) January 17, /21Physics 114A - Lecture 8

Example: Canoeing on the Lake Two canoeists start paddling at the same time and head toward a small island 1 km north in a lake. Canoeist 1 paddles at 1.35 m/s at an angle of 45 0 north of east and Canoeist 2 starts on the opposite shore, a distance of 1.5 km from Canoeist 1. (a) At what angle   relative to north must Canoeist 2 paddle to reach the island? (b) At what speed must Canoeist 2 have, if the canoes are to reach the island at the same time? January 17, /21Physics 114A - Lecture 8

Example: Cliff Diving Two boys, George and Sam, dive from a high overhanging cliff into a lake below. George (1) drops straight down. Sam (2) runs horizontally and dives outward. (a)If they leave the cliff at the same time, which boy reaches the water first? (b) Which boy hits the water with the greater speed? Since the vertical motion determines the time required to reach the water, both boys reach the water at the same. George reaches the water with only a vertical velocity, but Sam reaches the water with both horizontal and vertical velocity components. The vertical velocities are the same for both, so Sam’s speed on entering the water is greater than that of George. January 17, /21Physics 114A - Lecture 8

Example: Electron in Motion An electron in a cathode ray tube is traveling with a horizontal speed of 2.10 x 10 9 cm/s when it enters the gap between deflection plates that give it an upward acceleration of 5.30 x cm/s 2. (a) How long does it take for the electron to cover a distance of 6.20 cm? (b) What is the vertical displacement during this time? January 17, /21Physics 114A - Lecture 8

Example: Jumping a Crevasse A mountain climber encounters a crevasse in an ice field. The opposite side of the crevasse is 2.75 m lower, and the horizontal gap is 4.10 m. To cross the crevasse, the climber gets a running start and jumps horizontally. (a) What is the minimum speed v 0 needed for the climber to cross the crevasse? (b) Suppose the climber jumps at 6.0 m/s, where does he land? (c) What is his speed on landing? January 17, /21Physics 114A - Lecture 8

Before the next lecture on Thursday, read Walker, Chapter We will have Exam 1 on Friday. Homework Assignment #1 should be submitted using the WebAssign system by 11:59 PM tonight. Homework Assignment #2 is due on Thursday. End of Lecture 8 January 17, /21Physics 114A - Lecture 8