Mary Lee McJimsey (Physics BSc) High School Physics Teacher North Central High School Spokane, WA

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1 PHYSICIST PROFILE Mary Lee McJimsey (Physics BSc) High School Physics Teacher North Central High School Spokane, WA Attend Mary Lee McJimsey s high school physics class, and you may be surprised! Rather than delivering a long lecture at the board, she would be leading students in a hands-on exploration of how physics principles appear in everyday situations. In addition to teaching, Mary Lee works with a physics education researcher who collects data on how students learn in her classes. It s important for teachers to be well-versed in physics education research, she says. My understanding of this research informs the methods I use in my classroom. Mary Lee in action in her high school physics class!

2 Class 5 September 13, 2016 Learning Goals: Practice with constant acceleration problems No office hour Thursday

3 Equations for Constant Acceleration The following equations will solve most constant acceleration problems. x f x i = v i t + 1 a( t)2 2 v f v i = a t v 2 f v 2 i =2a (x f x i ) 3

4 A ball is thrown straight up into the air with an initial velocity v i = 5.00 m/s. (a) How high does the ball go above its starting point? (a) 1.27 m (b) 1.56 m (c) 2.05 m (d) 2.64 m (e) 3.00 m 4

5 A ball is thrown straight up into the air with an initial velocity v i = 5.00 m/s (a) How high does the ball go above its starting point? v f 2 v i 2 =2a (y f y i ) y f = y i v 2 i 2g =1.27 m 5

6 A ball is thrown straight up into the air with an initial velocity v i = 5.00 m/s. (b) How much time is required for the ball to go up and back down to its starting point? (a) 0.25 s (b) 0.51 s (c) 0.76 s (d) 1.02 s (e) 1.12 s 6

7 A ball is thrown straight up into the air with an initial velocity v i = 5.00 m/s (b) How much time is required for the ball to go up and back down to its starting point? v f v i = a t t up = v i =0.51 s g t up down =1.02 s The up and down motion are symmetric. You can solve the up problem by solving the down problem. 7

8 A car is parked on a street that descends at an angle of 5 below the horizontal. The car is initially at rest, but then the parking brake fails and the car begins to roll. Assume no friction and massless wheels and tires. What is the speed of the car after it travels 200 m to the end of the street? 8

9 Solution A car is parked on a street that descends at an angle of 5 below the horizontal. The car is initially at rest, but then the parking brake fails and the car begins to roll. Assume no friction and massless wheels and tires. What is the speed of the car after it travels 200 m to the end of the street? v 2 f v 2 i =2a x a = g sin p v f = p 2g sin 5 (200 m) = 18.5 m/s 9

10 A ball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and passes the top of the window 0.2 s later. The window is 2.0 m high from top to bottom. (a) What was the ball s initial velocity? (b) How far is the bottom of the window from the launch position? (c) How high does the ball rise above the launch position. 10

11 A ball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and passes the top of the window 0.2 s later. The window is 2.0 m high from top to bottom. (a) What was the ball s initial velocity? (b) How far is the bottom of the window from the launch position? (c) How high does the ball rise above the launch position. Getting Started: This is a problem with a ball being tossed up. It is totally equivalent to a problem in which the ball is dropped from a certain height. The height is unknown, but the ball s speed at the top of the window can be calculated from the 2 m and the 0.2 s. The rest of the problem is straightforward. 11

12 A ball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and passes the top of the window 0.2 s later. The window is 2.0 m high from top to bottom. (a) What was the ball s initial velocity? (b) How far is the bottom of the window from the launch position? (c) How high does the ball rise above the launch position. Devise Plan: Do the dropping problem. Analyze the motion past the window first. Analyze motion above the window next. Analyze the motion below the window last. 12

13 A ball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and passes the top of the window 0.2 s later. The window is 2.0 m high from top to bottom. (a) What was the ball s initial velocity? (b) How far is the bottom of the window from the launch position? (c) How high does the ball rise above the launch position. Execute Plan: Dropping ball by the window. v i is the speed of the ball at the top of the window. 13

14 A ball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and passes the top of the window 0.2 s later. The window is 2.0 m high from top to bottom. (a) What was the ball s initial velocity? (b) How far is the bottom of the window from the launch position? (c) How high does the ball rise above the launch position. Execute Plan: Above the window: 14

15 A ball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and passes the top of the window 0.2 s later. The window is 2.0 m high from top to bottom. (a) What was the ball s initial velocity? (b) How far is the bottom of the window from the launch position? (c) How high does the ball rise above the launch position. Execute Plan: Below the window: 15