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Physics 101: Lecture 6, Pg 1 Two Dimensional Dynamics Physics 101: Lecture 06 l Today’s lecture will cover Textbook Chapter 4 Exam I.

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Presentation on theme: "Physics 101: Lecture 6, Pg 1 Two Dimensional Dynamics Physics 101: Lecture 06 l Today’s lecture will cover Textbook Chapter 4 Exam I."— Presentation transcript:

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2 Physics 101: Lecture 6, Pg 1 Two Dimensional Dynamics Physics 101: Lecture 06 l Today’s lecture will cover Textbook Chapter 4 Exam I

3 Physics 101: Lecture 6, Pg 2 Brief Review Thus Far Newton’s Laws of motion:  F=ma Kinematics: x = x 0 + v 0 t + ½ a t 2 Dynamics Today we work in 2 Dimensions! 05

4 Physics 101: Lecture 6, Pg 3 2-Dimensions l X and Y are INDEPENDENT! l Break 2-D problem into two 1-D problems y x 7

5 Physics 101: Lecture 6, Pg 4 Position, Velocity and Acceleration l Position, Velocity and Acceleration are Vectors! l x and y directions are INDEPENDENT! x direction y direction 10

6 Physics 101: Lecture 6, Pg 5 Velocity in Two Dimensions 5 m/s 3 m/s A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s. What is the magnitude of the change in velocity? A) 0 m/s B) 2 m/s C) 2.6 m/s D) 3 m/s E) 5 m/s y x x-direction v ix = 5 m/s v fx = 3 m/s cos(25)  v x = 3cos(25)–5 =-2.28m/s y-direction v iy = 0 m/s v fy = 3 m/s sin(25)  v y = 3sin(25)=+1.27 m/s 15

7 Physics 101: Lecture 6, Pg 6 Acceleration in Two Dimensions 5 m/s 3 m/s A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s. What is the average acceleration? y x x-directiony-direction 18

8 Physics 101: Lecture 6, Pg 7 Kinematics in Two Dimensions l x = x 0 + v 0x t + 1/2 a x t 2 l v x = v 0x + a x t l v x 2 = v 0x 2 + 2a x  x l y = y 0 + v 0y t + 1/2 a y t 2 l v y = v 0y + a y t l v y 2 = v 0y 2 + 2a y  y x and y motions are independent! They share a common time t 21

9 Physics 101: Lecture 6, Pg 8 Train Demo Act A flatbed railroad car is moving along a track at constant velocity. A passenger at the center of the car throws a ball straight up. Neglecting air resistance, where will the ball land? A. Forward of the center of the car B. At the center of the car C. Backward of the center of the car correct 25 Demo - train x- direction ball and car start with same position and velocity, a=0, so always have same position

10 Physics 101: Lecture 6, Pg 9 ACT A flatbed railroad car is accelerating down a track due to gravity. The ball is shot perpendicular to the track. Where will it land? A. Forward of the center of the car B. At the center of the car C. Backward of the center of the car correct 31 y x x direction Ball mg sin(  ) = ma a = g sin(  ) x direction Cart mg sin(  ) = ma a = g sin(  ) Same acceleration gives same position

11 Physics 101: Lecture 6, Pg 10 Projectile Motion ACT One marble is given an initial horizontal velocity, the other simply dropped. Which marble hits the ground first? A) dropped B) pushed C) They both hit the ground at the same time 35 When ball hits depends on y only! y(t) = y 0 + v yo + ½ a y t Same for both balls!

12 Physics 101: Lecture 6, Pg 11 You are a vet trying to shoot a tranquilizer dart into a monkey hanging from a branch in a distant tree. You know that the monkey is very nervous, and will let go of the branch and start to fall as soon as your gun goes off. In order to hit the monkey with the dart, where should you point the gun before shooting? 1 Right at the monkey 2 Below the monkey 3 Above the monkey Monkey Pre-Flight correct Demo - monkey 39 69% 14% 17% No animals were harmed in the making of this demo

13 Physics 101: Lecture 6, Pg 12 x = x 0 x = x 0 yg y = - 1 / 2 g t 2 See text: 4-3 x = v 0 t x = v 0 t yg y = - 1 / 2 g t 2 Shooting the Monkey... 40

14 Physics 101: Lecture 6, Pg 13 Shooting the Monkey... yvg y = v 0y t - 1 / 2 g t 2 l Still works even if you shoot upwards! yg y = y 0 - 1 / 2 g t 2 Dart hits the monkey! 41

15 Physics 101: Lecture 6, Pg 14 Projectile Motion a x = 0 a y = -g  x = x 0 + v 0x t  v x = v 0x  y = y 0 + v 0y t - ½ gt 2  v y = v 0y – g t  v y 2 = v 0y 2 – 2 g  y Choose direction where you “know” information Solve kinematics in that direction. Use t from that direction as t in other direction 43

16 Physics 101: Lecture 6, Pg 15 Throw ball to Monkey You throw a ball to a monkey who is on a platform 12 meters above and 5 meters to the right of you. Determine the speed and angle you should throw it such that it “just reaches” the monkey. 5 m 12 m Y-direction v f 2 –v i 2 = 2 a  y v i = sqrt(2 9.8 12) = 15.3 m/s y = ½ (v f + v i ) t t = 1.56 s. X-direction v = d/t = 5 m / 1.56 s = 3.2 m/s 48

17 Physics 101: Lecture 6, Pg 16 Throw ball to Monkey You throw a ball to a monkey who is on a platform 12 meters above and 5 meters to the right of you. Determine the speed and angle you should throw it such that it “just reaches” the monkey. 5 m 12 m 48

18 Physics 101: Lecture 6, Pg 17 Summary of Concepts l X and Y directions are Independent! è Position, velocity and acceleration are vectors è “Share” t l F = m a applies in both x and y direction l Projective Motion è a x = 0 in horizontal direction è a y = g in vertical direction 50


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