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Two Dimensional Dynamics ,[object Object],Physics 101:  Lecture 06 Exam I Starting Wed Sept 15, W-F office hours will be in 32 Loomis. M office hours will continue in 236 Loomis
Brief Review Thus Far ,[object Object],[object Object],[object Object],[object Object]
2-Dimensions ,[object Object],[object Object],y x
Position, Velocity and Acceleration ,[object Object],[object Object],x direction y direction
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 x-direction v ix  = 5  m/s v fx  = 3  m/s  cos(25)  v x  = 3cos(25)–5 =-2.28 m/s   y-direction v iy  = 0  m/s v fy  = 3  m/s  sin(25)  v y  = 3sin(25)=+1.27  m/s   y x
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? x-direction y-direction y x
Kinematics in Two Dimensions ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],x  and   y   motions are   independent ! They share a common time t Must be able to identify variables in these equations!
Train Act/Demo 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 Demo - train ,[object Object],correct
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 x direction Ball mg sin(  ) = ma a = g sin(  ) x direction Cart mg sin(  ) = ma a = g sin(  ) Same acceleration gives same position correct y x
Projectile Motion ACT  ,[object Object],[object Object],[object Object],[object Object],When ball hits depends on y only! y(t) = y 0  + v yo  + ½ a y  t Same for both balls!
[object Object],[object Object],[object Object],[object Object],Monkey Pre-Flight  Demo - monkey 72% 14% 14% No animals were harmed in the making of this demo “ Poor monkey...I hope there's something soft for the little guy to land on.” correct
Most intriguing answer ,[object Object]
x  =  x 0 y   = - 1 / 2  g  t 2  See text:  4-3 x   =  v 0  t y   = - 1 / 2  g  t 2  Shooting the Monkey...
Shooting the Monkey... y   =  v 0y   t -  1 / 2  g  t 2 ,[object Object],y   =  y 0  -  1 / 2  g  t 2 Dart hits the   monkey!
Projectile Motion a x  = 0  a y  = -g   ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Throw ball to  Monkey  ,[object Object],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
Summary of Concepts ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

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Lecture06

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  • 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 x-direction v ix = 5 m/s v fx = 3 m/s cos(25)  v x = 3cos(25)–5 =-2.28 m/s y-direction v iy = 0 m/s v fy = 3 m/s sin(25)  v y = 3sin(25)=+1.27 m/s y x
  • 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? x-direction y-direction y x
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  • 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 x direction Ball mg sin(  ) = ma a = g sin(  ) x direction Cart mg sin(  ) = ma a = g sin(  ) Same acceleration gives same position correct y x
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  • 13. x = x 0 y = - 1 / 2 g t 2 See text: 4-3 x = v 0 t y = - 1 / 2 g t 2 Shooting the Monkey...
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Editor's Notes

  1. ACT A) 0 B) 2 C) 2.6 D)3 E) 5
  2. Do x(t) for cart and ball. Show they are the same.
  3. Could add y(t) equation here change to marble equation
  4. Vf2-vi2 = 2 a y v = d/t Vi = sqrt(2 9.8 12) = 3.2 m/s Vi = 15.3 m/s Vf = vi + a t T = 1.56 s