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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)โ5 =-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( ๏ฑ ) = ma a = g sin( ๏ฑ ) x direction Cart mg sin( ๏ฑ ) = ma a = g sin( ๏ฑ ) 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
ACT A) 0 B) 2 C) 2.6 D)3 E) 5
Do x(t) for cart and ball. Show they are the same.
Could add y(t) equation here change to marble equation
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