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Asst 22 seed 2 word
A 70 gram ball rolls off the edge of a table and falls freely to the floor 122 cm below. While in free fall it moves 40 cm in the horizontal direction. At the instant it leaves the edge it is moving only in the horizontal direction. In the vertical direction, at this instant it is moving neither up nor down so its vertical velocity is zero. For the interval of free fall:
• What are its final velocity in the vertical direction and its average velocity in the horizontal direction?
Vf^2 = Vo^2 + 2 a ‘ds. Vf^2 = 0 + 2(9.8 m/s^2) * 1.22 m = 23.9. Vf = 4.89 m/s. The vertical Vf = 4.89 m/s. Find the ‘dt by ‘dt = ‘ds / Vave; ‘dt = 1.22 m/ 2.44m/s = .5 s So if ball moves 40 cm in .5 seconds then it moves 80 cm in 1 sec so the velocity is . 8 meters in one sec so the horizontal Vave = .8 m/s because the acceleration is constant
• Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?
The horizontal component is .8 m/s and the vertical is 2.44 m/s
The ball will have achieved its final vertical velocity of 4.89 m/s. At the end of the fall it's not moving at its average vertical velocity.
• What are its speed and direction of motion at this instant?
The speed is the hypotenuse of 2.56 m/s but I suppose we don’t call it that so it’s the magnitude of the vector and the angle theta is -71.84 or 288.15 degrees.
Right, except that you used the wrong vertical velocity. Process is fine.
• What is its kinetic energy at this instant?
The KE is = ˝ mv^2 so it is .24 joules from my figures above KE = ˝*(.07kg)(2.56)^2
KE will be about 4 times as great if you use the final vertical velocity, rather than the average vertical velocity, to calculate the final speed.
• What was its kinetic energy as it left the tabletop?
The KE will be less but computed to .02 joules which is practically nothing which would stand to reason because the PE would be at its greatest on the table top.
• What is the change in its gravitational potential energy from the tabletop to the floor?
The PE is = m * g * y which is .07 kg * 9.8 m/s^2 * 1.22 m = .84 joules on the table top so the PE at the bottom would be zero.
• How are the initial KE, the final KE and the change in PE related?
They are related because they are somewhat opposites. If the PE is up the KE will be down and the units will change as the mass’s y dimension changes. So the PE will be going down and the KE will be going up as the ball is traveling downward.
• How much of the final KE is in the horizontal direction and how much in the vertical?
The horizontal KE = ˝ * .07 kg * (.8 m/s)^2 = .02 Joules
The vertical KE = ˝ * .07 kg * (2.44 m/s)^2 = .24 Joules
KE gain and PE loss will match if you correct the error mentioned above.
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30 minutes
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&#Good responses. See my notes and let me know if you have questions. &#