Phy 121
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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?
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The final velocity in the vertical direction is found by using a fundamental equation of motion. If we use vf^2= v0^2 +2a’ds we should be able to solve for this. I am going to assume that our acceleration is 9.8 m/s^2. So vf^2= 0 +2 (9.8m/s^2)(122 cm). To solve with the correct units, we can convert our 9.8 m/s^2 to 980 cm/s^2. When we solve for vf^2 it gives us, 239, 120 cm^2/s^2. When we take the square root it gives us 489 cm/s or 4.89 m/s as our final velocity.
I am assuming that we are to calculate the average horizontal velocity from the instant that the ball falls off the edge of the table. If this is so, then we need to solve for the final velocity first before we can solve for an average velocity. So we use vf^2= v0^2+2a’ds. When we plug in our given numbers it gives is vf^2= 2 (980cm/s^2)(40 cm).
40 cm is in the horizontal direction; 980 cm/s^2 is in the vertical direction.
The two quantities don't belong in the same equation.
Horizontal and vertical motion must be analyzed independently.
Vf^2= 78,400 cm^2/s^2. When we take the square root, it gives us 280 cm/s or 2.8 m/s. so knowing that we take 280 cm/s +0/2 and get 140 as our average velocity in the horizontal direction.
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• Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?
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If there is no acceleration, then does this mean that the fundamental equation of motion I used in the first question is still valid with the exception of plugging in a value for acceleration???
If so then that means that our vertical component is 2*120cm= 240cm and once we take the square root of this it gives us 15.49 cm or .15m.
The horizontal direction would be 2 (40 cm) = 80 cm. When we take the square root of this it gives us 9.48cm or 0.09 m.
This somehow doesn’t seem right to me though.
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• What are its speed and direction of motion at this instant?
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I’m not exactly sure how to determine speed here, but I think that the Pythagorean theorem and arctan will need to be used to solve for the direction of motion. Use c^2= a^2+ b^2 we use our given information and begin putting it in the formula. So c^2= .15m^2+ .09m^2. c^2= 0.0225 m +0.0081m. c^2= .0306m. When we taken the square root it gives us 0.17m
To solve for the direction of motion we take arctan (.09 m/.15) = .77 degrees From here I can’t remember if I would need to add 360 or 180. Since these numbers are so low, should I be sticking with cm instead of converting my numbers over to m???
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• What is its kinetic energy at this instant?
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I think that to find the kinetic energy we would use 1/2m a va. My a is supposed to be a subscript to represent the object.
KE is 1/2 m v^2
½ (70g)(2.8 m/s) = 98 J
I don’t think Joules is the correct unit here though.
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• What was its kinetic energy as it left the tabletop?
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I think that the kinetic energy would be the same as in the previous question.
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• What is the change in its gravitational potential energy from the tabletop to the floor?
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• How are the the initial KE, the final KE and the change in PE related?
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I think that the initial KE , the final KE and the change in PE are all related in the fact that PE and KE should be equal to one another. The potential energy that an object has is converted to the kinetic energy or the work that is actually done. The total KE before is also going to be the same KE as after.
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• How much of the final KE is in the horizontal direction and how much in the vertical?
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I’m not sure if the KE remains the same throughout or if you would need to calculate a KE amount for the object as it travels vertically downward and as it travels in the horizontal direction.
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45 minutes +
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I thought this was tricky!
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Solution
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