Phy 231
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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?
answer/question/discussion: The vertical velocity will be found by vf = - 'sqrt [2 (9.8 m/s^2) (1.22 m)] = - 'sqrt (23.91 m^2/s^2) = - 4.89 m/s, with negative sign indicating the direction is down. This tells us that the time to fall will be that of 1.22 m = 4.9 m/s^2 * 'dt^2; therefore 'dt^2= 0.25 s^2; 0.50 s = 'dt. This gives us an horizontal average velocity of 40 cm/ .5 s = 80 cm/s or .8 m/s = vAve.
Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?
answer/question/discussion: Vertical component is - 4.89 m/s, and horizontal component is .8 m/s.
What are its speed and direction of motion at this instant?
answer/question/discussion: It has a speed of v = 'sqrt [ (.8 m/s)^2 + (- 4.89 m/s)^2] = 'sqrt (24.65 m^2/s^2) = 4.96 m/s with a direction of arctan (-4.89/.8) = -80.71 degrees below the x-axis or if we rotate counterclockwise we find that the angle would be -80.71 + 360 = 279.29 degrees.
What is its kinetic energy at this instant?
answer/question/discussion: KE = ½ ( .07 kg) * (4.96 m/s)^2 = 0.861056 J
What was its kinetic energy as it left the tabletop?
answer/question/discussion: At the moment it left the table it had only the horizontal motion; therefore, we find KE = ½ ( .07 kg) * ( .8 m/s)^2 = 0.0224J.
What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: It went from having a PE of m * g * 'ds; therefore initial PE = .07 kg * (9.8 m/s^2) * 1.22 m = 0.83692 J
How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: The initial KE plus the initial PE is approximately equal to the final KE. This is shown by 0.0224 J + 0.83692 J = 0.85932, which is very close to our final KE value of .861056 J, and the difference is causes largely by rounding of values, and only causes about a .2% difference.
How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: 0.0224 J is in the horizontal, because there was no velocity change horizontally, which gives a KE final and initial being the same in the horizontal direction. The vertical direction has the remaining 0.8387 J, which it gained as a conversion of PE to KE during the drop.
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25 minutes
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Excellent.