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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?
First I solve for the final velocity of the vertical direction.
Vf^2 = v0^2 + 2*a*’ds
Vf^2 = 2*a*’ds
Vf^2 = 2*9.8m/sec^2*122cm
Vf= 48.889cm/sec
vAve = 24.449cm/sec
I need to find time to get the average velocity in the horizontal direction.
‘ds = vAve*’dt
122cm = 24.449cm/sec*’dt
‘dt = 4.989sec(this will be the same for horizontal motion) So
‘ds = vAve * ‘dt
40cm = vAve *4.989sec
vAve = 8.02m/sec
answer/question/discussion:
It doesn't take 4.9 + seconds to fall freely 122 cm, from rest.
Your procedure was good, but you didn't do the units calculations.
• Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?
Horizontal is 8.02m/sec and Vertical is the final velocity of 48.889m/sec
answer/question/discussion:
• What are its speed and direction of motion at this instant?
I need to take the square root of these sums.
Final v = sq. root (Vx^2 + Vy^2)
Final v = sq. root ( (8.02m/sec^2)^2 + (48.889m/sec^2))
Final v = sq. root (2454.455m^2/sec^2)
Final v = 49.54m/sec
Direction = tan y/x
Tan 48.889/ 8.02
80.684 degrees
answer/question/discussion:
Your units don't match up. Do the units calculations and be sure all units reconcile.
• What is its kinetic energy at this instant?
KE = 1/2mv^2
KE = ½(70kg)*(49.54m/sec^2)^2
KE = 85897.41 J
answer/question/discussion:
• What was its kinetic energy as it left the tabletop?
KE = ½ mv^2
KE = ½ *70kg * (8.02m/sec)^2
KE = 2251.214 J
answer/question/discussion:
• What is the change in its gravitational potential energy from the tabletop to the floor?
• PE = mgh
• PE = 70kg*9.8m/sec^2* 122cm
• PE = 83692 J
answer/question/discussion:
• How are the the initial KE, the final KE and the change in PE related?
The change in PE is equal to the difference of final KE minus initial KE.
answer/question/discussion:
• How much of the final KE is in the horizontal direction and how much in the vertical?
I am not sure how to solve that one.
What is the component of the velocity in each directions? What therefore is the KE in each direction.
Check your results by making sure they add up to the KE you calculated before.
Then think about how the vector nature of velocity ensures that this will be the case.
answer/question/discussion:
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20 minutes
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You are making a lot of errors with units, and the results you obtain when this happens tend to be unreasonable for the situation.
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