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Phy 241
Your 'cq_1_13.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A ball rolls off the end of an incline with a vertical velocity of 20 cm/s downward, and a horizontal velocity of 80 cm/s. The ball falls freely to the floor 120 cm below.
• For the interval between the end of the ramp and the floor, hat are the ball's initial velocity, displacement and acceleration in the vertical direction?
v0=20cm/s , `ds=120cm, a=980cm/s^2
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• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
vf^2=v0^2+2a`ds=(20cm/s)^2+2*980cm/s^2*120cm=235600cm^2/s^2 so vf=+-485.4cm/s
`dv=465.4cm/s ; vAve=252.7cm/s
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• What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
`dt=`dv/a=465.4cm/s / 980cm/s^2 = 0.5s ; v0=80cm/s a=0cm/s^2 since it has left the ramp and is no longer speeding up in that direction.
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• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
vf=`dt a+v0=.5s*0m/s^2+80cm/s = 80cm/s ; `dv=0cm/s ; vAve=80cm/s
`ds=vAve`dt=80cm/s*.5s=40cm
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No, the forces of friction, gravity, and the loss of energy from the other aspects of the impact will change the acceleration.
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• Why does this analysis stop at the instant of impact with the floor?
Since acceleration is no longer uniform, it becomes much harder for us to measure and calculate the various aspects of the balls travel without more knowledge and better equipment.
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Very good responses. Let me know if you have questions.