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phy121
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.
** CQ_1_13.1_labelMessages **
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?
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• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
In the vertical direction we have v0 = 20 cm/sec, a = 980 cm/sec^2, and `ds = 120 cm.
Vf^2 = v0^2 + 2*a*`ds
= 20^2 + 2*980*120
Vf = 485.39 cm/sec
vAVe = (485.39+20)/2 = 252.70
`dv = 465.39 m/sec
v = `ds/`dt
252.70 = 120/`dt
`dt=.47 seocnds
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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?
In the horizontal direction we have constant velocity of v=80 cm/second, `dt = .47 second
`ds = vAve * `dt
= 80 * .47 = 37.6 cm
V0 = 80 cm/sec
Acceleration would be 0
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• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
`ds = 37.6 cm, vf = 80 cm/sec, `dv = 0cm/sec vAVe = 80 m/sec
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
Upon impact with the floor, the ball and floor will exert equal and opposite force. I would therefore say that the acceleration of the ball changes with time and would not be constant
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• Why does this analysis stop at the instant of impact with the floor?
Because acceleration would no longer be uniform.
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&#Very good responses. Let me know if you have questions. &#