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Phy 121
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?
If the downward direction is positve...
v0 = 20cm/s
`ds = 120cm
a = 9.8m/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
vf^2 = 235600cm^2/s^2
vf = +- 485cm/s approx. Use the positive value
`ds= 120cm
`dv = 485cm/s - 20cm/s = 465cm/s
vAVe = (20cm/s + 485cm/s)/2 = 253 cm/s approx
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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?
a = 0cm/s^2
After it leaves the ramp it is not accelerating in the horizantal direction
v0 = 80cm/s
We base `dt off the time it take the ball to hit the ground.
vAve = `ds/`dt
`dt = `ds/vAve
`dt = 120cm/253cm/s = .47s
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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 = 80cm/s
vAve = (80cm/s + 80cm/s)/2 = 80cm/s
`dv = 0cm/s
`ds = vAve *`dt = 80cm/s *.47s = 38 cm approx.
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
It will not be uniformly accelerated. It may bounce around a bit, but those changes in velocity with respect to time are unlikely to be uniform.
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Why does this analysis stop at the instant of impact with the floor?
I would say because at this point we are no longer using uniform acceleration.
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*#&!
&#Very good responses. Let me know if you have questions. &#