Phy 201
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.
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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?
* voy = -20 cm/s, 'dsy = 120 cm, ay = 980 cm/s^2
The initial vertical velocity, the vertical displacement and the vertical acceleration are all downward, therefore all have the same sign.
If you choose downward as the positive direction, then they can all be positive. If you prefer to choose downward as the negative direction, then they will all be negative.
What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
* vfy^2 = voy^2 + 2ay(y - yo) = sqrt(vfy) = -20 cm/s^2 + 2(980 cm/s^2)(120 cm) = 485 cm/s; 'ds = 120 cm, Dv = 485 cm/s - 20 cm/s = 465 cm/s; Vave = 485 cm/s + 20 cm/s = 505 cm/s
Your statement that
vfy^2 = voy^2 + 2ay(y - yo) = sqrt(vfy)
implies among other things that vfy^2 = sqrt(vfy). This is certainly not so.
The expression -20 cm/s^2 + 2(980 cm/s^2)(120 cm) isn't correctly expressed. It should be (-20 cm/s)^2 + 2(980 cm/s^2)(120 cm). You result of 485 cm/s is, in any case, probably correct.
The average velocity would be the average of the initial and final velocities; you would need to divide your stated average velocity by 2 in order to get the correct average velocity.
What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
* vox = 80 cm/s; Dt = vx / ax = .48s, ax = 80 cm/s / .48 s = 167 cm/s^2
The initial horizontal velocity is indeed 80 cm/s; however the acceleration in the horizontal direction is 0. There is no force in the horizontal direction, therefore no acceleration during the ball's fall.
What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
* Ds = vox * t + 1/2 at^2 = 38.4 cm + 40.1 cm = 78.5 cm; vAve = 80 cm/s / 2 = 40 cm/s; Dv = 0
After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
* yes, but in the opposite direction
This might be the case, for example if the ball bounces back into the air, but the main point is that the uniform acceleration assumption does not apply after the ball touches the floor.
Why does this analysis stop at the instant of impact with the floor?
the ball is no longer in free fall
Good. Even better: the acceleration is no longer uniform and the analysis assumes uniformity of acceleration.
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20 minutes
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Your reasoning is on the whole sound, but there are couple of errors in detail.
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