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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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Seed Questions 13.1
Query Add comments on any surprises or insights you experienced as a result of this assignment.
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, what are the ball's initial velocity, displacement and acceleration in the vertical direction?
v0= 20 cm/s, ds = 120 cm, a= 9.8 m/s^2
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• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
Given: v0= 20 cm/s, ds = 120 cm, a= 9.8 m/s^2
Determine the final velocity:
vf^2 = v0^2 + 2 * a * ds
vf = +- sqrt(v0^2 + 2 a ds)
vf = +- sqrt( (20 cm/s)^2 + 2 * 9.8 m/s^2 * 120 cm)
vf = +- sqrt(400 cm^2 / s^2 + 2352 m^2 / s^2)
@& You have units errors in this step and the next.
m/s^2 * cm is not m^2 / s^2.
And 400 cm^2 / s^2 + 2352 m^2 / s^2 would not be 2752 cm^2 / s^2. The units cm^2 / s^2 and m^2 / s^2 are not the same, and the two quantities are therefore not like terms.
You are going to have to represent all your quantities in cm and seconds, or in meters and seconds.*@
@& Your procedures are correct, and you have done well to include the units. The error would have been difficult to spot without the units. All you need to do is fix the discrepancy with the units.*@
vf = +- sqrt(2752 cm^2 / s^2)
vf = +- 52.5 cm/s
vf = + 52.5 cm/s
Change in velocity = 52.5 cm/s - 20 cm/s = 32.5 cm/s
vAve = (vf + v0) / 2 = (52.5 cm/s + 20 cm/s) / 2 = 36.25 cm/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?
Horizontal: v0 = 20 cm/s, ds = 120 cm, a = 9.8 m/s^2, vf =52.5 cm/s, dt = 3.32 s
vf = v0 + a * dt
dt = vf -v0 / a
dt = (52.5 cm/s - 20 cm/s) / 9.8 m/s^2
dt = 32.5 cm/s / 9.8 m/s^2 = 3.32 s
@& Neither 52.5 cm/s nor 20 cm/s is a horizontal velocity.
All the force is in the vertical direction. There is no force in the horizontal direction, and hence zero acceleration in the horizontal 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?
Horizontal: v0 = 20 cm/s, ds = 120 cm, a = 9.8 m/s^2, vf =52.5 cm/s, dt = 3.32 s
ds = vAve * dt
ds = 20 cm/s * 3.32 s
ds = 66.4 cm
vf = v0, which is 80 cm/s
vAve = (vf + v0) / 2 = 80 cm/s + 80 cm/s / 2 = 80 cm/s
Change in velocity is 80 cm/s - 80 cm/s = 0
@& 20 cm/s and 9.8 m/s^2 are respectively the initial vertical velocity and the vertical acceleration. Neither applies to the horizontal motion, nor does the final vertical velocity.*@
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
Difficult question, I don’t have enough information to answer this question.
@& While falling the ball was accelerated by a nice, constant gravitational force.
Do other forces act on it when it hits the floor?
If so, what are some of these forces?
Gravity will still be acting on the ball, but do you expect that the net force on the ball (i.e., the total of all the forces) will be equal to the gravitational force?
Do you therefore expect that the acceleration of the ball will still be the acceleration of gravity?
If the acceleration changes, then it is not uniform. Do you think the acceleration is uniform between the instant the ball begins falling and some instant after it has hit the floor?*@
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• Why does this analysis stop at the instant of impact with the floor?
Difficult question, I don’t have enough information to answer this question.
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