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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Assignment 13 Seed Question 1
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?
We can just identify these quantities by the numbers given in the equation; v0= 20 cm/s, ‘ds= 120 cm, and a= gravity (9.8 m/s^2, or 980 cm/s^2).
• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
In order to find the vf, we need to first find ‘dt. In order to do so, we use the variables we know to plug into the equation y= (1/2)(a)(‘dt^2) + (v0)(‘dt) and solve for ‘dt. We need to do this using the quadratic equation and getting a time of .473 seconds. (since the negative time we get does not make sense). Then, we can use a= (vf-v0) / ‘dt and plug in what we know, now using the ‘dt= .473 sec, a= 9.8 m/s/s, and the v0= 20 m/s, we can solve for vf = 24.635 m/s. Now that we know the vf, we can say ‘dv= vf- v0 = 24.635 m/s – 20 m/s = 4.635 m/s. And, we can say that vAve = (vf + v0 ) / 2 = 22.32 m/s.
Everything but your final vertical velocity makes sense, and the procedure is very good.
For the vertical motion we choose the downward direction as positive, in which case we have v0 = 20 cm/s, a = 980 cm/s and `ds = 120 cm.
With v0, a and `ds we can start with either the third or fourth equation of motion to find vf and `dt. The fourth equation is simpler to use. Solving for vf we have
vf = +- sqrt(v0^2 + 2 a `ds) = +- sqrt( (20 cm/s)^2 + 2 * 980 cm/s * 120 cm) = +- sqrt( 240,000 cm^2/s^2) = +- 490 cm/s, approximately.
The downward direction is positive and the ball is clearly moving in the positive vertical direction, so we reject the negative solution.
This gives us an average vertical velocity of (490 cm/s + 20 cm/s) / 2 = 250 cm/s, approx., and a time of fall of about .48 sec.
The change in velocity is about 490 cm/s - 20 cm/s = 470 cm/s.
• What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
Initial velocity was told to be 80 cm/s. (Although, the units missed the “s” part, and just said cm). There is never any acceleration in the horizontal direction for a free falling object.
• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
To ifnd the horizontal displacement, we multiply velocity by time, so 80cm/s * .473 sec = 37.84 cm. The horizontal velocity will not change. Thus, it remains at 80cm/s for the final velocity and for the change in velocity. The average velocity is therefore 80 cm/s as well.
• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
Yes, because the vertical component of the velocity is constantly accelerating at a constant rate because only gravity is acting on the object.
• Why does this analysis stop at the instant of impact with the floor?
The analysis stops because velocity and acceleration, at the instant that the object hits the floor, are both zero.
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It took me about 45 minutes to do this assignment.
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Excellent work, but be sure to see my note about one result that doesn't seem right. I suspect a units error, though you didn't show all the details of that calculation and I can't tell for sure.