Phy 231
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.
1.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?
(a)We have v0 = 20 cm/s, 'ds = 120, and acceleration is 980 cm/s^2
2.What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
(a) Time elapsed will be calculated with only the vertical component and the vertical displacement with the acceleration of gravity being treated as positive and the initial height as negative since we have already assumed down as positive for velocity. This means that 0 cm = -120 cm + (20 cm/s) * 'dt + (980 cm/s^2) 'dt^2; therefore, with the quadratic equation we can find 'dt, if we have a = 980, b = 20, and c = -120. We then have the equation { -20 +- 'sqrt[ 20^2 – 4(980)(-120)]}/ 2(980) = [- 20 +- 'sqrt( 400 + 470400)]/ 1960 = [ -20 +- 'sqrt(470800)]/ 1960 = (- 20 +- 686.15)/ 1960 = 666.15/ 1960 & -706.15/1960 = 0.340 & -0.360, and since we are only concerned with the positive time value this gives us 'dt = .340. Now we can determine its vf = 20 cm/s + 980 cm/s^2 (.340s) = 20 cm/s + 333.2 cm/s = 353.2 cm/s; therefore, 'dv = 353.2 cm/s, and the vAve = (353.2 cm/s + 20 cm/s) / 2 = 186.6 cm/s.
3.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)Since we are neglecting air resistance there is no acceleration in the horizontal direction, 'dt = .340 s, which is the same as for the vertical, but with a different v0 = 80 cm/s
4.What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
(a) 'ds = (80 cm/s) * (.340 s) = 27.2 cm. The vf = 80 cm/s at the instant of impact, but it will then intantly become 0 cm/s since motion will cease upon hitting the ground.
5.After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
(a)Yes, because there will be no motion; therefore, there will be no velocity, and consequently no change in velocity to provide an acceleration value; however, if this question is implying that there will be motion then the assumption of uniform acceleration cannot be made with the available information.
The ball will inevitably compress a bit after the instant of initial impact, as will the nearby region of the floor; during this (likely brief) period acceleration forces will vary greatly and acceleration will not be uniform. If the ball quickly comes to rest, then after coming to rest the acceleration will as you say be uniform.
6.Why does this analysis stop at the instant of impact with the floor?
(a)The analysis stops with the impact of the floor, because for the information given we cannot predict at this point in our studies how the ball may behave after the moment of impact.
The one certainty is that the acceleration will cease to be uniform.
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30 minutes
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&#Good responses. See my notes and let me know if you have questions. &#