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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.
** 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?
In the vertical direction, initial velocity is 20cm/s.
Displacement is 120cm.
Acceleration is 980cm/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
vf=sqrt[(20cm/s)^2 + 2(980cm/s^2)(120cm)]
vf=sqrt(400cm^2/s^2 + 235200cm^2/s^2)
vf=sqrt(235600cm^2/s^2)
vf=+-485cm/s
Since we've defined down as the positive direction, vf=485cm/s
vertical displacement is 120cm
change in velocity equals 485cm/s-20cm/s = 465cm/s
average velocity equals (20cm/s+485cm/s)/2 = 252.5cm/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?
The ball should not be accelerating in the horizontal direction, because we assume the force causing the acceleration down the ramp is gravity, acting in the vertical direction. So, horizontal acceleration = 0
Initial horizontal velocity = 80cm/s
To find the change in clock time, we need to go back to the vertical direction:
vAve=`ds/`dt
252.5cm/s=120cm/`dt
`dt=.48s
Change in clock time is .48s.
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What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
If we assume horizontal acceleration is constant at 80cm/s, then:
80cm/s = `ds/.48s
38.02cm = `ds
final and average velocity both equal 80cm/s. Change in velocity is 0.
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
Not necessarily-- we cannot assume that impact with the floor will cause uniform acceleration.
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Why does this analysis stop at the instant of impact with the floor?
From dropoff until impact, we assume that fNet is generally constant, the force of gravity on the mass of the ball (perhaps lessened by the force of air resistance, etc.). But we don't know the mass of the ball, so we can't calculate the force of impact with the floor, which would propel the ball upwards with an initial velocity that we don't know, at which point gravity would begin to be the primary force again.
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10 minutes
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&#Very good responses. Let me know if you have questions. &#