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Phy 121
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?
v0= 20 cm/s
'ds= 120 cm
a= 980 cm/s^2
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What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
vf = sqrt( v0^2 + 2 a 'ds )
vf = sqrt( (20 cm/s)^2 + 2 * 980 cm/s^2 * 120 cm)
vf = sqrt( 400 cm^2/s^2 + 235200 cm^2/s^2)
vf = sqrt( 235600 cm^2/s^2 ) = 485.39 cm/s
'ds = 120 cm
'dv = 485.39 cm/s - 20 cm/s = 465.39 cm/s
vAVe = (485.39 cm/s + 20 cm/s)/2
vAve = 252.7 cm/s
'dt = 'ds / vAve
'dt = 120 cm / 252.7 cm/s
'dt = 0.47 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?
v0 = 80 cm/s
a = 0
'dt = 0.47 s
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What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal
direction during this interval?
'ds = v0 * 'dt + 0.5 a * 'dt^2
'ds= v0 * 'dt
'ds= 37.6 cm
vf = sqrt( (80 cm/s)^2)
vf = sqrt( 6400 cm^2/s^2)
vf = 80 cm/s
vAve = 80 cm/s
'dv=80 cm/s - 80 cm/s
'dv=0
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No the ball will stop.
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Why does this analysis stop at the instant of impact with the floor?
Once the ball hits the floor it is no longer falling with acceleration.
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** **
30 min
** **
@&
&#Very good responses. Let me know if you have questions. &#
*@
#$&*
Phy 121
Your 'cq_1_13.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_13.1_labelMessages **
A ball rolls off the end of an incline with a vertical velocity of 20 cm/s downward, and a horizontal velocity of 80 cm/s. The ball falls freely to the floor 120 cm below.
For the interval between the end of the ramp and the floor, hat are the ball's initial velocity, displacement and acceleration in the vertical direction?
v0= 20 cm/s
'ds= 120 cm
a= 980 cm/s^2
#$&*
What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
vf = sqrt( v0^2 + 2 a 'ds )
vf = sqrt( (20 cm/s)^2 + 2 * 980 cm/s^2 * 120 cm)
vf = sqrt( 400 cm^2/s^2 + 235200 cm^2/s^2)
vf = sqrt( 235600 cm^2/s^2 ) = 485.39 cm/s
'ds = 120 cm
'dv = 485.39 cm/s - 20 cm/s = 465.39 cm/s
vAVe = (485.39 cm/s + 20 cm/s)/2
vAve = 252.7 cm/s
'dt = 'ds / vAve
'dt = 120 cm / 252.7 cm/s
'dt = 0.47 s
#$&*
What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time,
during this interval?
v0 = 80 cm/s
a = 0
'dt = 0.47 s
#$&*
What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal
direction during this interval?
'ds = v0 * 'dt + 0.5 a * 'dt^2
'ds= v0 * 'dt
'ds= 37.6 cm
vf = sqrt( (80 cm/s)^2)
vf = sqrt( 6400 cm^2/s^2)
vf = 80 cm/s
vAve = 80 cm/s
'dv=80 cm/s - 80 cm/s
'dv=0
#$&*
After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No the ball will stop.
#$&*
Why does this analysis stop at the instant of impact with the floor?
Once the ball hits the floor it is no longer falling with acceleration.
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** **
30 min
** **
&#Very good responses. Let me know if you have questions. &#