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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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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=20cm/s vertical,
'ds=120cm vertical,
a=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+(2*a*'ds),
vf=+-sqrt(v0^2+(2*a*'ds)),
vf=+-sqrt((20cm/s)^2+(2*980cm/s/s*120cm)),
vf=+-485.4cm/s
We will use +485.4cm/s because down will be positive.
'ds = 120cm,
'dv = 485.4cm/s - 20cm/s = 465.4cm/s.
'dt = 'dv/a = 465.4cm/s / 980cm/s/s = .47s,
vAve = 'ds/'dt = 120cm/.47s = 255cm/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?
There is no acceleration in the horizontal,
v0=80cm/s,
'dt horizontal = 'dt vertical.
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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='dt*v0,
'ds=.47s(80cm/s)=37.6cm,
vf = 0 cm/s,
'dv = -80 cm/s,
vAve = 37.6cm/.47s = 80 cm/s
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@& 0 final velocity is not achieved during the uniform-acceleration phase. This might well be the final result, after the object hits the ground and comes to rest, but that is not relevant to this analysis.
During the uniform-acceleration phase horizontal acceleration is 0. Since average horizontal velocity is zero, the initial and final velocity are equal to the average velocity.*@
After the instant of impact with the floor, can we
expect that the ball will be uniformly accelerated?
No
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Why does this analysis stop at the instant of impact
with the floor?
The floor has acted upon the ball, interupting its
acceleration and travel with an additional force. This
force will cause the ball to bounce, it will continue to
move in the general direction it was traveling but part
of its velocity and its acceleration will have been
dissipated by the impact with the floor.
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*#&!*#&!
&#This looks good. See my notes. Let me know if you have any questions. &#