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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.
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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, what
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^2 = v0^2 + 2a(y-y0)
vf^2 = 20^2 + 2* 980 *(120)
vf = 485.39 cm/s
ds = 120 cm
change in velocity = vf -v0
= 485.39 - 20 = 465.39 cm/s
v_Ave = (vf + v0)/2
v_Ave = (485.39+20)/2
v_Ave = 252.70 cm/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?
From the verticle find t
vf = v0 + a * t
485.39 = 20 +980 * t
t = 0.47 s
v0 = vf = 80 cm/s
So a = 0
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What therefore are its displacement, final velocity, average
velocity and change in velocity in the horizontal direction
during this interval?
vf = 80 m/s
x-x0 = 0.5 * (v0 +vf) * t
x-0 = 0.5 * (80 +80) * 0.47
x = 37.6 cm
v_Ave = 80 m/s
change in velocity = 0
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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?
At that instant the ball will experience more forces that will
change it trajectory and speed.
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*#&!
&#Your work looks very good. Let me know if you have any questions. &#