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.
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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 , vf = 0 , `ds = 120cm, `dv = -20cm/s, `dt = `ds/vAve = 120cm / -10cm/s = 12s and a =`dv/`dt = -10cm/s / 12s = .83cm/s^2
You assumed final vertical velocity zero; with this assumption your analysis would be correct, but this assumption is not compatible with uniformity of acceleration.
During its uniform acceleration to the floor the ball attains a downward vertical velocity of nearly 500 cm/s. Upon contact with the floor acceleration becomes nonuniform and is not included in the analysis.
What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction? vf = 0, `ds = 120cm, vAve = 20cm/s +0 /2 = 10cm/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 = 80cm/s , vAve = 80cm/s / 2 = 40cm/s, `dt = 120cm*40cm/s = 4.8s
The final velocity in the horizontal direction, for the uniform-acceleration interval being analyzed, is not zero.
What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval? vf = 0, vAve = 40cm/s, `dv = 80cm/s, a = `dv/`dt = 80cm/s/4.8s = 16.7cm/s^2
After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated? yes
Why does this analysis stop at the instant of impact with the floor? Velocity reaches 0
velocity might or might not reach zero soon after impact (vertical velocity probably will, horizontal velocity not necessarily), but at the instant of impact with the floor, it has not done so
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25min
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cq1_13_1 solution and discussion
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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?
What therefore are its final velocity, displacement, change in velocity and
average velocity in the vertical direction?
The vertical motion is independent of the horizontal.
The vertical motion is characterized by a downward acceleration of 980
cm/s^2, a downward displacement of 120 cm and an initial downward velocity of 20
cm/s. Thus for the vertical motion, if we choose the downward direction to
be positive, we have
v0 = 20 cm/s, a = 980 cm/s^2 and `ds = 120 cm.
We cannot reason anything directly from this information.
Both the third and fourth equations of uniformly accelerated motion
include the three variables v0, a and `ds. So the third equation could
be solved for `dt, or the fourth for vf.
It is not recommended that students without a strong mathematical
background solve the third equation for `dt; the equation is quadratic in `dt
and the solution is algebraically challenging. University Physics
students must have a strong mathematical background and should be able to
solve the quadratic and interpret the solutions.
So we use the fourth equation vf^2 = v0^2 + 2 a `ds. We solve this
equation for v0, obtaining
v0 = + - sqrt(v0^2 + 2 a `ds).
Substituting our values for v0, a and `ds we have
vf = +- sqrt( (20 cm/s)^2 + 2 * 980 cm/s^2 * 120 cm) = +- 480 cm/s.
The final velocity is clearly in the downward direction, so we discard
the solution -480 cm/s and accept the solution + 480 cm/s.
Initial velocity v0 and final velocity vf, with uniform acceleration,
imply average velocity
vAve = (v0 + vf) / 2 = (20 cm/s + 480 cm/s) / 2 = 250 cm/s
so that the time required for the fall is
`dt = `ds / vAve = 120 cm / (250 cm/s) = .48 sec.
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 horizontal motion of an ideal projectile is characterized by zero
acceleration (since the only force acting on an ideal projectile there the
vertical gravitational force, it experiences no force in the horizontal
direction, implying zero acceleration in that direction).
The initial horizontal velocity is 80 cm/s, and since acceleration is zero
this velocity remains unchanged. Thus the average horizontal velocity is
also 80 cm/s and we conclude that the horizontal displacement is therefore
-
`ds = vAve * `dt = 80 cm/s * .48 sec = 48.4 cm.
After the instant of impact with the floor, can we expect that the ball will
be uniformly accelerated?
Upon impact with the floor the ball and the floor both begin compressing,
exerting equal and opposite forces on one another.
-
The constant gravitational force is, starting at that instant, not the
only force acting on the ball.
With increasing compression the force between the floor and the ball will
increase.
-
The net force on the ball therefore changes with time, so that the
acceleration of the ball changes with time and is no longer constant.
Why does this analysis stop at the instant of impact with the floor?
This analysis assumes uniform acceleration, and acceleration is no longer
uniform.