Phy 201
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A 70 gram ball rolls off the edge of a table and falls freely to the floor 122 cm below. While in free fall it moves 40 cm
in the horizontal direction. At the instant it leaves the edge it is moving only in the horizontal direction. In the
vertical direction, at this instant it is moving neither up nor down so its vertical velocity is zero. For the interval of
free fall:
What are its final velocity in the vertical direction and its average velocity in the horizontal direction?
answer/question/discussion:
We use vf^2=v0^2+2a`ds to find the vf,vf=sqrt(2(980cm/s^2)*122cm,vf=489cm/s. We now use vf=v0+a`dt to find the time which is
`dt=(489cm/s-0cm/s)980cm/s^2,`dt=.5s. We now take the `ds/`dt=vAve, vAve=`40cm/.5s,vAve=80cm/s
Since horizontal velocity is constant the vAve=80cm/s and vertical vf=489cm/s
Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the
instant before striking the floor?
answer/question/discussion:
y componentt=489cm/s and the x component=80cm/s
What are its speed and direction of motion at this instant?
answer/question/discussion:
We use the pythag. theorem to find speed wich is spd=sqrt((489cm/s)^2+(80cm/s)^2),spd=496cm/s. We now take the arctan(y/x)
which is tan^-1(489cm/s/80cm/s)=81deg=direction
What is its kinetic energy at this instant?
answer/question/discussion:
KE=.5mv^2 so we have the speed of the ball at 4.96m/sand mass at .07kg we get KE=.5(.07kg)(4.96m/s)^2,KE=.86joules
What was its kinetic energy as it left the tabletop?
answer/question/discussion:
-.86joules
KE cannot be negative. You might be saying that the difference in the kinetic energies is -.86 J, but the question asked for kinetic energy, not the change or difference in kinetic energy. (The difference in KE is close to the value give, but see my next comment).
What was the speed of the ball as it left the tabletop?
What therefore was its kinetic energy as it left the tabletop?
What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion:
Since PEgrav=mgh we can say PEgrav=.07kg*9.8m/s^2*1.22m,PEgrav= .84joules
From the tabletop to the floor the force exerted on the ball by gravity is in the same direction as the displacement. The work done on the ball by gravity is therefore positive. The work done by the ball against gravity is therefore negative.
The change in PE is equal to the work done by the object against conservative forces, so that in this case the change in gravitational potential energy is negative.
This -.84 J change in gravitational potential energy is equal to the change in the KE (from about .02 J to about .6 J).
How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion:
`dke is equal and opposite to `dpe.
How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion:
The vertical KE would be .84joules since the `dke is = and opposite of the `dpe and the horizontal KE would be the remaining
.02joules of the total .86joules of KE
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25min
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