Phy 201
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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?
The intial velocity is 20cm/s,and the displacement is 120cm and acceleration is 980cm/s^2
What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
First we need to find `dt. We use equation `ds=v0`dt+.5a`dt^2. We rearrange to find `dt=sqrt(((2`ds)/a)/vo)). We
The solution to this equation is not `dt=sqrt(((2`ds)/a)/vo))/ Among other things, the units of this solution do not match the units of `dt. That is, the units of sqrt(((2`ds)/a)/vo)) are not seconds.
The equation is quadratic in `dt and to solve that you would either need to complete the square or use the quadratic formula; the latter is easier, though the former demonstrates why the quadratic formula works.
Alternatively, you could use the fourth equation of motion to find vf, then use this information to find `dt. This would allow you to bypass the quadratic formula.
have`dt=sqrt(((2*120cm)/980cm/s^2)/20m/s)),`dt=sqrt((240cm/980cm/s^2)/20m/s),`dt=sqrt(.245/20m/s),`dt=.11s.We now know `dt we
As mentioned before, those units don't work out.
can find the rest.We now find vf=v0+a`dt,vf=20m/s+980cm/s^2*.11s,vf=127.8cm/s. `dv=vf-vo,`dv=127.8cm/s-
20m/s,`dv=107.8cm/s.vAve=(vf+v0)/2,vf=(127.8cm/s+20cm/s)/2,vAve=73.9cm/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?
Intial velocity is 80cm/s and the `dt is .11s and acceleration is 980cm/s^2
What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction
during this interval?
vf=v0+a`dt,vf=80cm/s+980cm/s^2*.11s,vf=187.8cm/s. `dv=187.8cm/s-80cm/s=107.8cm/s.vAve=(187.8cm/s+80cm/s)/2,vAve=133.9
980 cm/s^2 is the acceleration in the vertical direction and does not affect the horizontal direction.
The horizontal and vertical motion do share the time interval, which however was miscalculated in your work.
After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
no
Why does this analysis stop at the instant of impact with the floor?
Because the ball changes direction and we have no other information
The key reason is that the acceleration of the ball changes, so our assumption of uniform acceleration must and at this instant.
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30min
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You have made some errors in this analysis. I don't think you'll have any trouble correcting them.
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