Phy231
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=20cm/s 'ds=120cm 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=+-'sqrt(v0^2+2a'ds)=+-'sqrt((20cm/s)^2+2*980cm/s^2*120cm)=+-(400cm^2s^2+235200cm^2/s^2)=+-'sqrt(235600cm^2/s^2)=485.386cm/s
'dv=(vf-v0)=485.386cm/s-20cm/s=465.386cm/s vAve=(20cm/s+485.386cm/s)=252.693cm/s
'dt=vAve/'ds=252.693cm/s/120cm= 2.106s
252.693cm/s/120cm= 2.106 s^-1, not 2.106 s.
This calculation gives you the reciprocal of the time interval, not the time interval.
If you're careful with your units calculations, you'll be much more likely to catch errors of this type.
You could also drop something from a height of a little over a meter, and see clearly that it reaches the floor in less than 2 seconds. Of course for much greater heights this isn't feasible.
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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?
v0=80cm/s 'dt=2.106s a=0cm/s^2
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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=v0+.5a'dt=80cm/s+0=80cm/s vAve=80cm/s 'dv=80cm/s-80cm/s=0cm/s
'ds=vAve*'dt=80cm/s*2.106s=168.48cm
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
Yes because it will not acceleration in either direction. The motion will be over.
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Why does this analysis stop at the instant of impact with the floor?
Because the ball will not travel anymore using these terms.
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20 min
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5/1/10 2pm
Your calculation of the time interval was incorrect, but all the steps leading up to that point, and all the steps after were correct (except of course that some of the steps after used the incorrect time interval).
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