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phy121
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.
** CQ_1_13.1_labelMessages **
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?
Vf= 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+2*a*’ds)
Vf= sqrt( 20cm/s^2+2*980cm/s^2*120cm)
Vf= sqrt(235,600cm/s)
Vf= 485.39cm/s
‘ds= 120cm
‘dv= 485.39cm/s=20cm/s= 465.39cm/s
vAve= (485.39cm/s+20cm/s)/2= 252.7cm/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?
A= 0
V0= 80cm/s
‘dt= ‘ds/vAve= 120cm/252.7cm/s= 0.47s
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• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
‘ds= v0*’dt+0.5*a*’dt^2
‘ds= 80cm/s*0.47s+0.5*0*0.47s^2
‘ds= 37.6cm
Vf= 0cm/s
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No, once the ball impacts the floor, it will no longer be uniformly accelerating.
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• Why does this analysis stop at the instant of impact with the floor?
The analysis stops at the instant of impact because the ball is no longer free falling.
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More generally, the analysis assumed uniform acceleration, and the acceleration changes at the instant the ball hits the floor.
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&#Good work. See my notes and let me know if you have questions. &#