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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, what are the ball's initial velocity, displacement and acceleration in the vertical direction?
initial velocity: 20cm/s
displacement : 120 cm
acceleration in the vertical direction: 120cm / 20cm/s = 6s
20cm/s / 6s =3.33cm/s^2
@& The 20 cm/s velocity is the initial vertical velocity. The ball falls in the vertical direction, and 980 cm/s^2 is the vertical acceleration of the falling ball. The ball speeds up, it doesn't continue moving at 20 cm/s. So is takes much less than 6 seconds to fall. If you drop a ball from 120 cm (about 4 feet) it takes less than a second to fall.
What you know is the initial vertical velocity, the vertical displacement and the vertical acceleration. How do you find vf and `dt when you know `ds, a and v0?
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• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
Final velocity: vf= v0 + a’ dt
Vf = 20cm/s +3.33cm/s^2 (120cm) = 419. 6 this doesn’t make sense
@& 120 cm is not `dt. 120 cm is a displacement and `dt is a time interval.*@
Displacement: 120cm
Change in velocity = acceleration? 3.33cm/s^2
Average velocity in vertical direction = 20cm/s
@& This is the initial vertical velocity, not the average vertical velocity.*@
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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?
Ball’s acceleration 3.33cm/s^2
Initial velocity - 0 cm/s
6 seconds
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• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
Displacement: 120 cm
Final velocity: 80cm/s
Average velocity: (80cm/s + 20cm/s) /2 = 50 cm/s
Change in velocity in horizontal direction
@& There is no horizontal force on the falling ball and hence zero acceleration in the horizontal direction.
80 cm/s is the velocity at the beginning of the interval.
20 cm/s is not a horizontal velocity; 20 cm/s is the initial velocity in the vertical direction and does not contribute to the horizontal motion.
The time of fall is the same for vertical and horizontal motion; however the 6 second result needs to be modified.*@
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No because the ball has stopped
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• Why does this analysis stop at the instant of impact with the floor?
Because there is a force, the group which is stopping it
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20min.
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I am very confused. Where do I start to solve this problem...
self-critique rating
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