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Phy 121
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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 ball’s initial vertical velocity is -20cm/s because the ball is moving downward. The ball’s displacement in the vertical direction is -120cm and the acceleration in the vertical direction is -980cm/s^2 because of gravity.
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• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
The ball’s final vertical velocity is 0cm/s because it is hitting the floor below. The vertical displacement is -120cm. The change in velocity in the vertical direction is 20cm/s because 0cm/s- -20cm/s = 20cm/s. The average velocity in the vertical direction is -10cm/s.
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&&&&-484.95 is the ball’s final velocity because the ball is traveling downward.
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The ball's final velocity will not be zero. The analysis ends at first contact with the floor, since at that instant acceleration ceases to be constant.
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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?
The ball’s initial horizontal velocity is 80cm/s. The ball’s change in clock time during this interval is as follows:
vf = v0+a*’dt
0cm/s = 20cm/s+980cm/s^2*’dt
-20cm/s=980cm/s^2*’dt
.02s = ‘dt
Given that the change in time is .02s and with an initial horizontal velocity of 80cm/s and final velocity of 0cm/s because the ball hits the ground, we can now find the acceleration in the horizontal direction to be -4000cm/s^2 because:
vf = v0+a*’dt
0cm/s = 80cm/s+a*.02s
-80cm/s = a*.02s
-4000cm/s^2 = a
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&&&&
The time the ball takes to fall to the ground is .48s because the average velocity for the vertical direction is 250cm/s and the displacement is 120cm, so 120cm/250cm/s = .48s.
Given that the clock time is .48s we can now find the displacement as follows:
‘ds = vAve*’dt
‘ds = 80cm/s * .48s
‘ds = 48.4cm
The acceleration in the horizontal direction is 0cm/s.
• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
The final horizontal velocity is 0cm/s because it is hitting the ground. The average velocity is 40cm/s because 0cm/s+80cm/s=80cm/s and 80cm/s / 2 = 40cm/s. The change in velocity is -80cm/s because 0cm/s-80cm/s=-80cm/s. The displacement is as follows:
‘ds = (vf+v0)/2*’dt
‘ds = (0cm/s+80cm/s)/2*.02s
‘ds = 80cm/s / 2 * .02s
‘ds = 40cm/s*.02s
‘ds = .8cm in the horizontal direction.
&&&&The displacement is as follows:
Given that the clock time is .48s we can now find the displacement as follows:
‘ds = vAve*’dt
‘ds = 80cm/s * .48s
‘ds = 48.4cm
The final velocity is also 80cm/s therefore the change in velocity is 0cm/s and the average velocity is 40cm/s. &&&&
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No we cannot expect that the ball will be uniformly accelerated.
&&&&As soon as the ball hits the floor the gravitational force is no longer the only force acting upon them. &&&&
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• Why does this analysis stop at the instant of impact with the floor?
This analysis stops the instant of impact because we do not have the required knowledge to further this analysis.
&&&&This analysis stops at impact because at that moment the acceleration is no longer constant. &&&&
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*#&!
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Good.
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