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Phy 241
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?
initial velocity in vertical direction is 20 cm/s downward
displacement in the vertical direction is 120 cm
acceleration in the vertical direction is 981 cm/s^2
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What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
vf^2 = v0^2 + 2 a `ds
vf = sqrt[(20cm/s)^2 + 2*981cm/s*120cm]
= sqrt[(400cm^2/s^2) + 235440)
= 485.6 cm/s
`ds = 120 cm
'dv = 485.6 cm/s - 20 cm/s = 465.6 cm/s
v_ave = (485.6 cm/s + 20 cm/s) / 2
= 252.8 cm/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?
acceleration in horizontal direction = 0 cm/s^2
initial velocity in horizontal direction = 80 cm/s
change in clock time is the same for horizontal as is for vertical direction
`ds = (v0 + vf) / 2 * `dt
'dt = 2'ds / (v0 + vf)
= 2*120cm / (20cm/s + 485.6 cm/s)
= 0.475 s
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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 in horizontal direction
v0 `dt + .5 a `dt^2
`ds = 80 cm/s * 0.475 s + 0.5*(0 cm/s^2)*dt^2
= 38 cm
final velocity in horizontal direction
vf = v0 + a * `dt
= 80 cm/s + (0 cm/s^2)*0.475s
= 80 cm/s
average velocity in horizontal direction
v_ave = 80 cm/s
change in velocity in horizontal direction
'dv = 0 cm/s
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No. because other forces come into effect. As the ball drops, only gravitational forces act on the ball. Once the ball impacts the floor, the force of deflection from
the floor and the force of gravity immediately after impact once the ball bounces back up, come into effect.
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Why does this analysis stop at the instant of impact with the floor?
See above question. I am assuming that constant acceleration is due to only one force. Gravity.
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&#Very good responses. Let me know if you have questions. &#