Phy 231
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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?
• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
• What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
• Why does this analysis stop at the instant of impact with the floor?
VERTICAL DIRECTION
For the interval between the end of the ramp and the floor, v0 = 20 cm/s, d = 120 cm, and a = 980cm/s^2.
vf = 485.39 cm/s , d = 120 cm, dv = 465.39 cm/s, vAve = 252.7 cm/s
HORIZONTAL DIRECTION
a = 0 cm/s^2
v0 = 80 cm/s
t = 0.515, based on the equation d = v0*t + 1/2 * a * t^2
D = vt = 80*0.525 = 41.2 cm
vf = 80 cm/s, vAve = 80cm/s, and dv = 0 m/s since it has no horizontal acceleration.
AFTER IMPACT
After impact with the floor, the acceleration will change drastically because it will come to a sudden stop. The analysis stops at the instant of impact with the floor, because after the impact, the trends of the velocity/acceleration will not hold true. During impact, the velocity becomes 0, and the acceleration becomes 0.
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25 min
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Very good work. You might pick up a little more information by checking the link below, but that's just FYI; the part about needing a revision doesn't apply.
&#Please compare your solutions with the expanded discussion at the link
Solution
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