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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?
V0 = 20 cm/s
‘ds_v_ = 120 cm
A = 980 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 = [sqrt] 20^2 + 2(980)(120)
Vf = + - 485.4 cm/s
V f = 485.4 cm/s
‘ds_v_ = 120 cm
Change in Velocity = 465.4 cm/s
vAve = 485.4 + 20 cm/s / 2 = 252.7 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?
A = 0 cm/s^2
V0_x_ = 80 cm/s
‘dt = x = v0^2(t) + ½(a)(t^2)
‘dt = + - 0.5 s
‘dt = 0.5 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?
“dv = 80 cm/s as acceleration is constant at 0 cm/s^2
‘ds = 80 cm/s * 0.5 s = 40 cm
vAve = 80 cm/s
Vf = 80cm/s
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
Yes, because the horizontal motion is uniform motion.
That's only so while in free fall. The interaction of the ball with the floor is very likely to influence the horizontal acceleration during the time of contact between ball and floor. However the influence on horizontal motion might be small.
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• Why does this analysis stop at the instant of impact with the floor?
Because the floor is the impact of vertical motion, thus ending accelerated motion.
that ends the uniformly accelerated motion
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1 hour
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10/5 4 pm
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