Seed Question 13-1

course Phy 201

This seed question was posted but wasn't graded of commented on

I am confused about the last couple of questions. It tells us the horizontal velocity is 80cm/s. I don't understand, is that initial or average, also how do I find the horizontal acceleration?

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=20cm/s, a = 980cm/s, `ds = 120cm,

• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?

using vf^2 = v0^2 + 2a`ds

vf^2 = (20cm/s)^2 + 2*980cm/s*120cm

vf^2 = 235600cm^2/s^2

vf = 485cm/s

`dv = vf – v0 = 485cm/s – 20cm/s = 465cm/s

vAve = (vf+v0)/2 = 505cm/s /2 = 252.5cm/s

• What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?

v0=80cm/s, `dt = `dv/a = .47s

The only force acting in the horizontal direction is air resistance. At this speed the air resistance isn't high enough to significantly accelerate a steel ball, so its acceleration in the horizontal direction is effectively zero. This means its horizontal velocity can be regarded as constant.

• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?

`ds = .47s*80cm/s = 37.6cm

• 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?

Once the ball encounters the floor, various nongravitational forces act on it and there is no reason to expect that its acceleration will remain uniform.

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