cq_1_222

Phy 231

Your 'cq_1_22.2' 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 70 gram ball rolls off the edge of a table and falls freely to the floor 122 cm below. While in free fall it moves 40 cm in the horizontal direction. At the instant it leaves the edge it is moving only in the horizontal direction. In the vertical direction, at this instant it is moving neither up nor down so its vertical velocity is zero. For the interval of free fall:

• What are its final velocity in the vertical direction and its average velocity in the horizontal direction?

answer/question/discussion:

Answer: Starting just using vertical: a=9.8m/s/s. ds=.122m, v0=0m/s. vf^2=v0^2+2ads = 0+2(9.8m/s/s)(.122m) = 2.4m^2/s^2, vf= 1.5m/s. ave vel=(v0+vf)/2 = (1.5m/s + 0m/s)/2 = .75m/s. .75m/s = .122m/dt, dt = .122m/.75m/s = .2s.

Good solution, but note that it should have been 1.22 m, not .122 m.

horizontal ave vel = ds/dt = .040m/.2s = .2m/s.

40 cm is .40 m so the result would be 2 m/s.

• Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?

answer/question/discussion:

Answer: sin(theta) = opp/hyp = .122m/.128m = .95. Where sin equals .95 is at 288deg past horizontal. Fy=1.5m/s*cos(288deg)=.46. Fx=.2m/s*sin(288deg)=-.2.

• What are its speed and direction of motion at this instant?

answer/question/discussion:

Answer: speed=sqrt((1.5m/s^2)+(.2m/s^2)) = 1.52m/s. Direction: 288degrees past horizontal.

• What is its kinetic energy at this instant?

answer/question/discussion:

Answer: KE=.5mv^2 = .5*.070kg*(1.52m/s)^2 = .08J.

• What was its kinetic energy as it left the tabletop?

answer/question/discussion:

Answer: KE=.5(.070kg)(.2m/s)=virtually zero.

• What is the change in its gravitational potential energy from the tabletop to the floor?

answer/question/discussion:

Answer: +.08J.

• How are the the initial KE, the final KE and the change in PE related?

answer/question/discussion:

Answer: The ball should lose potential energy as it falls because it this energy should be KE. The ball starts with a big PE and small KE, and finishes with a small PE and a big KE.

• How much of the final KE is in the horizontal direction and how much in the vertical?

answer/question/discussion:

Answer: 1.5m/s / .2m/s = 7.5. .008/7.5=.001J are in the horizontal direction. .007J in the vertical direction.

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20 minutes

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cq1_22_2 solution and discussion

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A 70 gram ball rolls off the edge of a table and falls freely to the floor 122

cm below. While in free fall it moves 40 cm in the horizontal direction. At the

instant it leaves the edge it is moving only in the horizontal direction. In the

vertical direction, at this instant it is moving neither up nor down so its

vertical velocity is zero. For the interval of free fall:


What are its final velocity in the vertical direction and its average velocity

in the horizontal direction? </p>

 



We analyze the vertical motion first:

The vertical velocity of the ball is initially zero, its vertical

displacement is 122 cm downward and its acceleration in the vertical

direction is 980 cm/s^2 in the downward direction.  Choosing the

downward direction as positive we therefore have v0 = 0, a = 980 cm/s^2 and

`ds = 122 cm.

We use the fourth equation of unif accel motion:

vf^2=v0^2+2a`ds so that

vf=+-sqrt(2(980cm/s^2)*122cm) = +- 489cm/s. The final velocity is

clearly downward, which is in the positive vertical direction, so we

discard the negative solution.

Using the first equation of motion:

vf=v0+a`dt so that

`dt = (vf - v0) / a = (489cm/s-0cm/s) / (980cm/s^2) =.5s.

Note that `dt could have also been reasoned out from the definition of

acceleration.

We now analyze the horizontal motion:

Acceleration in the horizontal direction is zero, so horizontal velocity

doesn't change. 

Thus, for the horizontal motion,

 

Assuming zero acceleration in the horizontal direction, what are the vertical

and horizontal components of its velocity the instant before striking the floor?

What are its speed and direction of motion at this instant?

The x and y components of the velocity relative to a right-hand coordinate

system in which y is vertically upward are respectively 80 cm/s and -490 cm/s. 

Thus the final velocity has magnitude 

final resultant velocity magnitude= sqrt( (80 cm/s)^2 + (-490 cm/s)^2 )

= 495 cm/s, approx.

and is directed at angle

theta = arcTan(-490 cm/s^2 / (80 cm/s^2) ) = -84 deg,

i.e., at angle 84 deg below horizontal.



What is its kinetic energy at this instant? </p>

</p>

answer/question/discussion:</p>

</p>

</p>

 

The .07 kg mass is moving at 495 cm/s, which is 4.95 m/s, so its KE is

</p>

</p>

</p>

What was its kinetic energy as it left the tabletop? </p>

</p>

answer/question/discussion:</p>

</p>

 

As it left the tabletop is was moving at 80 cm/s = .8 m/s so its KE was 

KE at tabletop = 1/2 * .07 kg * (.8 m/s)^2 = .02 Joules, approx.



<h3>




</h3>

</p>

</p>

What is the change in its gravitational potential energy from the tabletop to

the floor? </p>

 

From the tabletop to the floor the force exerted on the ball by gravity

is in the same direction as the displacement. The work done on the ball by

gravity is therefore positive. The work done by the ball against gravity is

therefore negative.

The change in PE is equal to the work done by the object against conservative

forces, so that in this case the change in gravitational potential energy is

negative.

This -.84 J change in gravitational potential energy is equal to the change in

the KE (from about .02 J to about .6 J).

Note how energy conversion works here.  From tabletop to floor, KE

increases from .02 J to .86 J while PE decreases by .84 J.  We have assumed

that the only force is the gravitational force, so that no force is exerted by

nonconservative forces, and thus `dW_NC_ON = 0. 

Since `dW_NC_ON = `dKE + `dPE, we expect that `dKE + `dPE = 0. 

This is easily verified: 

`dKE = .86 J - .02 J = .84 J, and

`dPE = -.84 J, so

`dKE + `dPE = +.84 J - .84 J = 0.