phy201
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.
** **
seed 22.2
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: ->->->->->->->->->->->-> :
m= 70g = .07kg
'ds= 122cm= 1.22m
v0= 0
a= 9.8m/s^2
40cm= .04m (horizontally)
Vf^2= v0^2 + 2a 'ds
vf^2= 0 + 2(9.8m/s^2) 1.22m
vf^2= 23.912m/s
vf= 4.89m/s (vertically)
'ds= vave * 'dt
1.22m= (4.89m/s/2) 'dt
1.22m= = 2.44m/s 'dt
'dt= .5s (vertically)
HOrizontally
vave= 'ds/'dt
= .04/.5= .08m/s
40 cm is .4 m, not .04 m; otherwise OK
#$&*
Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
HOrizontal = .08m/s , .04m/s
x= .04m cos(.08m/s) = .039m/s
.08 m/s is a velocity, not an angle
vertical= 4.89m/s 1.22m
y= 1.22m sin( 4.89m/s) = .104m/s
#$&*
What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
magnitude = sqrt(.039^2 + .104^2)
= sqrt.001521 + .010816
= sqrt.012337
= .111m/s
direction = tan-1 .104m/s/.039m/s= 69.44 degrees
#$&*
What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
Ke= .5m v^2
= .5(.07kg)(.111m/s)^2
= .00043J
#$&*
What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
kE= .5(.07kg) (.08m/s)^2
KE= .000224J
#$&*
What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
Fnet= m *a
fnet= 9.8m/s^2 .07
nfet= .686N
'dw= fnet * 'ds
'dw= .686N * 1.22m = .837J
KE= .837J
PE= -.837J
#$&*
How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
.00043J + .000224J - .686J= -.685
#$&*
How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
Vertical= .00043J- .686J= -.686J
horizontal= .000224J
#$&*
** **
About an hour
** **
&#Please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified. &#
phy201
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.
** **
seed 22.2
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: ->->->->->->->->->->->-> :
m= 70g = .07kg
'ds= 122cm= 1.22m
v0= 0
a= 9.8m/s^2
40cm= .04m (horizontally)
Vf^2= v0^2 + 2a 'ds
vf^2= 0 + 2(9.8m/s^2) 1.22m
vf^2= 23.912m/s
vf= 4.89m/s (vertically)
'ds= vave * 'dt
1.22m= (4.89m/s/2) 'dt
1.22m= = 2.44m/s 'dt
'dt= .5s (vertically)
HOrizontally
vave= 'ds/'dt
= .04/.5= .08m/s
40 cm is .4 m, not .04 m; otherwise OK
#$&*
Assuming zero acceleration in the horizontal direction, what are the vertical and horizontal components of its velocity the instant before striking the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
HOrizontal = .08m/s , .04m/s
x= .04m cos(.08m/s) = .039m/s
.08 m/s is a velocity, not an angle
vertical= 4.89m/s 1.22m
y= 1.22m sin( 4.89m/s) = .104m/s
#$&*
What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
magnitude = sqrt(.039^2 + .104^2)
= sqrt.001521 + .010816
= sqrt.012337
= .111m/s
direction = tan-1 .104m/s/.039m/s= 69.44 degrees
#$&*
What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
Ke= .5m v^2
= .5(.07kg)(.111m/s)^2
= .00043J
#$&*
What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
kE= .5(.07kg) (.08m/s)^2
KE= .000224J
#$&*
What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
Fnet= m *a
fnet= 9.8m/s^2 .07
nfet= .686N
'dw= fnet * 'ds
'dw= .686N * 1.22m = .837J
KE= .837J
PE= -.837J
#$&*
How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
.00043J + .000224J - .686J= -.685
#$&*
How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
Vertical= .00043J- .686J= -.686J
horizontal= .000224J
#$&*
** **
About an hour
** **
&#Please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified. &#