#$&*
phy121
Your 'cq_1_22.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_22.2_labelMessages **
A 70 gram ball rolls off the edge of a table and falls freely to the floor 122 cm below. While in free fall it moves 40 cm in the horizontal direction. At the instant it leaves the edge it is moving only in the horizontal direction. In the vertical direction, at this instant it is moving neither up nor down so its vertical velocity is zero. For the interval of free fall:
• What are its final velocity in the vertical direction and its average velocity in the horizontal direction?
answer/question/discussion: ->->->->->->->->->->->-> :
122cm= 1.22m
Y:
Vf= sqrt(0^2+2*9.8m/s^2*1.22m)
Vf= 4.89m/s
vAve= 4.89m/s /2
vAve= 2.45m/s
‘dt= 1.22m/2.45m/s
‘dt= 0.5s
X:
40cm= 0.4m
V= ‘ds/’dt
V= 0.4m/0.5s
V= 0.8m/s
#$&*
• 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: ->->->->->->->->->->->-> :
70g= 0.07kg
Horizontal:
0.07cos(0.8)= 0.07m/s
0.07sin(0.8)= 9.77m/s
@&
You would take a sine or cosine of an angle, but you wouldn't take a sine or cosine of a velocity.
*@
@&
The velocity at impact will have components of 4.89 m/s in the negative y direction and .8 m/s in the x direction.
What are the magnitude and angle of a vector having these components?
*@
Vertical:
0.07cos(2.45)= 0.07m/s
0.07sin(2.45)= 0.003m/s
Not sure what to do from here, or even if im doing it properly or in the right direction.
#$&*
• What are its speed and direction of motion at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
c= sqrt(-4.89m/s^2+0.8m/s^2)
c= 4.96m/s
arctan(-4.89/0.8)= -80.7 degrees before the horizontal or 279.3 degrees above the horizontal.
@&
Good. This answers my previous question.
A little clarification:
The angle would be -80.7 degrees, which is 80.7 degrees clockwise from the positive x axis, or equivalently 279.3 degrees as measured counterclockwise from that axis.
*@
#$&*
• What is its kinetic energy at this instant?
answer/question/discussion: ->->->->->->->->->->->-> :
KE= 0.5*m*v^2
KE= 0.5*0.07kg*4.96m/s^2
KE= 0.86J
@&
KE= 0.5*0.07kg*(4.96m/s)^2
This is what you calculated, but it's not what you actually said when you wrote
KE= 0.5*0.07kg*4.96m/s^2.
*@
#$&*
• What was its kinetic energy as it left the tabletop?
answer/question/discussion: ->->->->->->->->->->->-> :
KE= 0.5*0.07kg*0.8m/s^2
KE= 0.02J
#$&*
• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion: ->->->->->->->->->->->-> :
gPE= m*g*y
gPE= 0.07kg*-9.8m/s^2*1.22m
gPE= -0.84J
#$&*
• How are the the initial KE, the final KE and the change in PE related?
answer/question/discussion: ->->->->->->->->->->->-> :
They are related by to one another by W= ‘dKE+’dPE, which means they are directly related.
W= (0.86-0.02)+(-0.84)= 0
#$&*
• How much of the final KE is in the horizontal direction and how much in the vertical?
answer/question/discussion: ->->->->->->->->->->->-> :
The final KE is almost in the vertical direction as shown above.
#$&*
** **
30 minutes
** **
@&
Very good, but do check my notes.
*@