Phy 201
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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:
ds=v0dt+.5adt so dt=sqrt(2ds/a)=sqrt(2*1.22m/9.8m/s^2)=.50s
dvx=ds/dt=.40m/.50s=.8m/s
vfx=dv+v0=.8m/s+0=.8m/s
vxavg=vf+v0/2=.8m/s+0/2=.4m/s
dvy=9.8m/s^2*.50s=4.9m/s
vfy=4.9m/s+0=4.9m/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:
horizontal=.8m/s
vertical= 4.9m/s
• What are its speed and direction of motion at this instant?
answer/question/discussion:
v=sqrt(40^2+122^2)=128 m/s
40 cm/s is a velocity; 122 cm is not a velocity and there is no 122 cm/s velocity.
The units of the (40 cm/s)^2 would be cm^2 / s^2, and the square root would result in units of cm/s, not m/s.
arctan(4.9/.8)=80.7 degrees
• What is its kinetic energy at this instant?
answer/question/discussion:
KE=.5mv^2=.5*70kg*(128m/s)^2=573440J
Your calculation is correct for mass 70 kg moving at 128 m/s. However the mass is 70 grrams, not 70 kg. And the final velocity is about 5 m/s , not 128 m/s.
• What was its kinetic energy as it left the tabletop?
answer/question/discussion:
KE=.5mv^2=.5*70kg*(0m/s)^2=0J
• What is the change in its gravitational potential energy from the tabletop to the floor?
answer/question/discussion:
initial gpe=mgh=70kg*9.8m/s^2*1.22m= 836.92
final gpe=mgh=70kg*9.8m/s^2*0m= 0
change in gpe=-836.92
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1 hour
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You used the right ideas but your numbers and units don't work out. No need to submit anything if you understand everything, but check out the link:
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#