060226_notes15questions

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00:10:39 Why do we expect the velocity attained by a ball on a ramp to be proportional to the square root of the vertical position change of the ball?

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RESPONSE --> PE=KE the solve for v v=srt(2*g*'dy)

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00:10:47 If the object was released from rest and and allowed to fall freely through a downward distance `dy equal to the vertical distance traveled on the ramp, its gravitational potential energy will convert to kinetic energy. In this case, setting the potential energy decrease equal to the kinetic energy increase (i.e., `dKE = -`dPE) gives.5 m v^2 = m g `dy. We solve to obtain v = `sqrt(2 g `dy). This demonstrates that v is proportional to`sqrt(`dy).

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RESPONSE --> right

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00:12:29 Why do we expected distance traveled by a ball after being projected horizontally off of a ramp and falling a fixed distance to be proportional to the velocity with which the ball leaves the ramp?

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RESPONSE --> not sure

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00:16:23 If v^2 is proportional to �dy, then for some constant k we have v^2 = k * dy. From experimental results it appears that �dx^2 is also proportional to �dy, so that �dx is emprircally proportional to �sqrt(�dy). Since `dy is proportional to v^2, it follows that sqrt(`dy) is proportional to v, and we finally conclude that �dx is proportional to the velocity v with which the ball is projected horizontally from the ramp.

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RESPONSE --> so v^2 = 'dy which would make since that the srt of 'dy would equal v

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00:17:51 Why is the distance traveled by the projectile in this situation less than that predicted from the velocity v, where v is determined by setting 1/2 m v^2 equal to the potential energy loss on the ramp? Why is the distance still less even if frictional losses are taken into consideration?

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RESPONSE --> since it is on a ramp it is gaining KE

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00:18:55 This situation is due mostly to the fact that in addition to attaining a velocity v the ball also rolls down the ramp and therefore gains kinetic energy associated with its rotational motion. As a ball rolls (without slipping) down a smooth ramp, it will turn out that 2/7 of its PE loss converted to rotational KE. This means that the KE associated with its velocity v on the ramp, which is called translational kinetic energy, theoretically, should only be 5/7 of the PE loss. Frictional losses further reduce v.

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RESPONSE --> 2/7 of its PE loss converted to rotational

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You've completed most of the assignment and your work is good. I'm not sure what might be going on with the runtime error, but if it doesn't occur again don't worry about it. If it does, and if the error gives you a more specific message, try to make note of it and let me know what it says.

Let me know if you have questions.