SCIENCE HELP ASAP

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zmcarry
Posts: 1
Joined: Tue Nov 09, 2010 5:52 pm
Occupation: Student: 9th grade
Project Question: Does the initial starting point's height, on a rollercoaster track, affect the distance a car will travel around a loop at the bottom of the track?
Project Due Date: ASAP
Project Status: I am finished with my experiment and analyzing the data

SCIENCE HELP ASAP

Post by zmcarry »

I am doing a project entitled "Rise Vs Run". I was hoping that you could help me by answering a fw simple questions.
Question #1 : Would the initial starting point's height, affect the distance a marble will travel on a slope?
Question #2 : Does the starting point of a rollercoaster affect whether the car will ake it around a loopty-loop at the bottom of the slope?
Question #3 : Can you explain rise vs rn and the relationship between that and potential and kinetic energy?
deleted-71588
Former Expert
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Re: SCIENCE HELP ASAP

Post by deleted-71588 »

Answer 1: Maybe - change in height would be a change in potential energy, so it has the potential to effect the outcome; however, unless that potential energy difference is converted into kinetic energy, it won't change the result.
Answer 2: Maybe - same basic question #1 so same answer.
Answer 3: Yes, if and only if I make all sorts of assumptions not stated in your question.

True understanding is not obtained without effort. You need to do some research/reading on potential energy and kinetic energy and Newton's 3 laws of motion. There are some fundamental equations that come from F = m a (Force = mass times acceleration) and its derivaties and integrals that relate distance, velocity, acceleration, force, gravitational constant, gravity, etc. One of these is F = 0.5 * m * v * v which is commonly referred to as a momentum force.

If you have a straight line track, then rise vs run is simply a relationship between the change in height and the horizontal distance travelled along the track.
-Craig
deleted-71709
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Occupation: Engineer - Product & Technical Development Executive Director
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Re: SCIENCE HELP ASAP

Post by deleted-71709 »

You have asked several interesting questions here.

There are some ScienceBuddies experiments that should help you understand the concepts of potential and kinetic energy better. Try these links:

https://www.sciencebuddies.org/science- ... p036.shtml

https://www.sciencebuddies.org/science- ... p017.shtml

https://www.sciencebuddies.org/science- ... p071.shtml

Have fun with your experiments!
Ed Neu
Buffalo, MN
deleted-71360
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Re: SCIENCE HELP ASAP

Post by deleted-71360 »

Sorry Craig, you are off a little on the terminology.

The sentence "One of these is F = 0.5 * m * v * v which is commonly referred to as a momentum force."
That is the formula for kinetic energy, which is the same energy as the mass times the change in elevation for potential energy. In the absence of nuclear effects, the potential and kinetic energy transformations are alway bidirectional.

Momentum is simply mass times velocity. It too is conserved and the combination of momentum and energy will define the relative motions asd speeds of objects before and after collisions.

As to "momentum force", that is a new one for me, please explain. Are you referring to the force to accelerate an object and give it some momentum and energy?

Bob
deleted-71588
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Re: SCIENCE HELP ASAP

Post by deleted-71588 »

robertreavis wrote:Sorry Craig, you are off a little on the terminology.
Funny thing about terminology, it evolves, has evolved, and can even evolve such that what was old terminology is born again and becomes new terminology as a new generation things about something in an old and forgotten way. I've definitely dated myself and exposed my engineering perspective to terminology. Engineers working on modeling complex systems of forces often find it useful to classify (name) forces as to where they originate from:

momentum force (left side of equation) = kinetic energy (right side of equation)
F = 0.5 * m * v * v

Your right hand side perspective and terminology calling it kinetic energy is in total agreement with my left hand side terminology perspective of momentum force.
-Craig
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