How do we get to the concept of kinetic energy?

[Robin04]

Hi,

I'm reading a high school textbook about mechanics. It's amazing how the author draws up the problems and solves them by introducing a new consistent concept.

Now I'm reading about collisions. He writes the conservation of momentum but the problem is that we have two unknowns in one equation so another one is needed. He defines beautifully the concept of elasticity by analyzing the bouncing of a ball dropped from a certain height (with geometric sequence) thus solving the problem, we have the second equation. With some simple math he changes it to a form which is very close to the conservation of kinetic energy (only the 1/2-s are missing) then he states that by taking the half of mv^2, so 1/2 mv^2 we get a new notion which is the kinetic energy. I don't really get this part, I feel he's missing something or there's another way to get to the kinetic energy. He doesn't say anything about why we have to multiply the terms (that we got from a very logical thought) with 1/2.

Thanks for you help! :)

 

[Opuscroakus]

Okay, I think I know where you're wanting to head, but I'm going to need more clarification before I answer in full.

First, a few things.

These aren't mere notions, they're proven theories, so proper terminology will help. :)

Are you simply asking how he's able to go from Conservation of Momentum to the Conservation of Energy? Or is it more complex and you're just not sure how to articulate that? It sounds like he's discussing the Conservation of Energy in an Elastic Collision. Elastic collisions conserve kinetic energy as well as momentumIt would really help if you could at least give us the equations you're seeing.

Conservation of Momentum (Elastic collision): m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

Conservation of Energy (Elastic Collision): ½m₁(v_1i)² +
½m₂(v_2i)² = ½m₁(v_1f)² + ½m₂(v_2f)²

It's possible he may be doing just a straight substitution, but I have no way of knowing until you provide further information. Do these equations look familiar, or close to what you're seeing?

 

[jtbell]

there's another way to get to the kinetic energy.

There is, by associating the change in an object's kinetic energy with the work done on it by an external force, via the work-energy theorem.

If you require that W = KE_final - KE_initial, you get the 1/2 in the KE equation automatically.

http://faculty.wwu.edu/vawter/physicsnet/topics/Work/WorkEngergyTheorem.html

Every uinversity-level intro physics textbook covers this; don't know about high school textbooks. It's been a long time since I used one of those.

 

[Opuscroakus]

Yep--that's exactly what I was waiting for her to confirm with a new reply before showing her the step-by-step derivation from the Definite Integral, but needed to make *sure* that's where she was headed.

Eh, well. NO need for it now.

 

[Robin04]

Thank you for your answers, Opuscroakus and jtbell! :)

 

[Qwertywerty]

Every uinversity-level intro physics textbook covers this; don't know about high school textbooks. It's been a long time since I used one of those.

Many do , if not all , I believe .