1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Distinction between Kinetic Energy and Momentum

  1. Mar 6, 2016 #1
    Hello, I would like to provide my thoughts on the distinction between KE and p and would like to know if this reasoning is sound and objective.
    I will give an example to start: A bowling ball is dropped from some height into a pool of honey. It is observed that the bowling ball has a velocity V that it falls into the honey with. The time it takes for the bowling ball to reach a velocity VF is observed to be the time t. The distance traveled in the honey before the ball reaches VF is d.

    What one would observe to be momentum is the force exerted by the honey onto the bowling ball, to slow it down to VF, during the time interval t.
    Written mathematically as: [tex] \vec {P} = \int \vec {F}{dt} = {m}\vec {v} = m \int \vec {a}{dt} [/tex]

    What one would observe to be the [change in] kinetic energy or work is the force exerted by the honey onto the bowling ball, to slow it down to VF, through the distance d.
    Written mathematically as: [tex] W = \Delta {KE} = \int \vec {F} \cdot {d} \vec {x} = m \int \vec {a} \cdot {d} \vec {x}[/tex]
    Newton did not distinguish between momentum and kinetic energy (leibniz did, and his views were not supported).
    I don't see why they didn't make a distinction (Newton and his contemporaries, that is). I suppose they did not regard distance as importantly as time since they held the view of absolute time?

    edit: If I were to change the velocity of the bowling ball to V+1, (assuming weight is the only force downward) would the time it took to reach VF be t+1? Consequently, would distance traveled then be d2?
     
    Last edited: Mar 6, 2016
  2. jcsd
  3. Mar 6, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi
    Can you provide a reference ? It seems to me the ball should drop to the floor of the pool and only then stop.

    What is your question ? There is no contradiction between ##W = \displaystyle \int \vec F \cdot d\vec s## and ##\vec p = \displaystyle \int \vec F dt##

    No, a momentum is not a force. Same with work. The presence of time c.q. distance is essential.
     
  4. Mar 6, 2016 #3
    Oh you're right. I'll fix the example give me a minute.

    If you read the rest of that statement it is the force over the time period t and the integral yields momentum.
    I'm not proving any contradictions. I'm merely proving the distinction between the two ideas. I am looking for the approval of this reasoning so that I can move on in my textbook.
     
  5. Mar 6, 2016 #4
    Alright I fixed it.

    The bowling ball would only stop if the buoyant force of the honey was equal to the bowling ball's weight, right? Then the net displacement and time it took to reach zero velocity would be the description I had earlier. I changed it to a specific velocity VF since I want to keep the example as simple as possible.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Distinction between Kinetic Energy and Momentum
Loading...