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!

Conservation of energy in bouncing ball

  1. May 21, 2015 #1
    Throwing a ball which is (m)kg on the ground
    If the velocity of the ball just before it reaches the ground is (v)m/s
    Assume it is a elastic collision
    The velocity of the ball when it leaves the ground should be (-v)m/s
    According to conservation of momentum,the ground has a change of momentum = 2mv(pointing downward)
    Then the total KE in the system =
    The ball's KE+the ground's KE
    The total KE after the collision > The total KE before the collision

    Is there something I missed?
    And sorry for the grammar...
     
  2. jcsd
  3. May 21, 2015 #2
    Nope you are *almost correct* but the kinetic energy of the ground will be negligible (2mv)^2/2M where M is the mass of ground. In real world if we take into account the KE of the ground then the ball will not have -v velocity after the collision but just abit smaller so the difference in the kinetic energy of the ball give us the kinetic energy of the ground.
     
  4. May 21, 2015 #3
    Oh....thanks a lot
     
  5. May 21, 2015 #4
    Background:
    I want to define something, and use your example to answer your question. First, I assume the ground has 0 momentum just before the ball crashes on it, you are the observer standing on the ground. Neglect gravity effect because the whole process happens in very short time. Assume upward as positive.

    Analysis:
    When the ball crashes on the ground with momentum -mv. Then when it bounces upward, it will have mv momentum. However, remember one thing, you are an observer standing on the ground. What it means is that, the mv is actually mv'(v' is a vector pointing upward) if you are observer in definite stationary state. However, because the ground also moves downward, with momentum Mv''(v'' is a vector pointing downward), so, you also move downward with v'' if you are observer standing on the ground. The mv is actually related to relative velocity,
    v=v'-v''.

    Execution:
    Formula we have: v = v' - v'' -> v = v' + v'' ; Mv'' - mv' = mv + 0
    Evaluate (1/2)Mv''2 + (1/2)mv'2.
    Tricks in evaluation: (1/2)Mv''2 = (1/2)(Mv'')(v''). By Mv'' = mv' + mv and v'' = v - v',
    (1/2)(Mv'')(v'')
    = (1/2)(mv'+mv)(v-v')
    = (1/2)(mv'v-mv'2+mv2-mvv') = (1/2)(mv2) - (1/2)(mv'2)

    Back to original formula:

    (1/2)Mv''2 + (1/2)mv'2 = (1/2)mv2 -(1/2)mv'2 + (1/2)mv'2 = (1/2)mv2 (original kinetic energy of the falling ball)

    Conclusion: Energy is conserved!
     
    Last edited: May 21, 2015
  6. May 21, 2015 #5
    it is the problem of observer. the v will not decrease in reality even if the ground is low in mass.
     
  7. May 21, 2015 #6
    But then if we choose an observer in a stationary frame (not in the ground but in a frame where the ground is stationary before the collision) then for that observer the kinetic energy will not be conserved.
     
  8. May 21, 2015 #7
    I have already calculated that. for both observer, energy is conserved. please look at my post above
     
  9. May 21, 2015 #8
    To tell you the truth i ve a problem following your post, but for the observer i am saying, it will be

    Kinetic Energy before collision =1/2mv^2

    Kinetic Energy after collision=1/2mv^2+(2mv)^2/2M.

    As i see KE is not conserved for that observer . Where am i wrong?
     
  10. May 21, 2015 #9
    if you use the observer on the ground to think, you will conclude that after the ball crashes on the ground and bounces upward, there is no energy transferred into the ground because you see no change in velocity. If you use observer in definite stationary, you will get 1/2mv'^2 + 1/2Mv''^2 as kinetic energy after collision, which equals to 1/2mv^2. And this value is the same as kinetic energy before collision. The problem in your analysis is, you use observer on the ground to describe the energy of the ball, but you use observer in stationary state to describe the energy of ground, which makes problems. you can only use one observer for the whole event.
     
  11. May 21, 2015 #10

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    I don't think you are wrong, If the observer is non-inertial during the collision (because he is fixed to the accelerated Earth), kinetic energy isn't conserved in his rest frame.
     
  12. May 21, 2015 #11
    I don't think you are correct. the earth always has 0 kinetic energy if you are moving with the earth all the time. you only know the KE of the earth if you are in definite stationary state
     
  13. May 21, 2015 #12

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    I never claimed otherwise.
    There is no such thing.
     
  14. May 21, 2015 #13
    you don't know what is imaginary experiment?
     
  15. May 21, 2015 #14

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    I think in post #2 he used the inertial frame where the Earth is initially at rest for the whole event, which is perfectly valid.
     
  16. May 21, 2015 #15
    yup, then why would he say there is KE for the earth? if you use observer on the ground, there is no KE for the earth.
     
  17. May 21, 2015 #16

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Because the Earth is only initially at rest in that inertial frame. After collision it moves. Since the frame is inertial total KE is conserved (assuming perfectly elastic collision).

    The frame where the Earth is always at rest throughout the collision, is not inertial, so kinetic energy isn't conserved in that frame.
     
  18. May 21, 2015 #17
    you are making the same mistake. If you are observer on the earth, you won't see the earth moving, there is no KE for the earth. The earth is always at rest because observer will not see any speed for the earth. then why would KE be not conserved? you should think, the earth will not gain KE because you are observer on the earth! don't use different observers in same event to create troubles!
     
  19. May 21, 2015 #18

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    I never said otherwise, so I don't know what mistake you are referring to.
     
  20. May 21, 2015 #19
    Mistake about using stationary frame to describe the earth, using ground frame to describe the ball.
     
  21. May 21, 2015 #20
    actually I said that on the above posts :D but maybe you didn't see those posts.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Conservation of energy in bouncing ball
Loading...