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Conservation of momentum

  1. Dec 10, 2009 #1
    Does a free falling ball drops on earth violate this theory?

    If we consider the ball and earth in an isolated system then the ball should continue to drop over time.

    If the impulse given is extremely negligible that it doesn't cause any significant change in the path of earth, then the whole momentum must be conserved by the ball and it should bounce back with the same velocity.

    Which is the missing point?
     
  2. jcsd
  3. Dec 10, 2009 #2

    Doc Al

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    Staff: Mentor

    The total momentum of the 'ball + earth' system remains unchanged. Whether the ball bounces back with the same speed depends on the nature of the collision, not on momentum conservation. Regardless of the speed of the bouncing ball, the total momentum of the system remains unchanged.
     
  4. Dec 10, 2009 #3
    The change of momentum of the earth isn't exactly negligible. It is exactly the same change in momentum as for the ball. However, you are right that the change in momentum of the earth corresponds to a negligible change in velocity of the earth due to its large mass.
    Usually a ball won't bounce back with the full velocity, since it will transfer momentum to particles in the earth (which will scatter in all directions). This could create a tiny shock wave in the earth, but again due to the spread it is not detectable.
     
  5. Dec 11, 2009 #4
    @pf mentor
    My point is m(e)*v(e)+m(b)*v(b) must be unchanged after colission. Since m(e)*v(e) is equal to zero and there is no (negligible) change in m(b), the final velocity before colission must be equal to the initial velocity after colission. If the initial velocity while bouncing is same as the final velocity at the time of drop then there can not be any change in displacement. Yes if the material is more elastic then it will help the object to restore the kinectic energy and it will bounce back better, but lets talk abt only momentum.
    @gerenuk
    Do you think the momentum transfered to the particles should be considered? Why cant we just consider these simply as two objects? Lets forget earth consider a big stone.
     
  6. Dec 11, 2009 #5
    I'm not sure what you want to point out. The earth is a big object only?
    Maybe I make another comment. Momentum and energy is always perfectly conserved if only you take all objects into account.

    You are right if you mean that, provided the earth does not absorb any energy and does not deform, then energy and momentum conservation will dictate that the ball will have the same magnitude of velocity when it reaches a point that it was at before.
     
  7. Dec 11, 2009 #6

    Doc Al

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    Staff: Mentor

    That's true.
    Not true. Don't confuse having the same speed with having the same velocity. If the ball bounces back with the same speed but in the opposite direction, its momentum is not the same.
    Let's work out the details:
    Initial velocity of ball: - v(b) (it's moving towards the earth)
    Initial momentum of ball: -m(b)v(b)
    Initial velocity of earth: 0 (assume it's zero)
    Initial momentum of earth: 0

    Final velocity of ball: + v(b) (assume it bounces back with same speed, so now it's moving away from earth)
    Final momentum of ball: +m(b)v(b)

    Change in momentum of ball = final - initial = +m(b)v(b) - -m(b)v(b) = +2m(b)v(b)

    Thus momentum conservation allows us to conclude that the final momentum of the earth is now -2m(b)v(b). The earth is now moving away from the ball. The speed of the earth might be tiny, but you cannot neglect its momentum.
     
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