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A general question regarding classical mechanics.

  1. Feb 25, 2013 #1
    Ok, so I know that law of conservation of linear momentum holds in a system in a particular direction, provided no net external force is acting in that direction. So, if we drop a ball on the earth surface from a height much less than earth's radius and then to analyze its momentum, we take ball+earth as the system. Now in the system no net external force is acting as gravity has become its internal force. Thus conservation of linear momentum will hold. Now Initial momentum of ball+earth: mu+0=mu ,where u is the velocity gained by ball in time t1. Final momentum of ball+earth system=mv+0=mv at time t2.

    Now conserving linear momentum:

    mv=mu=> v=u ????

    Something is absurd. How can velocity of ball remain constant in spite of gravity acting on it?
  2. jcsd
  3. Feb 25, 2013 #2
    Initial momentum -> (mass of the ball)*(velocity of the ball) + (mass of earth)*(velocity of earth)

    Final momentum -> (mass of earth and ball combined)*(velocity of combined earth/ball object)
  4. Feb 25, 2013 #3
    Oh !! Thanks ModusPwnd !! Since, I am taking earth+ball as the system, I must account for its velocity also. But wait !! I am applying the law of conservation of linear momentum in vertical direction, I cannot see the momentum of earth in vertical direction. What's your point ?

    Also, another question: If we take earth + a body as a system, then friction can also be treated as internal force (no net external force in that direction). Then why was I not able to apply law of conservation of linear momentum?
  5. Feb 25, 2013 #4
    You can apply conservation of momentum. Hold a ball up, what is the total momentum of the earth and ball in your frame of reference? (Remember that you are approximating that you and the earth are "still" and not accelerating) velocity of the ball = 0, velocity of the earth = 0 --> initial momentum equals zero. Now drop the ball and let it hit the earth. After it hits the earth the velocity of the earth/ball object is 0, thus the final momentum is zero. Momentum is conserved. How about when the ball is falling, after you drop it but before it impacts? Then the earth and ball each have nonzero velocity. At this point the momentum is the sum of both, MV + mv. (M and V for the earth, m and v for the ball). Since we know that momentum is conserved we know this must be equal the momentum before and after the drop, it must equal zero. Thus during the drop, MV + mv = 0. (Note that V and v are changing in time, M*V(t) + m*v(t) = 0)
  6. Feb 25, 2013 #5
    Ok, initially total momentum of the system equals to zero. Now before striking the ground, ball has velocity downwards.

    mv+MV=0, as per conservation law of momentum.

    This must imply that earth has its small velocity upwards. Is this really possible?
  7. Feb 25, 2013 #6
    Of course it is. Its not just possible, its impossible for it to be otherwise. How can the earth and the ball have a force attracting each other and the earth not move and the ball does move? There is no cosmic tether holding the earth in place. If you push or pull it, it moves.

    For a fun exercise, solve for "V", plug in some realistic numbers for M,m and v and then figure out how fast the earth is moving. Probably not very fast, right?
  8. Feb 25, 2013 #7
    Yes, too slow.
    Ok thanks. I was doing an exercise and was making this same silly mistake of not accounting for earth's velocity. But such types of questions, textbook do not frame. We have to consider external force anyway.

    Now can you answer my other question in whic friction was involved in post #3 ?

    Thanks once again.
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