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A rubber ball thrown to a wall

  1. Jul 14, 2007 #1
    Can anybody explain how the momentum is conserved in the case of a rubber ball (bounced back),when it was throwned to a wall?
  2. jcsd
  3. Jul 14, 2007 #2


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    If you throw the ball at a wall with velocity v, so that its momentum is mv, then, in a perfectly elastic collision with the wall, it bounces back with momentum -mv (in reality, slightly less) and so have momentum change 2mv. That is "balanced" by the change in momentum of molecules in the wall.
  4. Jul 14, 2007 #3

    Doc Al

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    If you chose to look at the ball itself, then momentum is obviously not conserved. Why should it be? After all, there's an external force acting on it.

    But if you look at the ball + "wall & attached Earth" as a composite system, then there are no external forces and momentum is conserved. As HallsofIvy explained, the change in the ball's momentum is -2mv (assuming an elastic collision), while the change in momentum of the "wall & attached Earth" must be +2mv.
  5. Jul 14, 2007 #4
    you need to realise that the force is impulsive as well.
    Therefore on the ball alone there is basically a force acting. This is resulting in a increase in the momentum of the ball.
  6. Jul 14, 2007 #5
    * decrease
  7. Jul 14, 2007 #6
    obviously you dont observe the earth moving in the other direction with a momentum change of 2mv because of massive mass of earth
  8. Jul 16, 2007 #7
    let Mass of "wall " is M(consider only gravity is holding the wall in position) ;Initial and final velocity of wall is zero:if mass of ball is m,and let the ball was thrown with a velocity u and let v be the velocity with which the ball bounce back.
    substittuting the values in the equation,

    m1u1 + m2u2 = m1v1 + m2v2(conservation of momentum)

    we will get M*0 + m*u = M*0 + m*v

    considering elastic collision, v should be equal to u;but in this case
    v = -u
    so how the momentum is conserved?
  9. Jul 16, 2007 #8
    who says walls final velocity is 0??
    m1u1 + m2u2 = m1v1 + m2v2
    v2=m1(u1 - v1)/m2 (here u1 and v1 are in opposite direction, so they ll add up, remember vector addition??)
    now put m2 = mass of earth and get v2, is it even noticable???
  10. Jul 16, 2007 #9
    elastic collision means the conservation of energy and momentum. it does not say that initial and final velocities are same
  11. Jul 16, 2007 #10
    no it is not.
    Ok let us try to get rid of this earth.
    A huge iron ball is falling vertically down.a cotton ball of same volume hit it horizontally with a velocity u and bounce back with a velocity -u.
    can you please explain the momentum conservation in this case
  12. Jul 16, 2007 #11
    is this what you are talking about????:confused::confused:

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  13. Jul 16, 2007 #12
    yes can you please explain and please consider the iron ball is (not falling) but moving with uniform velocity
  14. Jul 16, 2007 #13
    ok if you consider that the cotton ball strikes head on with the heavy ball, n bounces back along its initial line of motion withits change in momentum = 2McUc.
    the heavy iron ball ll alter its line of motion. it ll now move at an angle to the vertical direction so that its horizontal component is 2McUc(plz note that Vi_horizontal ll be very small, ie. Vi_horizontal=2McUc/Mi ) and vertical momentum remains unchanged(of the iron ball ofcourse)

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  15. Jul 16, 2007 #14

    Doc Al

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    OK. Assume the heavy wall is resting on frictionless ice, if you like. (Gravity doesn't hold the wall in position--the ice does; but there's no horizontal force from the ground on it, unlike the actual case where the wall is attached to something. No big deal.)

    The final velocity of the wall is not exactly zero, just close to it. But realize that a huge mass M multiplied by a small speed is still a finite momentum.

    You assume the final momentum of the wall is 0. Wrong! The final momentum of the wall equals 2mv (to an very good approximation).

    Try this: Apply both conservation of energy and momentum and solve for the final speed of the wall.
  16. Jul 16, 2007 #15


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    The wall is effectively connected to the earth. The final velocity of the wall and earth will be changed very little by a bouncing ball.
  17. Jul 16, 2007 #16


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    If at all - the energy has to transfer through the wall to the floor and due to the large mass and inelasticity of the wall, most of the energy would likely just be absorbed and converted to heat at the impact site....

    ....none of which in any way contradicts the laws of motion.
  18. Jul 16, 2007 #17
    ok ,thanks .agreed.Can you please tell me what is the force exerted by cotton ball on iron ball?
  19. Jul 16, 2007 #18
    duh.... force is the rate of change of momentum.
    do you see momentum of cotton ball changing???
    if yes, that is the force on the iron ball
  20. Jul 16, 2007 #19
    rate of change in one second or rate of change during contact period?
    If contact period,how can i get the contact period(just assumption or?)?
    what is the time interval ,in which the cotton ball ,attained its final velocity ?
    Last edited: Jul 17, 2007
  21. Jul 17, 2007 #20
    well obviously it is the rate of change during the contact period. and it depends upon the physical properties of the ball like the extent of deformation they undergo when they collide(elasticity)[but that is a complex case, because if you consider that part, you ll also have to take into account the energy lost in the deformation], for simplicity and considering it as an ideal case, the reversal of velocity is considered instantaneous
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