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Conservation of Momentum of a ball

  1. Nov 29, 2014 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown against a fixed wall where it bounces elastically. The mass of the ball is M and the velocity just before it hits the wall is U. Ignore the force of gravity in this question.
    a) Does the Principle of Conservation of Momentum apply to this situation?
    b) Obtain an expression for the change in momentum of the ball in terms of M and U

    2. Relevant equations
    momentum before = momentum after
    change in momentum = m(v-u)
    change in momentum= force x time

    3. The attempt at a solution
    I think that the situation does not follow the principle of conservation of momentum, because the situation is just before the ball hits the wall, so there has not been a transfer of momentum yet, but I am still not sure.
     
  2. jcsd
  3. Nov 29, 2014 #2

    ShayanJ

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    The momentum of any system not subject to a net force, is conserved. Here it seems we have such a system!
    Note that its regardless of the fact that the wall gets some of the ball's momentum or not.
     
  4. Nov 29, 2014 #3
    Thank you so much! Do you have any idea how to do the second question? I'm guessing that the initial and final velocity will be the same because the ball bounces elastically, but I don't really know.
     
  5. Nov 29, 2014 #4

    ShayanJ

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    You don't need to write any equations. The wall is fixed and so gets no momentum and momentum is conserved. So its only the ball that can have all the momentum after the collision.
     
  6. Nov 29, 2014 #5
    Thank you!
     
  7. Nov 29, 2014 #6

    PeroK

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    Are you sure about that?

    The question asks about the change in momentum of the ball. So, you don't have to worry about the wall's momentum.
     
  8. Nov 29, 2014 #7

    ShayanJ

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    I don't understand your objection but I'm sure about what I said.
     
  9. Nov 29, 2014 #8

    PeroK

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    The question (part a) is ambiguous, as it's not clear what is meant by "this situation". If you take the system to be the ball and the wall, then momentum is not conserved.
     
  10. Nov 29, 2014 #9

    ShayanJ

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    The problem says we should neglect the Earth's gravity!
     
  11. Nov 29, 2014 #10

    PeroK

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    Gravity is out of the equation.

    If the ball changes direction and the wall does not move (which I how I interpreted the question), then is momentum conserved?

    It would be a better question if it asked for an explanation of why momentum is conserved, or not.
     
  12. Nov 30, 2014 #11

    Orodruin

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    This would mean the ball would continue through the wall.
     
  13. Nov 30, 2014 #12

    ShayanJ

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    Consider mass m with velocity v hitting a stationary object with mass M. After the collision, the speed of m is u and the speed of M is U. Conservation of momentum and conservation of kinetic energy(collision is elastic), give us:
    <<Moderator note: Full expressions removed>>
    Which in the limit [itex] M \rightarrow \infty [/itex], give [itex] U=0 [/itex] and [itex] u=-v [/itex].
     
    Last edited by a moderator: Nov 30, 2014
  14. Nov 30, 2014 #13

    Orodruin

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    The velocity of the heavy object goes to zero. Its momentum goes to 2mv.
     
  15. Nov 30, 2014 #14

    PeroK

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    In which case, momentum is only conserved if ##\infty \cdot 0 = 2mv##
     
    Last edited by a moderator: Nov 30, 2014
  16. Nov 30, 2014 #15

    ShayanJ

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    Yeah...didn't see that coming!!! What does it mean to have an stationary object with momentum???
     
  17. Nov 30, 2014 #16

    Orodruin

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    The point is it is not stationary. For every M you have a non-zero velocity, it is just the limit that is zero.

    Another observation of interest for this situation is: The final momentum of the wall will increase with its mass.

    Edit: You could just as well ask what it means to have an infinitely massive object.
     
  18. Nov 30, 2014 #17

    ShayanJ

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    Well, this is how I see it: We throw the ball against the wall. But the wall is rigidly attached to earth and so can only move with earth. This means we're effectively hitting the ball with the earth itself. Comparing the mass of the ball with the mass of the earth, we can easily see what it means to have an infinitely massive object.
    So this all means that when we do such an experiment, the earth gains a momentum equal to 2mv. But its velocity is 2 (m/M) v which is effectively zero because of the very very very small mass ratio in the formula.
     
  19. Nov 30, 2014 #18

    Orodruin

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    Yet it is not zero and the Earth mass is not infinite, so the Earth velocity does change although by a very minuscule and undetectable amount. This is the thing with limits. All limits are going to depend on the conditions. If an object has a given momentum, its velocity will naturally be lower if its mass is larger.

    However, I suspect the intention of the problem is to consider the ball only, which is subject to an external force and thus momentum non-conservation.
     
  20. Nov 30, 2014 #19

    ShayanJ

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    Of course but there is always an amount of largeness or smallness that resembles the limit very well.(Assuming continuity!). That was what I meant.

    The problem tells us to ignore gravity so there is no external force and the momentum is conserved!
     
  21. Nov 30, 2014 #20

    Orodruin

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    There is a force from the wall on the ball when it bounces, which is when the momentum of the ball is not conserved. The origin of this force is electromagnetic and has nothing to do with gravity.

    Edit: I removed the full expressions from post #12 as this is part of the problem that should be solved by the OP.
     
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