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Bouncing Ball

  1. May 2, 2008 #1
    A ball rolls off a table with initial velocity u (say 1m/s). Every time it hits the ground its velocity is halved. Find how many times it has bounced when it has traveled a distance x (say 1m) from the table.

    I am almost sure that you can do this without knowing the height of the table but I can't for the life of me get it out. Any ideas?
     
  2. jcsd
  3. May 2, 2008 #2

    Borek

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    Time it travels seems to be a sum of geometric series. This suggests that the distance travelled is limited and there is no general solution for any x.
     
  4. May 2, 2008 #3

    pam

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    It does depend on the height of the table. For a very high table, it will travel 1m before the first bounce.
     
  5. May 8, 2008 #4
    You would have to know the height of the table, otherwise there is no telling because as 'pam' said, the roll will make it travel a certain distance first, then the bounces will carry it further. Also, the higher the table the higher a longer it will bounce.
     
  6. May 8, 2008 #5

    Danger

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    Actually, every time it hits the ground its velocity is reversed and reduced. :rolleyes:
     
  7. May 8, 2008 #6

    Borek

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    Only vertical component is reversed.
     
  8. May 8, 2008 #7
    Wouldn't it be possible to solve this by putting it in a ratio to an unknown variable?
     
  9. May 8, 2008 #8

    Danger

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    But velocity is a vector, not a scalar. The entire velocity, therefore, is reversed and reduced.
     
  10. May 9, 2008 #9

    Borek

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    So how come ball doesn't change its horizontal direction?
     
  11. May 9, 2008 #10

    Shooting Star

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    When the coefficient of restitution between the ball and the floor is e, the magnitude of the vertical component of the velocity changes according to:
    final velo/initial velo = vy2/vy1 = e.

    In this idealized scenario, the horizontal component of the velocity remains constant, just as in projectile motion. This is what the OP possibly wanted to say. The horizontal component remains equal to the speed it had at the point of rolling off the table.

    Here e = ½, and we know exactly where the ball is at any time, and the horizontal distance covered will be given by x= vx*t. Given x, we can find the number of bounces using elementary algebra, and x is independent of the initial height.

    The velocity vector is not reversed and reduced before and after collisions. The magnitude of the velocity is, of course, reduced.

    Note that in this idealized case, the ball bounces to infinity. The height of a bounce tends to zero as number of bounces go to infinity.

    For a nice animation, see this.
     
  12. May 9, 2008 #11
    Shooting Star got what I meant, motion in the horizontal and vertical directions carry on independantly of each other, which made me think you might be able to figure out the number of bounces after a given distance since you know the initial velocity in the horizontal direction and so u*t = x and figuring out time it takes, t, you could find the number. Well okay looks like you can't, (though I could swear I remember seeing an old exam question where they did it...)

    Here's what I have anyway:

    Velocity it has in the vertical when it hits the floor: v0 = [tex]\sqrt{2gh}[/tex]. (There's one of those mysterious twos that turn up in physics.)

    Time it takes to hit the floor the first time: v0/g

    Time of flight of a projectile with vertical velocity component v: 2v/g

    So total time the ball has spent moving after k bounces is:

    v0/g + 2(ev0)/g + 2(e[tex]^{2}[/tex]v0)/g + 2(e[tex]^{3}[/tex]v0)/g + ... + 2(e[tex]^{k}[/tex]v0)/g

    = (v0/g)*(1 + 2e(1 + e + ... + e[tex]^{k}[/tex]) = T

    So x = uT
    If you let it bounce an infinite number of times the sum of the series is 1/(1-e) and for e = 0.5 the ball goes 3v0/g. So that depends on height through v0 (as a square root so you'd need a high table to get the ball to go far).

    After k bounces the sum of the series is (1-e[tex]^{k+1}[/tex])/(1-e) so
    x = (uv0/g)(1+2e((1-e[tex]^{k+1}[/tex])/(1-e)))
    So the number of bounces k depends on v0 (After taking logs and so forth).

    Of course you need to know the initial height, if you think about a real ball rolling off a table it's totally obvious. This comes from too much maths and too little physics on my part.
     
  13. May 9, 2008 #12

    Shooting Star

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    You are right -- the initial height will determine the number of bounces, through the initial vertical component of the velocity, even if the horizontal component remains the same. If the height and therefore the speed is higher, the ball will spend more time in the air and will cover a larger horizontal distance per bounce compared to being dropped from a lesser height.
     
  14. May 9, 2008 #13

    Borek

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  15. May 9, 2008 #14

    Shooting Star

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    I believe I have said that before.

    It is obvious from the context that I was referring to the number of bounces for a finite horizontal distance. Which portion have you not understood?
     
    Last edited: May 9, 2008
  16. May 9, 2008 #15

    HallsofIvy

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    Why are you asserting this? What reason do you have for believing this is true? As has been pointed out repeatedly here, the horizontal component is not affected by either gravity or the floor (ignoring friction) and so does not change.
     
  17. May 9, 2008 #16

    Borek

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    Sorry, missed that. That happens when you watch Goblet of Fire on TV and answer PF posts in commercial breaks :rofl:
     
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