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Homework Help: Bouncing ball and momentum

  1. Jul 25, 2006 #1
    Hi:

    I am working through this problem regarding a bouncing ball. I have gotten most of the way through, but can't figure out the time interval between bounces. Can someone give me some guidance? What am I not seeing?

    A 32 g steel ball bounces elastically on a steel plate, always returning to the same maximum height h = 20.2 m.
    a) With what speed does the ball leave the plate?
    vup = 19.9m/s
    b) What is the magnitude of the total change in momentum of the ball with each bounce?
    |Dpball| = 1.27
    c) What is the magnitude of the total momentum the ball transfers to the plate at each bounce?
    |Dpplate| = 1.27
    d) What is the time interval between bounces?
    Dt = ???

    I tried to simply t=d/v=20.2/19.9 and multiplying that by 2-- wrong. I can't think of any other way to solve this. And the next question builds off of it:

    e) Averaged over a long time interval, what is the magnitude of the rate at which momentum is transferred to the plate?
    |Dp/Dt| =
    f) What is the magnitude of the average force exerted by the ball on the plate?
    Favg =

    Thank you...
     
  2. jcsd
  3. Jul 25, 2006 #2

    Office_Shredder

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    Time interval between bounces...

    knowing the laws of kinematics, can you calculate how long it takes for an object to move a set distance, given the acceleration?
     
  4. Jul 25, 2006 #3
    I know
    dx=vo*t+(a*t^2)/2
    But that doesn't give me the right answer
    I got 0.84s ---
    ??
     
  5. Jul 25, 2006 #4
    Is there something I am completely missing here?
    dx=vo*t+(a*t^2)/2
    20.2 = 19.9t + 9.81/2*t^2
    solving for t = 0.84 or -4.9
    2*t = 1.68s. Wrong.

    Please, does anyone have any suggestions?
    Thank you so much.
     
  6. Jul 26, 2006 #5
    20.2=9.81/2*t^2,=>t=2.03
    or 20.2 = 19.90792t - 9.81/2*t^2
     
  7. Jul 26, 2006 #6
    [tex]v = v_0 + at[/tex]

    [tex]v - v_0 = at[/tex]

    [tex]t = \frac{v - v_0}{a}[/tex]

    v - velocity of the ball when reaches the maximum height
    v0 = vup = 19.9m/s
    a = g = - 9.80 m/s2
    t - That's the time the ball will take to reach the maximum height.

    But the time interval between two bounces let's call tb is equal to 2t.
    Then:

    [tex]t_b = 2(\frac{v - v_0}{a})[/tex]
     
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