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Dynamics with bounces

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    [PLAIN]http://img686.imageshack.us/img686/8532/phyf.jpg [Broken]
    It is observed that the ball always bounces exactly once on each step.

    2. Relevant equations
    s = ut + 1/2at^2

    3. The attempt at a solution
    [PLAIN]http://img28.imageshack.us/img28/6678/abccz.jpg [Broken]
    So after resolving, horizontal velocity must be always constant as it bounces exactly once on each step and must be 0.30/t ms^-1
    So, u*cos(theta) = v*cos(phi) = 0.30(1/t)
    And to find e, we form the equation e*u*sin(theta) = -v*sin(phi)
    I also find time t, for which the ball moves -0.20m vertically and 0.30m horizontally. Upwards as positive. Using s = ut + 1/2at^2
    -0.20 = v*sin(phi)*t + 1/2(-9.81)(t^2)
    But I tried manipulating around and I just can't find e.

    Is something wrong with my method? Do I need to use momentum to calculate this question?
    Thank you.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 6, 2010 #2
    Re: Dynamics

    Just consider the vertical component of the motion to find the value of e.
    The ball falls a distance 0.4m and bounces up a distance 0.2. That's a ratio of 2 to1.
    The value of e given, 0.71, is the square root of 1/2. This should provide a big clue.
    Use v2=u2+2gs for the fall and the bounce, with the values of s I mentioned, to find the ratio of the squares of the velocities.
     
    Last edited: Apr 6, 2010
  4. Apr 7, 2010 #3
    Re: Dynamics

    But why will the ball fall a distance 0.4m? What's the rationale behind it?
     
  5. Apr 7, 2010 #4
    Re: Dynamics

    It's what's (apparently) given in the question. In the diagram the ball bounces up to be level with the top of the previous step. It then falls down a height of 2 steps. Then it's motion down the steps is identical all the way down. I assume, in other words, that the diagram is giving this information.
     
    Last edited: Apr 7, 2010
  6. Apr 8, 2010 #5
    Re: Dynamics

    Okay... Thanks anyway :D
     
  7. Apr 8, 2010 #6
    Re: Dynamics

    As a bit of further information here. I've had a closer look at this question just to see if there is something deeper going on that I've missed.
    There is no unique solution to the question if you don't assume the ball bounces up 0.2m and then falls down 0.4m.
    You have to assume some initial height of bounce.
    For example, you could solve the problem for a bounce up height of 0.1m and fall down of 0.3m. This would give you the same effect of bouncing down the steps, albeit with a flatter trajectory, and would mean (if you do the maths) that e=0.58 and the horizontal component of velocity =0.77m/s
    In other words, there are an infinite number of solutions with different values of e and the horizontal component unless you assume the bounce up to be 0.2m
    This then gives the values required in the question.
    There's no other way of doing it with the information supplied.
     
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