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Ball collision when one is dropped and the other thrown up

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Ball A at rest is dropped from a height H and ball B at the same time is thrown up from the ground. When they collide at height h, the velocity of ball A is twice the velocity of ball B. What is h/H?


    2. Relevant equations
    1D Kinematics

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    3. The attempt at a solution
    I have no idea which of these equations to use, even when I list all of my knowns and what I can infer from this. I just randomly pick every single equation and try to plug everything in because it all looks about the same to me.

    So for ball A:
    (Initial velocity) V_0 = 0 m/s
    (Final velocity @ collision) V_A = -(2V_B) (negative velocity because it's going downwards)
    g = -9.8 m/s^2
    (displacement) x - x_0 = h - H

    ball B:
    (Initial velocity)V_0 = ?
    (Final velocity @ collision) V_B = |(V_A)/2| (positive velocity because going upward)
    g = -9.8 m/s^2
    x - x_0 = h

    And t_A = t_B since the same amount of time elapses when they both collide since they were released at the same time.

    I tried plugging in for the equations and got all these nonsensical statements that I have no idea what they mean.

    I got stuff like V_0 of B = -9.8t but I don't know if that's useful. Also -19.6t = 19.6h + 19.6H and a bunch of other stuff I have no clue if is meaninful or not.
     
  2. jcsd
  3. Sep 18, 2012 #2
    Plugging values into random equations is as far from good physics as it can possibly get. No wonder you get nonsensical results.

    Before using the equations, think about the physics of the problem. You have observed, correctly, that the time of motion for the two balls is equal. Can you find this time from what you know about any of the balls? Can you then find the velocities of the balls?
     
  4. Sep 18, 2012 #3
    Well, all but one of the equations use have time, so how do I know which one to pick? I can plug it into all of them, and they will all pretty much have at least 2 unknowns in the end. How can you tell which equation(s) to use?
    And how do you know that you need to solve for time first? Why not some other arbitrary variable?

    If I knew that solving for time would allow me to solve for velocity afterwards, of course I would have gone and done that. But I didn't know that I have to solve for velocity, and in order to do that I need to solve for time first. How do you go about knowing what to solve for, in order?
     
  5. Sep 18, 2012 #4
    For the top ball,
    v2=2a(H-h)

    Taking on magnitude for bottom ball,

    1/4(2a(H-h)) =u2-2a(H)

    2aH-2ah=4u2- 8aH

    10aH-2ah=4u2

    10a -2ah/H=4u2/H
     
    Last edited: Sep 18, 2012
  6. Sep 18, 2012 #5
    Consider ball A. What do you NOT know about it that would prevent you from solving for its velocity at h and the time it takes to get there?
     
  7. Sep 18, 2012 #6
    Well, since you told me to consider ball A, if I look at equation 2-15, in order to solve for time I'd need h and H. If I want the final velocity of it at h, I can use 2-11 but I need time for that. So to find the final velocity and the time, I need either one because they seem to depend on each other.

    As a side question, why consider ball A first? If I had not been told to, then I might have never known to. I need to know the mindset of solving problems, not the solution steps. I have a worked solution handbook that shows every single step to find the answer, but I don't know why the solution requires to do those steps, or why those steps were chosen to arrive at the answer.
     
  8. Sep 18, 2012 #7
    The mindset, as I said, is to think less about the repertoire of the equations you have in front of you, and to think more about the real physics. In the end, you need only a couple of equations, everything else can be derived from them.

    For ball A, falling down from rest, everything depends solely on the original height H and height h it is at any given time. Once this is clear, you can look at your equations. From 2.15 you obtain the time. From 2.11 you obtain its velocity. Both will still depend on the unknown h.

    Then consider ball B. For this ball, everything depends on its initial velocity and the time elapsed. Since we already know the time, and also know the final velocity, we obtain its initial velocity. Check what equation could be used for this. The result will again depend on the unknown h.

    So we can express this height h as the distance traveled by ball B (what equation?), and that will give you an equation, from which h can be found.
     
  9. Sep 18, 2012 #8
    These concepts are not clear to me. What does it mean that "everything depends on" these components?
     
  10. Sep 18, 2012 #9
    This is the concept of determinism. Every aspect of motion of a mechanical system depends solely on (or is determined by) its initial state (positions + velocities of all its constituent parts) and time elapsed.
     
  11. Sep 18, 2012 #10
    So when you say: "For ball A, falling down from rest, everything depends solely on the original height H and height h it is at any given time."

    You say that everything is dependent on the original height and final height, disregarding whatever time has passed or whatever speed it has travelled or is travelling at. Why is this?
     
  12. Sep 18, 2012 #11
    The initial conditions for A are its initial height H and its initial velocity, which is zero. So everything depends just on H and the time elapsed. Everything, including h at any given time. But that equally means that at any given h, time can be found from h and H. So we can exclude the time and express any aspect of A's motion as dependent on h and H.
     
  13. Sep 18, 2012 #12
    And all of this, you analyze in your head and it goes through in a few seconds?
     
  14. Sep 18, 2012 #13
    Well, I can't say I analyze this really. I analyzed this quite a while ago and I just KNOW now that, for example, I can trade one kinematic variable of a system for another (t for h in this case). Another thing I immediately take note of is that A has ALL the initial conditions given, while for B we know only the initial height. So it makes sense to start with A. Yes, all of this takes a little bit of practice.

    There is also another method, which is less physical and more mathematical. It usually is more roundabout, but sometimes, when you are stuck, you might consider it. Every uniformly accelerated 1D motion is fully described by two equations: [tex]x = x_0 + v_0t + \frac {at^2} 2 \\ v = v_0 + at [/tex]So you just express your problem in terms of these equations - for every part of the system. Physics mostly ends here, and mathematics begin: you just solve the resultant system of equations. For this problem, for example, you have two balls, so you have four equations: [tex]
    x_A = x_{A0} + v_{A0}t + \frac {at^2} 2 \\ v_A = v_{A0} + at
    \\ x_B = x_{B0} + v_{B0}t + \frac {at^2} 2 \\ v_B = v_{B0} + at
    [/tex]Plugging here all the known parameters, we end up with [tex]
    h = H - \frac {gt^2} 2 \\ v_A = - gt
    \\ h = v_{B0}t - \frac {gt^2} 2 \\ v_B = v_{B0} - gt
    [/tex] We further know that ## v_A = 2v_B ##, so we just add this to our system, and end up with five equations and five unknowns: ## h, \ t, \ v_A, \ v_{B0}, \ v_B ##, so we should be able to solve this system purely algebraically without any more physical considerations. Well, not completely, as we have quadratic equations there, so we may have to choose from two different roots, and that typically needs some physical consideration.
     
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