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A ball thrown up in an elevator.

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data

    An elevator goes down with a constant acceleration a. Someone throws the ball upward from the floor with an initial velocity u relative to the elevator. If the height of the elevator is h=2,25m, the ball has an initial speed relative to the elevator u=15 m/s and it hits the ceiling of the elevator in an elastic way, find the time needed to reach the ceiling and that one spent to get to the floor again.

    2. Relevant equations
    Let's call a' the relative acceleration of the ball with respect to the elevator.Then:
    a'=g-a
    h=ut-1/2 a' t2

    Would someone tell me explicitly the reasoning, please?
     
  2. jcsd
  3. Sep 12, 2011 #2
    It seems you are on the right track, where are you stuck?
     
  4. Sep 13, 2011 #3
    Troubles comes when I try to find t (the request of the exercise).
    To do that I need at least the final velocity so that I can solve a system of two unknowns of this kind:

    h=ut-1/2 a' t2
    vf-u=-a't

    where vf is the relative velocity of the ball when it hits the ceiling (with respect to the elevator reference frame).
    Since I don't have vf, I can't solve the system and it doesn't come to my mind anything else to handle that velocity or the acceleration a' (or a).

    Then: suggestions? :D
     
  5. Sep 13, 2011 #4

    NascentOxygen

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    You know h. You know u. You know a'. So determine t.
     
  6. Sep 13, 2011 #5
    The text doesn't give a' (nor a), how can you say I know it?
     
  7. Sep 13, 2011 #6

    NascentOxygen

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    'Cause you wrote back at the start:
    Sorry, I wasn't drawing a distinction between knowns that you know, and knowns that you don't know. :smile:
    But if you aren't given numerical data, then acceleration will have to remain algebraic. But once you are given a data value for a you will then be able to determine t.

    I see where you have balked. It seems the problem specification lacks the detail necessary for you to arrive at a numerical value for t and vf. I guess you'll have to leave them as expressions.
     
    Last edited: Sep 13, 2011
  8. Sep 13, 2011 #7
    I don't know if this is correct, but since the ball is moving up against gravity and also in the opposite direction to that of the lift, should not a' = -g -a or -(g+a)?
     
  9. Sep 13, 2011 #8
    @NascentOxygen:

    Then, there's no way for solving the problem without writing t in function of a' (or a), isn't it? (That was my doubt)

    @Quantum Mind:
    My one is the right relation because you have to keep in mind the reference frame we look from: in this case the elevator one.
    The ball from an external reference frame has an acceleration g while the elevator from the same external reference frame has an acceleration a (both signs supposed to be positive in the downward direction). Then, if you are on the elevator, moving with a, you see the ball moving with its acceleration minus that one you move: a'=g-a.

    It's like the relation of relative velocities of inertial reference frame, but in this case the elevator frame isn't inertial so you can differentiate v'=V-v to get the result (works just for a rectilinear motion ;) )
     
  10. Sep 14, 2011 #9

    NascentOxygen

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    There is no solution when essential data is missing. :frown: :frown: :frown:

    Are you sure than in summarising the problem to post it on PF, you haven't omitted some essential detail?

    If you aren't told a, then you'll just have to leave the solution in algebraic terms.
     
  11. Sep 14, 2011 #10

    phinds

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    As for the return of the ball to the floor, I can't see how you can possibly answer that without knowing specifics of the elasticity of the ball. For example one of those super-bouncy balls is going to get back faster than a bean-bag would.
     
  12. Sep 14, 2011 #11

    NascentOxygen

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    Elastic means that there in no energy loss in the collision. But a complication here is that during the interaction of the ball with the ceiling, the ceiling is delivering added impetus due to its continual acceleration.
     
  13. Sep 14, 2011 #12

    phinds

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    OOPS ... thanks for the explanation; I clearly had the wrong concept on the word.
     
  14. Sep 15, 2011 #13
    No, guys. I'm completely sure about the text ;)
    As for the total time, it should be 2 times that spent going up (or just down). If you take in account the elevator frame of reference, the ball has a constant acceleration downward, then we have a normal accelerated motion with its parabolic equation y(t), where the derivative is y'(t1)=-y'(t2). So same velocity magnitude but opposite sign; it's easy to show that the time t1-t0=t3-t2 (t0 at the floor, t3 at the floor again).Finally, we know the velocity after the collision is v2=-v1.
    Obviously, I assumed that during the collision the velocity of the elevator, whose mass is much greater than the ball one, doesn't change appreciably.
    Anyway, I'm just going to start my second university year and I would be pleased if someone could tell me whether my reasoning is right or not.
    As for the problem, my trouble was handling the acceleration to give a numerical answer, but since it's impossible I'm fine now.
    Thanks ;)
     
  15. Sep 16, 2011 #14

    NascentOxygen

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    You believe it takes the same amount of time to bound as rebound? That might be true, but have you a textbook that confirms this, or where did this figure come from?

    If I get time I'll take a closer look next week. You might be right, but I'm surprised.
    ah huh
    Is this the velocity relative to the elevator, or relative to earth? Regardless, I think you might be wrong, though just how wrong may be difficult to say.

    It's a pity the problem is incompletely specified, as it's an interesting one I would have liked to run to ground. Did you get this from a book, or from your teacher?
     
    Last edited: Sep 16, 2011
  16. Sep 16, 2011 #15
    v1=-v2 comes from y'(t1)=-y'(t2) that is relative to the elevator frame (the text consists just of what I wrote unfortunately)

    I'll tell you the truth.
    Actually, it was a point of an exercise of the exam I did a few days ago, the unique thing I didn't do and the unique error (and I want to underline the point I'm sure about the text).
    Meanwhile I did the oral and I asked for the solution, but I won't give you the teacher answer not to bias you.

    As for that "an huh", what does it mean? =)
     
  17. Sep 16, 2011 #16
    Forgetting the elevator for a second, if we throw a ball at a moving wall the rebound velocity of the ball will depend on whether the wall is moving away from or towards the ball. Won't this same effect come into play with the elevator ceiling to some extent?
     
  18. Sep 16, 2011 #17
    Mmh, it may be seen in this way.
    When you look at elevator and ball from the outside, you see a ball with an acceleration g and an elevator with an acceleration a (both constant). Since the elevator is going down, second by second the ball moves up and the elevator moves down, so when the ball reaches the ceiling it has a displacement D and the elevator a displacement H; then the ball goes down with a certain unknown velocity (from our outside view) and needs to move a distance H+D to reach the floor.
    Now, I think that from the outside the rebound velocity is different (greater in magnitude and downward), because we have the impulse given by the the accelerated motion of the elevator. That to say the ball for an outside observer takes the same time to cover the distance D and D+H.
    All that may be described giving the ball another acceleration a'=g-a, that's the relative to the elevator ref. frame. So, we may get to my "same-time reasoning". :D
     
  19. Sep 21, 2011 #18

    NascentOxygen

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    Bouncing off the ceiling is a whole separate issue, a problem complete in itself. I have had a look at it, to determine the speed at which the ball will bounce off the ceiling. (For this I assume the ceiling is a perfect solid body, and the collision is elastic.) A golf ball is a good choice for these experiments--it is designed for rough treatment and has surface pitting that minimizes air resistance.

    To derive a general solution I'll first examine horizontal motion, for example, a golf ball bouncing off the front of an oncoming locomotive. Let their masses be m1 and m2, and their closing velocities v1 and v2. Considering both momentum and energy to be conserved, write the equations and solve them simultaneously. After a perfectly elastic collision--where no energy is lost--the ball rebounds with velocity

    [tex]\color{Blue}{V{_{1}}^{'}\;=\;\frac{2\,V_2+(1-m)V_1}{m+1}}\;\;\small {where\;\;\;mass\;\;ratio\;\;m=\frac{m_1}{m_2}}[/tex]
    so when m2 is massive, the mass ratio m tends to 0 and the expression simplifies to

    [tex]V{_{1}}^{'} =\; 2\, V_2+V_1[/tex]
    What interesting things can these equations tell us? Let's consider the special case of equal masses colliding, i.e., where m=1. Applying the formula, we find that after the collision mass m1 has a speed unrelated to its pre-collision speed. Amazing! Have you ever witnessed this? Sure you have--those executive toys where a couple of shiny steel balls are suspended on thin nylon line and allowed to bounce back and forth hitting each other. When the moving one collides with the stationary one, the moving one stops dead and the other one bounces away. For equal masses colliding head-on, the speed of each is entirely determined by the speed of the other. In fact, they swap their speeds (meaning, they swap kinetic energies).
     
    Last edited by a moderator: Sep 22, 2011
  20. Sep 24, 2011 #19

    NascentOxygen

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    Example: A building's elevator starts from rest and accelerates downwards at 1.8m/sec2. After 2.5 secs a ball is thrown upwards from the lift floor at 15m/sec relative to the lift. The height of the elevator cage is 2.25m.

    Taking all relative to a fixed reference, the speed of the elevator after 2.5 secs is 4.5 m/sec.

    At time=2.5 secs, ball is thrown upwards and strikes the ceiling t seconds later. During its flight, the ball "falls" a further (-15+4.5)*t +0.5*9.8*t2 metres. In the same interval, the elevator drops by another 4.5t + 0.5*1.8*t2 metres. These two distances differ by 2.25 metres in order that the ceiling overtakes the ball.

    Solving for flight from floor to ceiling, t=0.1565 secs.

    Immediately before impact, velocity of elevator is 4.782 m/sec, and of ball -8.966 m/sec. Using my earlier derivation, after momentary interaction with ceiling, ball rebounds with velocity 2*4.782 + 8.966 = 18.529 m/sec. Speed of elevator immediately after impact is assumed to be 4.782 m/sec.

    t seconds after impacting the ceiling, ball impacts floor. During this time t, elevator has travelled distance d, and ball has fallen by distance d+2.25 metres.

    Solving, yields time for ball's flight from ceiling to floor, t=0.1565 secs.

    Thus, maCrobo's contention that the ball takes equal times to travel up as down, is supported.
     
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