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Math Help - Bodies with potential energy

  1. #1
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    Bodies with potential energy

    Hi
    I am struggling with this question. It goes like this:

    A rubber ball is dropped on to flat ground from a height of 2m.
    (a) Calculate how long it takes for the ball to first hit the ground.

    (b)The ball loses 10% of its kinetic energy at each bounce. Calculate the time taken for the ball to come to rest.

    I am able to solve the first part using equation of motion.
    S=UT +0.5aT^2

    I am not sure of the second part.
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  2. #2
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    Quote Originally Posted by gbenguse78 View Post
    (b)The ball loses 10% of its kinetic energy at each bounce. Calculate the time taken for the ball to come to rest.
    Are you sure this is the entire question?
    Loosing 10% of kinetic energy each bounce will never give 0 kinetic energy, so it will never stop (unless there is a threshold given).
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  3. #3
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    Yes. The other condition is ignore air resistance.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    No, that still means that the ball will never have 0 kinetic energy. Or, does the question mean '10% of the original kinetic energy at the before the first bounce'. This will mean that the ball will lose a constant amount of energy after each bounce and solve this problem.
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  5. #5
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    Re: Bodies with potential energy

    The ball will be at rest (for an instant) when it reaches its highest point between bounces.
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  6. #6
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    Hi unknown008
    Thats how the question is phrased. I have checked it. Its a past question.
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  7. #7
    MHF Contributor Unknown008's Avatar
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    SammyS' point might be what the question is looking for
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  8. #8
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    Quote Originally Posted by Unknown008 View Post
    SammyS' point might be what the question is looking for
    Thanks, Unknown008.

    After the first bounce, conservation of energy tells us that the ball will reach a height of 80% of 2 meters, which is 1.6 meters.

    So, rewording the question: "How long does it take from the time the ball is dropped for it to bounce off the floor then reach a height of 1.6 meters, at which point it has zero KE?"
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  9. #9
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    Re: Bodies with potential energy. The ball does remain at rest in a finite amount of

    Quote Originally Posted by SammyS View Post
    The ball will be at rest (for an instant) when it reaches its highest point between bounces.
    Upon further reflection, I do believe that the question is indeed referring to how long it takes the ball to come to rest and remain at rest on the ground.

    While it's true that the ball will bounce an infinite number of times, (in this idealized situation) the ball will bounce for only a finite amount of time.

    Let t_0 be the time it takes the ball to reach the floor the first time it falls.

    It loses 10% of its Kinetic Energy (KE) at each bounce, so it keeps 90% of its KE. |\vec{v}\,| = \alpha\sqrt{KE} (Velocity is proportional to square root of KE.) Time between successive bounces is proportional to rebound velocity, and time between successive bounces is double the time the ball takes to fall from it's highest point between those two bounces.

    Therefore, the time between the 1st & 2nd bounces is: 2t_0\sqrt{0.90}\,.

    The time between the 2nd & 3rd bounces is: 2t_0\sqrt{0.90}\cdot\sqrt{0.90}\,.

    The time between the 3rd & 4th bounces is: 2t_0\sqrt{0.90}\cdot\sqrt{0.90}\cdot\sqrt{0.90}\,.
    ...

    The time between the Nth & (N+1)th bounces is: 2t_0(\sqrt{0.90})^N\,.

    The total time the ball bounces is:

    \displaystyle T=t_0+2t_0\sum_{n=1}^{\infty} (\sqrt{0.90})^n\,.
    The sum is a geometric series which converges to: \displaystyle {{1}\over{1-\sqrt{0.90}}}={{1+\sqrt{0.90}}\over{1-0.90}}=10\left(1+\sqrt{0.90}\right)

    Therefore, \displaystyle T=t_0+20t_0\left(1+\sqrt{0.90}\right)\,.
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