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Math Help - tennis ball example/coefficient of restitution

  1. #1
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    Question tennis ball example/coefficient of restitution

    a tennis ball is hit from a height h above the ground with a speed V and at an angle alpha to the horizontal. it hits a wall at a horizontal distance d away . Air resistance is negligible

    i was told to find expressions for the height and velocity of the ball for which i did shown below

    Vx = Vo*cos(alpha)

    Where :
    Vo is the initial velocity
    alpha is the launch angle
    Vx is the x component of velocity

    Vy = Voy*sin(alpha) - g*t

    From the pythagorean theorem V = sqrt(Vx^2+Vy^2)

    X = Vo*cos(alpha)*t
    Y = Vo*sin(alpha)*t - 1/2*g*t^2

    Assuming that i started at (x,y) = (0,0) i will then end upat the point (D,h)

    that was the first part of the quesiton...but the part im stuck on is ...auusuming that the coefficient of restitution between the tennis ball and the wall is e, i need to write down an expression for te the velocity of the ball immediately after impact......is there anyone that can help ???thankz
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by dopi
    a tennis ball is hit from a height h above the ground with a speed V and at an angle alpha to the horizontal. it hits a wall at a horizontal distance d away . Air resistance is negligible

    i was told to find expressions for the height and velocity of the ball for which i did shown below

    Vx = Vo*cos(alpha)

    Where :
    Vo is the initial velocity
    alpha is the launch angle
    Vx is the x component of velocity

    Vy = Voy*sin(alpha) - g*t

    From the pythagorean theorem V = sqrt(Vx^2+Vy^2)

    X = Vo*cos(alpha)*t
    Y = Vo*sin(alpha)*t - 1/2*g*t^2

    Assuming that i started at (x,y) = (0,0) i will then end upat the point (D,h)

    that was the first part of the quesiton...but the part im stuck on is ...auusuming that the coefficient of restitution between the tennis ball and the wall is e, i need to write down an expression for te the velocity of the ball immediately after impact......is there anyone that can help ???thankz
    I think it will end up at (D,-h).
    Let t_1 be the total time of flight.
    You can use your equation:Y = Vo*sin(alpha)*t - 1/2*g*t^2
    If t= t_1, then Y=-h.
    Rest you can do yourself.

    Keep Smiling
    Malay
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  3. #3
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    Question expression/coefficient of restittuon

    Quote Originally Posted by malaygoel
    I think it will end up at (D,-h).
    Let t_1 be the total time of flight.
    You can use your equation:Y = Vo*sin(alpha)*t - 1/2*g*t^2
    If t= t_1, then Y=-h.
    Rest you can do yourself.

    Keep Smiling
    Malay
    i dont understand how that will help me to find an expression for the velocity of the ball immediately after the impact, when the coefficient of restitution between the ball and the wass is e.????
    thankz
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by dopi
    a tennis ball is hit from a height h above the ground with a speed V and at an angle alpha to the horizontal. it hits a wall at a horizontal distance d away . Air resistance is negligible

    i was told to find expressions for the height and velocity of the ball for which i did shown below

    Vx = Vo*cos(alpha)

    Where :
    Vo is the initial velocity
    alpha is the launch angle
    Vx is the x component of velocity


    X = Vo*cos(alpha)*t
    You have done good work, I will use two of your equations(see quote).
    To find the velocity after impact, we need to know two things:
    coefficient of restitution(which we know)
    velocity before impact(which is our focus now)

    Velocity before impact can be divided into two components:
    horizontal component(which you know V_ocos\alpha)
    vertical component--- for it you need to know the total time of flight[t1](since it depends on t)

    You could find t1 using the equation:
    X = Vo*cos(alpha)*t
    you get t_1=\frac{D}{V_0cos\alpha}

    I hope you have got your answer. Feel free to say.

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    Malay
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dopi
    i dont understand how that will help me to find an expression for the velocity of the ball immediately after the impact, when the coefficient of restitution between the ball and the wass is e.????
    thankz
    The definition for the coefficient of restitution is:
    KE_f = eKE_0

    So if you know the KE before collision (KE0) you can find the KE after collision (KEf). This also works for multibody collisions where the KE's in the formula are the total KE for the system.

    By the way, you say you are looking for the velocity after collision. That isn't correct. You can only get the speed after collision using this equation. What's going to happen to the velocity is anyone's guess because we don't know what effect the collision will have on the individual components of velocity. (The only example I can think of where you CAN get the final velocity is if the collision is "head-on." Then the velocity after impact will be directly opposite the velocity before impact.)

    -Dan
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  6. #6
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    Quote Originally Posted by topsquark
    The definition for the coefficient of restitution is:
    KE_f = eKE_0
    Not according to my sources, which seem to think it is the ratio of the after
    to before velocity component normal surface.

    RonL
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack
    Not according to my sources, which seem to think it is the ratio of the after
    to before velocity component normal surface.

    RonL
    (sigh) Different books, different definitions. Gotta love it.

    -Dan
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  8. #8
    Super Member malaygoel's Avatar
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    Heelo Topsquark!
    You said: KE_f=eKE_o
    this implies that e is the ratio of squares of final and initial velocities. Do you want to say that?

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    Malay
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by malaygoel
    Heelo Topsquark!
    You said: KE_f=eKE_o
    this implies that e is the ratio of squares of final and initial velocities. Do you want to say that?

    Keep Smiling
    Malay
    Well, that's how I learned it anyway, I think. (Sigh) I went back to the book I thought I had gotten it from (Serway's intro Physics book) and couldn't even find it! Seeing as I can't find it in my Graduate Mechanics book either, I obviously learned it from some out of book notes. I could simply be remembering it wrong.

    -Dan
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack
    Not according to my sources, which seem to think it is the ratio of the after
    to before velocity component normal surface.

    RonL
    I propose we do an experiment. Let us throw a ball gently on the patio so
    it bounces and has forward motion. If Topsquark is right the ball will stop
    bouncing with no residual velocity along the patio. If I am right it will continue
    rolling along the patio with approximately the same speed as the original
    horizontal component of the velocity (some loss to friction and conversion
    of translational to rotational energy), yes?

    I've done this experiment, and so know the answer

    RonL

    PS my spell checker wants to change TopSquark to pipsqeak!
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack
    I propose we do an experiment. Let us throw a ball gently on the patio so
    it bounces and has forward motion. If Topsquark is right the ball will stop
    bouncing with no residual velocity along the patio. If I am right it will continue
    rolling along the patio with approximately the same speed as the original
    horizontal component of the velocity (some loss to friction and conversion
    of translational to rotational energy), yes?

    I've done this experiment, and so know the answer

    RonL

    PS my spell checker wants to change TopSquark to pipsqeak!
    Good point. ("pipsqueak" Shakes his head in despair!)

    -Dan
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  12. #12
    Super Member malaygoel's Avatar
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    Quote Originally Posted by CaptainBlack
    I've done this experiment, and so know the answer
    I haven't performed the experiment, but let me know the answer.

    Malay
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by malaygoel
    I haven't performed the experiment, but let me know the answer.

    Malay
    The point is, using my definition, the ball would eventually stop even over a frictionless surface. (e is always less than 1 in reality) If the ball had a horizontal component of velocity Netwon's 1st says it should always be in motion. A contradiction, so my definition can't be correct.

    -Dan
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  14. #14
    Super Member malaygoel's Avatar
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    Quote Originally Posted by topsquark
    The point is, using my definition, the ball would eventually stop even over a frictionless surface. (e is always less than 1 in reality) If the ball had a horizontal component of velocity Netwon's 1st says it should always be in motion. A contradiction, so my definition can't be correct.

    -Dan
    It means that coefficient of restitution gives the ratios of final and initial normal components of velocities.
    Right???

    Malay
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  15. #15
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by malaygoel
    It means that coefficient of restitution gives the ratios of final and initial normal components of velocities.
    Right???

    Malay
    Yes. According to CaptainBlack (and I no longer have any reason to doubt his definiton) the coefficient of restitution is a ratio between the final and initial "vertical" components of the velocity. (Vertical referring to a normal to the surface of impact.)

    -Dan
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