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Is this book wrong? Tennis ball rebound speed

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  1. Dec 14, 2015 #1
    I am reading a book abut tennis physics and I have seen a result that I feel is wrong.
    We have a tennis ball and a racket, and several cases:
    1.- Ball hitting stacionary racquet that is sustained with your hand.
    (I agree with the results of the book here, it is just to show how they are reasoning)

    I assume that if the racquet wouldn't have any support, the ball would transfer the energy to the racquet that would fly backwards and the ball would rest quiet after the impact. *Doubt 1, Is this correct?


    These are the variables:

    ball speed (straigh line perpendicular to the racquet)= 60 mph
    racquet speed=0
    "Rebound power"=0.4, and it is defined as the ratio of the exit speed to the incoming speed of the ball for that racquet
    Relative speed between ball and racquet=60 mph
    Rebound speed= Relative speed x Rebound power= 24 mph
    Exit speed of the ball=Rebound speed = 24mph

    In this case I think the impact absorbs part of the energy and obviously the exit speed is lower than the ball speed before the contact, it seems correct to me.

    2.- Raquet and ball hitting each other with speed.

    ball speed (straigh line perpendicular to the racquet)= 20 mph
    racquet speed=40 mph
    "Rebound power"=0.4, and it is defined as the ratio of the exit speed to the incoming speed of the ball for that racquet
    Relative speed between ball and racquet=60 mph
    Rebound speed= Relative speed x Rebound power= 24 mph
    Exit speed of the ball=Rebound speed + Racquet speed= 40+24=64 mph

    In this case there is an impact that absorbs part of the energy of the incoming ball, but they add the racquet speed, I have some doubts:

    *Let's imagine that we choose a coordinate system that is in the center of gravity of the tennis ball, and let's repeat the calculation:

    ball speed (straigh line perpendicular to the racquet)= 0 mph, it is the origin of our system
    racquet speed=40 mph + 20 mph = 60 mph
    "Rebound power"=0.4, and it is defined as the ratio of the exit speed to the incoming speed of the ball for that racquet
    Relative speed between ball and racquet=60 mph
    Rebound speed= Relative speed x Rebound power= 24 mph
    Exit speed of the ball=Rebound speed + Racquet speed= 24+60=84 mph, I get a different result to the original which is absurd right? And I don't understand how a racquet moving at 60 mph could give a speed of 84mph to the ball which is higher. Is this possible?

    *Let's imagine that we choose a coordinate system that is in the center of gravity of the racquet, and let's repeat the calculation:

    ball speed (straigh line perpendicular to the racquet)= 60 mph, now the racquet is quiet and this is the sum of ball and racquet speeds.

    racquet speed=0 mph, it is the origin

    "Rebound power"=0.4, and it is defined as the ratio of the exit speed to the incoming speed of the ball for that racquet
    Relative speed between ball and racquet=60 mph
    Rebound speed= Relative speed x Rebound power= 24 mph
    Exit speed of the ball=Rebound speed + Racquet speed= 24+0=24mph, Now I get a different result because in the coordinate system of the racquet all the movement is on the ball which is the same as case 1.

    What am I doing wrong? Why do I get different results with different coordinate systems?

    3.- Ball quiet and racquet hitting the ball.

    ball speed (straigh line perpendicular to the racquet)= 0 mph
    racquet speed= 60 mph
    "Rebound power"=0.4, and it is defined as the ratio of the exit speed to the incoming speed of the ball for that racquet
    Relative speed between ball and racquet=60 mph
    Rebound speed= Relative speed x Rebound power= 24 mph
    Exit speed of the ball=Rebound speed + Racquet speed= 24+60=84 mph

    That is the third case of the book. How is that possible? I mean, a racquet at 60mph generating 84mph looks absurd to me. In this case there is not any rebound because the ball was quite, I guess the exit speed is just 60mph like the racquet, or a lower speed because some energy will be lost in the impact, but a HIGHER speed looks weird. Is this possible?

    Thanks for your time!
     
  2. jcsd
  3. Dec 14, 2015 #2
    Are these all examples from the same book? Or your solutions to questions from the book?

    It may be that you are using this coefficient of rebound improperly.
    By definition it is the ratio between the ball's speeds when bouncing from a stationary rocket.
    It is not obvious that the same value will apply when you have the ball stationary and the rocket moving.

    In order to use this coefficient you can always move to a frame where the rocket is at rest.
    But not to one where the ball is at rest.
     
  4. Dec 14, 2015 #3
    Cases 1, 2 and 3 are a copy from the book. The reasoning with the coordinate systems is mine.

    Do you think is it possible to get a 84mph speed with a 60mph racquet speed?
     
  5. Dec 14, 2015 #4
    So yes, the first 3 seem to be OK.
    But for the next one you should use the rebound power of the ball (and not of the racket) which is not the same 0.4.
     
  6. Dec 14, 2015 #5
    So I understand you agree with the last one?

    I mean racquet at 60 mph on serve gets a ball speed of 84 mph.
     
  7. Dec 14, 2015 #6

    A.T.

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    Gold Member

    It's OK, if the racket + arm have more mass than the ball:
    https://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_Newtonian

    You seem to confuse speed and energy.
     
  8. Dec 14, 2015 #7
    Thanks, yes I was confusing the speed with the energy.
     
  9. Dec 28, 2015 #8
    and dont forget the rotational energy that is transfered into motion when it hits the ground with a lot of "top spin"
     
  10. Dec 28, 2015 #9
    Thanks, the book talks about that at the end.
     
  11. Dec 28, 2015 #10

    phinds

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    Gold Member
    2016 Award

    First, this is not a "doubt" it is a question: http://www.physicsforums.com/showthread.php?t=607274

    Second, no it is not right. You are forgetting the inertia of the racket. If you tossed the racket upwards with no rotation and managed to hit it just at the top of its arc when it is motionless and unsupported, the ball would NOT just fall directly to the ground. How it would act would depend on its speed and weight and the weight of the racket.
     
  12. Dec 29, 2015 #11
    Third, thanks for your answer.

    Fourth, this is an international forum with people from around the world. It is completely logical to expect finding expressions that are not written in perfect English. That makes the forum better since there are more opinions, QUESTIONS, threads etc.

    Fourth, How many languages do you know a part from English? Maybe one day in the future you will have questions about something and you won't find an English forum that helps you and instead is in German, Will you be able to ask QUESTIONS there in perfect german? Maybe not, but for sure people there will be kind and will understand your issues.
     
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