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Why is it more difficult for to hit back a slow ball

  1. Oct 11, 2009 #1
    Why is it more difficult , in terms of generating speed, to hit back a ball traveling slowly than one traveling fast?
    For example, in cricket, we often hear that the batsmen have to "generate" pace (cricket language for speed) when the ball is coming at them slowly.
    In tennis, I know from experience that just by blocking a fast ball, it's possible to send it back to the other side of the court. However, just putting the racket in front of a slow ball won't even give the ball enough speed to clear the net.

    I'm trying to analyze this through physics but I'm not having much success.
    For example, if the racket is just "blocking" the ball, then in both instances it would be exerting the same force on the ball. The same force would cause the same change in momentum and that would mean that a slow ball would travel back faster...
  2. jcsd
  3. Oct 11, 2009 #2
    not so. force IS different....

    one way to think about momentum and an opposing force is from FT = MV considerations....
    let's say a tennis ball is traveling fast...v is larger...so holding a racket up means either a greater F or a greater T is needed to return the ball....likely both happens: there is a bit greater force on your wrist as the ball hits the racket while at the same time the racket is deformed a bit more, increasing the time of contact T....
  4. Oct 11, 2009 #3


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    Hi mahela007! :wink:

    hmm … that's interesting …

    ah, I think it's conservation of energy

    a stationary racket will send the ball back at the same speed because that's also the same energy, so no work needs to be done,

    but sending the ball back faster than it came requires added energy, and that comes from work done by the racket, which is force times distance. :smile:
  5. Oct 11, 2009 #4


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    In tennis, when the ball starts to make contact with the racket stringing, both ball and stringing start to deform. I've seen high speed camera footage, and the deformation is huge. I thought for sure the ball would burst, but miraculously it doesn't. In modern professional play rackets the string tension is very large, but there is enough elasticity to store elastic energy.

    In the deformation of ball and racket strinning some of the energy is dissipated, but a lot of it gets reconverted to kinetic energy on the rebound.

    I assume that is why blocking a fast ball can send it way back to the opponent's baseline, whereas blocking a slow ball can at best result in some sort of drop shot. (I suppose that a true drop shot is where the player removes pace from the ball while still controlling it.)

    In baseball:
    As far as I know baseballs are very elastic. The elasiticity is subject to moisture content. There was an episode of Mythbusters in which influence of moisture content on baseball bounciness was investigated. A baseball that is dropped from height will bounce up very high, much higher than a tennis ball (if I remember correctly)

    I think that when a bat hits a ball elastic energy is also stored in bending of the bat, and returned. A baseball bat can bend allright. In the same Mythbusters episode Adam Savage swings a bat, and with the tip he accidently clips the top of a workbench. Such a hit causes the bat to vibrate violently, which hurts like hell.

    Along the length of the bat there is a socalled 'sweet spot'. If you hit the ball with the sweet spot then a maximum of energy is transferred to the ball. The further away from the sweet spot, the larger the loss of energy to causing vibration of the bat. The sweet spot is the point along the length where there is a node of the bat's vibration (when it vibrates.) If you hit the ball with the sweet spot you don't induce vibration in the bat.

  6. Oct 11, 2009 #5
    One way to make all of this seem more intuitive, is to think of the bat (or tennis racket) as a wall...

    if you throw a tennis ball at a wall, do you not expect the distance the ball travels away from the wall to vary depending on the speed at which you throw it? think in the extreme cases where the tennis ball is traveling very slow, and very fast.
  7. Oct 12, 2009 #6
    Thanks for all your answers.
    As for the wall analogy, that's what I tried doing. If you throw a ball at a wall hard, it will bounce back hard. If you throw it slowly, it will bounce back slowly.
    That would mean that the wall exerted a different force at each instance.. a large force when the ball was thrown fast and a small force when it was throw slowly.
    Why would a wall do that? What decides the amount of force the wall generates? In other words, why doesn't the wall just apply the same force on the fast ball as the slow ball and simply send it back slowly?

    (I think I'm confusing everyone here as well... ):wink:
  8. Oct 12, 2009 #7
    well again, so long as we are answering in analogies:

    one way to answer this i suppose is... the wall doesn't make any decisions... instead it is like a defined reaction to a force being applied to it's surface...

    when you get down to the physics of it, the wall and the tennis ball are both made of molecules, and the force put on the tennis ball and the wall is created from molecular repulsions created from a coulomb potential. This is the "elastic energy storage" that was being discussed . The wall's molecules aren't "thinking" about what force to apply to something when it hits it, rather it is just sitting there... and depending on the incident velocity, the ball and the wall will reach a point of higher potential energy, which is rapidly converted back into mechanical energy in the opposite direction.

    if you choose your system correctly, and assume the ideal case of no frictional forces and no sound energy etc... both momentum and energy is being conserved in this case... (the wall and the earth have some momentum now in the same direction that the ball was traveling, but the velocity is incredibly tiny because the mass is so large)
  9. Oct 12, 2009 #8


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    OK, you do feel that your intuition is leading you astray, but where?

    I think in the first place you shouldn't attribute things to the wall. Let me give a comparison. A pogo stick has a long shaft, inside a coil spring, and at one end a small shaft sticks out. Imagine you are jumping around with that pogo stick on a concrete surface.

    Each time you land on the ground the two objects exerting a force upon each other are the pogo stick and the concrete covered Earth. But only one object undergoes deformation: the coil spring in the Pogo stick undergoes elastic deformation, so that is where kinetic energy gets converted to elastic potential energy.

    Deformation of the concrete surface is negligable, so the surface doesn't store elastic energy. In that sense the surface is passive, not doing work.

    If you put concrete under compressive stress then the amount of compression is exceedingly small. Still, during the impact the end of the pogo stick does put the concrete that it lands on under compressive stress. The concret doesn't budge, and subsequently the coil spring of the pogo stick is compressed.

    Imagine what would happen if you try to pogo jump on clay soil. On impact the clay undergoes plastic deformation. The clay fails to push back on the end of the pogo stick, and then the spring of the pogo stick won't be compressed.

    I probably could have said all this more economically.
    My point is that in the case of a tennis ball bouncing against a wall the wall is instrumental, but the wall is not doing work. The work that is done goes into elastic deformation of the tennis ball. On the rebound you get all that energy back.

    The question then becomes:
    "When a fast tennis ball impacts a wall, more elastic energy is stored in the ball than when a slow tennis ball impacts. Why?"
    Well, a faster ball is faster! The faster ball undergoes more elastic deformation because it impacts harder.

  10. Oct 13, 2009 #9


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    Quote:Well, a faster ball is faster! The faster ball undergoes more elastic deformation because it impacts harder.
    You could add that it has more kinetic energy. This KE is transferred to elastic energy which is then transferred back to KE. The momentum will be conserved- mening that the massive wall plus the Earth will move away VERY slowly and the ball will go back fast, getting (virtually) all the KE back.
  11. Oct 13, 2009 #10
    I said a "how does the wall do that" only as a figure of speech.. of course it can't decide on which force to exert :wink:...
    Anyway.. it seems pretty clear when energy is considered. faster ball = more KE. = more energy on ball on rebound.

    But I (stubbornly) want to get back to forces,.
    Let's think of it this way.
    Let's assume that the ball (fast or slow) comes to rest when it hits the wall. This would mean that the faster ball required a greater force to stop it. In other words, it has to achieve a greater change in momentum than the slow ball does in coming to rest.

    But how would one explain the forces involved if the rebounding of the ball is considered?
    You can't say that the faster ball exerts a larger force on the wall because is "has" to rebound faster..(right?) How would you explain why a faster ball exerts more force on the wall?
    P.S o:)
  12. Oct 13, 2009 #11


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    Impulse = Change of Momentum = Force times Time

    In the simplest model - regarding the ball as a mass on a spring -, the time constant of the oscillation is independent of displacement (simple harmonic motion) so the rebound time will be about the same, however hard it hits the wall. That implies that the (average) force on the ball must be proportionally larger, the bigger the momentum change.
    For a ball, what happens when it hits the wall is a bit more complicated than a mass on a spring and, as the displacement increases, I think the restoring force goes up faster than for a spring. So I think the average force will be even higher because the contact time may be shorter. But that is surmise; The time won't be very different for different ball speeds, in any case, so the force must be bigger.
    How's that?
  13. Oct 13, 2009 #12
    yes.. but why is a larger force generated? Why isn't a force similar to that generated when a slow ball hits the wall generated? To me, it seems paradoxical that a larger force would be generated. If the ball comes to rest, we could then argues that the acceleration is greater and hence a larger force is required / generated.
    However, the ball doesn't come to rest.. it rebound (fast). So it seems that the large change in momentum (in the case of the fast ball) is brought about by a large force. The large force is generated/required because of the large change in momentum... the large change in momentum occurs because of the large force and... round and round we go... (hope that was clear)
  14. Oct 13, 2009 #13


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    I don't see your problem, I'm afraid.
    When the ball hits the wall there is a reaction force. During the time that it is slowing down, the force gets bigger and bigger. The strain in the ball then produces a force against the wall which pushes the ball away. As I pointed out, the time for this process is independent of the approach speed so the force will be proportional to the initually momentum.
    The forces and momentum changes are similarly related when a mass bobs up and down on a spring. Energy is transfered from KE to strain and back again and momentum is reversed. For an elastic collision the process is reversible.
  15. Oct 14, 2009 #14
    here's my problem... I can't figure out what causes the large force.
    I'll try to rephrase it..
    Why is a large force exerted on the ball? You could say that the force acted in order to change the momentum of the ball. However, why couldn't a small force act on the ball and just bring it to rest?
  16. Oct 14, 2009 #15
    On second thought.. let's take it this way.
    The fast ball hit's the wall. The wall then exerts a force (F) for a time (t). This will change the momentum of the ball and make it head back in the direction it came from (at a speed of V).
    My question is... why couldn't F be smaller and thus send the ball back with a velocity less than V?
  17. Oct 14, 2009 #16
    because energy must be conserved, hence momentum must be conserved. That measn for a perfectly elastic collision, the ball will rebound at the same velocity as it hit the wall. In reality, some energy will be lost during the impact (sound, material inelasticicty causing heat as the ball is compressed, etc) and the ball will always return slower than when it hit the wall.
  18. Oct 14, 2009 #17
    AH.... Now i understand. many thanks to everyone who posted.
  19. Oct 14, 2009 #18


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    For a perfectly elastic collision, the ball could be very soft and take a long time to rebound. The average force would be small.
    For a steel ballbearing hitting a steel walk, the rebound would take a short time and the force would be enormous.
    The rebound speed would be the same in each case.

    And, as an afterthought, if the wall just brought the ball to a halt, it would be an inelastic collision.
    Last edited: Oct 14, 2009
  20. Oct 14, 2009 #19
    No, I don't think that is true. But I think you figured that out in your next statement:

  21. Oct 14, 2009 #20


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    If the collision is ELASTIC it has to be true because that's what elastic means. In practice, it maybe hard to find a matrial with a low modulus which actually approaches perfect elasticity but we were talking theoretically.
  22. Oct 15, 2009 #21
    Indeed, this is true, sorry, I misread your examples, i thought you were comparing elastic (with the rubber ball) and inelastic (with the steel ball bearing).
  23. Oct 15, 2009 #22


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    A steel ball on a thick steel plate is as near elastic as I think you'll get. Newton's cradle does quite well innit?
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