1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tennis ball-racket collision: Deformation and duration

  1. Nov 13, 2005 #1
    I've been presented to the following problem:

    A tennis ball can achieve a speed of 100 km/h. During the contact between tennis racket and ball the ball's dimension in the direction of movement is halved. Estimate how long ball and racket are in contact with each other. It can be presumed that the force between ball and racket is constant during the collision.

    ------

    It isn't wholly clear, if we are to involve the deformation in our calculations, since the sentence "During the contact between tennis racket and ball the ball's dimension in the direction of movement is halved." might be just a statement about normal racket-ball collisions. Then we'd just have to estimate a force for a normal collision - a quite loose estimate, if we just guess. Or it might mean that we'd have to include the deformation x = r/2 - where r is the ball radius.

    But how? We can't use work, since we'd just get W = Fs <=> F = W / (r/2) - if we assert that only the ball is elastic, which isn't true, and then we'd have to estimate the work instead of the force (since work doesn't come into the other relevant equations).

    Using elastic potential energy U[el], we'd have to estimate a spring constant, which is even more loose than estimating force.

    Is there an easy/right/good way? Or somewhere, at least, that I can find values of racket force or ball speeds immediately following collisions?
     
    Last edited: Nov 13, 2005
  2. jcsd
  3. Nov 14, 2005 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Interesting question. You could try using a pressure approach. If the volume of the ball is halved, what is the pressure inside the ball? Assume an adiabatic compression (no heat lost) because it happens so fast. That should give you the potential "spring" energy stored in the ball which is then transferred as kinetic energy of the ball when the air in the ball expands (adiabatically).
    You will have to assume that the force on the racquet stops when the ball is fully compressed.

    But there is an easier way to do this. If you assume that the force is constant, F, for a time [itex]\Delta t[/itex] the impulse [itex]F\Delta t[/itex] is equal to the momentum of the ball. Also, [itex]Fd/2 = mv^2/2[/itex] where d is the diameter of the ball.

    Combine the two equations to get rid of m to find the time.

    AM
     
    Last edited: Nov 14, 2005
  4. Nov 15, 2005 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Further to my last reply, it occurred to me that in order to solve the problem, we have to know either the magnitude of the force or the distance over which the force is applied. I think the question assumes that distance to be 1/2 the diameter of the ball. But is that really a reasonable assumption?

    There are two things that add to the distance: the motion of the racquet after impact and the stretching of the racquet strings. I would think that racquets are designed so that the racquet strings stretch on impact but return to normal while in contact with the ball.

    AM
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Tennis ball-racket collision: Deformation and duration
  1. Collision of two Balls (Replies: 1)

  2. Deformation gradient (Replies: 0)

Loading...