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Work done on changing direction of a tennis ball

  • Thread starter Chris0101
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Homework Statement


[/B]
A tennis player strikes an incoming tennis ball of mass m and speed v such that the ball leaves the racket at speed v as well. How much work was done by the tennis player to reverse the direction of the ball? Explain. Did the tennis player do any negative work?

Homework Equations



##W = Fx = 1/2 mv^2##

The Attempt at a Solution



Here is my attempt with the use of calculus and differential equations:

##F = ma = mv dv/dx##
##Fdx = mvdv##

Integrating both sides, where the left integral is integrated from 0 to x and the right integral is integrated from v to -v, yielding:

##Fx = 1/2 m(-v)^2 - 1/2mv^2##
##Fx = 1/2mv^2 -1/2mv^2##
##Fx = 0##

Mathematically, it appears that the work done by the tennis player is zero if the magnitude of velocity is equal to that of the starting velocity.

I would assume that this is correct because because work would be greater than zero if the ball leaves the racket at a greater speed than it arrived at the racket and alternatively, work would be negative if the velocity decreased leaving the racket.

I need some clarification with regards to this, any insight would be greatly appreciated.
 

Answers and Replies

  • #2
Simon Bridge
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Here is my attempt with the use of calculus and differential equations:
You don't need calculus - you already know the equations for momentum and work.

I would assume that this is correct because because work would be greater than zero if the ball leaves the racket at a greater speed than it arrived at the racket and alternatively, work would be negative if the velocity decreased leaving the racket.
... that's not too bad.
Work is the change in energy ... since the kinetic energy of the ball is unchanged during the collision we can conclude that the net work done on the ball to change it's direction is zero (not counting energy that changes the temperature of the ball, make that thwacking noise, swing the racquet etc.)
However, Work is also force times distance ... the ball clearly experiences a force (it's momentum changes) which acts over some distance (as the ball and racquet squash together)...
 
  • #3
haruspex
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the ball clearly experiences a force (its momentum changes) which acts over some distance (as the ball and racquet squash together)...
Not necessarily. If the collision is completely elastic then it must be the case that the racket was held stationary. If we take the racket to be rigid, the force applied by the racket does not move through any distance. If we allow for elasticity in the racket, it first does negative work then an equal quantity of positive work. The internal motion of the ball in squashing against the racket is its own affair.
 
  • #4
Simon Bridge
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Both the racquet and ball deform during the collision though. This is a necessary part of the operation of both objects: it's how they are designed to behave.
The extension to positive and negative work that results is what I was hoping the OP would come up with ;)

If it's a racket instead, then there will surely be substantial energy loss in making so much noise :D
 
  • #5
haruspex
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Both the racquet and ball deform during the collision though. This is a necessary part of the operation of both objects: it's how they are designed to behave.
Yes, but you wrote
the ball clearly experiences a force (it's momentum changes) which acts over some distance
To me, that implied the racquet doing net work on the ball.
If it is completely elastic, whether by virtue of the ball's or the racquet's elasticity or both, the racquet does no net work on the ball. There may well be no net change in position in space of point of contact during contact.
If it's a racket instead, then there will surely be substantial energy loss in making so much noise
Well hit, sir.
 

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