Velocity of ball after being hit by a racket


by pvm
Tags: ball, racket, velocity, velocity momentum
pvm
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#1
Nov22-13, 06:52 AM
P: 9
I'm trying to figure the velocity (and specifically the angle) of a squash ball (for example) after being hit by a racket.

Simple case: say the ball is initially stationary (hanging in the air!) and the racket hits the ball "flat". That is the velocity of the racket it normal (perpendicular) to the face of the racket. Then it seems simple - the ball will move away with a velocity also normal to the face of the racket, in line with the velocity of the racket.

More complex case: say the ball is initially stationary again, but this time the face is "open" or at an angle to the velocity vector of the racket. E.g. you are hitting it along a horizontal trajectory, but the face of the racket is not vertical, but tilted back, say 45 degrees. What angle does the ball come off the racket now? I'm not sure how to figure this out. The racket has strings - its not a flat surface.

Most complex case: as above, but now say the ball has in inbound velocity. I need its outbound velocity in the ground frame of reference. I'm not sure its a simple case of doing vector addition of its inbound velocity to what would happen had it been stationary when hit.

I also need the magnitude of the velocity. Will it be that of the racket + ball inbound - losses for heat etc.?

Thanks for your help!
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jedishrfu
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#2
Nov22-13, 06:56 AM
P: 2,493
Is this a homework assignment. If so you ou need to follow homework template and show some work. We are here to help but can't if you don't show what you've done so far.
pvm
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#3
Nov22-13, 07:30 AM
P: 9
No, not a homework assignment - I'm many years past that! :) Its a genuine problem I need to solve. I did A-level physics 30+ years ago... but I don't have enough knowledge to figure this out - unless its simpler than I am thinking - so I've nothing really to present in terms of work so far.

My thoughts are that if we look from the rackets frame of reference the ball is inbound with the racket velocity. Its also coming in at an angle if the face of the racket is open. I guess with simple "perfect" surfaces etc it would just bounce off with an equal and opposite angle. But with real strings and balls I think it will tend to come off much more towards the normal to the face. I was wondering if anyone could help me understand what this angle is likely to be...

xAxis
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#4
Nov22-13, 07:52 AM
P: 197

Velocity of ball after being hit by a racket


I guess you will have to assume that the racket is perfectly flat, and that the angles towards the normal are equal. Use the addition of velocity (v racket + v ball), so you can treat all three cases equally.
Then more complex case would be the ball having spin.
jedishrfu
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#5
Nov22-13, 07:54 AM
P: 2,493
It seems a quicker way to verify your assumption would be to video tape a ball bouncing off a stationary racket
to see what happens.

Here's a discussion on string tension in a racket (no math):

http://physicscentral.com/experiment...20080506030153

and here's a more detailed treatment:

http://www.physics.usyd.edu.au/~cros...iqueImpact.PDF
pvm
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#6
Nov22-13, 09:09 AM
P: 9
Two great references (esp. the paper).
Thanks folks...


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