Lab question on rebound

  • Thread starter sonya
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  • #1
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ok my physics class jst did a lab on studying rebound and last question in the report says:

An extrapolation of the data to a perfect bounce (ie. one that reaches to the height of the launch point) results in a bounce distance that is more than twice the horizontal distance of the point of impact from the launch point. Since the fall of the ball from the launch point is only half the trajectory of such a perfect bounce, this means that the horizontal velocity after the bounce is greater than it was before the bounce.

What is a possible reason for this??

i am stuck on this last question...ne help would be much appreciated!
 

Answers and Replies

  • #2
enigma
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If it were a snake, it woulda bit ya!

The ball is spinning. Friction with the floor during the collision gives it a forward momentum.
 
  • #3
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but wouldnt that happen with ne bounce then...not jst the "perfect" one?
 
  • #4
HallsofIvy
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With a "perfect bounce" (perfectly elastic), ignoring friction, the ball will come back up to the same height with the same speed. With less than a perfect bounce, it would have lower speed.

With friction, the rotational velocity can be converted to linear motion. If the bounce is "perfect", that can result in greater linear speed that initially. If the bounce is not perfectly elastic, that might not happen. The increase in linear speed due to the rotation might not make up for the reduction due to the inelastic bounce.
 

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